Consider the perturbed 1D infinite well, V(x) = 0 for 0 < x < a/4 and 3/4 < x <a,

V(x) = V_{0} for a/4 < x < 3a/4,

V(x) infinite everywhere else.

Solve for the energy eigenvalues of excited state with E >> V_{0} 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

∫_{cylce}pdx = ∫_{cylce}ħkdx = (n - ½)h, n = 1, 2, ... .

Here V(x) is infinite everywhere except in some finite region and ∫_{cylce}pdx = ∫_{cylce}ħkdx = nh. - Details of the calculation:

For stationary bound states we want ∫_{xmin}^{xmax}pdx = nh/2, n = 1, 2, ... .

For this problem k^{2}= k_{1}^{2}= (2m/ħ^{2})E for 0 < x < a/4 and 3/4 < x <a,

and k^{2}= k_{2}^{2}= (2m/ħ^{2})(E - V_{0}) for a/4 < x < 2a/4.

∫_{0}^{a}ħkdx = nh/2 = nπħ. ∫_{0}^{a/4}(2mE)^{½}dx + ∫_{3a/4}^{a}(2mE)^{½}dx +∫_{a/4}^{3a/4}(2m(E-V_{0}))^{½}dx = nπħ.

(2mE)^{½}a/2 + (2m(E - V_{0}))^{½}a/2 = nπħ.

E/4 + (E - V_{0})/4 + (E(E - V_{0}))^{½}/2 = n^{2}π^{2}ħ^{2}/(2ma^{2}) = n^{2}E_{0}.

E/2 - V_{0}/4 + (E/2)(1 - V_{0}/E)^{½}≈ E/2 - V_{0}/4 + E/2 - V_{0}/4 = E - V_{0}/2 = n^{2}E_{0}.

since E >> V_{0}.

E = n^{2}E_{0}+ V_{0}/2, keeping only first order terms in the expansion.

Suppose the potential energy between an electron and a proton had a term U_{0}(a_{0}
/r)^{2} in addition to usual electrostatic potential energy -e^{2}/r,
where e^{2} = q_{e}^{2}/(4πε_{0}).

To the first order in U_{0}, where U_{0} = 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' = U_{0}(a_{0}/r)^{2}. - Details of the calculation:

The ground state wave function of the hydrogen atom is ψ(**r**) = (πa_{0}^{3})^{-½}exp(-r/a_{0}).

The first order correction to the ground state energy is

E_{0}^{1}= [4πU_{0}a_{0}^{2}/(πa_{0}^{3})] ∫_{0}^{∞}dr r^{2}_{ }exp(-2r/a_{0}) (1/r^{2}) = 2U_{0}._{ }The ground state energy is shifted up by 0.02 eV.

Consider a system with a 4-dimensionl state space. The Hamiltonian of
the system is H_{0}. The normalized eigenbasis of H_{0} is
{|1>, |2>, |3>, |4>}. In that basis the matrix of H_{0} 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 E_{0}is 3-fold degenerate. - Details of the calculation:

H = H_{0}+ H'.

The eigenvalues of H_{0}are E_{0}with eigenvectors |1>, |2>, |3> and -3E_{0}with eigenvector |4>.

The eigenvectors |2> and |4> are also eigenvectors of H with eigenvalue E_{0}ad -3E_{0}respectively.

We now diagonalize the perturbation H' in the subspace spanned by |1> and |3>.

γ^{2}- ε^{2}= 0, γ = ±ε. The first order corrections to the eigenvalues are εE_{0}and -εE_{0}.

The corresponding eigenstates are 2^{-½}(|1> + |3>) and 2^{-½}(|1> - |3>) respectively.

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 = p^{2}/(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> > E_{0}.

We choose

ψ(x) = N x exp(-cx), N^{2}∫_{0}^{∞}x^{2}exp(-2cx) dx = N^{2}/(4c^{3}) = 1, N^{2}= 4c^{3}.

Here c is the adjustable parameter.

<H> = N^{2}∫_{0}^{∞}x exp(-cx)[(-ħ^{2}/2m)(∂^{2}/∂x^{2}) + bx]x exp(-cx) dx

= N^{2}(-ħ^{2}/2m)∫_{0}^{∞}(-2cx + c^{2}x^{2}) exp(-2cx) dx + N^{2}b∫_{0}^{∞}x^{3}exp(-2cx) dx

= N^{2}(ħ^{2}/4mc - ħ^{2}/8mc + 6b/(2c)^{4}).

d<H>/dc = 0 --> c^{3}= 3bm/(2ħ^{2}).

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)}.