## Assignment 10, solutions

#### Problem 1:

Find the ground state energy of the He atom using first-order perturbation theory.

Useful information:
ψ1s(r) = (1/πa03)½exp(-r/a0) for the hydrogen atom.
1s|(1/r)|ψ1s> = 1/a0.
1s|∇21s> = -1/a02.
∫∫d3r d3r'|ψ1s(r)|21s(r')|2(1/|r - r'|) = 5/(8a0).

Solution:

• Concepts:
First-order perturbation theory, scaling
• Reasoning:
We consider the electron-electron interaction to be the perturbation.
The eigenfunctions of the unperturbed Hamiltonian are products of hydrogenic wave functions with Z = 2.
• Details of the calculation:
H = P12/(2m) + P22/(2m) - 2e2/r1 - 2e2/r2 + e2/r12 = H1 + H2 + U12 = H0 + H'.
e2 = qe2/(4πε0) in SI units.
H0 = H1 + H2,  H' = U12.
ψ(r1,r2) = Φ100(r1100(r2) = (1/πa3)exp(-(r1 + r2)/a), with a = a0/2,  E00 = -8EI.
Here Φ100(r) = (8/πa03)½exp(-2r/a0) is the ground state wave function for a hydrogenic ion with Z = 2.
<ψ|H'|ψ> = (e2/(π2a6))∫∫d3r1d3r2 exp(-2(r1 + r2)/a)/|r- r2| = 5e2/(8a) = 5e2/(4a0) = (5/2) EI.
EI = e2/(4a0) = 13.6 eV is the ionization energy of the hydrogen atom.
E0 = E00 + E10 = -8EI + (5/2)EI = -74.8 eV.

#### Problem 2:

Find the ground state energy of the He atom using the variational method.
Useful information:
ψ1s(r) = (1/πa03)½exp(-r/a0) for the hydrogen atom.
1s|(1/r)|ψ1s> = 1/a0.
1s|∇21s> = -1/a02.
∫∫d3r d3r'|ψ1s(r)|21s(r')|2(1/|r - r'|) = 5/(8a0).

Solution:

• Concepts:
The variational method, scaling
• Reasoning:
We are asked to find the ground state energy of the He atom using the variational method
• Details of the calculation:
H = P12/(2m) + P22/(2m) - 2e2/r1 - 2e2/r2 + e2/r12 = H1 + H2 + U12.
e2 = qe2/(4πε0) in SI units.
Let us use as a trial function  ψ(r1,r2) = (1/πa3)exp(-(r1 + r2)/a).
We are guided by the ground state wave function of hydrogen, ψ(r) = (1/πa03)½exp(-r/a0).
The adjustable parameter in our trial function is a = a0/Z'.
ψ(r1,r2) is a product function, ψ(r1,r2) = Φ1(r12(r2).
Φ1(r1) = (1/πa3)½exp(-r1/a),  Φ2(r2) = (1/πa3)½exp(-r2/a).
<ψ|p21/(2m)|ψ> = <Φ1|p21/(2m)|Φ1> = -ħ2/(2m)<Φ1|∇(1)21> = ħ2/(2ma2).
<ψ|p22/(2m)|ψ> = <Φ2|p22/(2m)|Φ2> = -ħ2/(2m)<Φ2|∇(2)22> = ħ2/(2ma2).
<ψ|2e2/r1|ψ> = 221|1/r11> = 2e2/a.
<ψ|2e2/r2|ψ> = 2e22|1/r22> = 2e2/a.
<ψ|e2/|r-r2||ψ> = ∫d3r1d3r21(r1)|22(r2)|2(1/|r1 - r2|) = 5e2/(8a).
Therefore <H> = ħ2/(ma2) - 4e2/a + 5e2/(8a) = ħ2/(ma2) - 27e2/(8a).

d<H>/da = -2ħ2/(ma3) + 27e2/(8a2) = 0,  a = 16ħ2/(27me2).
<H> = Eground state(upper bound) = -(27/16)2me42 = -2.85*2*13.6 eV = -77.5 eV.

#### Problem 3:

Consider a quantum system with just three linearly independent states.
Suppose the Hamiltonian, in matrix form, is

H = V0

 1 - ε 0 0 0 1 ε 0 ε 2

.

where V0 is a constant and ε << 1.
(a)  Write down the eigenvalues and eigenvectors of the unperturbed Hamiltonian,  H0 (ε = 0).
(b)  Solve for the exact eigenvalues of H.
Expand each of them in a power series in ε up to second order.
(c)  Use first- and second-order non-degenerate perturbation theory to find the approximate eigenvalue for the state that grows out of the non-degenerate eigenvector of H0.
(d)  Use degenerate perturbation theory to find the first-order corrections to the initially degenerate eigenvalues.

Solution:

• Concepts:
Stationary perturbation theory for non-degenerate and for degenerate states
• Reasoning:
We are asked to find the exact eigenvalues of a Hamiltonian operator and to find its approximate eigenvalues using stationary perturbation theory.
• Details of the calculation:
(a)  The eigenvalues of H0 are V0 (two fold degenerate) and 2V0.
For the normalized eigenvectors with eigenvalue V0 we can choose

|a> =

 1 0 0

,   |b> =

 0 1 0

.

and the eigenvector with eigenvalue 2V0 is

|c> =

 0 0 1

.

(b)  The eigenvalues are λV0, where

 1 - ε - λ 0 0 0 1 - λ ε 0 ε 2 - λ

= 0.

λ1 is 1 - ε,  E1 = (1 - ε)V0, the corresponding eigenvector is |a>.
(1 - λ)(2 - λ) - ε2 = 0,   λ± = 3/2 ± (¼ + ε2)½ = 3/2 ± ½(1 + 4ε2)½.
λ2 = 3/2 - ½(1 + 4ε2)½ ≈ 1 - ε2,  E2 ≈ (1 - ε2)V0,
λ3 = 3/2 + ½(1 + 4ε2)½  ≈ 2 + ε2,  E3 ≈ (2 + ε2)V0.

(c)  H = H0 + H'.

H' = V0

 -ε 0 0 0 0 ε 0 ε 0

.

To find the first order corrections to the energy eigenvalues we diagonalize H'  in the subspace of degenerate states.
First order correction for the nondegenerate eigenvalue 2V0:  E31 = <c|H'|c> = 0.
Second order correction for this eigenvalue:  E32 = (|<a|H'|c>|2 + |<b|H'|c>|2 )/V0 = ε2V0.
E3 = V0(2 + ε2) to second order.

(d)  First order correction for the degenerate eigenvalue V0:

 -ε - λ 0 0 -λ

= 0.

λ = -ε or λ = 0.  The degeneracy is removed to first order.  The eigenstates to zeroth order are |a> with eigenvalue (1 - ε)V0 and |b> with eigenvalue V0.

#### Problem 4:

In one dimension, the potential energy of a particle of mass m as a function of x is given by
U(x) = (b2|x|)½.  Here b is a positive constant with units energy/lemgth½.
(a)  Use the WKB method to estimate the energy of the particle in the ground state.
(b)  Use the variational method to estimate the energy of the particle in the ground state.
(c)  Which estimate is closer to the true ground-state energy?

0exp(-x2) √x dx = Γ(3/4)/2 = 0.612708

Solution:

• Concepts:
The WKB and variational methods
• Reasoning:
The variational method often yields a very good estimate for the ground state energy of a system.
• Details of the calculation:
(a)  For this problem the WKB approximation requires that  ∫xminxmax pdx = ½(h/2),
or ∫xminxmax pdx = πħ/2, or ∫0xmax pdx = πħ/4.
p = (2m)½(E - b√x)½.  ∫0xmax pdx =  (2m)½0(E/b)^2 (E - b√x)½dx,
since E = b√(xmax).
Let u = b√x.  du = (b/2)dx/√x = (b2/2)dx/u.  dx = (2/b2)u du.
0xmax pdx = (2/b2)(2m)½0E (E - u)½u du = (2/b2)(2m)½ 4E5/2/15.
(2/b2)(2m)½ 4E5/2/15 = πħ/4.  E5/2 = 15πħb2/(32*(2m)½) = 1.041 (ħb2/m½).
E =  1.01632 (ħ2b4/m)1/5.

(b)  H = (-ħ2/2μ)(∂2/∂x2) + (b2|x|)½.
Choose Φ(x) = Aexp(-a2x2).  Here a is an adjustable parameter.
Φ(x) = 0 as x --> ±∞, and dΦ/dx is continuous at all x.
If we let a2 = ½mω/ħ, then Φ(x)  is the ground state wave function for a harmonic oscillator potential U(x) = ½mω2x2.
The normalization constant is A = (mω/(πħ))¼ = (2a2/π)¼,
and the expectation value of the kinetic energy is <T> = <p2>/(2m) = ħω/4 = ħ2a2/(2m).
<H> = <T> + <U>,  <U> = (2a2/π)½ 2∫0 exp(-2a2x2) b √x dx.
<U> = (2a2/π)½ 2∫0 exp(-2a2x2) b √x dx.
let x' = √2 a x,  then dx = dx'/(a√2) and √x = √x'/(2¼√a).  Therefore
<U> = (23/4b/π½) a0 exp(-x'2) √x' dx' = (23/4b/π½) a * X,  where X = 0.612708.
<H> =  ħ2a2/(2m) + (23/4b/π½) a * X.
d<H>/da2 = ħ2/(2m) - (1/4) (23/4b/π½) (a2)-5/4 * X = 0.
a2 = (mbX/(2¼ħ2π½))4/5 = (mb/ħ2)4/5 * 0.372116.
E = 0.186083 ħ2/5 m-1/5 b4/5 + 0.744332*ħ2/5 m-1/5 b4/5.
E =  0.930415 (ħ2b4/m)1/5.

(c)  The variational method yields an upper limit for the ground state energy.  Since this limit for E is lower than the value for E obtained using the WKB approximation, the variational estimate is closer to the true ground-state energy than the WKB estimate.

#### Problem 5:

The hydrogen atom states 2S½ and 2P½ are still degenerate, even after including the spin-orbit interaction.  However, they are split by a small "Lamb shift" energy ħω0 = 2πħ*1.06 GHz.
The 2P½ state has the lower energy.
Consider only the subspace spanned by these two states.  With the appropriate choice of the zero of the energy scale the Hamiltonian in the absence of an external field is

H0 = ħ

 ω0 0 0 0

.

(a)  Find the eigenvalues of H in an applied electric field E = Efield k.
(Use the matrix element <2P½|z|2S½> = 3a0, where the Bohr radius a0 ≈ 0.053 nm.)
Write down the low-field and the high-field approximations for the eigenvalues.
Sketch these eigenenergies vs. the field strength Efield.
What are the approximate low-field and high-field eigenstates of H?

(b)  How large should the electric field be for the energy shifts (i.e. Stark shifts) to become linear?
(c)  What is the value of the linear Stark shift (in MHz/(V/cm)) at large electric fields?

Solution:

• Concepts:
A two-state system
• Reasoning:
In the presence of an electric field H = H0 + H1 = H0 + qeEfieldz.

H1 = ħ

 0 U U 0

,   H = ħ

 ω0 U U 0

,

with U = 3a0qeEfield/ħ.
(The dipole moment operator qez only couples states of opposite parity, i.e. S and P.)

• Details of the calculation:
(a)  det(H - EI) = 0 yields
E± =  ħω0/2 ± ħ((ω0/2)2 + U2)½ = ½ħω0(1 ± (1 + 4U202)½).
Limiting cases:
|U| << ω0,  E± ≈ ½ħω0(1 ± (1 + 2U202)).
E+ = ħω0 + ħU20,  E- = -ħU20.
+> = a|S> + b|P>,  b ≈ aU/ω0 ≈ 0,  |ψ+> ≈ |S>.
-> = a|S> + b|P>,  b ≈ -a( ω0/U + U/ω0), |b| >> |a|,  | ψ-> ≈ |P>.

|U| >> ω0,  E± ≈ ½ħω0(1 ±  2U/ω0).
E+ = ħω0 + ħ|U|,  E- = ħω0 - ħ|U|.
+> = a|S> + b|P>,  b ≈ a,  |ψ+> ≈ (|S> + |P>)/√2.
-> = a|S> + b|P>,  b ≈ -a,  |ψ-> ≈ (|S> - |P>)/√2.
The Stark shift is quadratic at low fields and linear at high fields.

(b) E± changes behavior from quadratic to linear in U when 4U202 ≈ 1.
Then U = 3a0qeEfield/ħ  = ω0/2,  Efield = ħω0/(6a0qe) ≈ 1.4*104 V/m = 140 V/cm.

(c)  At large fields the shift is proportional to U =  3a0qeEfield/ħ.
ħU = hf.  f = ħU/h = U/(2π).
(2π)-1dU/dEfield = 3a0qe/h  ≈ 3.8 MHz/(V/cm).