1.1Problem 1 - Department of Physics and … = a,α 2,3 = −a,andβ 1 = −b, β 2,3 = b....

1

1.1 Problem 1(a) Solve the integral:

I =∞∫−∞

dxe−αx2

=√π

α

Hint: Consider solving the two-dimensional integral I2 =∞∫−∞

dxdye−α(x2+y2).

(b) Consider a particle in the stateψ(x) = 〈x|ψ〉 = e−αx

2

Calculate the expectation values 〈x2〉 and 〈p2〉. Take p = −i~∂x.

(c) Calculate the Fourier Transform of f(x) = e−µ|x| where µ = µ∗ and µ > 0.

Solution

(a) Convert the two-dimensional integral to polar coordinates.

I2 =∞∫∫

−∞

dxdye−αx2e−αy

2=∞∫

r=0

2π∫ϕ=0

e−αr2rdrdϕ = π

α→ I =

√I2

(b) Use parametric differentiation.

〈ψ|x2|ψ〉 =∞∫−∞

dxx2e−2αx2= −1

2∂

∂α

∞∫−∞

dxe−2αx2= 1

4

√π

2α3

Consider what it means for a wavefunction to be “bounded”. (It must vanish at the edges of the region of interest;draw some wavefunctions and consider what must happen for the wavefunction to be normalizeable.) This means thatψ|±∞ = 0. Use integration by parts:

〈ψ|p2|ψ〉 = −~2〈ψ|∂2x|ψ〉 = −~2ψ

dψ

dx

∣∣∣∣∞−∞

+ ~2∞∫−∞

dx

∣∣∣∣dψdx∣∣∣∣2 = ~2

∞∫−∞

dx

∣∣∣∣dψdx∣∣∣∣2

Note that for any bounded wavefunction ψ we may apply the above trick (the first term must always go to zero if thewavefunction is bounded). For this particular ψ the result is:

〈p2〉 = ~2√πα

2

(c) We deal with the absolute value by splitting the integral into positive and negative components.

1√2π

∞∫−∞

dxe−µ|x|e−ikx = 1√2π

0∫−∞

dxeµxe−ikx +∞∫

0

dxe−µxe−ikx

→ =√

2π

µ

µ2 + k2

1.2 Problem 2Consider two operators A and B, shown below, where a and b are real constants.

A =

a 0 00 −a 00 0 −a

B =

b 0 00 0 −ib0 ib 0

1

(a) Are these operators Hermitian? (A = A† – Self-adjoint).(b) Show that A and B commute.(c) Determine the eigenvalues and eigenvectors of A and B.(d) If two operators commute, it is always possible to find a set of simultaneous eigenvectors for those operators (ie

the operators can be diagonalized in the same basis). Find a set of orthonormal eigenvectors which are simultaneouseigenvectors of both A and B.

(e) Use the eigenvectors from part (d) to construct a rotation matrix R such that R†AR and R†BR are diagonal. Isthe rotation matrix R unitary? (For a unitary matrix U , we have U†U = UU† = I, where I is the identity matrix.)

Solution

(a) The matrices A and B are self-adjoint by inspection. (A∗ij = Aji and the same is true of B).(b) The commutator [A,B] = AB −BA. Do this out to show that AB −BA = 0.(c) Recall that the eigenvalues αi and βj are determined by solving the characteristic equation det(A− αI) = 0, and

the corresponding equation for B. We obtain α1 = a, α2,3 = −a, and β1 = −b, β2,3 = b. Since the matrix A is diagonalwriting the eigenvectors of A is trivial:

|α1〉 =

100

, |α2〉 =

010

, |α3〉 =

001

For B, we plug in each of the eigenvalues found above into B|βi〉 = βi|βi〉 in order to set up a system to solve for theelements of each vector |βi〉. One orthonormal choice of eigenvectors is:

|β1〉 = 1√2

01−i

, |β2〉 =

100

, |β3〉 = 1√2

0i−1

(d) Notice that the eigenvectors of A are the standard unit vectors. This makes things easier. We can see right away

that the pair of eigenvalues (α1, β2) = (a, b) share the same eigenvector (|α1〉 = |β2〉). The same logic can be applied toother pairs:

(−a, b) : |β3〉 = i√2|α2〉 −

1√2|α3〉 ∴ share |β3〉 (−a,−b) : |β1〉 = 1√

2|α2〉 −

i√2|α3〉 ∴ share |β1〉

(e) The matrix (which must perform a rotation from the original basis into the eigenbasis) is constructed using theshared eigenvectors as columns. Consider:

R = 1√2

√2 0 00 i 10 −1 −i

→ R† = 1√2

√2 0 00 −i −10 1 i

Evaluating with the original matrices gives the desired the result (simply perform the matrix multiplication):

R†AR =

a 0 00 −a 00 0 −a

R†BR =

b 0 00 b 00 0 −b

Such a rotation matrix must always be Unitary, and performing the matrix multiplication for R†R we find that this isindeed the case.

1.3 Problem 3Consider a Hamiltonian represented by the matrix shown below, in a state |ψ〉.

H =

1 0√

30 2 0√3 0 3

|ψ〉 =

12i−1

(a) Which eigenvalue of H is most likely to emerge from a measurement?(b) What is the average value of H in this state?

2

Solution

Applying the same process as in the preceding problem, we obtain the eigenvalues (0, 2, 4), and their correspondingeigenvectors:

|0〉 = 12

−√301

|2〉 =

010

|4〉 = 12

10√3

(a) The probabilities for each eigenvalue occurring are given by:

P (0) = | 〈0|ψ〉 |2

〈ψ|ψ〉= (1 +

√3)2

24 , P (2) = | 〈2|ψ〉 |2

〈ψ|ψ〉= 2

3 , P (4) = | 〈4|ψ〉 |2

〈ψ|ψ〉= (1−

√3)2

24

By inspection of the above results, P (2) is the most probable.(b) Two ways of calculating this result:

〈H〉 = 0 · P (0) + 2 · P (2) + 4 · P (4) = 2− 1√3

or

〈H〉 = 〈ψ |H|ψ〉 = 16(1,−2i,−1)

1 0√

30 2 0√3 0 3

12i−1

= 2− 1√3

1.4 Problem 4Consider a particle in a 1-dimensional infinite square well, described by the potential

V (x) =

0 x ∈ [0, L]∞ else

(a) Determine the normalized eigenfuctions of the Hamiltonian ψn(x)(b) Determine the probability of finding the particle in the range 0 < x < L

a . Show that the classical result is obtainedas n→∞.

(c) Determine 〈E〉 of a particle the the state |φ〉:

|φ〉 = 1√3|ψ1〉+

√23 |ψ3〉

(d) Compute 〈p〉t.

Solution

(a) Set up the Schrödinger equation and the boundary conditions ψ(0) = 0 = ψ(L).

H|ψn〉 = En|ψn〉 → − ~2

2m∂2ψn∂x2 = Enψn(x) → sin

[√2mEn~2 L

]= 0 → En = n2π2~2

2mL2

and ψn(x) = 〈x|ψn〉 =√

2L

sin(nπxL

)(b)

La∫

0

〈ψn|x〉〈x|ψn〉 dx =

La∫

0

ψ∗n(x)ψn(x)dx = 12

[2a− 1nπ

sin(

2πna

)]for n→∞, Solution → 1

a

(c) The average energy 〈E〉 in the state |φ〉 is given by 〈φ |H|φ〉. (Notice that each ψ is an eigenstate of H, so we canthrow out the cross terms right away.)

〈φ |H|φ〉 = 13 〈ψ1 |H|ψ1〉+ 2

3 〈ψ3 |H|ψ3〉 = E1

3 〈ψ1|ψ1〉+ 2E3

3 〈ψ3|ψ3〉 = 19π2~2

6mL2

3

This can also be calculated by evaluating:

〈φ |H|φ〉 = ~2

2m

L∫0

∣∣∣∣∂φ∂x∣∣∣∣2dx

(d) Let |Φ〉 be the time dependent version of |φ〉.

〈Φ |p|Φ〉 =L∫

0

(1√3ψ∗1e

iE1t~ +

√23ψ∗3e

iE3t~

)(−i~ ∂

∂x

)(1√3ψ1e

−iE1t~ +

√23ψ3e

−iE3t~

)dx = 0

This occurs because:L∫

0

sin(nπxL

)cos(mπx

L

)dx = 0

for n = m, and (n = 1,m = 3), and (n = 3,m = 1).

4

2

2.1 Problem 5Solve the integral using residue theory for n = 2 and n = 3:

∞∫−∞

dx1

(x2 + 1)n

Solution

The integrand has nth-order poles at z = ±i. Since the integrand goes to zero faster than |z|−1 we can close a semi-circularcontour at infinity without changing the value of the integral (the added part goes to zero), and use residue theory. Wechoose to close this contour in the upper half of the complex plane, and call this contour C. (The residues are thenevaluated as positive since we circle the poles counter-clockwise.) Then:

∞∫−∞

dx

(x+ i)n(x− i)n =∮C

dz

(z2 + 1)n = 2πiRes[

1(z2 + 1)n , z = i

]= 2πi

(n− 1)! limz→idn−1

dzn−1(z − i)n

(z2 + 1)n

= 2πi(n− 1)! limz→i

dn−1

dzn−11

(z + i)n

We evaluate this expression to obtain:n = 2 : π

2 n = 3 : 3π8

2.2 Problem 6We can see by dimensional analysis that the solution of the integral below will be of the form shown (for some constantλ):

∞∫−∞

dx1

(x2 + a2)3 = λ

a5

Given this result, compute the following in terms of a and λ without integrating.(a)

∞∫−∞

dx1

(x2 + a2)4

(b)∞∫−∞

dxx2

(x2 + a2)4

Solution

(a)d

d b2

∞∫−∞

dx(x2 + b2)−3 = −3∞∫−∞

dx(x2 + b2)−4 →∞∫−∞

dx(x2 + b2)−4 = −13d

db2λ

(b2) 52

= 5λ6b7

(b) Let x→ cx. Then:∞∫−∞

cdx

(c2x2 + b2)3 = λ

b5 →∞∫−∞

dx

(c2x2 + b2)3 = λ

cb5

d

dc2

∞∫−∞

dx(c2x2 + b2)−3 = −3∞∫−∞

x2dx

(c2x2 + b2)4 →∞∫−∞

x2dx

(c2x2 + b2)4

∣∣∣∣c=1

= −13d

dc2

(λ

(c2) 12 b5

)c=1

= λ

6b5

5

2.3 Problem 7Parts (a) - (d) show some familiar objects, properties, and operations from linear algebra expressed over a discrete basis,in Dirac notation. Write the analoguous statements for a continuous basis.

(a) Kronecker Delta δij .(b) Inner Product / Projection operator: 〈a|b〉 =

∑i a∗i bi =

∑i 〈a|i〉〈i|b〉. (Note

∑i |i 〉〈 i| = I).

(c) Matrix transformations: |a′〉 = M |a〉 → a′i =∑jMijaj ↔ 〈i|a′〉 =

∑j 〈i |M | j〉〈j|a〉.

(d) Unitary transformations: U†U = I →∑i U∗ijUik = δjk. Make sure you can derive the second expression shown

before writing the continuous form of this relationship. (Note that when this is true, for |a′〉 = U |a〉 and |b′〉 = U |b〉 wehave 〈a′|b′〉 =

⟨a∣∣U†U ∣∣ b⟩ = 〈a|b〉.)

(e) Consider how the Fourier Series and Fourier Transform relate to the ideas above. What property of the FourierTransform is suggested by Parseval’s theorem?

Solution

(Introduction to the Problem)Finite-Dimensional / Discrete Expressions Infinite-Dimensional / Continuous ExpressionsKronecker Delta function for two indices i, j

〈i|j〉 = δij

Dirac Delta function for two quantities over continuousvariables.

〈t2|t1〉 = δ(t2 − t1)Inner/Dot Product generalized to complex vectors:

〈a|b〉 = a† · b →∑i

〈a|i〉〈i|b〉 =∑i

a∗i bi

Inner Product for complex functions:

〈f |g〉 =∫dt 〈f |t〉〈t|g〉 =

∫dtf∗(t)g(t)

Matrix transformations: |a′〉 = M |a〉

a′i =∑j

Mijaj ↔ 〈i|a′〉 =∑j

〈i |M | j〉〈j|a〉

Integral Transformations: |f〉 = M |f〉 (M → Kernel)

f(τ) =⟨τ |f⟩

=∫dt 〈τ |M | t〉〈t|f〉 =

∫dtM(τ, t)f(t)

Unitary Transformations: 〈a|b〉 =⟨a∣∣U†U ∣∣ b⟩ = 〈a′|b′〉

True iff∑i U∗ijUik = δjk

〈a′|b′〉 =∑i

⟨a∣∣U†∣∣ i⟩ 〈i |U | b〉

=∑i

∑j

〈a|j〉⟨j∣∣U†∣∣ i⟩

(∑k

〈i |U | k〉〈k|b〉

)

=∑i

a′∗i b′i =

∑i

∑j

(Uijaj)∗∑k

(Uikbk)

=∑jk

a∗j bk∑i

U∗ijUik = 〈a|b〉 iff U is Unitary.

Unitary Transformations: 〈f |g〉 =⟨f∣∣U†U ∣∣ g⟩ =

⟨f |g⟩

True iff∫dτU(τ, t1)∗U(τ, t2) = δ(t2 − t1)

⟨f |g⟩

=∫dτ⟨f∣∣U†∣∣ τ⟩ 〈τ |U | g〉

=∫dτ

(∫dt1 〈f |t1〉

⟨t1∣∣U†∣∣ τ⟩)(∫ dt2 〈τ |U | t2〉〈t2|g〉

)=∫dτ

(∫dt1(U(τ, t1)f(t1))∗

)(∫dt2U(τ, t2)g(t2)

)=∫dt1dt2f

∗(t1)g(t2)∫dτU∗(τ, t1)U(τ, t2)

= 〈f |g〉 iff U is Unitary.

(e) Consider the following series and inverse Fourier Transform (g(k) is the Fourier Transform of g(x)):

Fourier Series: f(x) =∞∑

n=−∞Cne

2πinx/L Inverse Transform: g(x) = 1√2π

∫ ∞−∞

g(k)eikxdk

In each expression Cn and g(k) represent the “amount” of a given frequency needed to construct some function f(x) org(x). The constraint on x being finite for the series, however, makes the frequency space discrete, whereas in the unboundcase for the transform all frequencies are allowed. This is stated by the nature of Cn being dependent on an integer n,whereas k is a continuous variable. If we consider k ↔ 2πn

L we see that the the two expressions are related in a fashion verysimilar to all of the expressions considered in the previous parts of this problem. If we notate Cn = 〈n|C〉 and g(k) = 〈k|g〉this relationship is clearer. The Kernel for the Fourier transform is M(x, k) = e−ikx/

√2π for g(k) =

∫dxM(x, k)g(x).

6

Parseval’s Theorem states that for Fourier Transform⟨f |g⟩

= 〈f |g〉. The theorem is thereby essentially a statementthat the Fourier Transform is Unitary. (This can be shown using some integration tricks directly from the Kernel /definition of the transform.)

2.4 Problem 8In this problem we will show that in one dimension, bound states cannot be degenerate.

(a) Consider two degenerate states ψ1(x) and ψ2(x). Evalute the quantity ψ1(x)Hψ2(x) − ψ2(x)Hψ1(x) using theSchrodinger equation.

(b) Use the result from above to determine the relation between ψ1 and ψ2. (What do you expect if there is nodenegeracy?)

Solution

Solution from Prof. Das’ PHY 407 course. Since we are positing two degenerate states, they must have the same energy,by definition. This means Hψ1 = Eψ2 and Hψ2 = Eψ2. This implies:

ψ1Hψ2 − ψ2Hψ1 = E(ψ1ψ2 − ψ2ψ1) = 0

H = − ~2

2md2

dx2 + V (x) → − ~2

2m

[ψ1d2ψ2

dx2 − ψ2d2ψ1

dx2

]V (x)ψ1ψ2 − V (x)ψ2ψ1 = 0 → ψ1

d2ψ2

dx2 − ψ2d2ψ1

dx2 = 0

This last expression can be reduced to (where the prime is differentiation with respect to x):

d

dx(ψ1ψ

′2 − ψ2ψ

′1) = 0 → ψ1ψ

′2 − ψ2ψ

′1 = W = constant

Note that W is the Wronskian. We now apply the assumption that the states are bound. For a bound state, we must haveψ1, ψ2 → 0 as x→ ±∞. But if W is a constant and that constant is zero at ±∞, then that constant is zero everywhere.Therefore W = 0. If the Wronskian is zero however, that implies that ψ1 = λψ2 for some constant λ (i.e. ψ1 and ψ2 arelinearly dependent for all x). If that is the case, then ψ1 and ψ2, in fact, represent the same state. Therefore the solutionis unique and there is no degeneracy.

2.5 Problem 9Consider the Hamiltonian

H =p2x + p2

y

2m + V (x, y) → V (x, y) =

0 |x| < L2 , 0 < y < L

∞ else

(a) Determine the eigenvalues and normalized eigenfunctions of the system.(b) Set up and equation describing the degenerate states of the system.(c) Find some degenerage states.

Solution

(a) We can write down the answer based on previous derivations from workshop 1.

E = Ex + Ey = ~2π2

2mL2 (n2x + n2

y)

The wavefunction (not normalized) is given by:

ψ(x, y) = sin(nyπy

L

)sin(nxπx

L

)for nx even , ψ(x, y) = sin

(nyπyL

)cos(nxπx

L

)for nx odd

We may take the product of the x and y parts of the function because the Schrödinger equation can be solved by separationof variables in this case. The energies add cleanly because the Hamiltonian can be broken into H = Hx +Hy, where Hx

contains all the x dependence in the Hamiltonian, and Hy all the y, and [Hx, Hy] = 0. This separation of the Schrödingerequation as a differential equation, and ability to factor the Hamiltonian into commuting terms are, in fact, equivalentstatements.

7

The normalization condition can be expressed by:

L/2∫−L/2

dx

L∫0

dy|A|2ψ∗(x, y)ψ(x, y) = 1

Note that the integrals each in x and in y evaluate are the same as those for the one-dimensional box:

L∫0

sin2(nπxL

)dx = L

2 →L/2∫−L/2

dx

L∫0

dyψ∗(x, y)ψ(x, y) = L2

4 → A = 2L

ψ(x, y) = 2L

sin(nyπy

L

)sin(nxπx

L

)for nx even , ψ(x, y) = 2

Lsin(nyπy

L

)cos(nxπx

L

)for nx odd

(b) Degeneracy occurs when Enx,ny= En′x,n′y . Since E ∝ (n2

x+n2y), degeneracy arises when multiple integer coordinates

are on the circumference of a circle with radius proportional to√E.

(c) The following pair of coordinates (nx, ny) satisfy the relationship (there are many more): (5, 5) and (7, 1).

8

3

3.1 Problem 10Show the following using Dirac Notation, where Ω and Λ are some operators.

(a) Tr(ΩΛ) = Tr(ΛΩ).(b) (ΩΛ)† = Λ†Ω†. What happens if one takes the adjoint of 3 operators?

Solution

(a)Tr(ΩΛ) =

∑i

〈i |ΩΛ| i〉 =∑i

∑j

〈i |Ω| j〉〈j |Λ| i〉 =∑ij

〈j |Λ| i〉〈i |Ω| j〉 =∑j

〈j |ΛΩ| j〉 = Tr(ΛΩ)

(b) Method 1 (more formal):⟨i∣∣(ΩΛ)†

∣∣ j⟩ = (〈j |ΩΛ| i〉)† =∑k

(〈j |Ω| k〉〈k |Λ| i〉)† =∑k

(〈k |Λ| i〉〈j |Ω| k〉)† =∑k

⟨i∣∣Λ†∣∣ k⟩ ⟨k ∣∣Ω†∣∣ j⟩ =

⟨i∣∣Λ†Ω†∣∣ j⟩

Method 2: ⟨i∣∣(ΩΛ)†

∣∣ j⟩ = 〈ΩΛi|j〉 =⟨Λi∣∣Ω†∣∣ j⟩ =

⟨i∣∣Λ†Ω†∣∣ j⟩

Method two makes it clearer that (ABC)† = C†B†A†, and so forth for more operators.

3.2 Problem 11We apply a variational approach to a known problem. Consider the Hamiltonian for the one-dimensional SHO, and thetrial wavefuntion ψ.

H = p2

2m + mω2x2

2 ψ(x) = xe−α|x|

(a) Compute 〈H〉 as a function of α.(b) Minimize 〈H〉.(c) Which energy eigenstate of H most closely resembles the trial wavefunction?

Solution

(a)

〈H〉 =

~2

2m

∞∫−∞

dx(xe−α|x| d

2

dx2xe−α|x|

)+ mω2

2

∞∫−∞

dxx4e−2α|x|

∞∫−∞

dxe−2α|x|x2

=

~2

2m

∞∫−∞

dx[xe−α|x|

(−αx|x|

)+ e−α|x|

]2+ mω2

2

∞∫−∞

dxx4e−2α|x|

∞∫−∞

dxe−2α|x|x2

=

∞∫0

e−2αxdx

−1

×

~2

2m

∞∫0

(α2x2 − 2αx+ 1)e−2αxdx+ mω2

2

∞∫0

x4e−2αxdx

=[(

∂

∂(−2α)

)2 12α

]−1

×

[~2

2m

(α2(

∂

∂(−2α)

)2− 2α

(∂

∂(−2α)

)+ 1)

12α + mω2

2

(∂

∂(−2α)

)4 12α

]

=[2

(2α)3

]−1

×

[~2

2m

(α2

2

(2α)3 −2α

(2α)2 +1

2α

)+ 1

2mω2

24

(2α)512

(2α)2

]= ~2α2

2m + 32mω2

α2

9

(b) Minimize:d 〈H〉dα

= 0 → α =(

3m2ω2

~2

) 14

Plug that value in to get a measurement for the energy.

〈H〉 = ~2

2m

(3m2ω2

~2

) 12

+ 32mω

2(

~2

3m2ω2

) 12

=√

3~ω ≈ 1.73~ω

(c) This result is closest to the energy associated with the first excited state of the SHO, so we conclude that the trialwavefunction resembles that one the most.

3.3 Problem 12Recall the following properties of a Dirac delta function, and a generalized form of a Fourier Transform:

δ(k) = 12π

∞∫−∞

dxe−ikx f(x) =∫dx′f(x′)δ(x− x′) f(k) =

√|η|2π

∞∫−∞

f(x)e−ikxηdx

(a) Evaluate the expression∫dx′[∂x′δ(x− x′)]f(x′).

(b) Consider an operator D which differentiates a function, (i.e. 〈x |D| f〉 = ∂xf(x)). If this were written as anintegral transformation, what is the Kernel D(x, x′)? (The Kernel in an integral transformation is analoguous to thematrix element in a transformation.)

(c) Can we say anything about whether D is Hermitian? What about an operator P = −i~D?(d) What are the eigenfunctions of the operator P? (Let the eigenvalue be p. Find 〈x|p〉 such that P |p〉 = p|p〉.)(e) Write the quantity 〈p|ψ〉 using an integral over x. What do you notice about the relationship between 〈x|ψ〉 and

〈p|ψ〉? Can you use this to say something about the normalization of 〈x|p〉?(f) Evaluate 〈x |[x, P ]|ψ〉 for some |ψ〉, for the definition of P above. Do you get the expected result? Discuss.

Solution

(a) Assume a < x < b.

b∫a

dx′f(x′) d

dx′δ(x− x′) =

:0f(x′)δ(x− x′)

∣∣∣∣ba

−b∫a

dx′δ(x− x′)f ′(x′) = − dfdx

(b) Notice that ∂x′δ(x− x′) = −∂xδ(x− x′), and compare the following expansion with the result of part (a):

df

dx= 〈x |D| f〉 =

∫dx′ 〈x |D|x′〉〈x′|f〉 → 〈x |D|x′〉 = D(x, x′) = d

dxδ(x− x′) = δ′(x− x′)

(c)D†(x, x′) =

⟨x∣∣D†∣∣x′⟩ = 〈x′ |D|x〉∗ = D∗(x′, x) = δ′(x′ − x) = −D(x, x′)

D is naively anti-Hermitian. (There are some further restrictions on the normalizeability of the wavefunctions this actson for this to work a Hilbert Space, as we are interested in doing. Please ask questions in the event you would like furtherclarification on this point. Shankar’s “Principles of Quantum Mechanics”, on reserve at the POA, contains the morecomplete proof.)

P †(x, x′) = i~D†(x, x′) = −i~D(x, x′) = P (x, x′)

It is therefore reasonable to treat P as Hermitian (again given suitable functions consistent with a Hilbert space).(d)

〈x |P | p〉 = p 〈x|p〉 → −i~dp(x)dx

= p · p(x) → p(x) = Aeipx/~

(e)

〈p|ψ〉 =∞∫−∞

dx 〈p|x〉〈x|ψ〉 =∞∫−∞

dxp∗(x)ψ(x) = A

∞∫−∞

dxe−ipx~ ψ(x)

10

What if I choose the normalization A = (2π~)− 12 ?

p(x) = eipx/~√2π~

→ 〈p|ψ〉 = 1√2π~

∞∫−∞

dxψ(x) exp(ipx

~

)= ψ(p) for η = 1

~

Then 〈p|ψ〉 = ψ(p) is nothing but the Fourier transform of 〈x|ψ〉 = ψ(x). This implies:

〈p|p′〉 =∫dx 〈p|x〉〈x|p′〉 = 1

2π~

∞∫−∞

dxe−i(p−p′)x/~ = δ

(p− p′

~

)1~

= δ(p− p′)

So we have, in fact normalized 〈x|p〉. (Functions in a Hilbert space must be normalizeable to unity, as for bound functions,or to a Dirac delta, as is the case here and for unbound wavefunctions.)

(f) The following is an informal way of evaluating this. (You can consider the 〈x| on the left to denote the basis inwhich the operators and state are being written. It is not correct to view the bra as being pulled through the commutatorhowever; we would then be operating a matrix on a scalar. The more formal version illustrating this is shown furtherbelow.)

〈x |[x, P ]|ψ〉 = x

(−i~ ∂

∂x

)ψ(x) + i~

∂

∂x(xψ(x)) = i~ψ(x) = i~ 〈x|ψ〉 → [x, P ] = i~

This is the usual commutator between a coordinate and its conjugate momentum. We therefore conclude that a givenwavefunction written in a coordinate basis can be transformed to the momentum basis by the application of a Fouriertransform. For the sake of completion, a more formal version of part (f) would read as follows, where we note thatthe terms |x〉 are eigenstates of x, just as |p〉 denotes an eigenstate of p above, and that 〈x |D|x′〉 differentiates termscontaining x′.

〈x |[x, p]|ψ〉 =∫〈x |xp− px|x′〉〈x′|ψ〉 dx′ =

∫〈x |x|x′′〉〈x′′ |p|x′〉〈x′|ψ〉 dx′dx′′ −

∫〈x |p|x′′〉〈x′′ |x|x′〉〈x′|ψ〉 dx′dx′′

= −i~[∫

x′′ 〈x|x′′〉〈x′′ |D|x′〉〈x′|ψ〉 dx′dx′′ −∫〈x |D|x′′〉〈x′′|x′〉x′ 〈x′|ψ〉 dx′dx′′

]= −i~

[∫x′′δ(x− x′′)[∂x′′δ(x′′ − x′)]ψ(x′)dx′dx′′ −

∫[∂xδ(x− x′′)]δ(x′′ − x′)x′ψ(x′)dx′dx′′

]= i~

[∫[∂xδ(x− x′)]x′ψ(x′)dx′ −

∫x[∂xδ(x− x′)]ψ(x′)dx′

]= i~

[d

dx(xψ(x))− x d

dxψ(x)

]Note that this recovers the result above. The abstracted commutator (devoid of any particular basis or state), can bewritten [x, p] = i~I, or simply [x, p] = i~. A fun fact for further context: this applies to any set of coordinates and momentax and p related in classical terms by p = ∂xL, where L is the Lagrangian for a system. The Heisenberg uncertainty principlecan be derived from this commutator (this will be done in class), and thereby applies over any conjugate set of variablesin the same way. Getting from one variable to the other by Fourier transform as shown above “closes” a theoretical loopof sorts; the uncertainty principle can also be derived from Fourier transforms independent of any reference to quantummechanics. It is a direct consequence of the fact that we are working in mathematical system in which our variables mustbe related in the way derived above.

Note also that the integral transform definition of P is still consistent with all of this (just a nice check):

〈p |P |ψ〉 = −i~∞∫∫

−∞

dxdx′ 〈p|x〉〈x |D|x′〉〈x′|ψ〉 = − i~√2π~

∞∫−∞

dxe−ipx/~∞∫−∞

dx′ψ(x′)δ(x− x′)

= − i~√2π~

∞∫−∞

dxe−ipx/~dψ

dx= −i~

(ip

~ψ(p)

)= pψ(p)

〈p |P |ψ〉 =⟨p∣∣P †∣∣ψ⟩ = p 〈p|ψ〉 = pψ(p)

(I have used the identity for the Fourier Transform of the derivative of a function to simplify. I have also used the factthat P is hermitian, even though we have not proven this strictly above.)

A further note on the utility of this result: What this means is that if I expand a wavefunction in coordinate space (asyou are accustomed to doing), you use x = x and p = −i~∂x. You could however expand in the momentum basis and use

11

p = p and x = i~∂p. In some cases the integrations might be easier one way than the other, so it is worth being aware ofboth options.

〈x〉 →∫dxψ∗(x)xψ(x)↔

∫dpψ∗(p)

(i~∂

∂p

)ψ(p) and 〈p〉 →

∫dxψ∗(x)

(−i~ ∂

∂x

)ψ(x)↔

∫dpψ∗(p)pψ(p)

3.4 Problem 13Let |a′〉 and |a′′〉 be eigenstates of a Hermitian operator A with eigenvalues a′ and a′′ respectively. (a′ 6= a′′). TheHamiltonian is given below, where b is some real number.

H = b |a′ 〉〈 a′′|+ b |a′′ 〉〈 a′|

(a) |a′〉 and |a′′〉 are not eigenstates of the Hamiltonian. Do you see why this is true? Determine the eigenstates andcorresponding eigenvalues.

(b) Suppose the system is in a state |a′〉 at t = 0. Determine the state vector for t > 0.(c) Determine the probability for finding the system in |a′′〉 for t > 0 if the system is known to be in a state |a′〉 at

t = 0.

Solution

(a) Begin by evaluating H|a′〉 and H|a′′〉, noting that 〈a′|a′′〉 = 0 = 〈a′′|a′〉.

H|a′〉 = b (|a′ 〉〈 a′′|+ |a′′ 〉〈 a′|) |a′〉 = b|a′′〉 H|a′′〉 = b (|a′ 〉〈 a′′|+ |a′′ 〉〈 a′|) |a′′〉 = b|a′〉

It is then clear that H is swapping us between |a′〉 and |a′′〉. We proceed by constructing the matrix form of H in orderto find an eigenbasis:

H =(〈a′ |H| a′〉〈a′ |H| a′′〉〈a′′ |H| a′〉〈a′′ |H| a′′〉

)=(

0 bb 0

)→

∣∣∣∣ −λ bb −λ

∣∣∣∣ = 0 → λ = ±b

We see that the eigenvalues are ±b, and notate the associated normalized eigenstates by |b〉 and | − b〉. Solve these theusual way to obtain:

|b〉 = 1√2

(11

)| − b〉 = 1√

2

(1−1

)(b) Recall that the time evolution operator is of the form U = e−iHt/~. When applied to an eigenstate of the

Hamiltonian with energy E this results in a term e−iEt/~. We write the vector |a′〉 in terms of the eigenbasis.

|a′〉 =(

10

)= 1

2

(11

)+ 1

2

(1−1

)= 1√

2|b〉+ 1√

2| − b〉

Then we may apply the evolution operator to obtain:

|a′〉t = U |a′〉 = 1√2e−ibt/~|b〉+ 1√

2eibt/~| − b〉 = 1

2

(11

)e−ibt/~ + 1

2

(1−1

)eibt/~ =

(cos(bt~)

−i sin(bt~) )

(c) Then the probability of the system being in |a′′〉 for t > 0 after starting in |a′〉 is given by:

P = | 〈a′′|a′〉t |2 → |a′′〉 =

(01

)→ P = sin2

(bt

~

)

12

4

4.1 Problem 14Consider a particle subject to the potential V (x) = −V0δ(x) where V0 > 0.

(a) What are the boundary conditions? What are the units of V0?(b) Look for a bound state using the variational approach and a trial function φ(x) = Ae−α|x|. Give the energy and

normalized wave function of the bound state.(c) Show that the trial wave function in part (b) is an exact solution to the Schrödinger equation for this system.(d) Repeat part (b) with a different trial wavefunction shown below to determine the minimum expected energy 〈E〉.

(L is a varational parameter).

ψ(x) =

1 + x/L x ∈ [−L, 0]1− x/L x ∈ (0, L]

Compare with the results you found in part (b).

Solution

(a) We note that δ(x) has units of m−1 ([δ(x)] = [x]−1). Then for V to have units of Joules, [V0] = Jm. We requirethat the wavefunction be continuous across the boundary (consider what would happen to the probability current if thiswere not the case). We obtain a relation for the discontinuity in the derivative by integrating the Schrödinger equationover the delta function. We define a small region x ∈ [−ε, ε], and are interested in the limit as ε→ 0.

− ~2

2m∂2φ

∂x2 − V0δ(x)φ(x)− Eφ(x) = 0 → limε→0

ε∫−ε

[∂2φ

∂x2 + 2m~2 (V0δ(x) + E)φ(x)

]dx

This gives a discontinuity in the derivative according to:

limε→0

φ′(ε)− φ′(−ε) + 2mV0

~2 φ(0) = 0 → φ′2(0)− φ′1(0) = −2mV0φ(0)~2

(b)

〈H〉 =1

2m∫|pφ|2dx− V0φ(0)∫|φ|2dx

=~2α2

2m∫e−2α|x|dx− V0∫e−2α|x|dx

=~2α2m − V0

1α

= ~2α2

2m − V0α

We then minimize 〈H〉 and plug the value of α back in to obtain a measure of the energy.

∂ 〈H〉∂α

= 0 = ~2α

m− V0 → α = mV0

~2 → 〈H〉 = ~2

2m

(mV0

~2

)− V0

mV0

~2 = −mV2

02~2

We then normalize φ by evaluating 1 =∫|φ|2dx = A2

α . This gives A =√α.

φ(x) =√mV0

~2 exp[−mV0

~2 |x|]

(c) It will be useful to write |x| = xε(x) where ε is the alternating function (you may also know it as sgn(x)). Notethat ∂xε(x) = 2δ(x). Plug the solution into the Schrödinger equation.[

− ~2

2m∂2

∂x2 − V0δ(x)]

exp[−mV0

~2 |x|]

?= −mV2

02~2 exp

[−mV0

~2 |x|]

[− ~2

2m

(mV0

~2

)2− ~2

2m

(−mV0

~2

)2δ(x)− V0δ(x)

]exp

[−mV0

~2 |x|]

?= E exp[−mV0

~2 |x|]

The δ-functions on the LHS cancel, such that both sides are in fact equal, and we see that we have been using an exactsolution.

(d) As usual we require the continuity of the wavefunction at all points, thus ψ1(0) = ψ2(0), where region 1 is fornegative x and region 2 is for positive x.

ψ(x) =

1 + x/L x ∈ [−L, 0] → ψ11− x/L x ∈ (0, L] → ψ2

13

We obtain a relation for the discontinuity in the derivative by integrating the Schrödinger equation over the delta functionas in part (a), to obtain the following.

limε→0

ψ′(ε)− ψ′(−ε) + 2mV0

~2 ψ(0) = 0 → ψ′2(0)− ψ′1(0) = −2mV0ψ(0)~2

〈H〉 = 〈ψ |H|ψ〉〈ψ|ψ〉

=

~2

m

L∫0

∣∣∣∣∂ψ2∂x

∣∣∣∣2dx− V0ψ(0)

2L∫0

(1− x

L

)2dx

=~2

mL − V02L3

= 32

(~2

mL2 −V0

L

)

d 〈H〉dL

= 0 = −3 ~2

mL3 + 3V0

2L2 → L = 2~2

mV0

Then plugging L back into 〈H〉 gives:〈H〉 = −3mV0

8~2

We have an energy slightly closer to zero than the actual value with this approximation, but we see that the variationalmethod still gives us a relatively good result for this trial wavefunction. (By plotting the wavefunctions one may see thattrial function from part (d) and the exact solution in parts (b) and (c) are qualitatively quite similar.)

4.2 Problem 15Calculate the following Commutators:(a) [x, p] (b) [x2, p] (c) [x, p2] (d) [xn, p] (e) [x, pn] (f) [x1x2, p1p2] (g) [x1p2, x2p1]

Solution

(a) [x, p] = i~. See part (e) of previous problem.(b) [x2, p] = x[x, p] + [x, p]x = 2i~x.(c) [x, p2] = p[x, p] + [x, p]p = 2i~p.(d) [xn, p] = x[xn−1, p] + [x, p]xn−1 = x2[xn−2, p] + x[x, p]xn−2 + i~xn−1 = ... = i~nxn−1.(e) [x, pn] = ni~pn−1 by same logic.(f) [x1x2, p1p2] = i~(p1x1 + x2p2).(g) [x1p2, x2p1] = i~(x2p2 − x1p1).

4.3 Problem 16The 1D SHO can be expressed/solved in terms of the following operators:

H = p2

2M + 12Mω2x2 = ~ω

(a†a+ 1

2

); a =

√Mω

2~

(x+ ip

Mω

), a† =

√Mω

2~

(x− ip

Mω

)(a) Calculate the commutator [a, a†].(b) Given the operator N = a†a, where N |n〉 = n|n〉, compute En = 〈n |H|n〉.(c) Compute [a,H]|n〉 and [a†, H]|n〉. Show that H(a|n〉) = En−1(a|n〉) and H(a†|n〉) = En+1(a†|n〉).(d) Determine a|n〉 and a†|n〉.(e) Write out the matrix elements 〈n |x|m〉, 〈n |p|m〉,

⟨n∣∣x2∣∣m⟩, and ⟨n ∣∣p2

∣∣m⟩. Write out enough elements in matrixform to get a sense of how this looks.

(f) Sum matrices to obtain a matrix for H. Does your result make sense?(g) From part (e), compute ∆x∆p of the nth state.

14

Solution

(a)

[a, a†] = Mω

2~

[x+ ip

Mω, x− ip

Mω

]= Mω

2~i

Mω(−[x, p] + [p, x]) = i

2~ (−2i~) = 1

(b)

H = ~ω(a†a+ 1

2

)= ~ωN + ~ω

2 → 〈H〉 = En = ~ω(n+ 1

2

)for n = 0, 1, 2...

(c)

[a,H] = ~ω[a, a†a] = ~ω[a, a†]a = ~ωa → [a,H]|n〉 = ~ωa|n〉 → H(a|n〉) = (En − ~ω)(a|n〉)

Similarly:[a†, H]|n〉 = −~ωa†|n〉 → H(a†|n〉) = (En + ~ω)(a†|n〉)

Thus we see that a† raises us one energy level, and a lowers us one level.(d) We may then say that a|n〉 = C1|n − 1〉 and that a†|n〉 = C2|n + 1〉. We apply the definitions and commutator

results above to evaluate expectation values.⟨n∣∣a†a∣∣n⟩ = |C1|2 = n → a|n〉 =

√n|n− 1〉⟨

n∣∣aa†∣∣n⟩ = |C2|2 = n+ 1 → a†|n〉 =

√n+ 1|n+ 1〉

(e) Solve the system of given relations defining a and a† to write x and p in terms of the ladder operators, and thenapply these in calculating the matrix elements.

x =√

~2Mω

(a+ a†) p = i

√Mω~

2 (a† − a)

〈n |x|m〉 =√

~2Mω

(√mδn,m−1 +

√m+ 1δn,m+1

)〈n |p|m〉 = −i

√Mω~

2(√mδn,m−1 −

√m+ 1δn,m+1

)⟨n∣∣x2∣∣m⟩ = ~

2Mω

(√m(m− 1)δn,m−2 +mδn,m + (m+ 1)δn,m +

√(m+ 1)(m+ 2)δn,m+2

)⟨n∣∣p2∣∣m⟩ = −Mω~

2

(√m(m− 1)δn,m−2 −mδn,m − (m+ 1)δn,m +

√(m+ 1)(m+ 2)δn,m

)These may be equivalently represented by:

x =√

~2Mω

0 1 0 0 ...

1 0√

2 00√

2 0√

30 0

√3 0

.... . .

p = i

√Mω~

2

0 −1 0 0 ...

1 0 −√

2 00√

2 0 −√

30 0

√3 0

.... . .

x2 = ~2Mω

1 0

√2 ...

0 3 0√2 0 5...

. . .

p2 = −Mω~2

−1 0

√2 ...

0 −3 0√2 0 −5...

. . .

(f)

H = 12mp2 + mω2

2 x2 = ~ω

12 0 0 ...0 3

2 00 0 5

2...

. . .

15

This is a thoroughly unsurprising result, which can be obtained directly from H = ~ω(a†a + 12 ) as well. We are in the

energy eigenbasis, so H should be diagonal with the energy eigenvalues populating the diagonal. This is exactly what wefind.

(g)

〈n |x|n〉 = 0 = 〈n |p|n〉⟨n∣∣x2∣∣n⟩ = ~

2Mω(2n+ 1)

⟨n∣∣p2∣∣n⟩ = Mω~

2 (2n+ 1) → ∆x∆p = ~2 (2n+ 1) ≥ ~

2

4.4 Problem 17(a) Evaluate the following for some operator Ω.

d

dt〈Ω〉

(Hint: Consider using the time-dependent Schrödinger equation to rewrite terms containing |ψ〉.)(b) The result above is known as Ehrenfest’s theorem. What classical equation does the result remind you of? What

does the relationship suggest is a classical quantity analoguous to the commutators used in quantum theory? Discuss.(c) Prove the following relationship where X and P denote a coordinate and conjugate momentum (both expectation

values are time dependent):

md 〈X〉dt

= 〈P 〉

Solution

(a)d

dt〈Ω〉 = d

dt〈ψ |Ω|ψ〉 =

⟨ψ |Ω|ψ

⟩+⟨ψ

∣∣∣∣∂Ω∂t

∣∣∣∣ψ⟩+⟨ψ |Ω| ψ

⟩The time-dependent Schrödinger equation can be written:

i~d

dt|ψ〉 = i~|ψ〉 = H|ψ〉 → |ψ〉 = − i

~H|ψ〉 , 〈ψ| = i

~〈ψ|H

Plugging these results into the first relationship for dt 〈Ω〉, we obtain:

d

dt〈Ω〉 = i

~〈ψ |HΩ|ψ〉+

⟨ψ

∣∣∣∣∂Ω∂t

∣∣∣∣ψ⟩− i

~〈ψ |ΩH|ψ〉 = i

~〈ψ |[H,Ω]|ψ〉+

⟨x

∣∣∣∣∂Ω∂t

∣∣∣∣ψ⟩ = 1i~〈[Ω, H]〉+ 〈∂tΩ〉

If Ω has no explicit time dependence, this reduces to

d 〈Ω〉dt

= 1i~〈[Ω, H]〉

(b) This result looks an awful lot like the following result from classical mechanics, describing the time evolution ofsome phase-space parameter ω, where the curly braces denote Poisson brackets:

dω

dt= ω,H+ ∂ω

∂tIf ω 6= ω(t) → dω

dt= ω,H

It would then be reasonable to suppose that the commutator in the quantum mechanical equation behaves very muchlike the Poisson Bracket in classical mechanics. One can show that this is true and applies quite generally (although wedo not have the time to do so here): for an equation with a Poisson bracket a, b in a classical setting, substitutingin a commutator [A,B]/i~ will give a correct quantum-mechnical statement. For instance, in classical mechanics for aconjugate coordinate and momentum x and p, x, p = 1 becomes [x, p]/i~ = 1 → [x, p] = i~ in the quantum case, wherex and p now denote operators.

(c)d 〈X〉dt

= 1i~〈[X,H]〉 = 1

i~⟨[X,P 2/2m+ V (x)]

⟩= 1

2mi~ 〈2i~P 〉 = 〈P 〉m

16

5

5.1 Problem 18Consider a particle subject to the potential

V (x) =−V0δ(x− a) x > 0

∞ x ≤ 0where V0 > 0 and a > 0. Does a bound state exist for this system?

Solution

We note that ψ = 0 for all x ≤ 0, and that the solutions to the Schrödinger equation are exponential for a bound state(E < 0). We define ψ1 over x ∈ [0, a] and ψ2 over x ∈ [a,∞). Since we would like a function that decays as x → ∞ andwhich is zero at x = 0 we propose the following:

− ~2

2m∂2ψ

∂x2 − V0δ(x− a)ψ(x) = −|E|ψ(x) → k2 = 2m|E|~2 → ψ1 = A sinh(kx) and ψ2 = Be−kx

The boundary conditions are obtained the same way as in workshop problem 4.1, and are:

ψ1(a)− ψ2(a) = 0 ψ′2(a)− ψ′1(a) = −2mV0ψ(a)~2

A sinh(ka)−Be−ka = 0 − kBe−ka −Ak cosh(ka) = −2mV0Be−ka

~2

This system can be expressed as: (sinh(ka) −e−kak cosh(ka) e−ka

(k − 2mV0

~2

) )( AB

)= 0

For a solution to the system of equations to exist, we require that the determinant of the matrix be zero.∣∣∣∣ sinh(ka) −e−kak cosh(ka) e−ka

(k − 2mV0

~2

) ∣∣∣∣ = sinh(ka)e−ka(k − 2mV0

~2

)+ k cosh(ka)e−ka = 0

We reduce this equation down:(k − 2mV0

~2

)sinh(ka) = −k cosh(ka) → 2mV0

~2 = k(coth(ka) + 1)

And arrive at:mV0

k~2 = e2ka

e2ka − 1 → η ≡ 2ka → ~2η

2mV0a= 1− e−η

We plot each side of the transcendental equation to see if we have a solution.(a) (b)

In case (a), where the LHS equation passes above the RHS we only have an intersection corresponding to the trivialsolution η ∝ k ∝ |E|1/2 = 0. In case (b) however, we have a nontrivial solution for a bound state energy. We quantifythis; for a bound state to exist we must have:

d

dη(1− e−η)

∣∣∣∣η=0

= 1 > ~2

2mV0a→ Bound State when: V0 >

~2

2ma

17

5.2 Problem 19Two potentials are shown below, where V0 > 0, V2 > V1 > 0, and a > 0.

V1(x) =

0 x < 0V1 0 < x < aV2 x > a

V2(x) =

0 x < −a−V0 −a < x < 0V0 0 < x < a0 x > a

(a) Consider a particle incident from the left in system 1, with energy E > V2. Set up, but do not solve a system of

equations you can use to calculate the transmission and reflection coefficients for this system. Write expressions for thetransmission and reflection coefficients in terms of wave amplitudes.

(b) What are the transmission and reflection coefficients in system 1 if the particle is incident with an energy E < V2?(c) Repeat part (a) in system 2 for a particle incident from the left with energy E > V0.(d) What if the particle in part (c) had an energy 0 < E < V0?

Solution

(a) Number the regions in system 1 as 1-3, from left to right. Then make the following assignments for wave functions,which are solutions to the Schrödinger equation in each region.

ψ1(x) = eik1x +Be−ik1x ψ2(x) = Ceik2x +De−ik2x ψ3(x) = Feik3x

k1 ≡√

2mE~2 k2 ≡

√2m(E − V1)

~2 k3 ≡√

2m(E − V2)~2

We must have the wavefunction and its derivative continuous across each boundary. This results in the following equations:

ψ1(0) = ψ2(0) → 1 +B = C +D

ψ′1(0) = ψ′2(0) → ik1(1−B) = ik2(C −D)ψ2(a) = ψ3(a) → Ceik2a +De−ik2a = Feik3a

ψ′2(a) = ψ′3(a) → ik2(Ceik2a −De−ik2a) = ik3Feik3a

This can be rewritten more nicely as:−1 1 1 0k1 k2 −k2 00 eiak2 e−iak2 −eiak3

0 k2eiak2 −k2e

−iak2 −k3eiak3

BCDF

=

1k100

The reflection coefficient is R = |B|2 and the transmission coefficient is T = (k3/k1)|F |2. For those interested, one cansolve for the coefficients and obtain:

B = (k1 − k2)(k2 + k3) + e2iak2(k1 + k2)(k2 − k3)(k1 + k2)(k2 + k3) + e2iak2(k1 − k2)(k2 − k3)

C = 2k1(k2 + k3)(k1 + k2)(k2 + k3) + e2iak2(k1 − k2)(k2 − k3)

D = 2k1e2iak2(k2 − k3)

(k1 + k2)(k2 + k3) + e2iak2(k1 − k2)(k2 − k3)

F = − 4k1k2eia(k2−k3)

−(k1 + k2)(k2 + k3) + e2iak2(k2 − k1)(k2 − k3)(b) Since the wave function must now decay in region 3, there is no chance of the particle being transmitted to infinity.

Therefore R = 1 and T = 0.(c) Number the regions in system 2 as 1-4, from left to right. We then follow the same process as in part (a):

ψ1(x) = eik1x +Be−ik1x ψ2(x) = Ceik2x +De−ik2x ψ3(x) = Feik3x +Ge−ik3x ψ4(x) = Heik1x

18

k1 = k4 =√

2mE~2 k2 =

√2m(E + V0)

~2 k3 =√

2m(E − V0)~2

The boundary conditions are applied at x = 0,±a, to obtain:eiak1 −e−iak2 −eiak2 0 0 0k1e

iak1 k2e−iak2 −k2e

iak2 0 0 00 1 1 −1 −1 00 k2 −k2 −k3 k3 00 0 0 eiak3 e−iak3 −eiak1

0 0 0 k3eiak3 −k3e

−iak3 −k1eiak1

BCDFGH

=

−e−ik1a

k1e−ik1a

0000

The reflection coefficient is given by R = |B|2 and the transmission coefficient is given by T = |H|2.

(d) This change in the energy of the incident particle only affects region 3. We may replace the ψ3 in part (c) with

ψ3(x) = Fe−k3x +Gek3x

and then proceed in the same fashion. This does not imply that T → 0 as in part (b) however, since region 3 does notextend to infinity. There is a nonzero chance that the particle will tunnel through the classically forbidden region andcontinue onto region 4.

5.3 Problem 20Consider the 3-dimensional anisotropic oscillator described by the Hamiltonian:

H = p2

2m + 12mω

2(x2 + 4y2 + 9z2)

(a) Write down a general expression for the energy eigenvalues, and list the first eight energies. What is the degeneracyof each of these states?

(b) Write out the wavefunctions corresponding to the first three energies.

Solution

(a) Consider 12mω

2(x2 + 4y2 + 9z2) = m2 (ω2x2 + (2ω)2y2 + (3ω)2z2). Then:

E = ~ω(nx + 2ny + 3nz + 1

2 + 212 + 31

2

)= ~ω (nx + 2ny + 3nz + 3)

Energy (nx, ny, nz) No. of combinations3~ω all 0 14~ω (1,0,0) 15~ω (2,0,0),(0,1,0) 26~ω (3,0,0),(1,1,0),(0,0,1) 37~ω (4,0,0),(0,2,0),(2,1,0),(1,0,1) 48~ω (5,0,0),(3,1,0),(1,2,0),(2,0,1),(0,1,1) 59~ω (6,0,0),(4,1,0),(2,2,0),(0,3,0),(1,1,1),(0,0,2),(3,0,1) 710~ω (7,0,0),(5,1,0),(3,2,0),(1,3,0),(2,1,1),(1,0,2),(4,0,1),(0,2,1) 8

(b) We use the convention from Liboff (table 7.1), where ξ2i ≡ mωi

~2 x2i , and An = (2nn!

√π)− 1

2 . For the case above, wemay take ξ2

x = mωx2/~2, ξ2y = 2mωy2/~2, and ξ2

z = 3mωz2/~2. Then:

E000 = 3~ω → ψ(ξx, ξy, ξz) = A30e−(ξ2

x+ξ2y+ξ2

z)/2

E100 = 4~ω → ψ(ξx, ξy, ξz) =(A2

0e−(ξ2

y+ξ2z)/2

)·(

2A1ξxe−ξ2

x/2)

E010 = 5~ω → ψ(ξx, ξy, ξz) =(A2

0e−(ξ2

x+ξ2z)/2

)·(

2A1ξye−ξ2

y/2)

E200 = 5~ω → ψ(ξx, ξy, ξz) =(A2

0e−(ξ2

y+ξ2z)/2

)·(A2(4ξ2

x − 2)e−ξ2x/2)

19

5.4 Problem 21Consider the matrices

σx = 12

(0 11 0

)σy = 1

2

(0 −ii 0

)(a) Find a matrix σz such that [σy, σz] = iσx, and [σz, σx] = iσy, and Tr(σz) = 0.(b) Compute [σx, σy]. What does this say about σx, σy, and σz?(c) Compute σ2 = σ2

x + σ2y + σ2

z . What are the eigenvalues of σ2?(d) Determine the kets |± 1

2 〉, which are the eigen-kets of σz corresponding the eigenvalues ± 12 . Define σ± = σx± iσy,

and compute σ±| ± 12 〉. What do you notice?

Solution

(a) Set up a suitable system of equations, guess and check, or know the answer by some other means to obtain:

σz = 12

(1 00 −1

)(b) You will notice that for this choice of σz we have [σx, σy] = iσz, which means σx, σy, and σz are closed under

commutation.(c)

σ2 = 14

[(1 00 1

)+(

1 00 1

)+(

1 00 1

)]=( 3

4 00 3

4

)The eigenvalue is 3

4 with degeneracy 2.(d) The eigenvectors of σz are given by (this is trivial since σz is diagonal):

|+ 1/2〉 =(

10

)| − 1/2〉 =

(01

)

σ+ = 12

(0 11 0

)+ i

2

(0 −ii 0

)=(

0 10 0

)σ− = 1

2

(0 11 0

)− i

2

(0 −ii 0

)=(

0 01 0

)With these definitions, we can see the following:

σ+

(10

)= 0 , σ−

(10

)=(

01

), σ+

(01

)=(

10

), σ−

(01

)= 0

We conclude that σ± are raising and lowering operators, moving us on a ladder with only two states (raising above | 12 〉and lowering below | − 1

2 〉 both give zero.)

20

6

6.1 Problem 22The following are theorems that have been useful throughout the course, and will continue to prove useful as the semesterprogresses. Prove (a)-(e), then discuss (f). Then try to make connections between these and problems 6.2-6.3.

(a) A Hermitian operator has real eigenvalues.(b) Eigenvectors of a Hermitian operator with distinct eigenvalues are orthogonal.(c) The operator which transforms an orthonormal set of basis vectors into another is unitary.(d) If A is a Hermitian matrix, then there exists a unitary matrix U such that U†AU is diagonal.(e) If Ω and Λ are two commuting Hermitian matrices/operators with non-degenerate eigenvalues, then the eigenstates

of Ω are also eigenstates of the Λ.(f) Extend part (e) to account for the possibility of degeneracies. Discuss how one might prove that for any commuting

and Hermitian Ω and Λ there exists some set of shared eigenstates. (This is provable, but this question is asking for aninformal discussion rather than a formal proof, due to time constraints.)

Solution

(a) Let A be a Hermitian operator, such that A = A†. Let |a〉 be an eigenvector/eigenstate of A with eigenvalue a.Then:

A|a〉 = a|a〉 → 〈a |A| a〉 = a 〈a|a〉 and 〈a|A† = 〈a|A = a∗〈a| → 〈a |A| a〉 = a∗ 〈a|a〉

By subtracting the results of each expression we obtain (a− a∗) 〈a|a〉 = 0. Since the inner product of a vector with itselfis positive semi-definite (i.e. 〈a|a〉 ≥ 0), for the relationship to hold for any |a〉 and a, we must have a = a∗, meaning thata is real.

(b) Let A be a hermitian operator and |a1〉 and |a2〉 be eigenstates of A with eigenvalues a1 and a2, respectively. Sincethe eigenvalues are distinct, we require a1 6= a2. Then we may do the following:

A|a1〉 = a1|a1〉 → 〈a2 |A| a1〉 = a1 〈a2|a1〉 and A|a2〉 = a2|a2〉 → 〈a2|A = a2〈a2| → 〈a2 |A| a1〉 = a2 〈a2|a1〉

We again take the difference of the results, to obtain the equation (a1 − a2) 〈a2|a1〉 = 0. Since a1 6= a2 by assumption, wemust have 〈a2|a1〉 = 0.

(c) We define a set of orthonormal basis vectors |ei〉, and let U be the operator which takes it to another set oforthonormal basis vectors |ai〉. Then:

|ai〉 = U |ei〉 → 〈aj | = 〈ej |U† → 〈aj |ai〉 =⟨ej∣∣U†U ∣∣ ei⟩

Since both sets of vectors are orthonormal we have 〈aj |ai〉 = δij = 〈ej |ei〉. Then the above is true when U†U = I, or U isunitary.

(d) Let |ei〉 be the standard orthonormal basis, and |ai〉 be an orthonormal eigenbasis of the operator A. U is thematrix that transforms the standard basis to the eigenbasis. Then:

|ai〉 = U |ei〉 and A|ai〉 = ai|ai〉 → 〈aj |A| ai〉 = ai 〈aj |ai〉 = aiδij

=⟨ej∣∣U†AU ∣∣ ei⟩ = aiδij

This shows that U†AU is diagonal, with the eigenvalues of A as the matrix elements down the diagonal. Then Udiagonalizes A.

(e) Let Ω and Λ be commuting operators ([Ω,Λ] = 0), and |ωi〉 represent the complete set of eigenstates of Ωcorresponding to eigenvalues ωi. We further assume that neither Ω nor Λ have degenerate eigenvalues. Since [Ω,Λ] = 0,we may write (ΩΛ− ΛΩ)|ωi〉 = 0. From here we do the following:

ΩΛ|ωi〉 = ΛΩ|ωi〉 = ωi(Λ|ωi〉)

Since Ω(Λ|ωi〉) = ωi(Λ|ωi〉) we see that Λ|ωi〉 is also an eigenvector of Ω with eigenvalue ωi. This is only possible ifΛ|ωi〉 = λi|ωi〉. This means that all eigenstates of Ω are also eigenstates of Λ, and that the same matrix which diagonalizesΩ would diagonalize Λ.

(f) We now consider the possibility of degeneracy in the eigenvalues ωi. Let |ωi, j〉 be an Ni-fold degenerate eigenstateof Ω, where j = 1, 2, ...Ni, such that Ω|ωi, j〉 = ωi|ωi, j〉. Note that for a non-degenerate state we simply have Ni = 1 forthat particular i. For the sake of formality, we state that Ω is a linear operator over an N-dimensional vector space V.

21

There are a few results and observations we need to motivate this discussion. (1) A linear operator on V is diagonalizableiff V can be written as the direct sum of the eigenspaces of the operator, (2) all Hermitian operators are diagonalizable, asshown above, and (3) an operator which is Hermitian on V will also be Hermitian within each subspace Wi. This meansthat each eigenspace of Ω Wi (necessarily a subspace of V) has dimensionality Ni such that

∑iNi = N , or that all of the

states |ωi, j〉 are orthogonal and span V. To summarize, we are saying that:

〈ωi, k|ωi, j〉 = δjk ∀ i and W1 ⊕W2 ⊕ ...⊕Wimax= V or Wi ∩

∑i 6=j

Wj = 0 ∀ j

We have already shown that 〈ωk, j|ωi, j〉 = δik ∀ j in part (b). The result we are interested in follows reasonably well oncewe have digested the above. We proceed by noting something we used in workshop 1. Consider a linear combination of|ωi, j〉 for fixed i:

Ω

N∑j

αj |ωi, j〉

=N∑j

αjΩ|ωi, j〉 =N∑j

ωiαj |ωi, j〉 = ωi

N∑j

αj |ωi, j〉

It is therefore obvious that a superposition of degenerate eigenstates of Ω is still an eigenstate of Ω. Then we can say thefollowing as we did in part (e):

[Ω,Λ] = 0 → Ω

ΛN∑j

αj |ωi, j〉

= ΛΩ

N∑j

αj |ωi, j〉

= ωi

ΛN∑j

αj |ωi, j〉

This reduces the question down somewhat. Can we find sets of coefficients αj such that we construct a linear combinationof |ωi, j〉 within each subspace Wi which create Ni eigenvectors of Λ? Part (d) tells us yes, because if Λ is Hermitian in Vit must also be Hermitian within each subspace Wi. Therefore we conclude that we have the freedom to perform a changeof basis that diagonalizes Λ in each subspace, which solves our problem.

Note that parts (e) and (f) show that given two commuting Hermitian operators, each eigenbasis of Ω is also aneigenbasis of Λ iff the eigenvalues of the operators are non-degenerate. If there are degeneracies however, we will be ableto find a common set of eigenvectors, but not every eigenbasis of one operator will also be an eigenbasis of the other.Those wishing to see the same thought process explained differently may wish to readhttp://ocw.mit.edu/courses/physics/8-04-quantum-physics-i-spring-2013/study-materials/MIT8_04S13_OnCommEigenbas.pdf

which was referenced in creating this solution.This problem can also be understood by looking at the matrix representations of the operators Ω and Λ directly.

Clearly Ω is diagonal in its own eigenbasis, and Λ, which commutes with Ω, must then be block diagonal in an eigenbasisof Ω.

Ω =

ω1 0 00 ω1 00 0 ω2

→ Λ =

? ? 0? ? 00 0 λ2

Since ω1 has multiplicity 2 in the example above, the corresponding eigenspace is not necessarily already diagonalized,but can be in order to obtain a common basis. Since ω2 is non-degenerate, this portion of Λ is already diagonal and willalready share an eigenvector. A further example of how this works can be found in problem 6.3.

One final point to ensure that the utility of this result is clear. In homework problem 4.2 you were given a 2D SHO,and states |0, 0〉 with E = ~ω, |1, 0〉 and |0, 1〉 with E = 2~ω, where these are indexed according to |nx, ny〉. These arestates written in the energy eigenbasis (eigenbasis of H), and you were given two operators A and B which you had shownwere Hermitian and commuted with H. You were asked to find an eigenbasis common to H and A, and one common toH and B. You were lucky with A, because the eigenstates of H were already eigenstates of A. For B however, you foundthat |0, 0〉 was an eigenstate of B (which it had to be, since the energy eigenvalue ~ω was non-degenerate), but then forB, you had to take a linear combination of the two states in the 2~ω-subspace in order to diagonalize B.

6.2 Problem 23Consider a Hamiltonian given by the 3× 3 matrix

H =

3 0 00 0 50 5 0

22

(a) Determine the Energy Eigenvalues.(b) Does H form a CSCO?

Solution

(a)

det

3− λ 0 00 −λ 50 5 −λ

= 0 → λ = 3, 5,−5

(b) Since the eigenvalues are non-degenerate, H completely describes the system and is a CSCO.

6.3 Problem 24Consider a system with Hamiltonian H and operator A given by:

H =

−2 0 00 1 00 0 1

A =

5 0 00 0 20 2 0

(a) Do H and A commute? If yes, give a basis of eigenvectors common to H and A.(b) Which among the set of operators H, A, H,A, and H2, A form a CSCO?

Solution

(a) We find that [H,A] = 0. H has eigenvalues −2, 1, and 1 and eigenvectors:

| − 2〉H =

100

|1〉Ha =

010

|1〉Hb =

001

A has eigenvalues −2, 2, 5, with eigenvectors:

| − 2〉A = 1√2

01−1

|2〉A = 1√2

011

|5〉A =

100

We look for a set of eigenvectors common to both operators. We do this in the same way as in workshop 1, noting thatany linear combination of |1〉Ha and |1〉Hb will still be an eigenvector of H with eigenvalue 1.

| − 2〉A = 1√2|1〉Ha −

1√2|1〉Hb |2〉A = 1√

2|1〉Ha + 1√

2|1〉Hb

We therefore reindex our vectors according to |λH , λA〉, and propose the following for the shared set:

|1,−2〉 = 1√2

01−1

|1, 2〉 = 1√2

011

| − 2, 5〉 =

100

(b) H cannot be a CSCO on its own because it is degenerate. A is however a CSCO since its eigenvalues are

non-degenerate, and since [H,A] = 0 = [H2, A] these are also complete sets of commuting operators.

6.4 Problem 25Consider a particle subject to the Hamiltonian below, where ε is a constant.

H =p2x + p2

y

2m + 12mω

2x2 − εx+ V (y) V (y) =

0 y ∈ [−L/2, L/2]∞ else

Determine the energy spectrum, wavefunctions, and comment on parity and possibility of degenerate states.Hint: Consider an SHO with a linear transformation in x, namely x→ x− x0.

23

Solution

Consider an SHO with x→ x− x0

Hx = p2x

2m + 12mω

2(x− x0)2 = p2x

2m + 12mω

2(x2 − 2xx0 + x20)

We match this up with the given Hamiltonian, and conclude that ε = mω2x0. Then:

H =p2x + p2

y

2m + 12mω

2(x− ε

mω2

)2− 1

2mω2( ε

mω2

)2+ V (y)

Since ε2/2mω2 is a constant, this simply shifts the whole spectrum, but does not otherwise affect the problem. We canthen write the spectrum, where nx = 0, 1, 2... and ny = 1, 2, 3....

E = ~ω(nx + 1

2

)+n2yπ

2~2

2mL2 −ε2

2mω2

We can then also write the wavefunctions, where ξ follows the conventions in Liboff, and An is the SHO normalizationfrom Liboff table 7.1. H denotes a Hermite polynomial.

Let ξ2x = mω

~

(x− ε

mω2

)2→ ψ(x, y) =

√2LAnx

eξ2x/2Hnx

(ξx)

cos(nyπy

L

)ny odd

sin(nyπy

L

)ny even

The parity is determined by (−1)nx(−1)ny+1 = (−1)nx+ny+1.

For an arbitrary choice of m, ω and ε, the spectra in x and y will not necessarily line up and create degeneracies. Theycould however be tuned in such a way that particular states in each dimension end up combining to create degeneracies.

6.5 Problem 26In cartesian coordinates, the parity operator transforms x→ −x, y → −y, and z → −z.

(a) What does a parity transformation do in polar and spherical coordinates?(b) Find an H in spherical coordinates where [H,π] = 0.

Solution

(a) Parity works differently in an even number and odd number of dimensions. For polar coordinates, we let ϕ→ −ϕ,which mirrors the functions across the x-axis. (A mirroring through the origin in an even-numbered coordinate systemwould be equivalent to a single rotation, which is not useful).

For spherical coordinates we may mirror through the origin. This could naively be achieved by letting r → −r, but rcannot be negative. We can however achieve the same result by letting θ → π − θ and ϕ→ π + ϕ.

(b) Consider the following:

H = 12m

(p2r + p2

θ

r2 +p2ϕ

r2 sin2 θ

)+ V (r)

Since r2 = (−r)2 nothing happens to the kinetic energy under parity transformation. We leave V (r) only as a function ofr however, because r is the only coordinate that is unaffected under parity transformation.

24

7

7.1 Problem 27Consider an SHO where 〈H〉 = n~ω and n ≥ 2, in the state

|ψ〉 = A|n〉+B|n− 2〉

(a) Determine A and B.(b) Determine ∆H.(c) Show ∆x∆p ≥ ~/2 for all time t. (Hint: use parity arguments.)

Solution

(a) We use the fact that |A|2 + |B|2 = 1.

〈H〉 = |A|2~ω(n+ 1

2

)+ |B|2~ω

(n− 2 + 1

2

)= ~ω

(n+ 1

2 − 2|B|2)

= ~ωn

We then solve to obtain |B|2 = 14 and |A|2 = 3

4 . Therefore:

|ψ〉 =√

32 |n〉+ 1

2 |n− 2〉

(b)

⟨H2⟩ = 3

4

[~ω(n+ 1

2

)]2+ 1

4

[~ω(n− 3

2

)]2= (~ω)2

[34

(n2 +n+ 1

4

)+ 1

4

(n2 −3n+ 9

4

)]= (~ω)2

[n2 + 3

4

]We already know 〈H〉2 from the problem statement, so

∆H =(⟨H2⟩− 〈H〉2) 1

2 =√

32 ~ω

(c) Notice that |ψ〉 has either odd or even parity. In a coordinate basis, it is evident that x and p both have oddparity (x has odd parity, and taking the derivative of an odd or even function results in one of the opposite parity). Let|Ψ〉 = U |ψ〉 be the time dependent version of |ψ〉 where U is the time evolution operator. Then by parity arguments, thefollowing is evident because the overall expressions in these integrands are odd upon expansion over a coordinate.

〈x〉t = 〈Ψ |x|Ψ〉 = 0 〈p〉t = 〈Ψ |p|Ψ〉 = 0

Next we consider⟨x2⟩

t. Note that the commutator [a, a†] = 1 implies aa† + a†a = 2aa† − 1.

⟨x2⟩

t= ~

2mω

⟨Ψ∣∣∣a2 + a†

2 + aa† + a†a∣∣∣Ψ⟩ = ~

2mω 〈Ψ|[a2 + a†

2 + 2aa† − 1](√3

2 e−iEnt/~|n〉+ 12e−iEn−2t/~|n− 2〉

)= ~

2mω 〈Ψ|[√

32 e−iω(n+1/2)t

(√n(n− 1)|n− 2〉+ (2n+ 1)|n〉

)+ 1

2e−iω(n−3/2)t

(√n(n− 1)|n〉+ (2n− 3)|n− 2〉

)]= ~

2mω

[√3

2 e−iω(n+1/2)t

(√n(n− 1)

2 eiω(n−3/2)t + (2n+ 1)√

32 eiω(n+1/2)t

)+ ...

...+ 12e−iω(n−3/2)t

(√n(n− 1)

√3

2 eiω(n+1/2)t + 2n− 32 eiω(n−3/2)t

)]= ~

2mω

[√3

4√n(n− 1)

(e2iωt + e−2iωt)+ 3

4(2n+ 1) + 14(2n− 3)

]= ~

2mω

[√3n(n− 1)

2 cos(2ωt) + 2n]

= ∆x2

We perform a similar calculation to obtain⟨p2⟩

t. We shorten it greatly by noticing some similarities with the above.

x2 = ~2mω (a2 + a†

2 + a†a+ aa†) = ~2mω (a2 + a†

2 + 2aa† − 1) → p2 = mω~2 (2a†a− 1− a2 − a†2)

25

We see that we can simply change the sign on the last two terms and the calculation is otherwise identical. Therefore:

∆p2 = mω~2

[−√

3n(n− 1)2 cos(2ωt) + 2n

]

We may now verify against the uncertainty principle (recall that n ≥ 2):

∆x∆p = ~2

[4n2 − 3n(n− 1)

4 cos2(2ωt)] 1

2

≥ ~2

√4n2 − 3n(n− 1)

4 >~2


V (x) =−V |x| < a0 else

with V > 0. Look for an even-parity state with vanishingly small (or zero) energy.(a) Determine the wavefunctions in each region.(b) Apply boundary conditions to find a condition on V and a. What values of V are possible?(c) Show that the result from part (b) agrees with the result derived in class for bound states of even parity in the

finite well, namely:

ξ2 + η2 = 2mV a2

~2 , η = ξ tan ξ (Recall ξ = ka, η = κa)

Solution

(a) Let region 1 be for x < −a, region 2 for −a < x < a, and region 3 for x > a. We note that the solutions will beeigenfunctions of the parity operator, since the potential is symmetric.

− ~2

2m∂2ψ1,3

∂x2 = 0 → ψ1(x) = Ax+BEV EN→ ψ3(x) = −Ax+B

− ~2

2m∂2ψ2

∂x2 − V ψ2(x) = 0 → ψ2(x) = C cos(kx) +D sin(kx) for k =

√2mV~

We note that except for A = B = 0 this state will not be normalizable, but since it is on the border of not being bound,we accept this. We must however still consider that this is still a well, and that having a divergent probability density atx → ±∞, which dominates over finding the particle near the box does not appear physical. We require that A = 0 onthese grounds. (Having a constant is also consistent with an approach where we take the limit as k → 0 from the solutionsof the bound case.) We apply the usual continuity conditions to these simplified wave-functions.

ψ1 = ψ3 = B , ψ2 = C cos(kx) → B = C cos(ka) , 0 = kC sin(ka) →(

1 − cos(ka)0 k sin(ka)

)(BC

)=(

00

)Having sin(ka) = 0 implies:

ka = nπ → V = 12m

(nπ~a

)2for n = 1, 2, 3...

We now compare with the known result for the even-parity bound states of the finite square well. For E = 0:

ξ2 = k2a2 = 2mV a2

~2 η2 = κ2a2 = 0

See the following page for the plot of this result. We see that for η = 0, we find the solution exactly along the ξ-axis,which fall precisely on the roots of tan(ξ), which is where ka = nπ. Therefore the result we have obtained above agreeswith the previous work we have done on this potential in class.

26


V (x) = V1[δ(x+ b) + δ(x− b)] +−V0 |x| < a

0 else

with b ≥ a > 0, and V1, V0 > 0. Look for a state with vanishingly small or zero energy.(a) Determine the wavefunctions in each region.(b) Apply boundary conditions to derive

cot ζ = ζ

(b

a− 1− ~2

2maV1

)for ζ ≡ a

~√

2mV0

(c) Comment on the solutions of V0 for the following cases (look a graphical solutions)i) b

a > 1 + ~2

2maV1, ii) b

a < 1 + ~2

2maV1, iii) b

a 1

Solution

(a) We number the regions in the problem from left to right in increasing order. We note that in the finite square well,only the even state solutions will have the possibility of appearing for any choice of V and a, and that the delta functionsadded here make it more difficult to create bound states. (Look at the roots in the graphical solution from class; theE = 0 condition once again sets us on the ξ axis, but the odd solution with cotangent instead of tangent does not have aroot until ξ = π/2, meaning that the well can be “too shallow” for an odd bound state, even without the delta functions).We will once again work only with the even solutions in solving this problem, since they offer a more general look at thesystem. Following the same reasoning applied in problem 7.2, we write the following as the wavefunctions in each region:

ψ1(x) = A ψ2(x) = Bx+ C ψ3(x) = D cos(kx) ψ4(x) = −Bx+ C ψ5(x) = A k =√

2mV~

(b) We set up the usual boundary conditions for the coefficients. Note that since we have established certain symmetriesin the problem that we only need “one side” of the problem, so to speak.

ψ3(a) = ψ4(a) → D cos(ka) = −Ba+ C

ψ′3(a) = ψ′4(a) → −Dk sin(ka) = −B

ψ4(b) = ψ5(b) → −Bb+ C = A

b+ε∫b−ε

dx

(− ~2

2md2ψ

dx2 + V1δ(x− b)ψ(x))

= 0 → 0 +B = 2mV1A

~2

27

These can be summarized in the following homogeneous system of equations:2mV1~2 −1 0 00 1 0 −k sin(ka)1 b −1 00 a −1 cos(ka)

ABCD

= 0

As usual, when confronted with this type of homogeneous solution and a question where we do not care to solve for A,B, C, or D, we note that the determinant must be zero for a solution to exist, and examine the conditions under whichthis is true. This requires:

2mV1

~2 [(b− a)k sin(ka)− cos(ka)]− k sin(ka) = 0 → 2mV1

~2 [(a− b)k + cot(ka)] + k = 0

V1 = ~2

2m−k

cot(ka) + k(a− b) → cot(ka) + ka

(1− b

a

)+ ~2ka

2maV1= cot(ka) + ka

(1− b

a+ ~2

2maV1

)= 0

Which gives us the desired result:

cot(ka) = ka

(b

a− 1− ~2

2maV1

)→ cot ζ = ζ

(b

a− 1− ~2

2maV1

)(c) We may look at cases i) and ii) together:

The upper linear function is for ba > 1 + ~2

2maV1, and the lower linear function is for b

a < 1 + ~2

2maV1. We see that the

solutions for case i) are shifted to the right relative to those for case ii). This makes sense because it means that a deeperwell can bind a particle despite the delta functions being closer to the well. For case iii) we observe that:

b

a 1 → sin(ka) = 0 → ka = nπ for n = 1, 2, 3...

which is the same as in previous problem, which meets any expectations we might have had. (This is equivalent to sayingthat for very large slopes in the lines on the graphical solution, we will get intersections very near the asymptotes of cot ζ.)

28

8 Review for Midterm

Important Formulas and ResultsFive systems / general types of problems you want to be familiar with: (1) Infinite Square Well, (2) Simple HarmonicOscillator, (3) Delta-Function Potential, (4) Unbound State Barrier Problems (Transmission and Reflection), (5) FiniteSquare Well.

Important Concepts to Apply in these problems: (1) Calculating and Interpreting Expectation Values, (2) TimeEvolution (including time-dependent expectation values), (3) Heisenberg Uncertainty Principle, (4) Generalizations tohigher dimensions and degeneracies that may arise, (5) Superpositions of States, Probability, and Probability Amplitudes,(6) Operator Algebras (including matrix manipulation, use/role of Hermitian operators, ladder operators, Commutators,CSCOs, etc.), (7) Appropriate use of graphical solution techniques.

These topics are summarized on the equation sheet that has also been posted. You should know how to use everyresult on that sheet and qualitatively be able to explain where it comes from. If you are able to actually derive everythingon that sheet, then you are in very good shape!

Suggestion: Study the homework first and foremost, followed by the workshop problems. Problems labeled “practiceproblems and other” below are those I find particularly representative of the kinds of topics that might be covered.

Workshop Problems Organized by TypeMath Practice / Notation 1.1, 1.2, 1.3, 2.1, 2.2, 2.3, 3.1, 3.3, 3.4, 6.5Theoretical Foundations 1.3, 2.4, 3.1, 3.3, 4.4, 6.1, 6.2, 6.3Practice Problems and Other 1.4, 2.5, 3.2, 3.4, 4.1, 4.2, 4.3, 5.1, 5.2, 5.3, 5.4, 6.4, 7.1, 7.2, 7.3

8.1 Short Exercises1. Plot any of the following functions that you cannot sketch without a reference. Notice their roots, behavior at 0 and±∞, parity, and any other salient features.sinh(x), cosh(x), tanh(x), e−|x|, Θ(x) (Heaviside), sgn(x), The first 4 states of the SHO.Understand the derivatives of Θ and sgn in terms of their plots.

2. Define bound and unbound states.3. Define a CSCO.4. Write the time evolution operator. Why is it generally necessary to know some state in terms of energy eigenstates

to use it?5. What is the action of the parity operator on a function? Which functions are eigenfunctions of P and with what

eigenvalues? Construct operators which project out the even or odd part of any function.

Solution

1. Make the plots yourself! Mathematica has Hermite polynomials built in if you need it, and the derivatives of interestare proportional to dirac delta functions. The scaling factor on the delta functions is proportional to the size of thediscontinuity in the functions.

2. Bound states are those for which the particle is confined to some region of space. Confinement leads to quantizedenergies, each of which corresponds to a bound state(s). A bound state is normalizable if E < V at x = ±∞, whichmeans that the particle has no amplitude to tunnel to ±∞. An unbound state is one for which V < E at x =∞ orx = −∞, where the particle is away from any kind of local minimum / unconfined (which is what you are workingwith in a barrier/scattering problem). An unbound state cannot exist in a potential which tends towards positiveinfinity at both edges (in this case all states are bound).

3. CSCO means complete set of commuting operators or complete set of commuting observables. It is a set of operatorswhich are Hermitian, commute with each other, and thereby have shared eigenstates. The set is “complete” if theeigenvalues of all the operators in the set are able to completely distinguish between all of the states. (An operatorwith degenerate eigenvalues cannot be a CSCO alone – it needs another operator(s) with the same eigenstates butdifferent eigenvalues, such that the set of eigenvalues totally characterizes the state.)Examples: 2D SHO: Hx and Hy form a CSCO, as do H and either Hx and Hy. None of these alone are able todistinguish between states however.

29

4. U = exp[−iHt/~]. In order to substitute in a number for H, the operator must act on a state corresponding to asingle energy, namely an energy eigenstate.

5. Pf(x) = f(−x) in one dimension. Eigenfunctions are either even (f(x) = f(−x); eigenvalue 1), or odd (f(x) =−f(−x); eigenvalue -1). See Liboff, problem 6.18 for the projection problem (p. 180). If you are uncomfortable withthe parity operator you are equally encouraged to do problem 6.16, and review workshop problem 6.5 for furtherdetails and higher-dimensional cases.

8.2 Problem 30A particle in one dimension has a first excited state eigenfunction associated with energy eigenvalue E1 given by

ψ1(x) = xψ0(x),

where ψ0 is the ground state wavefunction with energy eigenvalue E0. The potential vanishes at x = 0.(a) Apply the Schrodinger equation to ψ0 and ψ1, and solve for the wavefunction ψ0(x).(b) Determine the potential V (x) in which the particle moves, and a numerical answer for the ratio E1

E0. What type of

system do you infer the above describes?

Solution

Problem from Prof. Das’ book.(a)

H = − ~2

2m∂2

∂x2 + V (x) → Hψ0 = E0ψ0 , Hψ1 = E1ψ1 = xE1ψ0

We expand these out, and reduce the equation containing E1:

d2ψ0

dx2 −2m~2 (V (x)− E0)ψ0 = 0 (8.1)

d2ψ1

dx2 −2m~2 (V (x)− E1)ψ1 = 0 → x

d2ψ0

dx2 + 2dψ0

dx− 2mx

~2 (V (x)− E1)ψ0 = 0 (8.2)

We multiply through (8.1) by x, and then subtract this modified form of (8.1) from (8.2).

xd2ψ0

dx2 −2mx~2 (V (x)− E0)ψ0 = 0 → 2dψ0

dx− 2mx

~2 (E0 − E1)ψ0 = 0

This is a first order, separable ODE for ψ0, which we can proceed to solve.

∂ψ0

∂x= m

~2 (E0 − E1)xψ0 → lnψ0 = m

~2 (E0 − E1)x2

2 + C → ψ0 = A exp[− m

2~2x2(E0 − E1)

](b) Next, obtain the equation below by differentiating (8.2), which is of the same form as the original ground state

equation (8.1) once the result for the single derivative above is plugged in.

d2ψ0

dx2 −m

~2 (E0 − E1)(ψ0 + x

dψ0

dx

)= 0 → d2ψ0

dx2 −m

~2 (E0 − E1)ψ0 −m2

~4 x2(E0 − E1)2ψ0 = 0

We infer that

V (x)− E0 = 12

(E0 − E1 + m

~2 (E0 − E1)2x2)→ V (x) = 3

2E0 −12E1 + m

2~2 (E0 − E1)2x2

We then apply the boundary condition that V (0) = 0, and obtain:

32E0 = 1

2E1 → E1

E0= 3 → V (x) m2~2 (E0 − 3E0)2x2 = 2mE2

0~2 x2 = 1

2mω2x2 for suitable ω

We have a quadratic potential, which is a form of the SHO, and have found that E1/E0 = 32~ω/

12~ω = 3. This is

self-consistent, and consistent with all previous results derived in this class.

30

8.3 Problem 31Consider a particle subject to the following potential, with V0, V1 > 0.

V (x) = V0δ(x)− V1[δ(x− a) + δ(x+ a)]

Comment on the existence of:(a) an odd parity bound state.(b) an even parity bound state.(c) a bound state for a = 0. Of which parity?(d) a bound state for a→∞. Parity?

This is a qualitative question, but your arguments should be founded on an understanding of how to solve the systemmathematically, and an intuition built up from the homework regarding how you expect the solutions to go.

Solution

In order to gain extra practice, you may wish to demonstrate each of the following points with more mathematical rigor,especially if you do not find the following arguments convincing at first glance.

(a) Note that any continuous odd function f(x) must satisfy f(0) = 0. Therefore an odd solution would not “feel” therepulsive delta function at x = 0, and only sees the two attractive ones at x = ±a. We know that for even one attractivedelta we have a bound state, and we will for two as well.

(b) An even parity state will feel both the attractive and repulsive potentials, and so the overall ability of the potentialto bind will depend on the relative strengths of V1 and V0. Solving the E = 0 case would be the simplest way to determinethe conditions under which a bound state might be possible.

(c) Here the delta functions are all on top of each other. As long as 2V1 > V0 this will be an overall attractive deltaand single even-parity bound state will exist.

(d) Sending a → ∞ essentially makes the effects of the attractive parts of the potential negligible, leaving only therepulsive portion at x = 0. No bound state can exist under these conditions.

8.4 Problem 32A canonical coherent state is an eigenstate of the 1D SHO annihilation operator, which you examined in HW problem 4.1.Let |α〉 be a canonical coherent state with eigenvalue α such that a|α〉 = α|α〉. Part (a) was completed on your homeworkin a slightly different way, and you then proceeded to determine that the distribution of energies to create such a statefollows a Poisson distribution, and the typical uncertainties for such a distribution. We now examine the dynamics of thissystem.

(a) Redo this if you want the practice, or move forward using the homework results. Write |α〉 as a linear combination∑n Cn|n〉 of states |n〉 of the 1D SHO. Find a recursion relation to determine the coefficients Cn in terms of C0, then

determine C0 by normalizing the state. Compare with results from the homework.(b) Let |ψα〉 be the time-dependent form of the state |α〉, obtained by using the time evolution operator. Write and

simplify |ψα〉 using the results from the previous parts.(c) Calculate the time-dependent expectation values 〈x〉t = 〈ψα |x|ψα〉 and 〈p〉t = 〈ψα |p|ψα〉. (Do not do the whole

calculation twice. Once you obtain one, you should be able to obtain the other with relatively little work.)(d) Find the wavefunction 〈x|α〉.(e) What do you notice about the results from parts (c) and (d) / How does the behavior of this system compare with

that of a classical SHO? How does this system move? How is the Heisenberg uncertainty principle satisfied (a qualitativeanswer based on a plot is sufficient)?

Solution

(a) We begin by expanding |α〉 in terms of the energy eigenstates of the SHO.

|α〉 =∑n

|n〉〈n|α〉 =∑n

Cn|n〉 → Determine Coefficients Cn = 〈n|α〉

31

We use the property that a|α〉 = α|α〉 to proceed.

a|α〉 =∑n

Cna|n〉 =∞∑n=1

Cn√n|n− 1〉 =

∞∑n=0

Cn+1√n+ 1|n〉

= α|α〉 = α∑n

Cn|n〉

We obtain the recursion relation αCn =√n+ 1Cn+1. If we begin with C0 and apply this repeatedly, we see that

Cn = αn√n!C0. Therefore:

|α〉 =∞∑n=0

αn√n!C0|n〉

We determine C0 by normalization.

〈α|α〉 = 1 =(∑

m

〈m| (α∗)m√m!

C∗0

)(∑n

αn√n!C0|n〉

)=∑n,m

|C0|2(α∗)mαn√

m!n!δnm = |C0|2

∑n

|α|2n

n!

→ |C0|2 =(∑

n

|α|2n

n!

)−1

= e−|α|2

(b) The evolution operator is U = e−iHt/~. We want to calculate |ψα〉 = U |α〉.

|ψα〉 = e−iHt/~∑n

αn√n!C0|n〉 = C0

∑n

αn√n!e−iEnt/~|n〉 = C0

∞∑n=0

αn√n!e−iωt(n+1/2)|n〉

(c) We calculate 〈x〉t first.

〈ψα |x|ψα〉 = |C0|2∞∑

m,n=0

(α∗)mαn√n!m!

eiωt(m−n) 〈m |x|n〉

The term 〈m |x|n〉 is of interest.

x =√

~2Mω

(a+ a†) → 〈m |x|n〉 =√

~2Mω

(〈m |a|n〉+⟨m∣∣a†∣∣n⟩) =

√~

2Mω(√nδm,n−1 +

√n+ 1δm,n+1)

We plug this result into the first expression above and simplify:

〈x〉t = |C0|2∞∑

m,n=0

(α∗)mαn√n!m!

eiωt(m−n)√

~2Mω

(√nδm,n−1 +

√n+ 1δm,n+1)

= |C0|2√

~2Mω

( ∞∑n=1

(α∗)n−1αn√n!(n− 1)!

eiωt((n−1)−n)√n+∞∑n=0

(α∗)n+1αn√n!(n+ 1)!

eiωt((n+1)−n)√n+ 1)

= |C0|2√

~2Mω

( ∞∑n=1

α|α|2(n−1)

(n− 1)! e−iωt +∞∑n=0

α∗|α|2n

n! eiωt

)

= |C0|2√

~2Mω

∞∑n=0

|α|2n

n!(αe−iωt + α∗eiωt

)=√

~2Mω

(αe−iωt + α∗eiωt

)We then note that 〈p〉t = M∂t 〈x〉t, such that we may write the following:

〈p〉t = i

√~Mω

2 (α∗eiωt − αe−iωt)

(d) We write the operator a in a coordinate representation, and then solve the resulting differential equation from therequirement that 〈x |a|α〉 = α 〈x|α〉. I’ll start writing the mass lower case again since there is no longer an index m toconfuse it with.

a =√mω

2~

(x+ ~

mω

∂

∂x

)→

√mω

2~

(x+ ~

mω

∂

∂x

)α(x) = α · α(x) →

[x

√mω

2~ − α+√

~2mω

d

dx

]α(x) = 0

32

dα

dx=√

2mω~

(α− x

√mω

2~

)α(x) =

(√2mω~

α− mω

~x

)α(x)

We propose a slightly modified Gaussian of the form f(x) = e−bx2+cx since ∂xf(x) = (−2bx+ c)f(x), which is of the same

form as the above.

b = mω

2~ , c = α

√2mω~

→ 〈x|α〉 = α(x) = A exp[−mω2~ x

2 + xα

√2mω~

]

This is a shifted Gaussian. We can make this more apparent and easier to work with by completing the square in theexponential:

〈x|α〉 = A exp[−mω2~ x

2 + xα

√2~mω

]→ −mω2~

(x2 − xα 2~

mω

√2mω~

)→ x2 − 2α

√2~mω

x

x2 − 2α√

2~mω

x = x2 − 2α√

2~mω

x+ 2α2~mω

− 2α2~mω

=(x− α

√2~mω

)2

− 2α2~mω

We absorb the extra constant into the normalization to obtain a function of the form:

〈x|α〉 = B exp

−mω2~

(x− α

√2~mω

)2

(e) We may rearrange the result in part (c) into real and imaginary parts to write it purely in terms of sinusoidalfunctions. In analogy with a classical system, the “spring” begins at its equilibrium position for a purely imaginary α,and begins at the end of its range of motion for a purely real α. We can thereby see that the real and imaginary partsof α determine the initial conditions of the system at t = 0, and that the expectation values of position and momentumthen oscillate as in a classical oscillator. This is however still a quantum system, and we can qualitatively see how theHeisenberg uncertainty principle is satisfied here; instead of two known values of position and momentum oscillatingin time (which we could represent with oscillating Dirac delta functions), we have two distributions (Gaussian) whosecentroids are oscillating as shown above. The width of these distributions tells us about the uncertainty in the positionand momentum of the particle at any given time.

Specifically, for the case where α is real:

〈x〉t =√

~2mωα(e−iωt + eiωt) = α

√2~mω

cos(ωt)

At t = 0 the expectation value of position is at is extreme (the “spring” is at its “most stretched”), which correspondsexactly to the shift observed in the peak of the Gaussian at t = 0 found in part (d). You can satisfy yourself that thiswill still work out as described above for complex α. We can infer all of the above with reasonable confidence withoutfinding the fully time dependent 〈x|ψα〉. All of this can be verified analytically with the time dependence along for theride, or you can evaluate the partial sums in terms of the SHO eigenstates with mathematica to confirm. (Use the Poissonuncertainties in H derived on the homework to choose where to truncate a sum for a given α!)

33

9

9.1 Problem 33Consider the first two energy levels of the 2D SHO.

(a) Write out the unnormalized eigenfunctions in polar coordinates.(b) Write out the eigenfunctions of Lz. What are the possible values of m for each level?

Solution

(a) We use notation such that ψnx,ny (r, ϕ) = 〈r, ϕ|nx, ny〉.

ψ00 ∝ e−αr2

ψ10 ∝ r cosϕe−αr2

ψ01 ∝ r sinϕe−αr2

(b) Recall the following about the angular momentum operator Lz in a coordinate basis:

Lz → −i~∂

∂ϕEigenfunctions: eimϕ with eigenvalues m

We can write the following:ψ00 ∝ e−αr

2→ m = 0

ψ10 + iψ01 ∝ re−αr2eiϕ → m = 1

ψ10 − iψ01 ∝ re−αr2eiϕ → m = −1

Notice that the states with m = ±1 have the same energy. You may wish to re-examine homework problems 3.3 and 4.2in order to put these observations in their proper context. The following suggests that the states we listed as 〈r, ϕ|nx, ny〉might be better expressed with a different set of eigenvalues appropriate for a polar representation rather than a sphericalone. Consider instead 〈r, ϕ|nr,m〉. This suggests a second CSCO composed of radial and angular operators.

9.2 Problem 34In Einstein notation A ·B = AiBi, where the sum over i is implicit. (You may find it convenient to denote a derivativein a particular coordinate by ∂i, where i can be x, y, or z.) The cross product in Einstein notation is written using theLevi-Cevita tensor. In particular if we have C = A × B, then that can be written Ci = εijkAjBk, where the index i isfixed since it appears on both sides of the equation, and there is an implicit sum over j and k. Recall that the Levi-Cevitatensor is given by

εijk = +1 ijk are cyclically permuted−1 ijk are anti-cyclically permuted0 i = j or i = k or j = k

(a) Write out ∇ ·C in Einstein notation.(b) Prove the following vector identity using Einstein notation:

∇ · (A×B) = (∇×A) ·B−A · (∇×B)

(c) Expand ∇ × [φ(r×A)] in Einstein notation, and then rewrite your result in vector notation. You may find theidentity εijkεklm = δilδjm − δimδjl useful.

(d) Using L = r× p, show that [Li, rj ] = i~εijkrk and [Li, pj ] = i~εijkpk.

Solution

(a) In general, the divergence of some vector quantity can be notated:

∇ ·C = ∂Ci∂xi

= ∂iCi

For C = A×B in particular, we may write:

∇ · (A×B) = ∂iεijkAjBk = εijk∂iAjBk = εijk(Aj∂iBk +Bk∂iAj)...

34

(b) which is continued below:

∇ · (A×B) = ∂iεijkAjBk = εijk∂iAjBk = εijk(Bk∂iAj +Aj∂iBk) = εkijBk∂iAj − εjikAj∂iBk= (∇×A) ·B−A · (∇×B)

(c)

∇ × [φ(r×A)]i = εijk∂j [φ(r×A)]k = εijk∂j [εklmφrlAm] = εijkεklm[φrl∂jAm + φAm∂jrl + rlAm∂jφ]= (δilδjm − δimδjl)[φrl∂jAm + φAmδjl + rlAm∂jφ]= φri∂jAj − φrj∂jAi + φAjδij − φAiδjj + riAj∂jφ− rjAi∂jφ

We note that since there is an implicit sum over j, δjj is effectively the trace of a 3 × 3 identity matrix (since we are inthree dimensions), and consequently δjj = 3. We convert the following expansion back into vector notation:

∇× [φ(r×A)] = φ[r(∇ ·A)− (r · ∇)A− 2A] + r(A · ∇φ)−A(r · ∇φ)

(d) Note that Li = εijkrjpk. Then we may do the following:

[Li, rn] = εijk[rjpk, rn] = −i~εijkrjδkn = i~εinjrj

[Li, pn] = εijk[rjpk, pn] = i~εinkpk

9.3 Problem 35Review some of the relationships between angular momentum operators discussed in class in parts (a)-(c).

(a) Using [Li, Lj ] = i~εijkLk, compute [L2, Li] where L2 = L2x + L2

y + L2z. What does this imply?

(b) Using Lz|l,m〉 = m~|l,m〉, evaluate [L±, Lz]|l,m〉 where L± = Lx± iLy. What happens when L± acts on a state?(c) Using L2|l,m〉 = l(l + 1)~2|l,m〉, show L±|l,m〉 = ~

√l(l + 1)−m(m± 1)|l,m± 1〉.

(d) Construct the matrix of Lx for l = 1. What are the possible values of Lx? Write out the eigenfunctions of Lx inthe Lz basis.

(e) Write out the functions 〈ϕ, θ|1,−1〉, 〈ϕ, θ|1, 0〉, and 〈ϕ, θ|1, 1〉, where the kets are of the form |l,m〉.

Solution

(a)[L2, Li] = [L2

j , Li] = i~εjikLjLk + i~εjikLkLj but εijkLjLk = −εijkLkLj → [L2, Li] = 0

We conclude that L2 and Li share some eigenbasis. Note that since the Li for different i do not commute, that theeigenbasis will have to be shared between L2 and only one of the three Li. We continue by defining eigenkets |l,m〉, whichare simultaneous eigenkets of both L2 and Lz.

(b) [L±, Lz] = ∓~L± using the above. But then it follows that:

∓~L±|l,m〉 = (L±Lz − LzL±)|l,m〉 → Lz(L±|l,m〉) = ~(m± 1)(L±|l,m〉)

We conclude that L± are ladder operators, which move us between different projections on a given state. In other words,L±|l,m〉 = C±|l,m± 1〉.

(c) Notice that since Li = L†i ∀ i we may say that L+ = L†− and vice versa. We now find the proportionality constantsC± defined in the line above.

〈l,m |L−L+| l,m〉 = |C+|2 = 〈l,m |(Lx − iLy)(Lx + iLy)| l,m〉 =⟨l,m

∣∣L2 − L2z − ~Lz

∣∣ l,m⟩where we have used (Lx − iLy)(Lx + iLy) = L2

x + L2y + iLxLy − iLyLx = L2 − L2

z + i[Lx, Ly]. Then:

|C+|2 = ~2[l(l + 1)−m2 −m] = ~2[l(l + 1)−m(m+ 1)]

We repeat this with L+ and L− reversed in order to determine |C−|2:

〈l,m |L+L−| l,m〉 = |C−|2 = ~2[l(l + 1)−m(m− 1)]

35

These results can then be summarized by:

L±|l,m〉 = ~√l(l + 1)−m(m± 1)|l,m± 1〉 = ~

√(l ∓m)(l ±m+ 1)|l,m± 1〉

(d) Notice that the definitions of L± can be added or subtracted to obtain:

Lx = 12(L− + L+) Ly = i

2(L− − L+)

We can construct matrices for L+ and L− with l = 1 according to the following: 〈11 |L±| 11〉〈11 |L±| 10〉〈11 |L±| 1− 1〉〈10 |L±| 11〉〈10 |L±| 10〉〈10 |L±| 1− 1〉〈1− 1 |L±| 11〉〈1− 1 |L±| 10〉〈1− 1 |L±| 1− 1〉

This allows us to determine:

L+ = ~√

2

0 1 00 0 10 0 0

L− = ~√

2

0 0 01 0 00 1 0

→ Lx = L+ + L−2 = ~√

2

0 1 01 0 10 1 0

We then find the eigenvalues and eigenvectors in the usual way. We find that the eigenvalues are 0,±~, and that thecorresponding eigenvectors are:

0→ 1√2

10−1

~→ 12

1√2

1

− ~→ 12

1−√

21

(e) Functions 〈ϕ, θ|l,m〉 (the coordinate representations of the angular momentum eigenstates) are given by the

spherical harmonics Y ml (ϕ, θ). Many can be found in Liboff, table 9.1, including the following:

Y 11 = −1

2

√3

2π sin θeiϕ Y 01 = 1

2

√3π

cos θ Y −11 = 1

2

√3

2π sin θe−iϕ

Note that this implies that the eigenvectors found in part (d) can also be written (See Liboff 9.100):1√2

(Y 11 − Y −1

1 ) 12(Y 1

1 +√

2Y 01 + Y −1

1 ) 12(Y 1

1 −√

2Y 01 + Y −1

1 )

9.4 Problem 36A system of two spin-1/2 particles is in the state

|ψ〉 = 1√5|+ +〉+ 2√

5|+−〉.

What are the possible values of J2 and Jz? With what probabilities?

Solution

First we need the CG coefficients for this system. These have been worked out in class, and we list them again below. (Workthese out by starting at the top or bottom of the ladder, raising and lowering, and using orthonormality considerations ifyou do not remember how to find these.) We are considering the system 1

2 ⊗12 = 1⊕ 0.

|11〉 = |+ +〉 |10〉 = 1√2

(|+−〉+ | −+〉) |00〉 = 1√2

(|+−〉 − | −+〉) |1− 1〉 = | − −〉

We need to make this change of basis because the states in the coupled basis above are the eigenstates of J2 and Jz. Thepossible values of J2 are 2~2 (j = 1) and 0~2 (j = 0). The possible values of Jz are ~ and 0~.

|ψ〉 = 1√5|+ +〉+ 2√

5|+−〉 = 1√

5|11〉+ 2√

51√2

(|10〉+ |00〉) =√

15 |11〉+

√25 |10〉+

√25 |00〉

We can read off the probabilities from the state in this form:

J2 =

2~2

0 with probabilities 3/52/5 Jz =

~0 with probabilities 1/5

4/5

36

10

10.1 Problem 37In workshop problem 9.4 you were asked to determine the possible values of J2 and Jz, and their corresponding probabil-ities, given the state below for a system of two spin- 1

2 particles.

|ψ〉 = 1√5|+ +〉+ 2√

5|+−〉.

Repeat the calculation for two spin-1 particles, letting + and − denote projections of m = ±1 instead of m = ± 12 . You

should begin by finding the appropriate Clebsch-Gordan coefficients.

Solution

We begin by finding the CG coefficients. The uncoupled representation will be notated with |±,±〉 as above, while thecoupled with |J,mJ〉. The uncoupled states will be notated with + (m = +1), · (m = 0), and − (m = −1). We start atthe top of the ladder and lower:

|22〉 = |+ +〉 → J−|22〉 = (j1− + j2−)|+ +〉√

2 · 3− 2 · 1|21〉 =√

1 · 2− 1 · 0| ·+〉+√

1 · 2− 1 · 0|+ ·〉 → |21〉 =√

12 |+ ·〉+

√12 | ·+〉

We repeat this continuing down the ladder with each subsequent result to obtain the rest of the |2,m2〉 states. I’ll omitthe algebra below.

J−|21〉 = (j1− + j2−)(√

12 |+ ·〉+

√12 | ·+〉

)→ |20〉 =

√16 |+−〉+

√23 | · ·〉+

√16 | −+〉

→ |2− 1〉 =√

12 | · −〉+

√12 | − ·〉 → |2− 2〉 = | − −〉

We determine |11〉 by considering orthnormality conditions. Specifically, |11〉 shares uncoupled states with |21〉, and wemake an orthogonal |11〉:

|11〉 =√

12 |+ ·〉 −

√12 | ·+〉

We apply the same lowering process as before to obtain the next two states down the ladder with J = 1:

→ |10〉 =√

12 |+−〉 −

√12 | −+〉 → |1− 1〉 =

√12 | · −〉 −

√12 | − ·〉

Finally, we note that our last state |00〉 must be orthogonal to both |20〉 and |10〉, which it will share uncoupled stateswith. The following satisfies this requirement:

|00〉 =√

13 |+−〉+

√13 | · ·〉+

√13 | −+〉

Recall that the CG coefficients are of the form 〈mj1 ,mj2 |J,mJ〉, and that since they are real, 〈mj1 ,mj2 |J,mJ〉 =〈J,mJ |mj1 ,mj2〉. We can expand them either way. We have two states of interest due to |ψ〉, namely | + +〉 and| + −〉, which we would like to rewrite in the coupled basis. Clearly | + +〉 = |22〉. We consider | + −〉. There are threecoupled states which could contribute to it:

|+−〉 =∑J,mJ

|JmJ〉〈JmJ |+−〉 → |+−〉 = |20〉〈20|+−〉+ |10〉〈10|+−〉+ |00〉〈00|+−〉

We are simply expanding the ket in the coupled basis over the uncoupled basis (both are complete orthonormal sets).

|ψ〉 =√

15 |22〉+

√25

(√16 |20〉+

√12 |10〉+

√13 |00〉

)=√

15 |22〉+

√215 |20〉+

√25 |10〉+

√415 |00〉

With the state rewritten in this way, we can now simply read off the answers.

J2 =

6~2

2~2

0with probabilities

1/32/54/15

Jz =

2~0 with probabilities 1/5

4/5

37

10.2 Problem 38The Runge-Lenz vector is given by:

K = 12me2 (L× p− p× L) + r

r

(a) Show that this is equivalent toK = 1

me2 (L× p− i~p) + rr

(b) To compute [H,K] you will find it necessary to compute [pi, 1/r]. Compute this commutator.(c) Does the result from part (a) hold if motion is constrained to the xy plane (i.e. in two dimensions)? Why?

Solution

(a) We expand one of the cross products being careful about order of operations.

p× Li = εijkpjLk = εijk([pj , Lk] + Lkpj) = εijkεjkmi~pm − (L× p)i= (δkk − 1)δimi~pm − (L× p)i = 2i~pi − (L× p)i

Then we may rearrange the L× p− p× L term in K to read:

L× p− p× L = 2L× p− 2i~p → K = 1me2 (L× p− i~p) + r

r

(b) We will calculate the commutator with the aid of a trial function to make it extra-clear:

[pi, 1/r]ψ = pi1rψ − 1

rpiψ = −i~

(ψ∂

∂rir−1 + 1

r

∂ψ

∂ri− 1r

∂ψ

∂ri

)= −i~

(− 1r2∂r

∂ri

)ψ = i~ri

r3 ψ

This can be summarized: [pi,

1r

]= i~ri

r3 ↔[p, 1r

]= i~r

r3

(c) The result from part (a) does not hold in two dimensions, where L = Lz and p and r only have two components.The clearest mathematical reason for this is that δkk evaluates to 2 in 2-dimensions rather than 3, which changes theresult somewhat. You are free to evaluate this out the long way with suitable simplifications however if you wish. Thechange in evaluating the delta results in:

K = 12me2 (2L× p− i~p) + r

r

for two-dimensions instead of the result derived above for 3. (In general you would clearly like to use the 3D result. Thispart of the problem is here simply to draw your attention to the fact that there is a difference, and that you can spot itif you are being careful about your use of Einstein notation.)

10.3 Problem 39Consider the system of three particles each with S2 = 2~2. The Hamiltonian is shown below, where λ is a constant.

H = λ∑i 6=j

Si · Sj

(a) Write out all the states of the system in the coupled representation. What are the possible values of S2total and

the degeneracies of each?(b) Write H in a form which is diagonal (i.e. change of basis). Determine the eigenvalues of H and the degeneracies

of each. Compute the average eigenvalue. Why should we expect this result? (Hint: compute the trace of H.)(c) Compute the average eigenvalue for a general sum on i and j (i can equal j). Does the result make sense?

38

Solution

(a) We have a system of 3 spin-1 particles. The coupling of the Hilbert spaces can be denoted by:

1⊗ 1⊗ 1 = (2⊕ 1⊕ 0)⊗ 1 = 3⊕ 2⊕ 1⊕ 2⊕ 1⊕ 0⊕ 1

The first expansion above denotes the coupling possibilities for the first two particles. There are 33 = 27 states, and wedistinguish states with |STotal, S1 + S2, Sz〉. We list them out clarify the above:

State Range of m No. of States|3, 2,m〉 −3→ 3 7|2, 2,m〉 −2→ 2 5|1, 2,m〉 −1→ 1 3|2, 1,m〉 −2→ 2 5|1, 1,m〉 −1→ 1 3|0, 1,m〉 0 1|1, 0,m〉 −1→ 1 3

Total 27

This implies that S2 = (12, 6, 2, 6, 2, 0, 2)~2 are (7, 5, 3, 5, 3, 1, 3)-fold degenerate.(b) We first write out H explicitly.

H = λ(S1 · S2 + S2 · S1 + S2 · S3 + S3 · S2 + S1 · S3 + S3 · S1)

We then note thatS2T = (S1 + S2 + S3)2 = S2

1 + S22 + S2

3 + 2S1 · S2 + 2S2 · S3 + 2S1 · S3

Since Si and Sj correspond to different particles, [Si,Sj] = 0 for i 6= j. (They exist in different Hilbert spaces.) Then Hcan be rewritten as follows:

H = λ(S2T − S2

1 − S22 − S2

3) = λ(S2T − 6~2) = λ(6, 0,−4, 0,−4,−6,−4)~2 are (7, 5, 3, 5, 3, 1, 3)-fold denegerate.

This indicates that average eigenvalue is 0, since we may evaluate the trace of the Hamiltonian in the basis in which it isdiagonal in order to obtain:

Tr(H) = ~2λ(6 · 7 + 5 · 0− 4 · 3 + 5 · 0− 4 · 3− 6− 4 · 3) = 0

Since the trace is invariant under unitary transformation, we may take the trace in the energy eigenbasis and expect itto be the same regardless of the basis in which we might be expressing the matrix. We expect the overall trace of theHamiltonian to be zero because the trace of any angular momentum is zero (we are summing over all possible projectionsm = −s → s, so then Tr(Sik) = 0). A remark on the notation here: I am using the first subscript to denote which spinthis is (i.e. i and j below are 1,2, or 3), and the second subscript to denote the coordinate of that particular spin (i.e. kbelow is x, y, or z). So any single case would read like S1x. The trace in this scheme can be separated by the first index(different Hilbert spaces), but not the coordinate.

Tr(H) = λ∑i 6=j

Tr(SikSjk) = λ∑i 6=j

Tr(Sik)Tr(Sjk) = 0

(We may separate the traces in the last step because each term corresponds to a different particle.) Since the originalHamiltonian must have trace zero, it would a problem if the average eigenvalue evaluated with this trace method (i.e.eigenvalue weighted by degeneracy) did not give zero. This is a good check of the solution.

(c) Let H0 be the original Hamiltonian.

H = λ(S21 + S2

2 + S23) +H0 → Tr(H) = λTr(S2

1 + S22 + S2

3) + Tr(H0) = λTr(6~2) + 0 = 27(6λ~2)

The average is then 6λ~2.

10.4 Problem 40(a) Find the eigenvalues of the spin operator S of an electron in the direction n which lies in the xz plane. (You should

be able to say what you expect for this before doing the problem. Do you find what you should?)(b) Given a particular projection measured at some angle w.r.t. z, comment on how you would obtain the probability

of obtaining a particular measurement along z itself. (I am not asking you to do the math in workshop. It is good practiceto actually compute the probabilities if you wish do so later however.)

39

Solution

(a) Consider S · n = Sx sin θ + Sz cos θ. We use the Pauli matrices for Sx and Sz.

S · n = ~2

(cos θ sin θsin θ − cos θ

)We determine that the eigenvalues of this matrix are ±~

2 in the usual way. (If we remove the factor ~2 in front, the

eigenvalues of the matrix itself are ±1, which we use below to simplify calculation of the eigenvectors). This is whatwe would expect, because in principle what we called the z axis could have been anywhere. If the possibilities for themeasurement were to change based on the direction of an arbitrarily defined set of coordinates, we would have a problem.

(b) Recall from workshop problem 5.4 that the eigenvectors of Sz are the most convenient eigenvectors, since wegenerally work in the Sz basis where it is a diagonal matrix. It is therefore simple to decompose any eigenvector intoeigenvectors of z, and then read off the probabilities. To solve the problem, we would find the eigenvectors associatedwith the eigenvalues found in part (a), and take the mod-square of the appropriate components.

40

11

11.1 Problem 41Consider a particle with charge q and mass m in an excited state of the two-dimensional harmonic oscillator. A photon isemitted when the particle de-excites. The spontaneous emission rate (up to first order) is given by the following expression:

R = 4ω3q2

3~c3 | 〈ψf |r|ψi〉 |2

ψi and ψf are the initial and final states, respectively, and ω is the frequency of the emitted photon.(a) Calculate the transition rate from the first excited state to the ground state.(b) Calculate the lifetime of the first excited state and the power radiated.

Solution

(a)

R = 4ω3q2

3~c3 | 〈ψf |r|ψi〉 |2 → 〈ψf |r|ψi〉 = 〈00 |r| 10〉 =

⟨00

∣∣∣∣∣√

~2mω′

(ax + a†xay + a†y

)∣∣∣∣∣ 10⟩

=√

~2mω′

(10

)Notice that we happen to have put the excitation in x above, and that of the four ladder operators shown, only the termwith ax is giving us an answer. We could equally well have put the excitation in y however, and obtained the same resultexcept for the vector, which will become irrelevant when we evaluate the mod-square anyway.

Since E1 − E0

~= ω → ω = ω′

We evaluate the above to obtain:R = 2ω2q2

3mc3

(b) The lifetime of a state is just the inverse of the transition rate, i.e.:

τ = 1R

= 3mc3

2ω2q2

Furthermore, the power emitted is given by the energy emitted times the rate of emission. The energy emitted in thistransition is E = ~ω. Therefore:

P = 2~ω3q2

3mc3

11.2 Problem 42The rotation operator U , which rotates an operator about the n axis through an angle ϕ is given by exp(−iJ · nϕ/~)where J is the angular momentum vector.

(a) For the case of spin-1/2, show

exp(− i2(σ · n)ϕ

)= cos

(ϕ2

)− i(σ · n) sin

(ϕ2

)(b) Determine a U that rotates Sx into −Sy and show that it works via the operation USxU† for spin-1/2.

Solution

(a) We note that (σ · n)2 = I (identity matrix). Expand the exponential in series form about ϕ0:

exp(− i2(σ · n)ϕ

)=∞∑n=0

1n!

(− i2(σ · n)ϕ

)n=

∞∑n=0,EV EN

(σ · n)n

n!

(− iϕ2

)n+

∞∑n=1,ODD

(σ · n)n

n!

(− iϕ2

)n

=∞∑

n=0,EV EN

1n!

(− iϕ2

)n+ (σ · n)

∞∑n=1,ODD

1n!

(− iϕ2

)n= cos

(ϕ2

)− i(σ · n) sin

(ϕ2

)

41

(b) We apply the above and ascertain that it in fact does what it is supposed to do. If we rotate by π2 about the z

axis that should move us from x to y. Thus we get from Sx to −Sy by trying the proposed scheme for n = −z and π2 = ϕ.

USxU† = ~

2

[cos(π

4

)+ iσz sin

(π4

)]σx

[cos(π

4

)− iσz sin

(π4

)]= ~

2(2)

[(1 + i 0

0 1− i

)(0 11 0

)(1− i 0

0 1 + i

)]= −~

2

(0 −ii 0

)= −Sy

11.3 Problem 43Use rotation operators and the properties of trace to prove the following:

(a) Tr(SiSj) = 0 for i 6= j where i and j are spatial directions.(b) Tr(S2

i ) = Tr(S2j ) (no sum, i and j are fixed). Combine this result with that of part (a) to find Tr(SiSj).

(c) Tr(SiSjSk) = 0 for any two indices alike.(d) Tr(SiSjSk) = −Tr(SjSiSk) for no two indices alike. Combine this result with that of part (c) to find Tr(SiSjSk).

Solution

(a) Let U rotate about j sending i→ −i.

Tr(SiSj) = Tr(USiU†USjU†) → Tr(−SiSj) = −Tr(SiSj) ∴ Tr(SiSj) = 0 for i 6= j

(b) Let U rotate about j sending i→ j.

Tr(SiSi) = Tr(USiU†USiU†) → Tr(S2i ) = Tr(S2

j )

We conclude that Tr(SiSj) = aδij for some number a. We find a as follows:

Tr(S2i ) = Tr(S2)

3 = ~2s(s+ 1)(2s+ 1)3 → Tr(SiSj) = ~2s(s+ 1)(2s+ 1)

3 δij

(c) Let us first consider Tr(SiSiSk), with a U which rotates about i sending k → −k.

Tr(SiSiSk) = Tr(USiU†USiU†USkU†) → Tr(SiSiSk) = −Tr(SiSiSk) = 0

Then let us consider Tr(SiSiSi) and a U which rotates about j 6= i sending i→ −i.

Tr(SiSiSi) = Tr(USiU†USiU†USiU†) → Tr(SiSiSi) = −Tr(SiSiSi) = 0

(d) Let U rotate about k sending i→ j and j → −i.

Tr(SiSjSk) = Tr(USiU†USjU†USkU†) → Tr(SiSjSk) = −Tr(SjSiSk)

We may therefore conclude that Tr(SiSjSk) = bεijk for some number b. We find b as follows:

Tr([Si, Sj ] + SjSi)Sk = Tr(i~εijlSlSk + SjSiSk) = i~εijlTr(SlSk) + Tr(SjSiSk)

→ 2Tr(SiSjSk) = i~3s(s+ 1)(2s+ 1)3 εijk → Tr(SiSjSk) = i~3

6 s(s+ 1)(2s+ 1)εijk

11.4 Problem 44(a) Determine the degeneracy of a general bound state of the hydrogen atom with principal quantum number n.

Include spin of an electron in the derivation. Use the summationm∑i=0

i = m(m+ 1)2

(b) Explain qualitatively the effects one might take into consideration to break this degeneracy, and the give the orderof magnitudes of these splittings relative to the energies themselves. (I am asking you to summarize things which havebeen covered in greater detail in class.)

42

Solution

(a) Per level n and orbital ` there is a degeneracy 2`+ 1. We sum over the possible orbitals for a given n:

n−1∑`=0

2`+ 1 = 2n−1∑`=0

`+ n = (n− 1)n+ n = n2 − n+ n = n2

This does not account for the spin of the electron yet. Since the spin is either up or down, the degeneracy including spinis 2n2.

(b) For further details concerning a derivation of these points, see your class notes and/or Liboff p. 589.The so called “fine structure” correction for Hydrogen breaks the degeneracy. This correction includes spin-orbit

coupling (interactions between the spin of an electron and its orbital angular momentum), and approximate relativisticeffects. These each change the energy levels by approximately one part in 105 relative to the usual spectrum where theenergy only depends on n.

There are also “hyperfine” corrections one can undertake to the spectrum, which are corrections several orders ofmagnitude smaller than those in the fine structure. These hyperfine corrections account for interactions due to the electricand magnetic fields of the atomic nucleus.

43

12

12.1 Problem 45This problem concerns the derivation and applications of the quantum-mechanical Virial theorem.

(a) Show that for any time-independent operator a the relationship ∂t 〈a〉 = 0 holds, where the expectation value istaken between eigenstates of the Hamiltonian.

(b) Consider the Hamiltonian:

H = p2

2m + kxn = T + V

Show that for the above system 2 〈T 〉 = n 〈V 〉. (Hint: Choose a = xp. This is the Virial Theorem.)(c) We extend the above into three dimensions. First, show the following. (What choice for the operator A makes

sense in analogy with the 1D choice for a above?)

[H,A] = i~(

r · ∇V− p2

m

)(d) Then compute 〈ψ |[H,A]|ψ〉 to show that 〈r · ∇V〉 = 2〈T 〉, which is a statement of the Virial theorem in 3

dimensions.(e) Use the Virial theorem to show 〈V 〉 = −2 〈T 〉 for the Coulombic potential.(f) Use the Virial theorem to show 〈V 〉 = 〈T 〉 for the SHO.

Solution

Note that we are sloppy about notating a with its hat below. a and a denote the same operator here.(a) We have H|ψ〉 = E|ψ〉 and notate 〈a〉 = 〈ψ |a|ψ〉. We invoke Heisenberg’s equation of motion, (see workshop

problem 4.4)da

dt= 1i~

[a,H] → d 〈a〉dt

= 1i~〈ψ |[a,H]|ψ〉 = 1

i~(〈aH〉 − 〈Ha〉)

and note that since H is Hermitian 〈ψ|H = 〈ψ|H† = E〈ψ|. This implies:d 〈a〉dt

= 1i~

(E 〈a〉 − E 〈a〉) = 0 → ∴ ˙〈a〉 = 0 = 〈[a,H]〉

(b) Let a = xp; we have established from part (a) that 〈[xp,H]〉 = 0. We consider the commutator in this expression:

[xp,H] = x[p,H] + [x,H]p = x[p, V ] + [x, T ]p = x[p, kxn] + 12m [x, p2]p = kx(−ni~xn−1) + 1

2m (2i~p)p = i~(2T − nV )

Therefore, using the result from part (a):

〈[xp,H]〉 = i~ 〈2T − nV 〉 = i~(2 〈T 〉 − n 〈V 〉) = 0 → 2 〈T 〉 = n 〈V 〉

(c) Consider the general Hamiltonian H = T + V where T = p2/2m and V is only spatially dependent.Let A = r · p = xipi.[

p2j

2m + V, xipi

]= 1

2m [p2j , xi] + xi[V, pi] = pj

m[pj , xi]pi + xi

(i~∂V

∂xi

)= − i~

mpjpiδij + i~xi

∂V

∂xi

= i~(xi∂V

∂xi− p2

i

m

)= i~

(r · ∇V− p2

m

)(d) Once again we have [H,A] = 0, so...

i~⟨

r · ∇V− p2

m

⟩= 0 → 〈r · ∇V〉 =

⟨p2

m

⟩= 2 〈T 〉

(e) ⟨r ·(d

dr

−1r

)r

⟩=⟨

r · r(

1r2

)⟩=⟨

1r

⟩= −〈V 〉 → 〈V 〉 = −2 〈T 〉

(f) In one dimension, simply let n = 2 in part (b), and the result 2 〈T 〉 = 2 〈V 〉 is obtained trivially. In three dimensions,the result is the same:

r · ∇V = xi∂x2

i

∂xi= 2x2

i = 2V → 〈V 〉 = 〈T 〉

44

12.2 Problem 46(a) Prove the Hellmann-Feynman Theorem (λ is a variational parameter):

dE

dλ= 〈ψ(λ) |∂λH|ψ(λ)〉

(b) Using part (a) and E for the Hydrogen atom, show the Virial theorem. (Hint: pick the right variational parameter.)(c) Using part (a) and E for the SHO, show the Virial theorem.

Solution

(a) |ψ〉 is an eigenstate of H, ie H|ψ(λ)〉 = E|ψ(λ)〉.

dE

dλ= d

dλ〈H〉 = d

dλ〈ψ(λ) |H|ψλ〉 =

⟨∂ψ

∂λ|H|ψ(λ)

⟩+⟨ψ(λ)

∣∣∣∣∂H∂λ∣∣∣∣ψ(λ)

⟩+⟨ψ(λ) |H| ∂ψ

∂λ

⟩= E

(⟨∂ψ

∂λ|ψ⟩

+⟨ψ|∂ψ∂λ

⟩)+ 〈∂λH〉 = E

d

dλ〈ψ|ψ〉+ 〈∂λH〉 = 〈∂λH〉

(b) For the H-atom, we may use e2 as our variational parameter:

H = p2

2m −e2

r& E = − me4

2~2n2 → dE

de2 = −me2

~2n2 =⟨−1r

⟩It then follows that

〈V 〉 = 2 〈E〉 = 2(〈V 〉+ 〈T 〉) → 〈V 〉 = −2 〈T 〉

(c) For the SHO, we will choose m as our variational parameter:

H = p2

2m + 12mω

2x2 & E = ~ω(n+ 1

2

)→ dE

dm= 0 =

⟨1m

(− p2

2m + 12mω

2x2)⟩

= 1m

(−〈V 〉+ 〈T 〉)

which then shows that 〈V 〉 = 〈T 〉.

12.3 Problem 47The magnetic moment of an electron moving in a loop is µ = q(r × v)/2. In the presence of a magnetic field B theinteraction energy is V = −µ ·B.

(a) Determine |µ|. (Hint: write µ in terms of orbital angular momentum.)(b) Determine the energy levels of the 2p state of the Hydrogen atom in the presence of B = Bz.

Solution

(a)

µ = q(r× v)2 = q

2m (r× p) = qL2m → |µ| = |q|~2m

√`(`+ 1)

(b)

H = p2

2m −e2

r− µ ·B = H0 −

qB

2mLz → Since [H0, Lz] = 0 → E = − me4

2~2(2)2 −q~B2m

−101

This can be summarized by the diagram:

45

12.4 Problem 48Consider an electron with Hamiltonian H = λSz. At t = 0, the electron is in a spin-up state along the y-axis. Determinethe probability that at t > 0 the spin will be reversed.

Solution

Note that we notate the probabilities according to p(x|I), which reads the probability of x given I. We are interested inseeing if a state |+y〉 can evolve into a state |−y〉 over time. This is expressed as follows, where we assume that the statesare normalized:

p(−y|+y) = | 〈−y |exp[−iHt/~]|+y〉 |2

We can write out the states |+y〉 and |−y〉 in vector form. They are the eigenvectors of the operator Sy for spin one-half.

σy =(

0 −ii 0

)→ |+y〉 = 1√

2

(1i

)|−y〉 = 1√

2

(i1

)Then the probability above can be calculated explicitly:

p(−x|+x) = 14

∣∣∣∣(−i, 1)[cos(λt

2

)I− iσz sin

(λt

2

)](1i

)∣∣∣∣2 = 14

∣∣∣∣(−i, 1)(e−iλt/2

ieiλt/2

)∣∣∣∣2 = sin2(λt

2

)Note that we have made use of the identity derived in problem 11.2 in the context of a rotation operator. We may useit here since the same arguments allowing us to remove a dot product from the taylor expansion in that problem applyhere. (σ2

z = I, which allows the spin operator to be removed from the summations).

46

13

13.1 Problem 49Consider an infinite square well extending from x = 0 to x = L. Calculate the energy of the nth excited state using firstorder perturbation theory for the following two perturbations (λ 1).

(a) H1 = λV0 sin(πx/L).(b) H1 = λV0δ(x− L/2).

Note that you can compare the solution of part (b) to the solution of HW problem 6.3 after workshop if you wish. Whatbehaviors of the system does the first order approximation capture?

Solution

Problem and solution from Zettili, Quantum Mechanics Concepts and Applications (2nd Ed.). Recall that the usualenergy and wavefunction are given by:

En = ~2π2n2

2mL2 ψn(x) = 〈x|n〉 =√

2L

sin(nπxL

)We are in a non-degenerate system, so the energy correction is calculated as follows:

En = ~2π2n2

2mL2 + E(1)n → E(1)

n = 〈n |H1|n〉 = 2L

L∫0

sin2(nπxL

)H1(x)dx

(a) We will derive a useful relation before proceeding with the actual calculation, namely:∫cos(nx) sin(mx)dx = 1

4i

∫(einx + e−inx)(eimx − e−imx) = 1

4i

∫(ei(m+n)x − e−i(m+n)x + ei(m−n)x − e−i(m−n)x)dx

= 12

∫[sin ((m+ n)x) + sin ((m− n)x)] dx = −1

2

[cos((m+ n)x)

m+ n+ cos((m− n)x)

m− n

]Then we calculate E(1)

n .

E(1)n = 2λV0

L

L∫0

sin2(nπxL

)sin(πxL

)dx = λV0

L

L∫0

(1− cos

(2nπxL

))sin(πxL

)dx

= λV0

π

[− cos

(πxL

)+ cos[(1− 2n)πx/L]

2(1− 2n) + cos[(1 + 2n)πx/L]2(1 + 2n)

]L0

= 2λV0

π

4n2

4n2 − 1

Then the corrected energy spectrum up to first order is:

En = n2(

~2π2

2mL2 + 8λV0

π(4n2 − 1)

)(b)

E(1)n = 2λV0

L

L∫0

sin2(nπxL

)δ(x− L/2)dx = 2λV0

Lsin2

(nπ2

)As in the homework problem in which you solved this problem exactly, the odd-parity states (even n) are unaffected,whereas the first order perturbation is giving us an upward shift in the energy levels of even parity (odd n).

En = ~2π2n2

2mL2 +

0 n even2λV0/L n odd

47

13.2 Problem 50Consider the 2D SHO, with Hamiltonian H0.

(a) Determine the energies of the three lowest states.(b) Now apply the perturbation H1 = δmω2xy where δ 1. Determine the energy eigenvalues to first order in δ for

the three lowest states.(c) Solve H = H0 +H1 exactly. (Hint: Rewrite H in terms of a 45 degree counter-clockwise rotation of the coordinates

and conjugate momenta.) Expand to first order in δ and compare with your result in part (b).

Solution

(a) E = ~ω, 2~ω, 2~ω as per results derived many times throughout the course.(b) We calculate the ground state submatrix (1 element), and the first excited state submatrix (2×2, in accordance

with the degeneracies of the associated eigenvalues). Liboff section 13.2 should be consulted for further details on whythis approach arises.

〈00 |xy| 00〉 = ~2mω

⟨00∣∣(ax + a†x)(ay + a†y)

∣∣ 00⟩

= 0

We see that this perturbation leaves the ground state unchanged. Next for the first excited states:(〈10 |H1| 10〉〈01 |H1| 10〉〈10 |H1| 01〉〈01 |H1| 01〉

)= δmω2 ~

2mω

(0 11 0

)= δ~ω

2

(0 11 0

)where each inner product is reduced via the same procedure that was applied to the ground state. We are interested inthe eigenvalues of this matrix, which will give us the corrections to the energies. We notate the eigenvalues of the H1submatrix as λ:

λ = ±δ~ω2 → E00 = ~ω E01 or 10 = 2~ω ± δ~ω2

(c) The rotation referred to in the hint can be expressed mathematically as follows, where the lower case variables arethe old ones, and the capital variables are the new:

PX,Y = (px ± py)/√

2 X,Y = (x± y)/√

2

You should satisfy yourself that upon making this change, the Hamiltonian shown below is equivalent to the original one:

H = P 2X + P 2

Y

2m + 12mω

2(1 + δ)X2 + 12mω

2(1− δ)Y 2

Since the variables are separated out now however, we can easily write an explicit form of the spectrum for this system:

E = ~ω√

1 + δ(n1 + 1/2) + ~ω√

1− δ(n2 + 1/2)

We expand this to O(δ) and see that this approximation of the exact result matched precisely the one given by perturbationtheory.

13.3 Problem 51Spin and isospin are similar. By convention, |p〉 = |1/2, 1/2〉 and |n〉 = |1/2,−1/2〉 where |p〉 and |n〉 are the isospin statesof a proton and neutron, respectively.

(a) Find the various states that can be constructed out of a two-nucleon system.(b) Now consider the particles |P 〉 = |1/2, 1/2〉 and |N〉 = |1/2,−1/2〉, where |P 〉 and |N〉 are distinct from |p〉 and

|n〉 by their masses. Assuming scattering only occurs in I = 1 states, compute the ratio of cross sections for the reactions

P + p→ P + p

P + n→ P + n

P + n→ N + p

(c) Repeat part (b) for scattering in I = 0 states.

48

Solution

(a) Have pp, pn, nn, and np. These can be notated in the fashion we are accustomed to for two spin- 12 particles.

12 ⊗

12 = 1⊕ 0 → |11〉 = |+ +〉 |1− 1〉 = | − −〉 |10〉 = 1√

2(|+−〉+ | −+〉) |00〉 = 1√

2(|+−〉 − | −+〉)

We make the following correspondences:

|pp〉 → |+ +〉 = |11〉 |pn〉 → |+−〉 = 1√2

(|10〉+ |00〉)

|np〉 → | −+〉 = 1√2

(|10〉 − |00〉) |nn〉 → | − −〉 = |1− 1〉

(b) Scattering cross-sections are proportional to the probability that a given interaction might occur. We may calculatethe ratios between them without much trouble, even though the calculation of the absolute values of the cross sections isconsiderably more complicated, we may find the ratios with relative ease using the relationships determined above. Forinstance, |Pp〉 is always in an I = 1 state. |Pn〉 only has a probability of 1

2 to be in an I = 1 state though, and an I = 1event will result in |Pn〉 half the time and |Np〉 half the time. Therefore:

σPp→Pp : σPn→Pn : σPn→Np = 1 : 14 : 1

4 → 4 : 1 : 1

(c) We apply the same logic, noting that for I = 0 nothing has changed except that no Pp events can occur in I = 0.Then the ratio of the cross-sections is:

0 : 1 : 1

13.4 Problem 52Consider the spin-orbit coupling Hamiltonian (this is the term in fine structure)

H = λ(L · S)

Determine the energy levels for a spin one particle with general l. Show that the sum is zero.

Solution

Use J = L + S to writeH = λ

2(J2 − L2 − S2)

Let’s start with the case ` = 0. We may write:

0⊗ 1 = 1 = j → j = s = 1 → E = λ

2 (J2 − S2) = 0

Now let’s consider some ` > 0, where ` ⊗ 1 = ` + 1 ⊕ ` ⊕ ` − 1. This gives us three possibilities for j. We begin byconsidering j = `+ 1.

E = λ

2 ((`+ 1)(`+ 2)− `(`+ 1)− 2)~2 = λ`~2

For j = ` we have:E = λ

2 (−2)~2 = −λ~2

For j = `− 1 we have:E = λ

2 ((`− 1)(`)− `(`+ 1)− 2)~2 = λ

2 (−2`− 2)~2 = −λ(`+ 1)~2

We check that this is reasonable by summing these up (taking the trace of H, effectively), noting that the degeneracy fora given j is given by 2j + 1:

(λ`~2)(2(`+ 1) + 1) + (−λ~2)(2`+ 1) + (−λ(`+ 1)~2)(2(`− 1) + 1) = 0

49

14

14.1 Problem 53A particle is initially in the state |ψi〉 (for t < 0). At time t = 0 a small perturbation H1 is applied to the system. Thefirst-order transition probability is given by

Pi→f = 1~2

∣∣∣∣t∫

0

eiωfit′〈ψf |H1|ψi〉 dt′

∣∣∣∣2

where ωfi = 1~ (Ef − Ei).

(a) Consider a 1D SHO in the ground state subject to a perturbation Ax3e−t/τ . Determine the possible final statesand the probabilities of each as t→∞.

(b) Why doesn’t the sum of the probabilities add up to one?

Solution

(a) We’ll begin by determining the states the SHO could transition to from the ground state. This means we areinterested in the expression

⟨n∣∣x3∣∣ 0⟩.

⟨n∣∣x3∣∣ 0⟩ =

(~

2mω

) 32 ⟨n∣∣(a†)2 + a†a+ aa† + a2)(a† + a)

∣∣ 0⟩ =(

~2mω

) 32 ⟨n∣∣(a†)3 + a†aa† + a(a†)2∣∣ 0⟩

=(

~2mω

) 32

(√

6δn3 + δn1 + 2δn1) =(

~2mω

) 32

(√

6δn3 + 3δn1)

We conclude that the possible states are n = 1 and n = 3. We’ll look at transitions from 0→ 1 first (noting that ωfi = ωin this case).

P0→1 = 9(

~2mω

)3A2

~2

∣∣∣∣∣∣∞∫

0

e−t/τeiωtdt

∣∣∣∣∣∣2

= 9A2~8(mω)3

∣∣∣∣e−t(1/τ−iω)

1τ − iω

∣∣∣∣∞0

∣∣∣∣2 = 9A2~8(mω)3

∣∣∣∣ −11τ − iω

∣∣∣∣2

= 9A2~8(mω)3

(1τ2 + ω2

)−1

To calculate the transition probability for the 0→ 3 transition, we note that ωfi = 3ω for this case, and that apart fromthis and the coefficient out front nothing changes about the calculation. Thus:

P0→3 = 3A2~4(mω)3

(1τ2 + 9ω2

)−1

(b) It is relatively apparent that P0→1 + P0→3 6= 1. This is alright however, because we are only considering the firstorder correction. Since this scheme is necessarily approximate, it is not surprising that we may have missed some aspectsof the system which would rectify this situation.

14.2 Problem 54Consider a perturbation due to linear Stark effect H1 = e(r ·E).

(a) Can a first-order perturbation due to a linear Stark effect exist between the degenerate states |n1, n2〉 and |n2, n1〉of the two dimensional infinite well? Assume the well is centered at the origin.

(b) Can a first order perturbation due to a linear Stark effect exist between degenerate states of the 3D SHO? Theenergy spectrum is E = ~ω(2nr + `+ 3/2), where ` determines the parity of the associated wave function. (The parity ofthe state |nr, `〉 when written in the coordinate basis is (−1)`.)

50

Solution

(a) Let E = Ex. Then the question is whether or not the statement 〈n1n2 |x|n2n1〉 is correct or not. Recall that theparity of the the eigenstates of the well goes as (−1)n+1 in each spatial dimension. Consider the following:

〈n1 |x|n2〉〈n2|n1〉 →((−1)n1+1(−1)(−1)n2+1) ((−1)n2+1(−1)n1+1) = (−1)n1+n2+3(−1)n1+n2+2 = −1

We conclude that for any combination of n1 and n2 the overall parity of the expression of interest is odd, and the matrixelement must therefore always be zero. Therefore there is no first-order perturbation due to the linear Stark effect in the2D infinite well.

(b) Let’s pick E = Ez this time. We try the same sort of argument as in part (a). Consider:

〈n′r`′ |z|nr`〉 → (−1)`′(−1)(−1)` = (−1)`+`

′+1

Therefore we require that `+ `′ be an odd number in order for a matrix element to be nonzero. But we have also specifiedthat we are looking at degenerate states, which means that

2n′r + `′ = 2nr + ` → 2(n′r − nr) = `− `′

Since the LHS of this last expression has a two in it, this implies that for degenerate states ` − `′ can never be odd.Therefore such a first-order perturbation is not possible.

14.3 Problem 55A Hydrogen atom, in the ground state, is placed between the plates of a parallel plate capacitor. A time-dependent, butspatially uniform electric field is applied as follows:

E =

0 t < 0E0e

−t/τ z t > 0

To first-order, can the atom excite to the 2s state? How about the 2p states?

Solution

Recall that the parity of a state Y m` is given by (−1)`. The initial state for the system is the ground state |100〉, andthe matrix elements of interest are of the form 〈n′`′m′ |z|n`m〉. We are considering the possibility of transitions to the 2sstate 〈200| and 2p states 〈21m′|. For the 2s state we observe that:

〈200 |z| 100〉 → (−1)0(−1)(−1)0 = −1 → 〈200 |z| 100〉 = 0

For the 2p state we see that:

〈21m |z| 100〉 → (−1)1(−1)(−1)0 = 1 → Possible so far

Let’s now consider which values of m would be possible, noting that [Lz, z] = 0.

〈n′`′m′ |[Lz, z]|n`m〉 = 0 = 〈n′`′m′ |Lzz − zLz|n`m〉 = (m−m′) 〈n′`′m′ |z|n`m〉

This last expression must equal zero one way or another, so if (m−m′) 6= 0 then 〈n′`′m′ |z|n`m〉 must be zero. Thereforethe only possible transition is to the state |210〉.

14.4 Problem 56Consider a Hydrogen atom subjected to the perturbation H1 = Axy (quadrupole Stark effect). Construct the first-ordern = 3 submatrix. Do not evaluate any integrals; identify which elements must be zero and then identify non-zero elementswith constants of your choosing.

51

Solution

First-order perturbation theory requires us to consider the matrix elements of the perturbing Hamiltonian calculatedwithin all combinations of degenerate eigenstates of the unperturbed Hamiltonian. Let’s begin by considering the per-turbing Hamiltonian H1 = Axy = A(r cos θ cosϕ)(r cos θ sinϕ). The parity of H1 is even, and the eigenfunctions of theunperturbed Hydrogen atom have parity that goes as (−1)`. Therefore any matrix elements for which ` − `′ is odd willinclude an integral in θ and ϕ that is overall odd, and will therefore go to zero. Matrix elements marked with a black zeroin the grid below have been discarded because of this consideration.

Furthermore the ϕ dependence of H1 can be rearranged as follows:

cosϕ sinϕ = 14i (e

iϕ + e−iϕ)(eiϕ − e−iϕ) = 14i (e

2iϕ − e−2iϕ)

We note that the integral below for integer p and q obeys:

2π∫0

e−ipϕeiqϕdϕ = 2πδpq

In calculating the matrix elements 〈3`′m′ |H1| 3`m〉 the spherical harmonics from each wavefunction will contributeei(m−m

′)ϕ to the integral. Combined with the ϕ dependence of H1, we conclude that only matrix elements with ∆m = ±2will survive integration with respect to ϕ. Elements discarded for this reason, which were not already discarded due toparity considerations, are marked below with a red zero.

The remaining matrix elements are marked with constants. We obtain a total of 6 matrix elements which need to becalculated, when we consider that the matrix must be Hermitian and mirror elements across accordingly.

nlm 300 31-1 310 311 32-2 32-1 320 321 322300 0 0 0 0 a 0 0 0 b31-1 0 0 0 c 0 0 0 0 0310 0 0 0 0 0 0 0 0 0311 0 c∗ 0 0 0 0 0 0 032-2 a∗ 0 0 0 0 0 d 0 032-1 0 0 0 0 0 0 0 e 0320 0 0 0 0 d∗ 0 0 0 f321 0 0 0 0 0 e∗ 0 0 0322 b∗ 0 0 0 0 0 f∗ 0 0

In matrix form this reads:

H1 =

0 0 0 0 a 0 0 0 b0 0 0 c 0 0 0 0 00 0 0 0 0 0 0 0 00 c∗ 0 0 0 0 0 0 0a∗ 0 0 0 0 0 d 0 00 0 0 0 0 0 0 e 00 0 0 0 d∗ 0 0 0 f0 0 0 0 0 e∗ 0 0 0b∗ 0 0 0 0 0 f∗ 0 0

52

15

15.1 Problem 57Consider the momentum eigenstates ψ(x) = e±ikx in one dimension of length L.

(a) What condition on k makes ψ(x) = ψ(x+ L)?(b) In the continuum, the number of states is determined by the expression below, where ρ(E) is the density of states.∫

ρ(E)dE = L

2π

∫dk

Determine ρ(E) for a free particle.(c) Determine dρ(E) for a free particle in three dimensions (in terms of the differential solid angle in k-space).

Solution

(a)eikx = eik(x+L) → kL = 2πn for n ∈ Z

(b) Note that this is all performed in the continuum, which means that we are looking at the (non-quantized) unboundstates of a free particle. The density of states tells us how many states we can expect to find within a certain interval ofenergies. ∫

ρ(E)dE = L

2π

∫dk → E = ~2k2

2m → dE = ~2

mkdk∫

ρ(E)dE = L

2π

∫m

~2kdE → ρ(E) = Lm

2π~√

2mE(c) We begin by generalizing the expression given above for one dimension into three. Note that dΩk is the differential

solid angle in k-space. ∫ρ(E)dE = L

2π

∫dk →

∫ρ(E)dE = L3

(2π)3

∫k2dkdΩk

The expression we found above relating dk and dE holds, so we do the following:∫ρ(E)dE = L3

(2π)3

∫k2dkdΩk = V

(2π)3

∫km

~2 dEdΩk → dρ(E) = mV√

2mE(2π)3~3 dΩk

Note that we have repeatedly used k =√

2mE/~ above, and that the differential volume k2dkdΩk in k-space is directlyanaloguous to d3x = r2drdΩ for regular spatial integrals.

Effectively what we are saying with this problem is that we are considering a very large box for V = L3, wherevery large means that the difference in energies between states has become so small that we can model the behaviorwith integrals rather than discrete sums. This is in general how we treat the continuum. (You may recall that similarapproaches are used regularly in statistical mechanics.)

15.2 Problem 58Consider the perturbation shown below, where Ω is a time-independent operator (we are looking at the transition proba-bility for a Harmonic Perturbation).

V (t) = Ωeiωt + Ω†e−iωt

(a) Compute the first-order transition probability (in terms of the undetermined spatial component). Disregard thecross-terms because they will not play a role in the next part.

(b) Take the limit of the transition rate (probability per unit time) for t→∞. Use the identity

lima→∞

sin2(ax)πax2 = δ(x)

(c) From part (b) what physical effect does e±iωt have on the transition rate?(d) How does this result change when transitions are to the continuum?

53

Solution

(a) This is effectively a statement to set up the problem using the first-order transition probability used in problem 1of workshop 14:

Pi→f = 1~2

∣∣∣∣∣∣t∫

0

〈ψf |Ω|ψi〉 eiωt′eiωfit

′dt′

∣∣∣∣∣∣2

+ 1~2

∣∣∣∣∣∣t∫

0

⟨ψf∣∣Ω†∣∣ψi⟩ e−iωt′eiωfit

′dt′

∣∣∣∣∣∣2

We are already neglecting the cross terms by taking the mod-square of each individual term.(b) The transition rate R is given by R = Pi→f/t. We are interested in lim

t→∞R. We’ll reduce the expression above and

then take the limit.

R = limt→∞

1~2t

(∣∣∣∣〈ψf |Ω|ψi〉 1− ei(ω+ωfi)t

i(ω + ωfi)

∣∣∣∣2 +∣∣∣∣⟨ψf ∣∣Ω†∣∣ψi⟩ 1− ei(−ω+ωfi)t

i(ωfi − ω)

∣∣∣∣2)

= limt→∞

1~2

[| 〈Ω〉fi |

2∣∣∣∣ei(ωfi+ω)t − 1

ωfi + ω

∣∣∣∣2 + |⟨Ω†⟩fi|2∣∣∣∣ei(ωfi−ω)t − 1

ωfi − ω

∣∣∣∣2]

= limt→∞

4~2

[| 〈Ω〉fi |

2 sin2[(ωfi + ω)t/2](ωfi + ω)2 + |

⟨Ω†⟩fi|2 sin2[(ωfi − ω)t/2]

(ωfi − ω)2

]= 2π

~[| 〈ψf |Ω|ψi〉 |2δ(Ef − Ei + ~ω) + |

⟨ψf∣∣Ω†∣∣ψi⟩ |2δ(Ef − Ei − ~ω)

](c) Let’s stop and interpret this result0 for a moment, because the two terms have two different meanings. In the

first, we see that the initial energy is necessarily higher than the final energy. In that case, we are looking at a systeminitially in an excited state, which de-excites and emits a photon of energy ~ω under the effect of the perturbation athand. This process is referred to as stimulated emission. The second term goes the opposite way: the perturbation causesthe absorption of a photon of energy ~ω leading to an excitation. So the terms e±iωt correspond to the absorption oremission of a photon of energy ~ω.

(d) In the continuum (unbound states) we must modify our expressions a bit. In particular, we want to consider thenumber of final states which exist between the energies Ef and Ef +dEf ; we use the density of states ρ, which we exploredin problem 1, which is defined such that Ef + dEf = ρ(Ef )dEf . We integrate our final result over the possible states inthe continuum case, instead of just leaving the delta functions in place for the quantized states above:

R =∫Pif (t)t

ρ(Ef )dEf

Integrating the delta functions will then use their sifting property over the density of states. We consider the absorbingand emitting cases below, notated as Ra and Re, respectively.

Ra = 2π~∣∣⟨ψf ∣∣Ω†∣∣ψi⟩∣∣2 ∫ δ(Ef − Ei − ~ω)ρ(Ef )dEf = 2π

~∣∣⟨ψf ∣∣Ω†∣∣ψi⟩∣∣2 ρ(Ef )

∣∣∣∣Ef =Ei+~ω

Re = 2π~|〈ψf |Ω|ψi〉|2

∫δ(Ef − Ei + ~ω)ρ(Ef )dEf = 2π

~|〈ψf |Ω|ψi〉|2 ρ(Ef )

∣∣∣∣Ef =Ei−~ω

These first order results treating transitions to or from the continuum to first order are often referred to as Fermi’s GoldenRule. Those wishing for further practice with this should repeat the derivation we have performed above to obtain

R = 2π~|〈ψf |H1|ψi〉|2 ρ(Ei)

in the case where the perturbation H1 is constant in time. (This derivation is similar to, and slightly easier than the onedone above, and I highly recommend it as practice.)

15.3 Problem 59Comment on the calculation of the integral ∫

zeik·re−r/a0d3x

54

Solution

Suppose we are interested in using the result of problem 2 to calculate a transition probability for the ionization of aHydrogen atom due to Stark effect. This corresponds to an excitation of a particular quantized state of the H-atom tothe continuum. We could reasonably model the continuum state (ψf ) with a plane-wave (as motivated in problem 1), andwould then use a Hydrogen wavefunction as the other state (ψi). This would give us a radial integral of the form shownabove.

Now let’s consider the following regarding the actual evaluation of such an integral:∫zeik·re−r/a0d3x = −i ∂

∂kz

∫eik·re−r/a0d3x

The integral left over after re-writing with the parametric differentiation is invariant under a change of direction of k.We’ll choose k along z then, so that when we re-write k · r = kr cos θ the angle θ will be the polar angle in sphericalcoordinates. We call the integral below f(k).

−i ∂∂kz

∫eikr cos θe−r/a0r2 sin θdrdθdϕ = −i ∂

∂kzf(k) = −if ′(k) ∂k

∂kz= −if ′(k) cos θk

Consider how this can reduce the amount of work involved in this sort of calculation.

15.4 Problem 60A Hydrogen atom is subjected to V = Axyz.

(a) Show that the n = 1 and n = 2 levels are unaffected up to first-order in A.(b) Construct the n = 3 first-order correction submatrix of H. Identify non-zero elements with constants of your

choosing.(c) Make the matrix block-diagonal. When doing so, you must conserve the property H = H†.(d) Determine the eigenvalues in terms of your constants.(e) Explain why all of the non-zero matrix elements can be expressed in terms of a single radial integral.

Solution

(a) Note that H1 is overall odd in this problem, and that xy creates a selection rule ∆m = ±2. The only way to get∆m = ±2 for the n = 0 and n = 1 submatrices are for the matrix elements with m′ = −1 and m = +1 or vice versa.But these are both ` = 1 states, so these matrix elements are zero due to parity. Therefore there are no non-zero matrixelements within the ` = 0 and ` = 1 regions of any submatrix. (And for n = 0 and n = 1 this is all there is.)

(b) We apply the same approach and labeling as in problem 14.4.

n`m 300 31-1 310 311 32-2 32-1 320 321 322300 0 0 0 0 0 0 0 0 031-1 0 0 0 0 0 0 0 c 0310 0 0 0 0 a 0 0 0 b311 0 0 0 0 0 d 0 0 032-2 0 0 a∗ 0 0 0 0 0 032-1 0 0 0 d∗ 0 0 0 0 0320 0 0 0 0 0 0 0 0 0321 0 c∗ 0 0 0 0 0 0 0322 0 0 b∗ 0 0 0 0 0 0

(c) Let’s number the columns of the matrix above 1-9. Then consider manipulating the columns so that they now goin the order (3,5,9,2,8,4,6,1,7) according to that numbering. Change the order of the rows in the same way in order topreserve H = H†. (Notice that the order in which we happened to write each state down the sides was arbitrary to beginwith so this should be no problem. Also, exchanging two rows or columns only affects the determinant of a matrix up to

55

a sign, so any such interchanges leave the eigenvalues unchanged.) This gives the block-diagonal matrix:

0 a ba∗ 0 0b∗ 0 0

0 cc∗ 0

0 dd∗ 0

0 00 0

(d) We can then take the eigenvalues of each block individually (the purpose of part (c) is to allow the eigenvalues to

be determined by hand without taking a 9× 9 determinant). Going down the blocks along the diagonal, we obtain:

First : λ = 0,±√|a|2 + |b|2 Second : λ = ±|c| Third : λ = ±|d| Fourth : λ = 0(×2)

Then the final result for all the energies may be notated:

E = E3 +

±√|a|2 + |b|2±|c|±|d|

0 (3 States)

where E3 is the unperturbed Hydrogen energy for n = 3.

(e) The functions ψn`m = 〈r, φ, θ|n`m〉 can be written ψn`m = Rn`Ym` . All of the non-zero matrix elements above

are of the form 〈31m |H1| 32m〉 or the complex conjugate of such a state. This means that the r integrand for each ofthese is completely defined by R31R32 (both of the solutions R to LaGuerre’s equation are real). Therefore evaluating∫∞

0 (R31R32)(r3)r2dr will deal with the radial dependence of all of the matrix elements of interest. (The term r3 is fromH1, and the r2 is part of the Jacobian to evaluate in spherical coordinates). The remaining angular parts are simple tocalculate for each non-zero matrix element.

Those wishing for extra practice may wish to look up the appropriate wavefunctions in Liboff, perform the integration(good practice of parametric differentiation), and then draw a splitting diagram of the results.

56

16 Review for Final

Problem 61Consider the following Hamiltonian, where B A:

H = AJz +BJy

(a) Determine the eigenvalues of H up to O(B2).(b) Determine the eigenvalues of H exactly.(c) Expand the result from part (b) to O(B2) and show that it agrees with part (a).

Solution

(a) We take H0 = AJz and H1 = BJy. The eigenstates of H0 are of the form |jm〉. We begin with the first ordercorrections:

E(1) = 〈jm |BJy| jm〉 = B

2i 〈jm |J+ − J−| jm〉 = 0

There are no first order corrections, so we proceed to second order. The general formula is shown below.

E(2)n =

∑m 6=n

|⟨n(0) |H1|m(0)⟩ |2E

(0)n − E(0)

m

We have E(0)n = An~ for the purposes of that notation, and wish to evaluate:

E(2)m = B2

4∑m6=m′

| 〈jm |J+ − J−| j′m′〉 |2

Am~−Am′~= B2~

4A

[|√j(j + 1)−m(m− 1)|2

m− (m− 1) +|√j(j + 1)−m(m+ 1)|2

m− (m+ 1)

]

= B2~4A [j(j + 1)−m(m− 1)− [j(j + 1)−m(m+ 1)]] = B2~

4A (2m) = B2~2A m

So the energy up to O(B2) is given by:

E = Am~ + B2~2A m

(b) Now let’s solve it exactly. Note that any Hamiltonian of the form H = CJ · n will have eigenvalues Cm~ whenthe projections are measured along the n axis. Consider:

H = AJz +BJy = CJ · n = C(Jx, Jy, Jz) · (0, B,A) 1√A2 +B2

→ C =√A2 +B2 → E =

√A2 +B2m~

(c) Now we have eigenvalues, and can expand assuming B A to check against part (a).

E =√A2 +B2m~ = m~A

√1 + B2

A2 ≈ Am~(

1 + B2

2A2

)= Am~ + B2~

2A m

We see that is in perfect agreement with the perturbative result to the same order.

Problem 62Consider the Hamiltonian

H = AJ2 +BJ2z

(a) Determine the eigenvalues and eigenvectors of H.(b) Consider a perturbation H1 = Cxz. Justify why the first-order perturbations vanish.

57

Solution

(a) The eigenvalues are easily determined in the coupled basis, and the basis states in the coupled basis are eigenstatesof H. Consider:

H|jm〉 = (AJ2 +BJ2z )|jm〉 = ~2(Aj(j + 1) +Bm2)|jm〉

That takes care of both the eigenvalues and eigenvectors.(b) Notice that H breaks up the degeneracy between the |jm〉 states that the eigenvalues J2 and Jz alone would

leave. Except for possible specific integer relationships between A and B, which we will neglect in this treatement, theonly degeneracy is between and ±m for a given j. In this case, we would in general a need to evaluate matrix elementsof the form where m′ = ±m.

〈jm′ |xz| jm〉

Note that xz is even, and implies a selection rule of ∆m = ±1. (See workshop problems 14.4 and 15.4 if you do notimmediately see why.) But when j and m are the same on both sides, the selection rule ∆m = ±1 would not be satisfied,and this expectation value is always zero. Likewise, when m 6= 0 and m′ = −m, the difference m′−m is at least two, alsoviolating the selection rule.

The part we are ignoring: There are almost certainly special cases choices of A and B for which states with a differentj and m could create the same energy. This is more difficult to deal with, and we will not consider this case.

Problem 63Consider an H-atom with an angular wavefunction

AY 02 +BY 1

2

where |A|2 + |B|2 = 1.(a) Determine the expectation of L2 and Lz.(b) Determine the expectation of Lx.(c) Determine the minimum value of energy that can be measured.

Solution

Let |ψ〉 = A|20〉+B|21〉 where the states are labeled according to |`m〉(a) In order to calculate

⟨L2⟩ we consider the following:⟨

L2⟩ = (A∗〈20|+B∗〈21|)L2 (A|20〉+B|21〉) = (A∗〈20|+B∗〈21|)6~2(A|20〉+B|21〉) = 6~2(|A|2 + |B|2) = 6~2

For Lz:〈Lz〉 = 〈ψ|Lz(A|20〉+B|21〉) = 〈ψ|(~B|21〉) = ~|B|2

(b) Recall that Lx = 12 (L+ + L−).

Lx|ψ〉 = 12(L+ + L−)(A|20〉+B|21〉) = ~

√6

2 [A(|2− 1〉+ |21〉) +B|20〉] +B|22〉

〈Lx〉 = ~√

62 (B∗A+A∗B)

(c) We have ` = 2, and ` ranges up to n− 1. So n = 3 is the smallest principle quantum number of a Hydrogen statefor which the above could be the angular wavefunction.

Problem 64Conisder four spin- 1

2 particles.(a) How many states does the system have?(b) Determine the possible values of S2 and Sz.(c) Determine a CSCO.

58

Solution

(a) Each particle has 2 states, and there are 4 particles. The number of states is given by 24 = 16.(b)

12 ⊗

12 ⊗

12 ⊗

12 = (1⊕ 0)⊗ 1

2 ⊗12 =

(32 ⊕

12 ⊕

12

)⊗ 1

2 = (2⊕ 1)⊕ (1⊕ 0)⊕ (1⊕ 0)

The possible values of S2 → ~2S(S + 1) are thus 6~2, 2~2, and 0. We check that above is expansion is consistent withpart (a), using the fact that there are 2S + 1 projections for each S. We see that 5 + 3 + 3 + 1 + 3 + 1 = 16, which agreeswith the above. The possible values of Sz are ±2~, ±1~, and 0.

(c) The simplest CSCO is S21 , S

22 , S

23 , S

24 , S1z, S2z, S3z, S4z, although many others are possible.

Problem 65Consider a system of ji = 1

2 with Hamiltonian

H = A(J1 · J2)(J3 · J4)

(a) Determine the eigenvalues of H and the degeneracies of each.(b) Show that the sum of the above is zero and explain why.(c) Determine a CSCO.

Solution

(a) Let J12 = J1 + J2 and J34 = J3 + J4.

H = A

4 (J212 − J2

1 − J22 )(J2

34 − J23 − J2

4 )→ A

4

(J2

12 −3~2

2

)(J2

34 −3~2

2

)Clearly, J12 and J34 commute with each other, and each can take on values of 0 or 1 (since each contains two spins). Eachvalue occurs in 1 and 3 states respectively in each of the two-spin subspaces. Thus we have the following combinations toconsider in putting in numbers for J12 and J34:

J12 = J34 = 1 (9 states) J12 = 1, J34 = 0 or J12 = 0, J34 = 1 (3 states each) J12 = J34 = 0 (1 state)

Then we have the following eigenvalues:

E = A

4

(2~2 − 3~2

2

)(2~2 − 3~2

2

)= A~4

16 → 9×

E = A

4

(0− 3~2

2

)(2~2 − 3~2

2

)= −3A~4

16 → 6×

E = A

4

(0− 3~2

2

)(0− 3~2

2

)= 9A~4

16 → 1×

(b) The weighted sum of the eigenvalues is:

A~4

16 (1 · 9− 3 · 6 + 9 · 1) = 0

This is expected because it is equivalent to taking the Trace of H and we have:

Tr(H) = ATr[(J1iJ2i)(J3kJ4k)] = ATr(J1i)Tr(J2i)Tr(J3k)Tr(J4k) = 0

(The second subscript indicates a particular spatial coordinate of each angular momentum operator. The operation ofsplitting up the trace is allowed for different angular momenta which are in separate subspaces, but not different spatialdirections corresponding to the same operator.)

(c) Once again J21 , J

22 , J

23 , J

24 , J1z, J2z, J3z, J4z works, but it is more interesting to consider a CSCO containing the

Hamiltonian. Note that since the Hamiltonian contains only dot products, it is a scalar and must therefore be rotationallyinvariant. Consider H,J2

1 , J22 , J

23 , J

24 , J

2, J12z, J34z.

59

Problem 66Consider three spin one-half particles with the Hamiltonian

H = A(S1 · S3 + S3 · S2)

(a) Determine the eigenvalues of H and degeneracies of each.(b) Show that the sum is zero.(c) Determine a CSCO.

Solution

(a) Let’s rewrite H a little bit. Let S = S1 + S2 + S3, and let S12 = S1 + S2.

H = A(S1 + S2) · S3 = AS12 · S3 = A

2((S12 + S3)2 − S2

12 − S23)

= A

2 (S2 − S212 − S2

3) = A

2

(S2 − S2

12 −3~2

4

)Possible values: S = (3/2, 1/2, 1/2) S12 = (1, 0) S3 = 1/2

Let S = 3/2, which requires that S12 = 1. 4 such states.

E = A~2

2

(154 − 2− 3

4

)= A~2

2

Let S = 1/2 and S12 = 1. 2 such states.

E = A~2

2

(34 − 2− 3

4

)= −A~2

Let S = 1/2 and S12 = 0. 2 such states.

E = A~2

2

(34 − 0− 3

4

)= 0

(b) We take the sum (weigted by degeneracy), to verify that the trace of H will be zero. (We haven’t found theeigenbasis of H, but we know that an eigenbasis in which H is diagonal must exist, and that this would be the trace inthat basis.)

Tr(H) = 4(A~2

2

)+ 2

(−A~2)+ 2(0) = 0

(c) We expect that we will require 6 operators to define a CSCO for this system. Again it is more interesting toconsider a CSCO including H. Consider:

H,Sz, S3z, S2, S2

12, S23

Problem 67Two protons are located on the z-axis, separated by a distance d, and subjected to a magnetic field B = B0z. Themagnetic moment of the ith proton is µi = 2µ0Si/~.

(a) Ignoring proton-proton interaction, find the energy levels and states of the system.(b) The dipole-dipole magnetic interaction energy between the protons is

H1 = 1r3

(µ1 · µ2 − 3(µ1 · r)(µ2 · r)

r2

)where r is the separation vector. Calculate the Energy levels.

(c) Determine a CSCO for parts (a) and (b) which includes H.

60

Solution

(a) Ignoring the proton-proton interaction, the Hamiltonian is given by

H0 = −µ ·B = −2µ0

~(S1 + S2) ·B = −2µ0

~B0 (S1z + S2z)

The states are those for the coupling of two spin- 12 objects, and the usual states |++〉, |+−〉, |−+〉, |−−〉 in the uncoupled

basis happen to be eigenstates of this Hamiltonian. (These are the states which are eigenstates of the z-components ofeach of the two individual spin operators.) The matching energies are:

E = −2µ0B0, 0, 0, 2µ0B0

(b) We now have a Hamiltonian H = H0 + H1 in order to account for the proton-proton interaction. Note that theseparation vector r = dz. Then:

H1 = 4µ20

d3~2 (S1 · S2 − 3S1zS2z) = 4µ20

d3~2

(S2 − S2

1 − S22

2 − 3S1zS2z

)We note that we have one term which is evaluated naturally in the coupled basis, and one term which is evaluated naturallyin the uncoupled basis.

H1|11〉 = H1|+ +〉 = 4µ20

d3~2

(12

[2~2 − 3

2~2]−[~2

]2)|11〉 = − 2µ2

0d3~2 |11〉

H1|1− 1〉 = H1| − −〉 = − 2µ20

d3~2 |1− 1〉

H1|10〉 = H11√2

(|+−〉+ | −+〉) = 4µ20

d3~2

(~2

4 + 3 · 2[~2

]2)|10〉 = 4µ2

0d3~2 |10〉

H1|00〉 = H11√2

(|+−〉 − | −+〉) = 0

Recombining this with H = H0 +H1 the complete spectrum for this problem can be effectively summarized as:

E =

〈11 |H| 11〉

〈1− 1 |H| 1− 1〉〈10 |H| 10〉〈00 |H| 00〉

= 2µ0B0

−1100

+ 2µ20

d3

−1−120

(c) H0 is not a CSCO due to denegeracy, but H has no degeneracy left once the two-particle interaction is taken into

account. Therefore H is a CSCO by itself in the coupled problem (part (b)).

Problem 68A particle is in the potential

V (r) =

0 a ≤ r ≤ b∞ else

(a) Determine the ground state wavefunction and energy.(b) Consider the perturbation (linear Stark effect) H1 = Az, which effectively transitions the particle to another state.

What are the smallest possible measurements of L2 and Lz.

Solution

(a) We begin with the radial Schrodinger equation, where U = rψR:

− ~2

2m

(∂2

∂r2 −`(`+ 1)r2

)U + V U = EU → Ground State ` = 0→ − ~2

2m∂2U

∂r2 = EU

61

This last expression is what we want to solve in the well region. Clearly the wavefunction is zero where the potential isinfinite. We have the following solutions for U :

U(r) =√

2b− a

sin(nπ(r − a)b− a

)→ n = 1 for ground state

And the spectrum:

E = ~2k2

2m = ~2

2m

(π

b− a

)2

In order to construct the complete ground state wavefunction we would need to add the angular distribution:

ψgs = ψRY0

0 = 1r

√2

b− asin(nπ(r − a)b− a

)Y 0

0 = 1r√

4π

√2

b− asin(nπ(r − a)b− a

)(b) The Stark effect Hamiltonian H1 ∝ z is odd, so we need ∆` = odd for a transition. Since z has no ϕ dependence in

spherical coordinates, it has a selection run ∆m = 0. We are starting in ` = 0 for the ground state, so the transition wouldhave to leave Lz = 0 unchanged, and the lowest possible value of L2 after this transition would be 2~2, corresponding to` = 1.

Problem 69Calculate the p4 relativistic correction to the energy spectrum of Hydrogen to first-order. Follow these steps:

(a) Write p4 in terms of H, V , and 1/r2

(b) Evaluate each of the terms using Virial theorem or HF theorem.(c) Write the correction in terms of the fine structure constant α = e2/~c ≈ 1/137.

Solution

(a) Recall that energy is given in relativity by E2 = p2c2 + (mc2)2. Kinetic energy would be what is left aftersubtracting the rest-mass energy, or T =

√(pc)2 + (mc2)2 −mc2. We expand this:

T = mc2

(√1 + p2

m2c2 − 1)≈ mc2

(1 + p2

2m2c2 −p4

8m4c4 − 1)

= p2

2m −p4

8m3c2 → H1 = − p4

8m3c2

We can take this second term containing p4 to be a first relativistic correction. We’re going to do something a littleunusual that will let us calculate expectation values of p4 in Hydrogen. consider the Hydrogen Hamiltonian:

H = p2

2m −e2

r→ p2 = 2m

(H + e2

r

)→ p4 = 4m2

(H2 +H

e2

r+ e2

rH + e4

r2

)We note that [p4, L2] = 0, that p4 does not affect m (the projection of `) in any way, and that p4 is an even operator.We conclude that in an expectation

⟨n`′m′

∣∣p4∣∣n`m⟩ that we must have ` = `′ and m = m′ to get a result. Then we are

effectively only looking at the diagonal of the first-order perturbative submatrix.(b) Let’s take expectation values using the above:

⟨n`m

∣∣p4∣∣n`m⟩ = 4m2(E2n + En

⟨n`m

∣∣∣∣e2

r

∣∣∣∣n`m⟩+⟨n`m

∣∣∣∣e2

r

∣∣∣∣n`m⟩En +⟨n`m

∣∣∣∣e4

r2

∣∣∣∣n`m⟩)The middle two terms contain

⟨e2/r

⟩= −〈V 〉. Recall from the Virial Theorem that for the Hydrogen atom 〈V 〉 = −2 〈T 〉,

−〈T 〉 = En, and 〈V 〉 = 2En. We apply this simplification to the above:

⟨n`m

∣∣p4∣∣n`m⟩ = 4m2(E2n + En(−2En) + (−2En)En +

⟨n`m

∣∣∣∣e4

r2

∣∣∣∣n`m⟩) = 4m2(−3E2

n +⟨n`m

∣∣∣∣e4

r2

∣∣∣∣n`m⟩)In order to deal with the term e4 ⟨1/r2⟩ we’ll use the HF theorem. Consider:

dE

dλ= 〈ψ |∂λH|ψ〉 → d

dλ

(− me4

2~2n2

)=⟨d

dλ

(p2r

2m + ~2`(`+ 1)2mr2 − e2

r

)⟩

62

We let λ→ ` and note that n = nr + `+ 1. Then:

d

d`

(− me4

2~2(nr + `+ 1)2

)=⟨

(2`+ 1)~2

2mr2

⟩→ me4

~2n3 = (2`+ 1)~2

2m

⟨1r2

⟩→

⟨1r2

⟩= 2m2e4

~4n3(2`+ 1)

Then using this result in our calculation of⟨p4⟩:

⟨p4⟩ = 4m2

(2m2e8

~4(2`+ 1)n3 − 3E2n

)= 4m2E2

n

(8n

2`+ 1 − 3)

= m4e8

~4n4

(8n

2`+ 1 − 3)

= α4m4c4

n4

(8n

2`+ 1 − 3)

(c) Given our expectation value for p4, we can return the relativistic correction we set out to calculate.

E(1)n` = − 1

8m3c2

⟨p4⟩ = −α

4mc2

8n4

(8n

2`+ 1 − 3)

= −α2|En|4n2

(8n

2`+ 1 − 3)

Note that since α ≈ 1/137, we can get the order of this correction. For E1 = −13.6eV , we would expect E(1)10 ≈ 10−3eV .

The total corrected energy (to first order) is given by E = En + E(1)n` .

Problem 70When placed in a magnetic field, the Hydrogen energy spectrum is given by

H = H0 +HFS +HB

where H0 is the unperturbed Hamiltonian, HFS is the fine structure correction (which includes a first order relativisticcorrection and spin-orbit interaction), and HB is the interaction from the magnetic field.

In the presence of a strong-field, HB HFS (Paschen-Back effect – or strong-field Zeeman effect), determine theenergy levels for n = 1 and n = 2.

Solution

Let’s consider the last term HB in the Hamiltonian. The orbital angular momentum and spin could both interact with amagnetic field. Note the two introduced on the spin term (this is the Landé g factor).

HB = −µL ·B− µs ·B = e

2mc (L ·B + 2S ·B) = eB

2mc (Lz + 2Sz)

We expect both this Zeeman term and the fine-structure term to have very small effects relative to the original HamiltonianH0, and will therefore treat them perturbatively. Specifically for the Paschen-Back effect, we are looking at a regime inwhich HB overwhelms HFS , and we will consequently neglect it. (If you wanted to include it, you would need the resultfrom the previous problem, along with a spin-orbit coupling term. It would be a good exercise to compute either if youare not sure you would be able to do so.) So we let H = H0 + HB for the remainder of the problem. We note that[HB , H0] = 0, and that we can therefore diagonalize them in a common set of states |n`m`ms〉.

H|n`m`ms〉 =[H0 + eB

2mc (Lz + 2Sz)]|n`m`ms〉 = E|n`m`ms〉 → E = − e2

2a0n2 + eB~2mc (m` + 2ms)

It is then easy to list out the energies for n = 1 (` = 0→ m` = 0, and ms = ± 12 ), and n = 2 (` = 0, 1→ m` = 0,±1, and

ms = ± 12 ).

Problem 71(a) Two spin-one half particles are in the state

|ψ〉 = A|+ +〉+B| −+〉

where |A|2 + |B|2 = 1. Determine the possible values of S2 and Sz and the probabilities of each.(b) Extend this to Hydrogen, where the angular wavefunction and state ket of the electron is,

AY 01 |−〉+BY 1

1 |+〉

No need to determine the probabilities this time.

63

Solution

(a) Recall the CG coefficients to get between the coupled and uncoupled basis. We plug those in to obtain:

|ψ〉 = A|11〉+ B√2

(|10〉 − |00〉)

Then the possible measurements and probabilities of the operators in question are as follows:

S2 =

2~2

0 With Probabilities: |A|2 + |B|2/2|B|2/2 Sz =

~0 With Probabilities: |A|2

|B|2

(b) This is not physically a two-particle system, but it works just like one. We have J = L + S, where the orbitalangular momentum and spin are coupled. Thus we can rewrite the wavefunction in question as:

AY 01 |−〉+BY 1

1 |+〉 = A|10〉|12 −12 〉+B|11〉|12

12 〉

The first term has Jz = −1/2 and the second Jz = 3/2. We then apply the same kind of arguments to measure J2 andJz as above.

Problem 72Consider a particle in the potential,

V = m

2(ω2

1x2 + ω2

2y2)

where ω1 6= ω2. What condition on ω1 and ω2 makes H a CSCO?

Solution

We can see that we can write an energy spectrum E1 in x and E2 in y, where the overall spectrum is E = E1 +E2. ThenE1 = ~ω1(n1 + 1/2) and E2 = ~ω2(n2 + 1/2). The key observation here is that H is not a CSCO when there is denegeracyin the system. Degeneracy occurs when:

~ω1(n1 − n′1) = ~ω2(n′2 − n2)

Thus we may writeω1

ω2= n1 − n′1n2 − n′2

where n1, n′1, n2, and n′2 are all integers. There is no possibility of degeneracy if we cannot satisfy this equation. Thus,if we require that one or both of ω1 and ω2 are irrational (such that their ratio is irrational), we will not get degeneratestates and H alone can be a CSCO.

64

17 New Problems

17.1(a) Show that for any time-independent operator Ω the relationship ∂t 〈Ω〉 = 0 holds, where the expectation value is

taken between eigenstates of the Hamiltonian.(b) Consider the Hamiltonian:

H = p2

2m + kxn = T + V

Derive the quantum Virial theorem for such a system, i.e. show that for the above system 2 〈T 〉 = n 〈V 〉. (Hint: ChooseΩ = xp.)

Solution

(a) We have H|ψ〉 = E|ψ〉 and notate 〈Ω〉 = 〈ψ |Ω|ψ〉. We invoke Heisenberg’s equation of motion,

dΩdt

= 1i~

[Ω, H] → d 〈Ω〉dt

= 1i~〈ψ |[Ω, H]|ψ〉 = 1

i~(〈ΩH〉 − 〈HΩ〉)

and note that since H is Hermitian 〈ψ|H = 〈ψ|H† = E〈ψ|. This implies:

d 〈Ω〉dt

= 1i~

(E 〈Ω〉 − E 〈Ω〉) = 0 → ∴ ˙〈Ω〉 = 0 = 〈[Ω, H]〉

(b) Let Ω = xp; we have established from part (a) that 〈[xp,H]〉 = 0. We consider the commutator in this expression:

[xp,H] = x[p,H] + [x,H]p = x[p, V ] + [x, T ]p = x[p, kxn] + 12m [x, p2]p = kx(−ni~xn−1) + 1

2m (2i~p)p = i~(2T − nV )

Therefore, using the result from part (a):

〈[xp,H]〉 = i~ 〈2T − nV 〉 = i~(2 〈T 〉 − n 〈V 〉) = 0 → 2 〈T 〉 = n 〈V 〉

17.2A canonical coherent state is an eigenstate of the 1D SHO annihilation operator. Let |α〉 be a canonical coherent statewith eigenvalue α such that a|α〉 = α|α〉.

(a) Write |α〉 as a linear combination of stationary states |n〉 of the 1D SHO.(b) Normalize the result from part (a).(c) Let |ψα〉 be the time-dependent form of the state |α〉, obtained by using the time evolution operator. Write and

simplify |ψα〉 using the results from the previous parts.(d) Calculate the time-dependent expectation values 〈x〉t = 〈ψα |x|ψα〉 and 〈p〉t = 〈ψα |p|ψα〉. (Do not do the whole

calculation twice. Once you obtain one, you should be able to obtain the other with relatively little work.)(e) What do you notice about the results from part (d) / How does the behavior of this system compare with that of

a classical SHO?

Solution

(a) We begin by expanding |α〉 in terms of the energy eigenstates of the SHO.

|α〉 =∑n

|n〉〈n|α〉 =∑n

Cn|n〉 → Determine Coefficients Cn = 〈n|α〉

We use the property that a|α〉 = α|α〉 to proceed.

a|α〉 =∑n

Cna|n〉 =∞∑n=1

Cn√n|n− 1〉 =

∞∑n=0

Cn+1√n+ 1|n〉

= α|α〉 = α∑n

Cn|n〉

65

We obtain the recursion relation αCn =√n+ 1Cn+1. If we begin with C0 and apply this repeatedly, we see that

Cn = αn√n!C0. Therefore:

|α〉 =∞∑n=0

αn√n!C0|n〉

(b) We determine C0 by normalization.

〈α|α〉 = 1 =(∑

m

〈m| (α∗)m√m!

C∗0

)(∑n

αn√n!C0|n〉

)=∑n,m

|C0|2(α∗)mαn√

m!n!δnm = |C0|2

∑n

|α|2n

n!

→ |C0|2 =(∑

n

|α|2n

n!

)−1

(c) The evolution operator is U = e−iHt/~. We want to calculate |ψα〉 = U |α〉.

|ψα〉 = e−iHt/~∑n

αn√n!C0|n〉 = C0

∑n

αn√n!e−iEnt/~|n〉 = C0

∞∑n=0

αn√n!e−iωt(n+1/2)|n〉

(d) We calculate 〈x〉t first.

〈ψα |x|ψα〉 = |C0|2∞∑

m,n=0

(α∗)mαn√n!m!

eiωt(m−n) 〈m |x|n〉

The term 〈m |x|n〉 is of interest.

x =√

~2Mω

(a+ a†) → 〈m |x|n〉 =√

~2Mω

(〈m |a|n〉+⟨m∣∣a†∣∣n⟩) =

√~

2Mω(√nδm,n−1 +

√n+ 1δm,n+1)

We plug this result into the first expression above and simplify:

〈x〉t = |C0|2∞∑

m,n=0

(α∗)mαn√n!m!

eiωt(m−n)√

~2Mω

(√nδm,n−1 +

√n+ 1δm,n+1)

= |C0|2√

~2Mω

( ∞∑n=1

(α∗)n−1αn√n!(n− 1)!

eiωt((n−1)−n)√n+∞∑n=0

(α∗)n+1αn√n!(n+ 1)!

eiωt((n+1)−n)√n+ 1)

= |C0|2√

~2Mω

( ∞∑n=1

α|α|2(n−1)

(n− 1)! e−iωt +∞∑n=0

α∗|α|2n

n! eiωt

)

= |C0|2√

~2Mω

∞∑n=0

|α|n

n!(αe−iωt + α∗eiωt

)=√

~2Mω

(αe−iωt + α∗eiωt

)We then note that 〈p〉t = M∂t 〈x〉t, such that we may write the following:

〈p〉t = i

√~Mω

2 (α∗eiωt − αe−iωt)

(e) We may rearrange this into real and imaginary parts to write it purely in terms of sinusoidal functions. In analogywith a classical system, the “spring” begins at its equilibrium position for a purely imaginary α, and begins at the endof its range of motion for a purely real α. We can thereby see that the real and imaginary parts of α determine theinitial conditions of the system at t = 0, and that the expectation values of position and momentum then oscillate as in aclassical oscillator. This is however still a quantum system, and with further analysis one can show that the Heisenberguncertainty principle is satisfied here; instead of two known values of position and momentum oscillating in time, we havetwo distributions (Gaussian) whose centroids are oscillating as shown above.

66

17.3Prove the following (Dirac notation recommended):

(a) A hermitian operator has real eigenvalues.(b) Eigenvectors of a Hermitian operator with distinct eigenvalues are orthogonal.(c) The operator which transforms an orthonormal set of basis vectors into another is unitary.(d) If A is a Hermitian matrix, then there exists a unitary matrix U such that U†AU is diagonal.(e) If Ω and Λ are two commuting matrices/operators, they can be simultaneously diagonalized.

Solution

(a) Let A be a Hermitian operator, such that A = A†. Let |a〉 be an eigenvector/eigenstate of A with eigenvalue a.Then:

A|a〉 = a|a〉 → 〈a |A| a〉 = a 〈a|a〉 and 〈a|A† = 〈a|A = a∗〈a| → 〈a |A| a〉 = a∗ 〈a|a〉By subtracting the results of each expression we obtain (a− a∗) 〈a|a〉 = 0. Since the inner product of a vector with itselfis positive semi-definite (i.e. 〈a|a〉 ≥ 0), for the relationship to hold for any |a〉 and a, we must have a = a∗, meaning thata is real.

(b) Let A be a hermitian operator and |a1〉 and |a2〉 be eigenstates of A with eigenvalues a1 and a2, respectively. Sincethe eigenvalues are distinct, we require a1 6= a2. Then we may do the following:

A|a1〉 = a1|a1〉 → 〈a2 |A| a1〉 = a1 〈a2|a1〉 and A|a2〉 = a2|a2〉 → 〈a2|A = a2〈a2| → 〈a2 |A| a1〉 = a2 〈a2|a1〉

We again take the difference of the results, to obtain the equation (a1 − a2) 〈a2|a1〉 = 0. Since a1 6= a2 by assumption, wemust have 〈a2|a1〉 = 0.

(c) We define a set of orthonormal basis vectors |ei〉, and let U be the operator which takes it to another set oforthonormal basis vectors |ai〉. Then:

|ai〉 = U |ei〉 → 〈aj | = 〈ej |U† → 〈aj |ai〉 =⟨ej∣∣U†U ∣∣ ei⟩

Since both sets of vectors are orthonormal we have 〈ωj |ωi〉 = δij = 〈ej |ei〉. Then the above is true when U†U = I, or U isunitary.

(d) Let |ei〉 be the standard orthonormal basis, and |ai〉 be an orthonormal eigenbasis of the operator A. U is thematrix that transforms the standard basis to the eigenbasis. Then:

|ai〉 = U |ei〉 and A|ai〉 = ai|ai〉 → 〈aj |A| ai〉 = ai 〈aj |ai〉 = aiδij

=⟨ej∣∣U†AU ∣∣ ei⟩ = aiδij

This shows that U†AU is diagonal, with the eigenvalues of A as the matrix elements down the diagonal. Then Udiagonalizes A.

(e) Let Ω and Λ be commuting operators ([Ω,Λ] = 0), and |ωi〉 represent the complete set of eigenstates of Ωcorresponding to eigenvalues ωi. Since [Ω,Λ] = 0, we may write (ΩΛ− ΛΩ)|ωi〉 = 0. From here we do the following:

ΩΛ|ωi〉 = ΛΩ|ωi〉 = ωi(Λ|ωi〉)

Since Ω(Λ|ωi〉) = ωi(Λ|ωi〉) we see that Λ|ωi〉 is also an eigenvector of Ω with eigenvalue ωi. This is only possible ifΛ|ωi〉 = λi|ωi〉. This means that all eigenstates of Ω are also eigenstates of Λ, and that the same matrix which diagonalizesΩ would diagonalize Λ.

17.4The following are properties, definitions, and relationships which apply to n-dimensional (discrete) vector spaces for vectors|a〉 and |b〉. Write the infinite-dimensional (continuous) vector space analog of each of the following for functions f(t) andg(t) / states |f〉 and |g〉.

(a) Projection over a basis: |a〉 =∑i |ei〉〈ei|a〉. (Hint: ask questions until you understand the notation f(x) = 〈x|f〉.)

(b) Inner Product: 〈a|b〉 =∑i a∗i bi =

∑i 〈a|ei〉〈ei|b〉.

(c) Matrix transformations: |a′〉 = M |a〉 → a′i =∑jMijaj ↔ 〈ei|a′〉 =

∑j 〈ei |M | ej〉〈ej |a〉.

(d) Unitary transformations: U†U = I →∑i U∗ijUik = δjk. (Note that for |a′〉 = U |a〉 and |b′〉 = U |b〉 this insures

that 〈a′|b′〉 =⟨a∣∣U†U ∣∣ b⟩ = 〈a|b〉.)

(e) Use parts (c) and (d) to relate the fourier series and transform (what is the kernel?), and Parseval’s theorem.What property of the Fourier transform does Parseval’s theorem imply?

67

Solution

Finite-Dimensional / Discrete Expressions Infinite-Dimensional / Continuous ExpressionsProjection over a discrete basis:

|a〉 =∑i

|ei〉〈ei|a〉

Projection over a continuous basis:

|f〉 =∫dt|t〉〈t|f〉 =

∫dtf(t)|t〉

Inner/Dot Product generalized to complex vectors:

〈a|b〉 = a† · b =∑i

a∗i bi

Inner Product for complex functions:

〈f |g〉 =∫dt 〈f |t〉〈t|g〉 =

∫dtf∗(t)g(t)

Matrix transformations: |a′〉 = M |a〉

a′i =∑j

Mijaj ↔ 〈ei|a′〉 =∑j

〈ei |M | ej〉〈ej |a〉

Integral Transformations: |f〉 = M |f〉 (M → Kernel)

f(τ) =⟨τ |f⟩

=∫dt 〈τ |M | t〉〈t|f〉 =

∫dtM(τ, t)f(t)

Unitary Transformations: 〈a|b〉 =⟨a∣∣U†U ∣∣ b⟩ = 〈a′|b′〉

True iff∑i U∗ijUik = δjk

〈a′|b′〉 =∑i

⟨a∣∣U†∣∣ i⟩ 〈i |U | b〉

=∑i

∑j

〈a|j〉⟨j∣∣U†∣∣ i⟩

(∑k

〈i |U | k〉〈k|b〉

)

=∑i

a′∗i b′i =

∑i

∑j

(Uijaj)∗∑k

(Uikbk)

=∑jk

a∗j bk∑i

U∗ijUik = 〈a|b〉 iff U is Unitary.

Unitary Transformations: 〈f |g〉 =⟨f∣∣U†U ∣∣ g⟩ =

⟨f |g⟩

True iff∫dτU(τ, t1)∗U(τ, t2) = δ(t2 − t1)

⟨f |g⟩

=∫dτ⟨f∣∣U†∣∣ τ⟩ 〈τ |U | g〉

=∫dτ

(∫dt1 〈f |t1〉

⟨t1∣∣U†∣∣ τ⟩)(∫ dt2 〈τ |U | t2〉〈t2|g〉

)=∫dτ

(∫dt1(U(τ, t1)f(t1))∗

)(∫dt2U(τ, t2)g(t2)

)=∫dt1dt2f

∗(t1)g(t2)∫dτU∗(τ, t1)U(τ, t2)

= 〈f |g〉 iff U is Unitary.

(e)

17.5(a) Consider an N -dimensional isotropic (all frequencies equal) SHO. Write down a general expression for the energy

eigenvalues, and list the first five energies and the degeneracies of each.(b) Consider the 3-dimensional anisotropic oscillator described by the Hamiltonian:

H = p2

2m + 12mω

2(x2 + 4y2 + 9z2)

Write down a general expression for the energy eigenvalues, and list the first eight energies and degeneracies of each.

Solution

(a)

E = ~ω

(N

2 +N∑i

ni

)We list the states below, and Cnk = n!

k!(n−k)! is the binomial coefficient which reads N choose k.

68

Energy Ways to obtain E Degeneracy (No. of Distinguishable states with E)~ωN2 all ni = 0 1 (Non-degenerate)

~ω(N2 + 1) (1,0...) N

~ω(N2 + 2) (2,0...),(1,1,0...) N + CN2 for N ≥ 2~ω(N2 + 3) (3,0...),(2,1,0...),(1,1,1,0...) N + 2CN2 + CN3 for N ≥ 3~ω(N2 + 4) (4,0...)(3,1,0...),(2,2,0...),(2,1,1,0...),(1,1,1,1,0...) N + 2CN2 + CN3 + CN4 for N ≥ 5

(b)

E = ~ω(nx + 2ny + 3nz + 3

2

)Energy (nx, ny, nz) No. of combinations

3~ω2 all 0 1

5~ω2 (1,0,0) 1

7~ω2 (2,0,0),(0,1,0) 2

9~ω2 (3,0,0),(1,1,0),(0,0,1) 3

11~ω2 (4,0,0),(0,2,0),(2,1,0),(1,0,1) 4

13~ω2 (5,0,0),(3,1,0),(1,2,0),(2,0,1),(0,1,1) 5

15~ω2 (6,0,0),(4,1,0),(2,2,0),(0,3,0),(1,1,1),(0,0,2),(3,0,1) 7

17~ω2 (7,0,0),(5,1,0),(3,2,0),(1,3,0),(2,1,1),(1,0,2),(4,0,1),(0,2,1) 8

17.6(a)

Solution

18 Old Problems (Retired)

18.1 Problem 2Consider a particle with E > 0 moving to the right subject to the potential V (x), where the constant V > 0.

V (x) =

0 x ∈ (−∞, 0]V x ∈ (0,∞)

Determine the transmission and reflection coefficients for the following cases:(a) E > V , (b) E V , (c) E < V .(d) Repeat part (a) for a particle moving to the left.

Solution

Let region 1 be for x < 0, and region 2 be for x > 0.(a) We define a ψ1 and ψ2 in each. (These are solutions to the Schrödinger equation for the the given potential. We

have already normalized the incoming probability amplitude by setting that coefficient to be one.)

ψ1(x) = eik1x +Be−ik1x ψ2(x) = Ceik2x

We require that the wavefunction be continuous and have a continuous derivative at the boundary, resulting in the followingboundary conditions. (We take the derivative to be continuous here, unlike with the delta function potential, because thediscontinuity in V is finite in this case.) We then solve the system of equations for the coefficients C and B.

1 +B = C , k1 −Bk1 = Ck2 → k1 −Bk1 = (1 +B)k2 → B = k1 − k2

k1 + k2and C = 2k1

k1 + k2

69

We then calculate the transmission coefficient T and reflection coefficient R from the coefficients.

T = k2

k1|C|2 = 4k1k2

(k1 + k2)2 R = |B|2 = (k1 − k2)2

(k1 + k2)2

A good check that we have not made a mistake is verifying that T +R = 1.(b) In the limit E V we have k1 ≈ k2. We see that T → 1 and R→ 0 in this case.(c) Region 2 has a decaying solution, which forces T = 0 and R = 1.(d) We now have:

ψ1(x) = Ce−ik1x and ψ2(x) = e−ik2x +Beik2x

We set the usual boundary conditions to obtain:

C = 1 +B − k1C = −k2 +Bk2

We note that these are the same conditions as in part (a), except that we have just interchanged k1 and k2. The resultsare therefore:

T = 4k1k2

(k1 + k2)2 R = (k2 − k1)2

(k1 + k2)2

70

1.1Problem 1 - Department of Physics and … = a,α 2,3 = −a,andβ 1 = −b, β 2,3 = b....

Documents

Transcript of 1.1Problem 1 - Department of Physics and … = a,α 2,3 = −a,andβ 1 = −b, β 2,3 = b....