Problem Set 6 Solution

Problem 1 Two dimensional Harmonic OscillatorGiven a spinless particle of mass m confined to move in the xy plane under the influence of harmonic oscillatorpotential

V (r) = V (x, y) =1

2mw2(x2 + y2) =

1

2mw2r2 ∀x, y ∈ R, r > 0

Infinite paraboloid (indpendent of polar angle θ)

(a) The Hamiltonian

H =p2

2m+ V (x, y) = − h2

2m∇2 + V (x, y) = − h2

2m

(∂2

∂x2+

∂2

∂y2

)+

1

2mw2(x2 + y2)

=

(− h2

2m

∂2

∂x2+

1

2mw2x2

)+

(− h2

2m

∂2

∂y2+

1

2mw2y2

)=

(− h2

2m∇2x +

1

2mw2x2

)+

(− h2

2m∇2y +

1

2mw2y2

)=

(p2x2m

+ Vx(x)

)+

(p2y2m

+ Vy(y)

)= Hx + Hy

where Hx =p2x2m

+ Vx(x) ∀x ∈ R and Hy =p2y2m

+ Vy(y) ∀y ∈ R.

The commutator between the Hamiltonian and angular momentum:

[H, Lz] = [Hx + Hy, Lz] = [Hx, Lz] + [Hy, Lz] = 0

since

[Hx, Lz] =

[p2x2m

+1

2mw2x2, Lz

]=

[p2x2m

, Lz

]+

[1

2mw2x2, Lz

]=

1

2m

[p2x, Lz

]+

1

2mw2

[x2, Lz

]=

1

2m

(px[px, Lz] + [px, Lz]px

)+

1

2mw2

(x[x, Lz] + [x, Lz]x

)= − ih

2m(pxpy + pypx)− ih

2mw2 (xy + yx)

= −ih(pxpym

+mw2xy

)

and in a similar manner [Hy, Lz] = ih

(pypxm

+mw2yx

). In the above working, we have use the results[

x, Lz

]= −ihy and [px, Lz] = −ihpy[

y, Lz

]= ihx and [py, Lz] = ihpx

proven in Problem Set 3 Question 1.

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(b) ClaimSuppose Hψ = Eψ. If H = Hx + Hy, then ψ = ψ(x, y) = ψnx

(x)ψny(y) = ψnx

ψnyis an eigenfunction of

H with eigenvalue E = Enx+Eny

where ψnx, ψny

are eigenfunctions of Hx, Hy with eigenvalues Enx, Eny

respectively.Proof of Claim

Hψ =(Hx + Hy

)ψnx

ψny= Hxψnx

ψny+Hyψny

ψnx= Enx

ψnxψny

+Enyψny

ψnx=(Enx

+ Eny

)ψnx

ψny= Eψ

As a result,

En = Enx,ny = Enx + Eny =

(nx +

1

2

)hw +

(ny +

1

2

)hw = (nx + ny + 1) hw = (n+ 1)hw

where n = nx + ny with nx, ny = 0, 1, 2, ...

(c) Energies and their degeneracy gn

E0 = E00 = hw (ground state) g0 = 1E1 = E10 = E01 = 2hw (first excited state) g1 = 2E2 = E11 = E20 = E02 = 3hw (second excited state) g2 = 3E3 = E30 = E03 = E21 = E12 = 4hw (third excited state) g3 = 4

(d) From part(c), the degeneracy for nth excited state gn = n+ 1.

(e) V (x, y) =1

2mw2(x2 + y2) =

1

2mw2r2 where r2 = x2 + y2.

The time independent Schrodinger equation in 2D polar coordinates:(− h2

2m∇2 +

1

2mw2r2

)ψ = Hψ = Eψ

=⇒(− h2

2m

[1

r

∂

∂r

(r∂

∂r

)+

1

r2∂2

∂θ2

]+

1

2mw2r2

)ψ = Eψ

=⇒ − h2

2m

[1

r

∂

∂r

(r∂ψ

∂r

)+

1

r2∂2ψ

∂θ2

]+

1

2mw2r2ψ = Eψ (1)

Multiply equation (1) throughout by r2 and rearranging we have(− h2

2m

)r∂

∂r

(r∂ψ

∂r

)+

1

2mw2r4ψ − Er2ψ =

h2

2m

∂2ψ

∂θ2(2)

Suppose we want the wavefunction to be variable separable at ψ = ψ(r, θ) = R(r)Θ(θ) = RΘ. Substituteinto equation (2) we have[(

− h2

2m

)rd

dr

(rdR

dr

)+

1

2mw2r4R− Er2R

]Θ =

h2

2mRd2Θ

dθ2

Divide throughout by RΘ[(− h2

2m

)rd

dr

(rdR

dr

)+

1

2mw2r4R− Er2R

]1

R=

h2

2m

1

Θ

d2Θ

dθ2= C (3)

where C is a constant. Equation (3) has to be equal to a constant as the left hand is a function of r onlywhile the right hand side is a function θ only. Hence[(

− h2

2m

)rd

dr

(rd

dr

)+

1

2mw2r4 − Er2 − C

]R = 0 (4)

[d2

dθ2− 2mC

h2

]Θ = 0 (5)

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(f) Solving for Θ(θ) in the angular equation (5)

d2Θ

dθ2− 2mC

h2Θ = 0 =⇒ Θ(θ) = Aeikθ +Be−ikθ

where k =

√2mC

hand A,B ∈ C are constants of integration. Since the potential V (r, θ) =

1

2mw2r2

is independent of θ, the wavefunction ψ and thus Θ(θ) must be θ independent also =⇒ Θ(θ) = 1 after

normalization.

(g) Equation (4) in part(e) is the radial equation that R(r) need to satisfy.

Alternative Solution for (e) and (f)

Recall that the harmonic oscillator energy eigenfunctions in cartesian coordinates are:

ψnx(x) =

e−x2/2x2

0√√π2nxnx!x0

Hnx

(x

x0

)and ψny

(y) =e−y

2/2y20√√π2nyny! y0

Hny

(y

y0

)

where x0 = y0 =

√h

mw= r0 and Hnx , Hny are the Hermite polynomials

Hnx(x) = (−1)nxex2 dnx

dxnxe−x

2

and Hny (y) = (−1)nyey2 dny

dynye−y

2

From the claim in part (b), we know that

ψnxny(x, y) = ψnx

(x)ψny(y) =

e−(x2+y2)/2r20

r0√π2nx+nynx!ny!

Hnx

(x

r0

)Hny

(y

r0

)= (−1)nx+nyex

2+y2 dnx

dxnx

dny

dynye−(x

2+y2) = (−1)ner2 dnx

dxnx

dny

dynye−r

2

= Rnxny(r)Θ(θ) = ψnxny

(r, θ)

where Rnxny (r) = (−1)ner2 dnx

dxnx

dny

dynye−r

2

and Θ(θ) = 1 .

Note The wavefunction should be independent of θ as the potential V (r) =1

2mw2r2 is θ independent.

Problem 2 Harmonic Oscillator + Infinite Square WellGiven a particle of mass m moves in the xy plane in the potential

Harmonic Oscillator and Infinite Square Well Potential

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(a) Time independent Schrodinger equation

Eψ = Hψ =

[− h2

2m∇2 + V (x, y)

]ψ =

[− h2

2m

(∂2

∂x2+

∂2

∂y2

)+

1

2mw2y2

]ψ

=

[− h2

2m

∂2

∂x2− h2

2m

∂2

∂y2+

1

2mw2y2

]ψ =

(Hx + Hy

)ψ

where

Hx = − h2

2m

∂2

∂x2for 0 ≤ x ≤ a (infinite square well of length a)

Hy = − h2

2m

∂2

∂y2+

1

2mw2y2 for all y (harmonic oscillator)

(b) From Problem 1(b) claim, we have

E = Enx + Eny =n2xπ

2h2

2ma2+

(ny +

1

2

)hw

ψ(x, y) = ψnx(x)ψny

(y) =

[√2

asin(nxπ

ax)][ e−y

2/2y20√√π2nyny! y0

Hny

(y

y0

)]

where nx = 1, 2, 3, ...;ny = 0, 1, 2, ...; y0 =

√h

mwandHny

(y) = (−1)nyey2 dny

dynye−y

2

is the Hermite

polynomials. Observe that

∫ψ(x, y)dA =

∞∫−∞

a∫0

ψnx(x)ψny

(y)dxdy =

a∫0

ψnx(x)dx

∞∫−∞

ψny(y)dy = 1

Hence ψ(x, y) is a normalised eigenfunction as ψnx(x) and ψny

(y) are also normalised eigenfunctions.

Problem 3 Given the radial momentum operator pr = −ih1

r

∂

∂rr.

[r, pr]ψ = (rpr − pr r)ψ =

(r

[−ih1

r

∂

∂rr

]−[−ih1

r

∂

∂rr

]r

)ψ = −ih

(∂

∂rr − 1

r

∂

∂rr2)ψ

= −ih(∂

∂r(rψ)− 1

r

∂

∂r

(r2ψ

))= −ih

(ψ + r

∂ψ

∂r− 1

r

[2rψ + r2

∂ψ

∂r

])= ihψ =⇒ [r, pr] = ih

Problem 4 Given a hydrogen atom in the energy eigenstate ψ4,3,3(r) i.e. n = 4 and l = m = 3.

(a) The hydrogen atom has energy E4 =E1

42where E1 = − me4

32h2π2ε20the ground state energy.

Recall that for each n, there are n possible values of l = 0, 1, 2, ..., n − 1 and for each l, there are 2l + 1possible values of m = −l,−l+1, ..., 0, ...l−1, l. Hence for a given energy state n, its degeneracy, i.e. totalnumber of possible (n, l,m) is

d =

n−1∑l=0

(2l + 1) = 2

n−1∑l=0

l +

n−1∑l=0

1 = 2

n−1∑l=1

l + n = 2

(n− 1

2

)(1 + n− 1) + n = n2

For n = 4 we have degeneracy d = 42 = 16 with degenerated states:

l = 0 (4, 0, 0)l = 1 (4, 1,−1), (4, 1, 0), (4, 1, 1)l = 2 (4, 2,−2), (4, 2,−1), (4, 2, 0), (4, 2, 1), (4, 2, 2)l = 3 (4, 3,−3), (4, 3,−2), (4, 3,−1), (4, 3, 0), (4, 3, 1), (4, 3, 2), (4, 3, 3)

Suppose we consider for each m, there are two possible spins (spin up and spin down), then the degeneracy

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d = 2n2

Therefore, the inner most orbit of the hydrogen atom (ground state) with n = 1 can hold up to 2 electrons.The next higher orbit with n = 2 can hold up to 8 electrons, etc.

(b) Recall L2(ψn,l,m) = l(l + 1)h2ψn,l,m. With l = 3, the magnitude of the orbital angular momentum of the

electron around the nucleus is L =√l(l + 1)h = 2

√3h.

(c) Recall Lz(ψn,l,m) = mhψn,l,m. With m = 3, the z-component of the angular momentum is Lz = mh = 3h.Thus, the angle between the orbital angular momentum vector and the z-axis is

cos θ =LzL

=|m|h√l(l + 1)h

=|m|√l(l + 1)

=3√

3(3 + 1)=

√3

2=⇒ θ = 300

From part(a), for n = 4 =⇒ l = 0, 1, 2, 3 and there are 16 states that have the same energy.

Case 1 n = 4, l = 0,m = 0 =⇒ cos θ =|m|√l(l + 1)

=0

0−→∞ =⇒ θ does not exist.

Case 2 n = 4, l = 1,m = −1, 0, 1

(i) m = 0 =⇒ cos θ =|m|√l(l + 1)

=0√2

=⇒ θ = 900

(ii) m = ±1 =⇒ cos θ =|m|√l(l + 1)

=1√2

=⇒ θ = 450

Case 3 n = 4, l = 2,m = −2,−1, 0, 1, 2

(i) m = 0 =⇒ cos θ =|m|√l(l + 1)

=0√6

=⇒ θ = 900

(ii) m = ±1 =⇒ cos θ =|m|√l(l + 1)

=1√6

=⇒ θ ≈ 65.90

(iii) m = ±2 =⇒ cos θ =|m|√l(l + 1)

=2√6

=⇒ θ ≈ 35.30

Case 4 n = 4, l = 3,m = −3,−2,−1, 0, 1, 2, 3

(i) m = 0 =⇒ cos θ =|m|√l(l + 1)

=0

2√

3=⇒ θ = 900

(ii) m = ±1 =⇒ cos θ =|m|√l(l + 1)

=1

2√

3=⇒ θ ≈ 73.20

(iii) m = ±2 =⇒ cos θ =|m|√l(l + 1)

=2

2√

3=⇒ θ ≈ 54.70

(iv) m = ±3 =⇒ cos θ =|m|√l(l + 1)

=3

2√

3=⇒ θ = 300

Hence possible angles for all the states with n = 4 are θ = 00, 300, 35.30, 450, 54.70, 65.90, 73.20, 900

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Problem 5 Given an electron in a hydrogen atom is in the energy eigenstate

ψ2,1,−1(r, θ, φ) = Nre−r/2a0Y1,−1 = R(r)Y (θ, φ)

where R(r) = Nre−r/2a0 is the radial wave function and Y (θ, φ) = Y1,−1 is the angular wave function.

(a) Normalisation

1 =

∫ψ∗ψdV =

∫|ψ|2dV = |N |2

2π∫0

π∫0

∞∫0

(r2e−r/a0 |Y1,−1|2

)r2 sin θdrdθdφ

= |N |2∞∫0

r4e−r/a0dr

2π∫0

π∫0

|Y1,−1|2sin θdθdφ = |N |2∞∫0

r4e−r/a0dr = |N |2∞∫0

(a0x)4e−xa0dx

= |N |2a50

∞∫0

x4e−xdx = |N |2a50Γ(5) = |N |2a504! = 24a50|N |2=⇒ N =1√24a50

(assuming N ∈ R).

where we have use the substitution r/a0 = x and the following results:

(i) The volume element in spherical coordinates is dV = r2 sin θdrdθdφ where θ and φ are the polar

and azimuthal angle respectively.

(ii) The Gamma function Γ(z) =

∞∫0

xz−1e−xdx and its property Γ(n) = (n− 1)!.

(iii) The spherical harmonics Yl,m are orthonormal functions, i.e.

2π∫0

π∫0

Y ∗l,mYn,k sin θdθdφ = 〈Yl,m|Yn,k〉 = δlnδml =

{1 if l = n and m = k0 otherwise

In general, given a wave function in spherical coordinates ψ(r, θ, φ) where we can separate the radialcomponent from its angular components, we can write:

ψ(r, θ, φ) = R(r)Y (θ, φ)

where R(r) is the radial wave function and Y (θ, φ) is the angular wave function. We usually capturethe normalization constant in the radial wave function. Observe that both the radial and angular wavefunctions can be complex functions. Normalization leads us to

1 =

∫|ψ|2dV =

2π∫0

π∫0

∞∫0

|R(r)|2|Y (θ, φ)|2r2 sin θdrdθdφ

=⇒∞∫0

r2|R(r)|2dr = 1 and

2π∫0

π∫0

sin θ|Y (θ, φ)|2dθdφ = 1

This is known as the normalization of the radial and angular wave function. Normalized angular wavefunction in hydrogen atom Y (θ, φ) = Yl,m are the spherical harmonics. The function

ρ(r) = r2|R(r)|2 where r > 0

is the probability density function (p.d.f) for the radial component and

P (a < r < b) =

b∫a

ρ(r)dr =

b∫a

r2|R(r)|2dr

gives the probability of finding the particle in between a and b.

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(b) Probability per unit volume function: |ψ|2= ψ∗ψ

|ψ2,1,−1|2= |Nre−r/2a0Y1,−1|2=r2e−r/a0

24a50|Y1,−1|2=

r2e−r/a0

24a50

(3

8πsin2 θ

)=r2 sin2 θe−r/a0

64πa50

where Y1,−1 =

√3

8πsin θe−iφ. Hence

|ψ2,1,−1(a0, 450, 600)|2=a20 sin2(450)e−a0/a0

64πa50=

1

128πa30e

(c) From above, probability per unit radial interval is

ρ(r) = r2|R(r)|2=r4e−r/a0

24a50

Thus finding the electron at a distance r = 2a0 per unit radial interval will be

ρ(2a0) =(2a0)4e−2

24a50=

2

3a0e2

Problem 6

Given an electron in a hydrogen atom ground state:

ψ100(r, θ, φ) =1√πa30

e−r/a0 =

(2√a30e−r/a0

)1√4π

= R10(r)Y00(θ, φ)

where R10(r) =2√a30e−r/a0 is the radial wave function. The angular wave function Y00(θ, φ) =

1√4π

is the

spherical harmonics and a0 is the Bohr radius.

(a) The most probable position for the electron is the value of r that maximize the radial probability densityfunction

ρ(r) = r2|R10(r)|2=4

a30r2e−2r/a0

Using product rule, the stationary points are

0 =dρ

dr=

8

a30re−2r/a0

(1− r

a0

)=⇒ r = a0 as 0 < r <∞

One can check by second order differentiation that r = a0 gives a maximum value for ρ(r). Hence themost probable position of the electron is at the Bohr radius a0.

(b) Average position of electron in hydrogen atom ground state.

〈r〉 =

∞∫0

rρ(r)dr =4

a30

∞∫0

r3e−2r/a0dr =4

a30

∞∫0

(a02x)3e−x

(a02

)dx =

a04

∞∫0

x3e−xdx =a04

Γ(4) =3!

4a0 =

3

2a0

where we have use the substitution x =2r

a0and the gamma function Γ(z) =

∞∫0

xz−1e−xdx and its property

Γ(n) = (n− 1)!.

Bohr radius is the most probable distance of an electron from the nucleus of a hydrogen atom in its ground state.

The value is a0 =4πε0h

2

me2≈ 5.29177× 10−11m where m and e is the mass and charge of electron respectively.

It is not the same as the average distance of the electron from the nucleus of a hydrogen atom in its groundstate.

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