## More Problems

#### Problem 1:

Consider the perturbed 1D infinite well, V(x) = 0 for 0 < x < a/4 and 3/4 < x <a,
V(x) = V0 for a/4 < x < 3a/4,
V(x) infinite everywhere else.
Solve for the energy eigenvalues of excited state with E >> V0 of this perturbed well using the WKB approximation.

Solution:

• Concepts:
The WKB approximation
• Reasoning:
In one dimension, if V(x) finite everywhere, then the WKB approximation requires
cylcepdx = ∫cylceħkdx = (n - ½)h, n = 1, 2, ... .
Here V(x) is infinite everywhere except in some finite region and ∫cylcepdx = ∫cylceħkdx = nh.
• Details of the calculation:
For stationary bound states we want ∫xminxmax pdx = nh/2, n = 1, 2, ... .
For this problem k2 = k12 = (2m/ħ2)E for 0 < x < a/4 and 3/4 < x <a,
and k2 = k22 = (2m/ħ2)(E - V0) for a/4 < x < 2a/4.
0a ħkdx = nh/2 = nπħ.  ∫0a/4 (2mE)½dx + ∫3a/4a (2mE)½dx +∫a/43a/4 (2m(E-V0))½dx = nπħ.
(2mE)½a/2 + (2m(E - V0))½a/2 = nπħ.
E/4 + (E - V0)/4 + (E(E - V0))½/2 = n2π2ħ2/(2ma2) = n2E0.
E/2 - V0/4 + (E/2)(1 - V0/E)½ ≈ E/2 - V0/4 + E/2 - V0/4 = E - V0/2 = n2E0.
since E >> V0.
E = n2E0 + V0/2, keeping only first order terms in the expansion.

#### Problem 2:

Suppose the potential energy between an electron and a proton had a term U0(a0 /r)2 in addition to usual electrostatic potential energy -e2/r, where e2 = qe2/(4πε0).
To the first order in U0, where U0 = 0.01 eV, by how much would the ground state energy of the hydrogen atom be changed?

Solution:

• Concepts:
First order perturbation theory for non-degenerate states
• Reasoning:
The ground state of the hydrogen atom is non-degenerate (neglecting spin).  We add the perturbation H' = U0(a0 /r)2.
• Details of the calculation:
The ground state wave function of the hydrogen atom is ψ(r) = (πa03)exp(-r/a0).
The first order correction to the ground state energy is
E01 = [4πU0a02/(πa03)] ∫0dr r2 exp(-2r/a0) (1/r2)  = 2U0.
The ground state energy is shifted up by 0.02 eV.

#### Problem 3:

Consider a system with a 4-dimensionl state space.  The Hamiltonian of the system is H0.  The normalized eigenbasis of H0 is {|1>, |2>, |3>, |4>}.  In that basis the matrix of H0 is

The system is perturbed and the Hamiltonian becomes

with ε << 1.
Use perturbation theory to find the first order corrections to the eigenvalues and the zeroth order eigenstates of H.

Solution:

• Concepts:
Stationary perturbation theory for degenerate states
• Reasoning:
The eigenvalue E0 is 3-fold degenerate.
• Details of the calculation:
H = H0 + H'.
The eigenvalues of H0 are E0 with eigenvectors |1>, |2>, |3> and -3E0 with eigenvector |4>.
The eigenvectors |2> and |4> are also eigenvectors of H with eigenvalue E0 ad -3E0 respectively.
We now diagonalize the perturbation H' in the subspace spanned by |1> and |3>.

γ2 - ε2 = 0, γ = ±ε.  The first order corrections to the eigenvalues are εE0 and -εE0.
The corresponding eigenstates are 2(|1> + |3>) and 2(|1> - |3>) respectively.

#### Problem 4:

Do a variational calculation.  A particle of mass M is subjected to a potential energy function of the form U = infinite for x < 0 and U = bx for x > 0
Let the trial wave function be ψ(x) = N x exp(-cx), where "c" is a variational parameter.  Using this function, construct a best estimate of the ground state energy.

Solution:

• Concepts:
The variational method
• Reasoning:
We are asked to use the variational method to estimate the ground state energy.
• Details of the calculation:
H = p2/(2m) + bx,  x > 0.
The eigenfunctions of H must be zero at x = 0 and x = ∞.  The derivative dψ/dx must be continuous at all x except x = 0.  Any function that satisfies these boundary conditions can be written as a linear combination of eigenfunctions of H and is an acceptable trial function giving <H> > E0.
We choose
ψ(x) = N x exp(-cx),  N20 x2 exp(-2cx) dx = N2/(4c3) = 1,  N2 = 4c3.
Here c is the adjustable parameter.
<H> =  N20 x exp(-cx)[(-ħ2/2m)(∂2/∂x2) + bx]x exp(-cx) dx
= N2(-ħ2/2m)∫0 (-2cx + c2x2) exp(-2cx) dx + N2b∫0 x3 exp(-2cx) dx
= N22/4mc  - ħ2/8mc + 6b/(2c)4).
d<H>/dc = 0 --> c3 = 3bm/(2ħ2).
Insert c back into the equation for <H>.
E = 1.96 b(2/3)ħ(2/3)/m(1/3).

Exact solutions:
E = 1.85 F(2/3)ħ(2/3)/m(1/3).