Three Dimensional systems

In three dimensions the time-independent Schrodinger Equation becomes[− ~2

2m∆r + V (r)

]ψ(r) = Eψ(r) ,

∆r =∂2

∂x2+

∂2

∂x2+

∂2

∂x2.

The easiest case one can imagine in 3D is when the potential energy is the sum of three independentterms V (r) = Vx(x) + Vy(y) + Vz(z). In this case we seek the solution as the product of threeindependent functions

ψ(r) = X(x)Y (y)Z(z) .

Substituting this anzats to the Schrodinger Equation and then dividing it by ψ we find

1

X(x)

[− 1

2mX ′′(x) + Vx(x)

]+

1

Y (y)

[− 1

2mY ′′(y) + Vy(y)

]+

1

Z(z)

[− 1

2mZ ′′(z) + Vz(z)

]= E

Each of the three terms on the l.h.s. is a function of a different variable, either x, or y, or z. Thisis only possible if each of them separately is a constant equal to E(x), E(y), and E(z), respectively.In other words, we are dealing with three separate 1D problems, and the total energy is the sumof three independent terms E = E(x) + E(y) + E(z).

The two cases which fall in this category are the parabolic potentials, V = (m/2)∑3

α=1 ω2αr

2α,

and the square box, V =∑

α V(sq.box)α (rα). [To simplify formulae, it makes sense to use notations

x = r1, y = r2, and z = r3 from time to time.] Solving them is no different from solving one-dimensional problems one-by-one. As an application, consider particles confined in the infinite 3Dsquare well with linear sizes x ∈ (0, Lx), y ∈ (0, Ly), z ∈ (0, Lz). We immediately jump to thefinal answers for the wavefunctions

ψn1,n2,n3 =∏α

√2

Lαsin

(πnαrαLα

), nα = 1, 2, 3, . . .

and spectrum (below n = (n1, n2, n3) is a vector with integer components)

En =k2

2m, kα =

~πLα

nα . (1)

Let us pretend that this is a model of some complex system where we replace interactionsbetween the particles with the mean-field potential which holds them together inside the volumeV = Lx×Ly×Lz. Let N of the particles are fermions with light mass m and spin 1/2. Accordingto the Pauli principle each energy level can be occupied only with two particles, one with spin up,and the other with spin down. We then ask the same question as before: ”What is the energy ofthe last occupied energy level, or Fermi energy εF , and total ground state energy of all fermionsin the box?”. We proceed along the same lines as in the discussion of 1D rings, and apply thesame approximations when replacing sums over integer index with momentum integrals for smoothfunctions. First, the condition of accommodating all particles is now (a factor of 2 is for spin index)

N = 2

En≤εF∑n1,n2,n3=1

≈ 2

∫ En≤εF(∏

α

θ(nα)

)dn ,

1

where the θ(x) function is filtering positive values of x only; it is unity for x > 0 and zero otherwise.We may allow for both positive and negative values of nα in this integral, but then it has to bedivided by 8 to have the same value

N = 21

8

∫ En≤εFdn = 2V

∫ k2/2m≤εF dk

(2π~)3.

In the last relation we went from integration over n to integration over k—the two are linearlyrelated to each other, see Eq. (1). Taking an integral we get

nf =N

V=

2 (4π)

(2π~)3k3F3, εF =

k2F2m

,

which defines the relation between the Fermi-momentum and fermion density

kF = ~(3π2nf )1/3 .

This relation happens to be protected by theorems to be interaction independent. Also, in thefinal form, this expression is valid for any shape of the well as long as the number of fermions isvery large.

States with energies Ek = εF form a ’Fermi-surface’ in momentum space. [It is a sphere here,but in condensed matter it can have other shapes, including ’plumber’s nightmare’ ones withnumerous ’valleys’, ’pockets’, ’necks’). The total energy of all fermions siting below the Fermisurface is

Etot = 2V

∫ k≤kF k2

2m

dk

(2π~)3=

V

π2~3k5F

10m=

3~2

10mV 2/3(3π2N)2/3N .

The average energy per particle is then

Etot/N =3k2F10m

=3

5εF

These are the core results for the ’Fermi gas’ model, no matter in what field of physics it happensto be a reasonable approximation.

In astrophysics, stability of white dwarf and neutron stars is based on counter-acting the gravitypull with the Fermi pressure

PF = −dEtot/dV = −~2(3π2)2/3

5m(N/V )5/3

In white dwarf stars the Fermi pressure is mostly due to electrons, in neutron stars, it is due toneutrons. In both cases this Fermi pressure has to balance the gravity pressure

PG = −dEgrdV

, Egr = −∫ R

0

G(4πR2dR ρ)(4πR3ρ/3)

R= −3

5

GM2

R= −3

5

(4π

3

)1/3 GM2

V 1/3,

where R is the star radius, ρ is its mass density, and M = ρ4πR3/3 = ρV is its total mass. Withthis

PG =1

5

(4π

3

)1/3 GM2

V 4/3.

2

From PF +PG = 0 one determines stability conditions, i.e. the radius of the star given its fermionnumber N and total mass M . In non-relativistic case PF increases faster than PG as volumeshrinks and thus there is always a solution for the radius. When kF /m becomes comparable withthe velocity of light one has to start using the relativistic dispersion relation which is less steepand thus leads to smaller Fermi-pressure for the same particle number and volume. This changesthe law from PF ∝ V −5/3 to PF ∝ V −4/3 in the relativistic limit and for massive enough stars(Chandrasekhar limit) leads to their collapse and black hole physics.

Problem 44. Spin and energy levels in nuclear physicsa). Consider a heavy nucleus consisting of Z = Np protons and N = Nn neutrons so that thetotal atomic number is A = Z +N . If the nucleons each have an elementary impenetrable volumeV0 and are packed so that they are ’just touching’, show that the volume of such a nucleus will goas V ∝ V0A, and thus the radius of the nucleus will be R = r0A

1/3. with r0 = 1.2 ÷ 1.4 fm thislaw works well for most nucleib). Instead of dealing with interactions between nucleons, model this system as A nucleons con-fined within the 3D infinite well of volume V = 4πR3/3. Using expressions from the notes find thetotal ground state energy of a system of A = Z + N nucleons, i.e. E(Z,N), recalling that bothprotons and neutrons are spin-1/2 fermions. For simplicity assume that protons and neutrons havethe same mass, mc2 = 940 Mev.c). For fixed A, minimize this energy, and show that equal numbers of neutrons and protons arefavored.d). Assuming that N − Z � A, show that the ground state energy can be approximated by anexpression of the form

E = Emin + Esym(N − Z)2/A+ . . . ,

and find an expression for Esym. Using the values above, show that the numerical value of Esym isroughly 12 Mev. Note: This contribution to the total nuclear energy (or rest mass) is often calledthe symmetry contribution and is part of the well-known semi-empirical formula of nuclear physics.

We now turn to another important case of spherically symmetric potentials V (r) = V (r). Todeal with it we introduce spherical coordinates z = r cos θ, x = r sin θ cosϕ, y = r sin θ sinϕ. TheLaplacian operator in spherical coordinates takes the form

∆r =1

r2

[∂

∂rr2∂

∂r+

1

sin θ

∂

∂θsin θ

∂

∂θ+

1

sin2 θ

∂2

∂ϕ2

]We will seek the solution in the form of a product ψ(r) = R(r)χ(θ)Φ(ϕ). Substituting this forminto the Schrodinger Equation and then dividing Schrodinger Equation by ψ we get

− ~2

2Mr2

[1

R(r)

∂

∂r[r2R ′(r)] +

1

sin θχ(θ)

∂

∂θ[sin θ χ ′(θ)] +

1

sin2 θΦ(ϕ)Φ ′′(ϕ)

]+ V (r) = E . (2)

Next we multiply it by r2 sin2 θ

− ~2

2M

sin2 θ

R(r)

∂

∂r[r2R ′(r)] + r2 sin2 θ[V (r)− E]− ~2

2M

sin θ

χ(θ)

∂

∂θ[sin θ χ ′(θ)]− ~2

2M

Φ ′′(ϕ)

Φ(ϕ)= 0 . (3)

3

Again, we observe that the first two terms are independent of ϕ and thus the last term must be aconstant

− ~2

2M

Φ ′′(ϕ)

Φ(ϕ)= const (4)

The solutions of this simple equation are simple exponential functions eikϕ subject to periodicboundary condition for the ϕ variable since point ϕ + 2π is the same as ϕ. Similarly to oursolution for the 1D ring, we must then have k = m=integer, i.e.

Φm(ϕ) =

√1

2πeimϕ , m = 0,±1,±2, . . .

Correspondingly, the value of the constant in Eq. (4) is ~2m2/2M (the notation for the particlemass was changed to avoid confusion with the quantum number m for the azimuthal motion).

With one variable less, we return to Eq. (3) and divide it by sin2 θ to get[− ~2

2MR(r)

∂

∂r[r2R ′(r)] + r2[V (r)− E]

]− ~2

2M

[1

sin θχ(θ)

∂

∂θ[sin θ χ ′(θ)]− m2

sin2 θ

]= 0 . (5)

Same story as before: the first square bracket depends only on r while the second depends onlyon θ. Thus both terms have to be equal to the same constant. Let us denote it by −~2λ/2M andtry to solve first

1

sin θχ(θ)

∂

∂θ[sin θ χ ′(θ)]− m2

sin2 θ= −λ . (6)

It can be simplified by introducing z = cos θ. The derivatives transform as (d/dθ)F (z) =− sin θ(d/dz)F (z) where F (z) = χ(θ). Now we have to solve(

λ− m2

1− z2

)F +

d

dz

((1− z2)F ′

)= 0 . (7)

Since θ ∈ (0, π) we have z ∈ (−1, 1) with 1/(1−z2) diverging at the interval boundary. Hence, it isexpected that F (z) tends to zero at the boundary. To understand how, we look at the equation inthe vicinity of the z = ±1 points. Let x = 1−z2 and f(x) = F (z) with (d/dz) = (dx/dz)(d/dx) =−2z(d/dx) ≈ ∓2(d/dx). Then, taking care of diverging terms,

−m2

xf(x) + 4

d

dx

(xf ′)≈ 0

whose solution is f(x) = x|m|/2 checked by elementary substitution and differentiation.This suggests that we use substitution F (z) = (1 − z2)|m|/2G(z) in Eq. (7) with expectations

that G(z) will be a regular function at z → ±1. The next step is a straightforward substitutionwith a lot of careful differentiation:

(1− z2) ddz

(1− z2)|m|/2G(z) = −z|m|(1− z2)|m|/2G(z) + (1− z2)|m|/2+1G′(z) ,

d

dz(1−z2)|m|/2

[−z|m|G(z) + (1− z2)G′(z)

]= −z|m|(1−z2)|m|/2−1

[−z|m|G(z) + (1− z2)G′(z)

]+

(1− z2)|m|/2[−|m|G(z)− z|m|G′(z)− 2zG′(z) + (1− z2)G′′(z)

].

4

The differential Equation to solve is then

(1− z2)G′′ − 2z(|m|+ 1)G′ + [λ− |m|(|m|+ 1)]G = 0 (8)

Let us introduce an auxiliary notation α = λ−|m|(|m|+1) to simplify things. Next, we employthe same trick we used to solve the harmonic oscillator, i.e we seek the solution as power seriesG =

∑∞n=0 anz

n which we substitute into the equation

∞∑n=0

n(n− 1)(1− z2)zn−2an −∞∑n=0

2zn(|m|+ 1)zn−1an +∞∑n=0

αznan = 0 ,

or, equating coefficients for each power

(n+2)(n+1)an+2−[n(n−1)+2n(|m|+1)−α]an = 0 −→ an+2 =n(n− 1) + 2n(|m|+ 1)− α

(n+ 2)(n+ 1)an .

The recursive relation can be also written as

an+2 =(n+ |m|)(n+ |m|+ 1)− λ

(n+ 2)(n+ 1)an . (9)

This immediately leads to the final solution because unless the recursion relation it terminated atsome finite n and all coefficients are zero afterwards, we would have G(z) ∼ (1− z2)−1 because itis the property of geometrical series in z2 to have an+2 ≈ an. Since divergent solutions are ruledout we must have

λ = (|m|+ n)(|m|+ n+ 1) , n = 0, 1, 2, . . .

which is equivalent to λ = l(l + 1) with l = |m|, |m|+ 1, . . . . Yet another, more canonical, way ofspecifying quantum numbers λ and m is to have

λ = l(l + 1) , with l = 0, 1, 2, . . . and − l ≤ m ≤ l .

For |m| = 0, we have Fl(z) = G(z) = Pl(z) where Pl(z) are the Legendre polynomials whichare conventionally normalized to have Pl(1) = 1 (the reason for this ’strange’ normalization is thesame as with hermitian polynomials—there is a generating function for which Pl(z) are related toTaylor-expansion coefficients). We easily get several functions from our recursion relations

P0 = 1 , P1 = z , P2 =1

2(3z2 − 1) , P3 =

1

2(5z3 − 3z) , P4 =

1

8(35z4 − 30z2 + 3) , etc.

The generating function is g(z, s) = (1− 2zs+ s2)−1/2 =∑∞

n=0 Pn(z)sn.For m 6= 0 we have

Fml (z) = (1− z2)|m|/2Gml (z) = (1− z2)|m|/2 d|m|

dz|m|Pl(z) ≡ P

|m|l (z) , (10)

which is called the Legendre polynomial of the second kind.

Problem 45. Legendre polynomialsCheck that (10) is indeed the solution of Eq. (7), or Eq. (8) for non-zero m (take for granted thatPl(z) solves this equation for m = 0). I suggest that you actually verify that the recursion relation

5

(9) is satisfied.

We have determined the full angle dependence of our solution at this point

χ(θ)Φ(ϕ) ∝ Y ml (θ, ϕ) ,

where Y ml (θ, ϕ) is called the spherical harmonic. With normalization factors in place it reads

(factors of (−1)m are a matter of convention, not principle)

Y ml (θ, ϕ) = (−1)m

√(2l + 1)(l −m)!

4π(l +m)!eimϕ Pml (cos θ) , for m ≥ 0 . (11)

For m < 0 we use Y ml (θ, ϕ) = (−1)m

(Y|m|l (θ, ϕ)

)∗. The normalization condition is such that∫ 2π

0dϕ

∫ π

0sin θdθ

(Y ml (θ, ϕ)

)∗Y m′l′ (θ, ϕ) = δl,l′ δm,m′ .

We proceed now with the formulation of the remaining radial equation for R(r) which is easysince all we have to do is to take Eq. (5) and equate the radial part to −λ~2/2M = −l(l+1)~2/2M .So we do just that and divide this equation by 1/r2 to have

− ~2

2M

1

r2∂

∂r[r2R ′(r)] +

[V (r) +

~2l(l + 1)

2Mr2

]R(r) = ER(r) . (12)

As before, we seek the solution in the form R(r) = u(r)/r to cast it in the form of a 1D SchrodingerEquation

u′′ + 2M

[E − V (r)− l(l + 1)

2Mr2

]u(r) = 0 , (13)

It is the same as for a 1D particle moving on the (0,∞) interval in the potential V (r)+l(l+1)/2Mr2

which includes the centrifugal-type piece. The normalization condition for the wavefunction ψ =R(r)Y m

l (θ, ϕ) is∫ ∞0

r2dr|R(r)|2∫ 2π

0dϕ

∫ π

0sin θdθ |Y m

l (θ, ϕ)|2 =

∫ ∞0

dr|u(r)|2 = 1 .

Angular momentumTo proceed we have to specify the potential. But before we do that, let us discuss a little thephysical meaning of quantum numbers m and l. Consider an operator which is a vector productof the coordinate and momentum, which in classical mechanics stands for the angular momentum

L = r× p = −i~ r×∇ .

In components α = 1, 2, 3 we have

Lα = −i~∑βγ

εαβγrβ∂

∂rγ,

6

where εαβγ in an anti-symmetric tensor such that ε123 = 1 and all other values are obtained fromthe property εβαγ = −εαβγ = −εβγα, i.e. it changes sign when we change any pair of indexesplaces. This, in particular, means that all indexes must be different to have a non-zero element.Let’s take α = 3. Then

L3 = Lz = −i~(x∂

∂y− y ∂

∂x

).

In spherical coordinates

∂

∂ϕ=∑α

∂rα∂ϕ

∂

∂rα= −y ∂

∂x+ x

∂

∂y=i

~Lz ,

meaning that (I will stop using ˆ on the operators)

Lz = −i~ ∂

∂ϕ.

With this we immediately check that

LzΦm(ϕ) = −i~ ∂

∂ϕ

√1

2πeimϕ = ~m Φm(ϕ) ,

i.e. Φm(ϕ) is an eigenfunction of the angular momentum projection on the z-axis with eigenvalue~m. This explains puzzling SG-machine experiments: no matter what axis z is selected by themagnet, one always finds that the measured angular momentum projections form equidistant linesat values ~m.

Next we look at the operator L2 = L · L =∑3

i=1 LiLi:

L2 = −~2∑ijkβγ

εijkεiβγrj∂

∂rkrβ

∂

∂rγ.

But∑

i εijkεiβγ = δjβδkγ − δjγδkβ because(i) j and k must be different from i and from each other, and(ii) β and γ must be different from i and from each other.Hence

L2 = −~2∑jk

rj∂

∂rkrj

∂

∂rk− rj

∂

∂rkrk

∂

∂rj=

−~2∑

jk

r2j∂2

∂r2k− rjrk

∂

∂rk

∂

∂rj+ (1− 3)

∑j

rj∂

∂rj

= −~2(r2∆r − r(r · ∇)∇− 2r · ∇

).

In spherical coordinates r · ∇ = r(∂/∂r) and we find

L2 = −~2(r2∆r − r2

∂2

∂r2− 2r

∂

∂r

)= −~2

(r2∆r − r2

∂2

∂r2− 2r

∂

∂r

)= −~2

(r2∆r −

∂

∂rr2∂

∂r

)=

= −~2r2(

∆r −1

r2∂

∂rr2∂

∂r

)= −~2

(1

sin θ

∂

∂θsin θ

∂

∂θ+

1

sin2 θ

∂2

∂ϕ2

).

7

In other words L2 is identical to the angular dependent part of the Laplace operator times −~2r2which has Y m

l (θ, ϕ) as eigenfunctions with eigenvalues l(l + 1)~2 [see Eqs. (2) and (5)]. Thus

L2Y ml (θ, ϕ) = l(l + 1)~2 Y m

l (θ, ϕ) .

With this we understand the origin of the L2/2Mr2 centrifugal potential in the radial equation.We also understand that the Hamiltonian symmetry is such that the particle angular momentumis conserved classically. Strangely enough, in quantum mechanics this translates into a counterin-tuitive law that the angular momentum projection on one, and only one (!) axis is conserved alongwith the angular momentum squared. This is because angular momentum operators for differentaxis form an algebra of non-commuting operators and thus can not assume well defined values atthe same time. We will deal with it later, after we finish with the radial equation for the hydrogenatom.

Hydrogen atomConsider the most important potential in Nature (for human beings at least) V (r) = −Ze2/r.

As usual we first introduce dimensionless variables (we look for bound states with negative energiesE = −|E|):

ρ =

√2m|E|~

r , γ = Ze2

√2m

~2|E|,

to bring this equation to the ’universal form’

u′′(ρ) +

[γ

ρ− l(l + 1)

ρ2− 1

]u(ρ) = 0 .

We demand that proper solutions go to zero as ρ→∞ and ρ→ 0 otherwise they cannot be eithernormalized or made with finite energy. Again, we check for the asymptotic behavior to factor itout. For ρ→ 0 we look at

u′′(ρ)− l(l + 1)

ρ2u(ρ) = 0 ,

whose solution is u(ρ) ∝ ρl+1 (we disregard the ρ−l solution). For large ρ we have

u′′(ρ)− u(ρ) = 0 ,

whose solution is u(ρ) ∝ e−ρ. Thus we propose to seek the solution in the form

u(ρ) = ρl+1 e−ρv(ρ) .

Next we have to differentiate this construction twice, plug it in, and attempt to find the solutionfor v as Taylor series expansion in ρ — same protocol as before for the harmonic oscillator andspherical functions. Following the drill

u′ = (l + 1)ρl e−ρv − ρl+1 e−ρv + ρl+1 e−ρv′

u′′ = (l+ 1)lρl−1 e−ρv − 2(l+ 1)ρl e−ρv + ρl+1 e−ρv + 2(l+ 1)ρl e−ρv′ − 2ρl+1 e−ρv′ + ρl+1 e−ρv′′ =

ρl+1 e−ρ{v′′ + 2

(l + 1)

ρv′ − 2v′

l(l + 1)

ρ2v + v − 2

(l + 1)

ρv

}.

8

With this small torture the equation for v reads

v′′(ρ) + 2

((l + 1)

ρ− 1

)v′(ρ) +

γ − 2(l + 1)

ρv(ρ) = 0 .

Next we employ v =∑

n anρn to get∑

n

[n(n− 1) + 2(l + 1)n]anρn−2 +

∑n

[−2n+ γ − 2(l + 1)]anρn−1 = 0

[(n+ 1)(n+ 2l + 2)]an+1 + [γ − 2(l + n+ 1)]an = 0 −→ an+1 =2(l + n+ 1)− γ

(n+ 1)(n+ 2l + 2)an .

Next we notice that if the series are not terminated then asymptotically an+1 ≈ (2/n)an whichis characteristic of the exponentially divergent solution e2ρ. Thus the only option we have is todemand that the series are terminated by having

γ = 2ν , with integer ν = l + 1, l + 2, l + 3, . . . (14)

The origin of the periodic table of elements is explained!We found that the spectrum of the hydrogen atom is determined by three quantum numbers:

Principal quantum number: ν = 1, 2, 3, . . .

Orbital Angular momentum: l = 0, 1, 2, . . . , ν − 1

Angular momentum projection: m = −l,−l + 1, . . . , l − 1, l

For this set of quantum numbers, the energy depends only on ν:

Ze2

√2m

~2|E|ν= 2ν −→ Eν = −Z

2e4m

2~2ν2,

reproducing the famous Balmer spectral line series. The degeneracies of the spectrum are suchthat for a given ν there are

ν−1∑l=0

(2l + 1) = ν(ν − 1) + ν = ν2

levels with the same energy which can accommodate Nν = 2ν2 electrons. We find

N1 = 2 , N2 = 8 , N3 = 18 , etc

In spectroscopy, ν = 1 orbitals are known as s-states (one can place 2 electrons on the s-orbital);ν = 2, l = 1 orbitals are known as p-states (one can place 2 electrons in the l = 0 state and8 − 2 = 6 electrons on the p-orbitals); ν = 3, l = 2 orbitals are known as d-states (one can place2, and 6 and electrons in l = 0, 1 states, and 18 − 10 on the d-orbital); other orbitals follow thealphabet f, g, h etc. It becomes evident that the periodic table is reflecting these magic numberscorresponding to filled energy orbitals. Of course, in reality electrons couple to each other by the

9

same Coulomb force and the picture is not that simple. It get scrambled in the many-electronsystem, but the basic structure of ’atomic energy shells’ remains valid. Thus in one stroke QMexplains the order among elements in Nature and many of their chemical properties by calculatingenergies of the last occupied states in atoms.

The dimensionless coordinate ρ is related to r by

ρ =

√2m|E|~

r =Z

ν

r

aB, where aB =

~2

me2is the Bohr radius

Let us now get familiar with the solutions a little closer. For l = 0 states with completespherical symmetry we have Y 0

0 (θ, ϕ) =√

1/4π, while for states with l = 1 the wavefunctionbecomes more elaborate

Y 01 (θ, ϕ) =

√3/4π cos θ , Y 1

1 (θ, ϕ) = −√

3/8π eiϕ sin θ , Y −11 (θ, ϕ) = −√

3/8π e−iϕ sin θ ,

The other way to present them is Y 01 ∝ z/r, Y ±11 ∝ (x ± iy)/

√2r. For large l and m = l the

spherical harmonic is sharply peaked in the θ = π/2 plane. The radial functions

R(r) = uνl(r)/r ∝ rle−Zr/νaB vνl(Zr/νaB) ,

are determined by the Laguerre polynomials studied by mathematicians well before quantummechanics (same with Hermite and Legendre polynomials). Just for the reference

vνl(ρ) = L2l+1ν−l−1(2ρ)

with generating function

F (z, s) = (1− z)−p−1e−sz/(1−z) =

∞∑n=1

Lpn(z)

(n+ p)!sn .

The normalized solutions for the hydrogen atom in terms of standard functions are

ψνlm =

[(2k)3(ν − l − 1)!

2ν[(ν + l)!]3

]1/2(2kr)l e−kr L2l+1

ν−l−1(2kr) Yml (θ, ϕ) , k =

Z

νaB, (15)

with ∫d3rψ∗νlm(r)ψ∗ν′l′m′(r) = δmm′δll′δνν′ .

For example:

Ground state ψ100 =

(Z3

πa3B

)1/2

e−Zr/aB

The expectation value of r, or the atom ’size’ is

〈r〉 =

(4πZ3

πa3B

)∫ ∞0

r3 e−2Zr/aBdr =aB4Z

∫ ∞0

x3 e−xdx = 3!aB4Z

=3

2ZaB .

Problem 46. Variational principle and the Hydrogen atom

10

a). Use the variational principle to estimate the ground state energy of the hydrogn atom by usingthe trial wavefunction

ψ(r, b) = e−r/b ,

where b is the variational parameter. [Note, the state as written is not normalized yet.] Youshould, of course, get the correct answer since this happens to be the right functional form of theexact solution.a). Repeat the calculation, but now use the gaussian form

ψ(r, b) = e−r2/2b2 ,

and find the fractional error made in the energy. Show (numerically) that the overlap of the trueground state wavefunction and the optimized variational estimate is roughly⟨

ψ100

∣∣ψ(r, bopt)⟩≈ 0.9782

Problem 47. Momentum space wavefunctions for Hydrogena). Calculate the momentum space wavefunctions for the two lowest-lying s-wave states of hydro-gen, ψνlm, i.e. for ν = 1 and ν = 2 and l = m = 0. When doing the 3D Fourier transform payattention that there is no preferred direction for s states and thus p-direction can be set to be thez-axis of the integration, i.e. p · r = pr cos θ. To check that you are on the right path compareyour answer to

|ψ100(p)|2 =64πk5

(p2 + k2)4, where k =

Z

aB.

(b) Do not forget to do ψ200(p).

Problem 48. Tritium decayThe nucleus of two neutrons and a single proton (hence with Z = 1) is called triton, t (by analogywith the deuteron, d), and the system consisting of a single electron and a triton is called a tritiumatom. The triton is unstable against radioactive β-decay via the process t→3 He+ e−+ νe where3He consists of one neutron and two protons (i.e. Z = 2). On an atomic time scale, the decayprocess and the ejection of the decay products e− and νe happens almost instantaneously. Aftersuch a decay, an electron initially in the ground state of the tritium atom suddenly finds itself ina Coulomb potential with twice the nuclear charge. Assuming that after decay the 3He nucleusremains statica). What is the probability that an electron that was originally in the ground state will remainin the ground state of the new system?b). In the n = 2s state?c). In some l = 1 state?Hint: this is essentially the same setup as in the ’moving walls’ problem you did before.

Problem 49. Wavefunction at the originConsider a particle bound to a fixed center by a spherically symmetric potential V (r).a). prove that

|ψ(0)|2 =(m

2π

)⟨dVdr

⟩11

for all s-states, ground and excited.Hint: Consider Schrodinger Equation for u(r), multiply by u′ and integrate over r ∈ (0,∞).b). Check this relation for the ground state of a three dimensional isotropic oscillator and thehydrogen atoms. Historically, this relation has been useful in guessing the form of the potentialbetween a quark and an antiquark.

Problem 50. Polarizability of the Hydrogen atomWhen an electric field E0 is applied to a system of charges, a deformation takes place which leadsto an electric dipole moment P = αEE0, where αE is called the linear electric polarizability. Thisleads to an energy shift

∆E = −1

2αEE

20 .

Actually, this relation may be used to define linear polarizability as αE = −(d2∆E/dE20)|0. This

is exactly what we will be doing in this problem.According to the second-order perturbation theory the energy shift of a hydrogen atom in a

constant electric field E0 oriented in the z direction is

∆E = −∑n6=1s

〈1s|eE0z|n〉〈n|eE0z|1s〉En − E1s

,

thus

αE = 2e2∑n6=1s

〈1s|z|n〉〈n|z|1s〉En − E1s

a). Calculate the leading (|2p〉) contribution to this answer and show that a lower bound is givenby

αE > 2e2|〈1s|z|2p〉|2

E2p − E1s=

219

311a3B .

b). Show that the upper bound is given by (use of completeness relation)

αE <2e2

E2p − E1s|〈1s|z2|1s〉 =

16

3a3B .

c). Demonstrate that (bonus part: extra 10 points to the homework if done correctly)

z |1s〉 = [H0, F ] |1s〉

where F = maB(r/2 + aB)z and H0 is the unperturbed hydrogen atom Hamiltonian. Note, thisis not an operator identity; this is equality valid when acting on a particular state.d). Use the proof established in c) to derive the exact result (bonus part: extra 10 points tothe homework if done correctly)

αE = 2e2〈1s|zF |1s〉 =9

2a3B .

Compare with the two bounds obtained in previous parts.

12

Three dimensional harmonic oscillator

Consider now an isotropic harmonic oscillator

V (r) =mω2

2r2 .

We have two ways to solve it—by considering three separate 1D problems for x, y, and z variablesseparately, as for the square box potential described at the beginning of this set of notes, or byusing spherical coordinates, as for the hydrogen atom. The former way is straightforward, andcan be done for the anisotropic oscillator easily too (different frequencies for different directions).The result for the spectrum is

En1,n2,n3 = ω

(n1 +

1

2

)+ω

(n2 +

1

2

)+ω

(n3 +

1

2

)= ω

(n1 + n2 + n3 +

3

2

), nα = 0, 1, 2, . . .

and for the wavefunctions we have the product of three harmonic oscillator solutions ψn1n2n3 =ψn1(x)ψn2(y)ψn3(z). We clearly see that excited states are degenerate since the spectrum dependsonly on the sum ν = n1 + n2 + n3 of three quantum numbers.

In spherical coordinates we will find that the angle-dependent part of solutions is given byspherical harmonics Y m

l (θ, ϕ), and the radial function R(r) = u(r)/r satisfies an equation (asbefore ξ =

√mωr and ε = 2E/ω)

u′′ +

[ε− ξ2 − l(l + 1)

ξ2

]u = 0 .

I will not go through yet another exercise of establishing an asymptotic behavior to motivate theform u(ξ) = ξl+1e−ξ

2/2v(ξ), and showing next that by looking for the power-series solution forv we arrive at the condition that power series must be truncated to have a meaningful solution.Ultimately this leads to the condition

ε = 2n+ 2l + 3 = 2ν + 3 −→ E = ω

(n+ l +

3

2

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

or

E = ω

(ν +

3

2

), with ν = 0, 1, 2, . . . ≥ l .

and additional rule that ν and l are both either even or odd.

Problem 51. Degeneracies in the 3D harmonic oscillatorCalculate degeneracy factors for the energy spectrum of the three dimensional harmonic oscillatorfrom two approaches, using

Eν = ω

(ν +

3

2

), with ν = n1 + n2 + n3

and integer non-negative nα, and using

Eνlm = ω

(ν +

3

2

),

13

with orbital momentum quantum numbers restricted by l ≤ ν and |m| ≤ l (pay attention toeven/odd rule for l). Make sure that the two calculations agree with each other.

Free particle in spherical coordinates

You may wonder why would one do a free particle with plane wave solutions using sphericalcoordinates? This is done (i) in preparation for the future discussions of scattering problems,when particles moving in free space are incident on the scattering center, and (ii) to deal withproblems which have free moving particles subject to certain spherically symmetric boundaryconditions. Thus instead of ψ(r) ∝ eik·r we seek solutions in the form ψ(r) ∝ Y m

l (θ, ϕ)u(r)/r with

u′′(r) +

(k2 − l(l + 1)

r2

)u(r) = 0 −→ u′′(ρ) +

(1− l(l + 1)

ρ2

)u(ρ) ,

where ρ = kr. This is the differential equation of the spherical Bessel ρjl(ρ) and Neumann ρnl(ρ)functions. In the limit of ρ→ 0 we have to deal with

u′′ − l(l + 1)

ρ2u(ρ) = 0 .

Substituting u = ρa we get a(a − 1) = l(l + 1) which solves by a = l + 1 and a = −l; the formercorresponding to the asymptotic behavior of ρjl(ρ) and the latter to the asymptotic behavior ofρnl(ρ). More specifically

jl(ρ) =2ll!

(2l + 1)!ρl +O(ρl+2) ρ� 1

nl(ρ) = −(2l)!

2ll!ρ−l−1 +O(ρ−l+1) ρ� 1

jl(ρ) ∼ 1

ρcos(ρ− (l + 1)

π

2

)ρ� 1

nl(ρ) ∼ 1

ρsin(ρ− (l + 1)

π

2

)ρ� 1

and more explicitly

j0 =sin ρ

ρ, j1 =

sin ρ

ρ2− cos ρ

ρ, j2 =

(3− ρ2) sin ρ

ρ3− 3 cos ρ

ρ2,

n0 = −cos ρ

ρ, n1 = −cos ρ

ρ2− sin ρ

ρ, n2 = −(3− ρ2) cos ρ

ρ3− 3 sin ρ

ρ2.

You immediately see that in the limit of large ρ the two functions can be combined into oneexponential

h(1)l = jl + inl ∼

1

ρei[ρ−(l+1)π/2] , ρ� 1 ,

h(2)l = jl − inl ∼

1

ρe−i[ρ−(l+1)π/2] , ρ� 1 ,

14

where h(1)l and h

(2)l functions are known as spherical Hankel functions of the first and second kind.

Three-dimensional square well: The Deuteron

As a practical application of free-particle-in-spherical-coordinates solutions we consider boundstates in the spherically symmetric potential well

V (r) =

{−V0 r ≤ a

0 r > a. (16)

Let k =√

2m(V0 − |E|), and κ =√

2m|E|, as usual. Then the solution in the well is that of afree particle which is regular at the origin

ψI(r) = Njl(kr)Yml (θ, ϕ) .

For the underbarrier region we need an exponentially decreasing solution with complex momentum,

which is h(1)l (iκr). Thus

ψII(r) = NAh(1)l (iκr)Y m

l (θ, ϕ) .

where N is the normalization factor. Now we match the solutions

jl(ka) = Ah(1)l (iκa) ,

djl(kr)

dr

∣∣∣∣r=a

= Adh

(1)l (iκr)

dr

∣∣∣∣r=a

,

to fix the allowed values of k and A.There are numerous physical examples of systems which are reasonably well described by this

setup. One of them is a deuteron which is a bound state of a proton and a neutron. They forma shallow bound state, |E| ≈ 2.236 Mev, which is spherically symmetric, i.e. l = 0 and m = 0. Ifwe approximate the potential between particles by Eq. (16) we can establish a relation betweenthe binding energy and strong force V0 (the relative motion of two particles is described by thereduced mass, which is, in this case, approximately 1/2 of the proton mass). For l = 0 we have

j0(z) =sin z

z, h

(1)l (z) = −ie

iz

z,

and the matching conditions read

sin ka

ka= A−ie−κa

iκa−→ sin ka = (−Ak/κ)e−κa

d(sin kr/kr)

dr

∣∣∣∣r=a

= −Ad(e−κr/κr)

dr

∣∣∣∣r=a

−→ cos ka = Ae−κa .

By dividing two equations we find the quantization condition

tan ka = −kκ,

which is exactly the same as for antisymmetric solutions in the 1D square well (!) (recall that herewe are looking for spherically symmetric solution in 3D). Thus, we can immediately state (fromnotes and homework) that in the limit of a deep well V0 � 1/ma2 the ground state energy is

E ≈ −V0 +π2

2ma2

(1− 1

2ma2V0

)2

.

15

If the well is shallow, i.e. V0 < Vc = π2/8ma2 then there are no bound states at all (!) in contrastwith square wells in 1D and in 2D. For a = 1.4 fm one finds Vc ≈ 50 Mev, meaning that the boundstate energy for a deuteron is much smaller than Vc and thus the value of V0 has to be slightlylarger, but close to Vc.This Problem is not part of Homework! Do it if you want to.Semiclassical quantization of the Hydrogen atomThe semiclassical picture of the hydrogen atom considers only planar orbits with variables r andϕ, and their corresponding momenta. The semiclassical quantization condition on the angularmomentum is that of free particle motion on a circle of length 2π:∫ 2π

0Ldϕ = 2πl or L = l

The quantization condition for the radial coordinate is∫ rmax

rmin

prdr =

∫ rmax

rmin

√2m

(E +

Ze2

r− l2

2mr2

)dr = πnr .

Evaluate the integral and show that the resulting answer is

E = − mZ2e4

(nr + l)2.

Hint: If you do it by residue analysis, do not forget the residue at infinity.

16