Assignment 10, solutions 

Problem 1:

In one dimension, the potential energy of an electron as a function of x is given by

U(x) = -30eV exp(-x2/(4Å2)).

Use the variational method to find the energy of the ground state in units of eV.

Solution:

Concepts, principles, relations that apply to the problem:
The variational method
Why do they apply?
The variational method often yields a very good estimate for the ground state energy of a system.
How do they apply?
The Hamiltonian is H = (-h2/2m)(2/x2) - V0exp(-x2/a2)).
Here V0 = 30 eV and a = 2Å and m = me.
Choose y(x) = Aexp(-x2/b2).  Here b is an adjustable parameter.
y(x) = 0 as x --> ±¥, and dy/dx must be continuous at all x.
A2ò-¥¥ exp(-2x2/b2)dx = 1,  A2 = (2/pb2)1/2.
<H> = A2ò-¥¥ exp(-x2/b2)((-h2/2m)(2/x2) - V0exp(-x2/a2))exp(-x2/b2)dx
= 2A2(-h2/2m)ò0¥ [-2/b2 + 4x2/b4]exp(-2x2/b2)dz - 2A2V0ò0¥ exp(-x2(2/b2 - 1/a2))dx
= 2A2[(-h2/2m)(2/b2)1/2 - V0(a2b2/(2a2+b2)1/2]ò0¥ exp(-x2)dx
- 2A2[(2h2/mb4)(b2/2)3/2ò0¥ x2exp(-x2)dx.
ò0¥ exp(-r2)dr = p1/2/2.  ò0¥  r2exp(-r2)dr = p1/2/4.
<H> = h2/(2mb2) - 21/2aV0/(2a2+b2)1/2.
d<H>/db = 0  -->  -h2/(mb3) + 21/2abV0/(2a2+b2)3/2 = 0.
Let V0' = V0ma2/h2.
Then a2/b2 - 21/2ab2V0'/(2a2+b2)3/2 = 0, a4/b4 = 2(a2b4)V0'2/(2a2 + b2)3,
a4/b4 = 2(a2/b2)V0'2/(2a2/b2 + 1)3.
Let y = a2/b2.
Then y2 = 2yV0'2/(2y + 1)3.  y(2y+1)3 = 2V0'2.
Here V0' = 30eV me a2/h2 = 17.5.

y = 2.6,  b2 = a2/2.6.
 E = 2.6h2/(2ma2) - V0(0.916) = -25.2 eV.
Details of the calculation:
None

Problem 2:

Using first order perturbation theory, find the shift in the ground state energy of the one-dimensional harmonic oscillator when the perturbation
(a)  V(x) = bx4
(b)  V(x) = bx3
is added to H = ½p2/m + ½mw2x2.

Solution:

Concepts, principles, relations that apply to the problem:
First order perturbation theory for non-degenerate states, the harmonic oscillator
Why do they apply?
We are asked to find the shift in the ground state energy of the one-dimensional harmonic oscillator using perturbation theory.  The ground state of the one-dimensional harmonic oscillator is non-degenerate.
How do they apply?
(a)  H = H0 + H’.  H0 = ½p2/m + ½mw2x2.  H' = bx4.
The ground-state wavefunction of the  harmonic oscillator is
y0(x) =  (mw/(ph))1/4exp(-mwx2/(2h)).
E10 = <y0|H'|y0> = b(mw/(ph))1/2ò-¥¥ x4 exp(-mwx2/h) dx.
E10 = b(mw/(ph))1/2(mw/h)-5/2ò-¥¥ x4 exp(-x2) dx = b(mw/(ph))1/2(mw/h)5/2(3/4)p1/2.
E10 = 3bh2/(4m2w2).
E10 is the first order energy correction for the ground state.

(b)  H = H0 + H’.  H0 = ½p2/m + ½mw2x2.  H' = bx3.
E10 = b(mw/(ph))1/2ò-¥¥ x3 exp(-mwx2/h) dx = 0,
from symmetry (odd-even).
To first order, the shift is zero.
Details of the calculation:
None

Problem 3:

It is known that the stable H- exists (two electrons bound to a proton).  Estimate the ground state energy of H- using the variational method.  Assume that each electron moves in a 1s orbit.  Neglect spin.

Useful information:
y1s(r) = (1/pa03)½exp(-r/a0)
< y1s |(1/r)| y1s > = 1/a0
< y1s |Ñ2| y1s > = -1/a02
òd3rd3r’|y1s(r)|2|y1s(r’)|2(1/|r-r’|) = 5/(8a0)

Solution:

Concepts, principles, relations that apply to the problem:
The variational method, the H- negative ion
Why do they apply?
Neglecting spin, the Hamiltonian for the H- negative ion is
H = P12/(2m) + P12/(2m) - e2/r1 - e2/r2 + e2/|r1-r2|.
We are asked to find the ground state energy using the variational method.
How do they apply?
Let us use as a trial function

(We are guided by the ground state wavefunction of hydrogen )

The adjustable parameter in our trial function is a.  The wavefunction is normalized.
y(r1,r2) is a product function, y(r1,r2) = f1(r1)f2(r2).

<y|p21/(2m)|y> = <f1|p21/(2m)|f1> = -h2/(2m)< f1|Ñ(1)2|f1 > = h2/(2ma2).
<y|p22/(2m)|y> = <f2|p22/(2m)|f2> = -h2/(2m)< f2|Ñ(2)2|f2 > = h2/(2ma2).
<y|e2/r1|y> = e2<f1|1/r1|f1> = e2/a.
<y|e2/r2|y> = e2<f2|1/r2|f2> = e2/a.
<y|e2/|r1-r2||y> = òd3r1d3r2|f1(r1)|2|f2(r2)|2(1/|r1-r2|) = 5e2/(8a).
Therefore <H> = h2/(ma2) - 2e2/a + 5e2/(8a) = h2/(ma2) - 11e2/(8a).
d<H>/da = -2h2/(ma3) + 11e2/(8a2) = 0,  a = 16h2/(11me2)
Eground state(upper bound) = -(121/256)me4/h2 = -0.945 * 13.6 eV
Details of the calculation:
This calculation would indicate that the H- ion is not stable, since the ground state energy is less negative than that of a free electron and a neutral hydrogen atom.  (H- is a stable negative ion.)

Problem 4:

Solve the perturbed 1-D particle in a box.  Follow the described setup precisely.  The 1-D box is of width 2a centered on x = 0.  The potential V0 = 0 for  |x| < a and V0 = ¥ for  |x| > a.
(a)  Write the Hamiltonian and solve for energies and wave functions.  (You must use the symmetric potential form specified.)
(b)  Let there be a small perturbing potential V’ = c - bx2 where b = c/a2.   Find the first order correction to the energy for the general n-th level.
(c)  Find the first order correction to the unperturbed wave functions.  Hint: In the expansion, non-zero coefficients may be left in integral form.

Solution:

Concepts, principles, relations that apply to the problem:
First order perturbation theory for non-degenerate states, the infinite square well
Why do they apply?
We are asked to find the first-order corrections to the eigenvalues and eigenfunctions of the infinite square well.  The eigenvalues of the 1-D infinite square well are not degenerate.
How do they apply?
(a)  H = (h2/2m)(2/x2) + V(x),  V(x) = 0 for  |x| < a and V(x) = ¥ for  |x| > a.
(2/x2)f(x) + (2m/h2)Ef(x) = 0 in the region |x| < a,  f(x) = 0 for |x| = a.
f(x) = Aexp(ikx) + Bexp(-ikx),   with E = h2k2/(2m).
Aexp(ika) + Bexp(-ika) = 0,  Aexp(-ika) + Bexp(ika) = 0.
Two solutions:
A = B: coska = 0,  ka = np/2, n = odd.
A = -B: sinka = 0,  ka = np/2, n = even.
En = n2p2h2/(8ma2),  fn(x) = Ncos(npx/(2a)), n = odd,  fn(x) = Nsin(npx/(2a)), n = even.
Normalization:  N2 = 1/a.
Details of the calculation:
(b)  n = odd:
En1 = N2ò-aa cos2(npx/(2a))(c - bx2)dx
= (2c/np)ò-np/2np/2 cos2(x)dx - (b/a)(2a/np)3ò-np/2np/2 cos2(x)x2dx
= c - c/3 + 2c/(np)2 = 2c/3 + 2c/(np)2.
n = even:
En1 = N2ò-aa sin2(npx/(2a))(c - bx2)dx = 2c/3 + 2c/(np)2.
The first order energy correction is 2c/3 + 2c/(np)2 for all n.

(c)  |yn1> = Sn'¹n bn'|fn> , where bn' = <fn'|V'|fn>/(E0n - E0n').
yn1(x) = Sn'¹n [<fn'|V'|fn>(8ma2)/((n2-n'2)p2h2)]fn(x)
is the first order correction to the wavefunction fn(x).
<fn'|V'|fn>  = 0 if n' is even and n is odd or if n' is odd and n is even.
<fn'|V'|fn>  = (1/a)ò-aa cos(n'px/(2a))cos(npx/(2a))(c - bx2)dx if n and n' are odd,
<fn'|V'|fn>  = (1/a)ò-aa sin(n'px/(2a))sin(npx/(2a))(c - bx2)dx if n and n' are even.

Problem 5:

Suppose the Hamiltonian of a rigid rotator in a magnetic field is of the form
 H = AL2 + BLz + CLy, if terms quadratic in the field are neglected.
Assuming B >> C, use perturbation theory to lowest non-vanishing order to get appropriate energy eigenvalues.

Solution:

Concepts, principles, relations that apply to the problem:
Stationary perturbation theory
Why do they apply?
A small correction term is added to a Hamiltonian H0, whose eigenstates and eigenvalues we can solve for exactly.  We are asked to find the lowest order non-vanishing corrections to the energy eigenvalues.

How do they apply?
 H = AL2 + BLz + CLy,  B >> C.
H = H0 + H', H0 =  AL2 + BLz.
The eigenstates of H0 are {|lm>}.  H0|lm> = (Al(l+1)h2 + Bmh)|lm>
The eigenvalues are not degenerate.
The first order corrections are E1lm = C<lm|Ly|lm> = -(iC/2)<lm|L+- L-|lm> = 0.
We have to find the second order corrections E2lm.
E2lm = Sl'm'¹lm C2|<l'm'|Ly|lm>|2/(E0lm - E0l'm')
Ly = -(i/2)(L+- L-)
L±|l,m> = [l(l+1)-m(m±1)]1/2h|l,m±1>.
<l'm'|L±|l,m> = [l(l+1)-m(m±1)]1/2h dl,l' dm',m±1.
 E2lm = (C2/4)[l(l+1)-m(m+1)]h2/(Bh(m-(m+1))) + (C2/4)[l(l+1)-m(m-1)]h2/(Bh(m-(m-1)))
= +C2mh/(2B).
The energy eigenvalues to second order are Al(l+1)h2 + Bmh + C2mh/(2B).

Details of the calculation:
None

Problem 6:

(a)  List the 8 possible states of the n = 2 manifold of the hydrogen atom in the common eigenbasis of  L2, Lz, S2, and Sz, and also in the common eigenbasis of L2, S2, J2, and Jz.
(b)  The 8 states of the n = 2 manifold would be degenerate if not for spin-orbit coupling, and hyperfine splitting.  Let us ignore the hyperfine splitting but treat the spin orbit coupling using first order perturbation theory.  Determine the energies of the 8 states of the n = 2 manifold under the perturbation V = lL×S, where L is the orbital angular momentum operator of the electron and S is the electron's spin operator.
(c)  What is the physical origin of the hyperfine structure?

Solution:

Concepts, principles, relations that apply to the problem:
The hydrogen atom, angular momentum coupling, perturbation theory.
Why do they apply?
We are asked use first-order perturbation theory to calculate the energies of the 8 states of the n = 2 manifold when spin-orbit coupling is not ignored.

How do they apply?
(a) n = 2; l = 1; m = 1, 0 -1; s = ½  ,-½   (6 states)
n = 2; l = 0; m = 0, s = ½  ,-½   (2 states)
These are the 8 states in the common eigenbasis of L2, Lz, S2 and Sz.
In the common eigenbasis of L2, S2, J2 and Jz we can enumerate the states in the following way.
n = 2; l = 1; s = ½ ; J = 3/2 , mj =  3/2, ½, -½, -3/2.  (4 states)
n = 2; l = 1; s = ½ ; J = 1/2 , mj =  ½, -½.  (2 states)
n = 2; l = 0; s = ½ ; J = 1/2 , mj =  ½, -½.  (2 states)
(b)  L×S = ½(J2 – L2 – S2)
First order perturbation theory:
We have to diagonalize the matrix of L×S in the subspace of the 8 degenerate states.
If we use the common eigenbasis of L2, S2, J2 and Jz, the matrix is already diagonal.
<l,s,j,mj| L×S |l,s,j,mj> = (h2/2)(j(j+1) - l(l+1) - s(s+1))
For the 4 states n = 2; l = 1; s = ½ ; J = 3/2 , mj =  3/2, ½, -½, -3/2
<l,s,j,mj| L×S |l,s,j,mj> = (h2/2)(15/4 – 2 – 3/4) = (h2/2).
The perturbed energies are –E0/4 + l(h2/2) with E0 = 13.6 eV.  (4 fold degenerate)
For the 2 states n = 2; l = 1; s = ½ ; J = ½, mj = ½, -½,
<l,s,j,mj| L×S |l,s,j,mj> = (h2/2)(3/4 – 2 – 3/4) = -h2.
The perturbed energies are –E0/4 - lh2.  (2 fold degenerate)
For the 2 states n = 2; l = 0; s = ½ ; J = ½, mj = ½, -½,
<l,s,j,mj| L×S |l,s,j,mj> = (h2/2)(3/4 – 0 – 3/4) = 0.
The perturbed energies are –E0/4. (2 fold degenerate)
(c)  The spin of the proton or nucleus results in a magnetic dipole, which can be aligned or counter-aligned with the magnetic dipole moment that is caused by the orbital angular momentum of the electron.

Details of the calculation:
None

Problem 7:

A particle of mass m is constrained to move in an infinitely deep, one-dimensional square well extending from -a to +a.  If this particle is under the influence of a perturbation H' = -Ad(x), where A is a constant and d(x) is a delta function at x, calculate the first order corrections to the energy levels.  What is the condition that A must satisfy if the corrected n* energy level is to have a negative energy?

Solution:

Concepts, principles, relations that apply to the problem:
First order perturbation theory for non-degenerate states, the infinite square well
Why do they apply?
The energy level of the 1D infinite square well are non-degenerate.
How do they apply?
For the unperturbed Hamiltonian we have:
H = (h2/2m)(2/x2) + V(x),  V(x) = 0 for  |x| < a and V(x) = ¥ for  |x| > a.
(2/x2)f(x) + (2m/h2)Ef(x) = 0 in the region |x| < a,  f(x) = 0 for |x| = a.
f(x) = Cexp(ikx) + Dexp(-ikx),   with E = h2k2/(2m).
Cexp(ika) + Cexp(-ika) = 0,  Cexp(-ika) + Dexp(ika) = 0.
Two solutions:
C = D: coska = 0,  ka = np/2, n = odd.
C = -D: sinka = 0,  ka = np/2, n = even.
En = n2p2h2/(8ma2),  fn(x) = Ncos(npx/(2a)), n = odd,  fn(x) = Nsin(npx/(2a)), n = even.
Normalization:  N2 = 1/a.

Let us calculate the first order energy corrections due to the perturbation.
n = odd:
En1 = -AN2ò-aa cos2(npx/(2a))d(x)dx = -(A/a)
n = even:
En1 = -AN2ò-aa sin2(npx/(2a))d(x)dx = 0.
En* = n2p2h2/(8ma2) - A/a if n = odd. We need A > n2p2h2/(8ma) for En* to be negative.
En* = n2p2h2/(8ma2) if n = even.  The perturbation does not change this energy level to first order.

Details of the calculation:
None