Consider the 1D infinite well, U(x) = 0 for 0 < x < L, U(x) infinite
everywhere else.

The eigenfunctions and eigenvalues of the Hamiltonian for a particle of
mass m trapped in this well are

ψ_{n}(x) = (2/L)^{½}sin(nπx/L),
E_{n} = n^{2}π^{2}ħ^{2}/(2mL^{2}),
respectively.

Now consider the perturbed 1D infinite well, U(x) = 0 for 0 < x < L/2,
U(x) = U_{0} for L/2 < x < L, U(x) infinite everywhere else.

Let E_{0} denote the ground state energy of the unperturbed
well.

Assume U_{0} = E_{0}/10 = 0.1*π^{2}ħ^{2}/(2mL^{2}),
i.e. U_{0} = 0.1 times the ground state energy of the
unperturbed well.

(a) Solve for the energy eigenvalues of the perturbed well without
using an approximation method.

(b) Solve for the energy eigenvalues of the perturbed well using the WKB
approximation.

(c) Solve for the energy eigenvalues of the perturbed well using 1st order
perturbation theory.

(c) Solve for the lowest energy eigenvalues of the perturbed well using
the variational method.

Solution:

- Concepts:

Comparing approximation methods - Reasoning:

We are asked to solve for the energy eigenvalues of the perturbed well using various approximation methods. - Details of the calculation:

(a) No approximation:

We have a piecewise constant potential.

Let 0 < x < L/2 define region 1 and L/2 < x < L define region 2.

The most general solution of the time independent solution in region 1 satisfying the boundary condition at x = 0 is

ψ_{1}(x) = A sin(k_{1}x)

The most general solution of the time independent solution in region 2 satisfying the boundary condition at x = L is

ψ_{2}(x) = B sin(k_{2}(L-x)).

Here k_{1}^{2}= (2m/ħ^{2})E and k_{2}^{2}= (2m/ħ^{2})(E - U_{0}).

The boundary conditions are that ψ(x) and (∂/∂x)ψ(x) are continuous at x = L/2.

A sin(k_{1}L/2) = B sin(k_{2}L/2).

k_{1}A cos(k_{1}L/2) = -k_{2}B cos(k_{2}L/2).

This yields tan(k_{1}L/2)/(k_{1}L/2) = -tan(k_{2}L/2)/(k_{2}L/2).

This equation can be solved numerically. We expect solutions for values of k_{1}L/2 = nπ/2 + ε_{1}and k_{2}L/2 = nπ/2 - ε_{2}, where ε_{1}and ε_{2}are small positive numbers.

Using Excel to find the ground state energy of the perturbed well E_{1}we obtain E_{1}= 1.04936 E_{0}.

E_{1}= E_{0}+ 0.4936 U_{0}.

(b) The WKB approximation:

In one dimension, if U(x) finite everywhere, then the WKB approximation requires

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

Here U(x) is infinite everywhere except in some finite region and ∫_{cylce}pdx = ∫_{cylce}ħkdx = nh.

∫_{0}^{L}ħkdx = nh/2 = nπħ. ∫_{0}^{L/2}(2mE)^{½}dx + ∫_{L/2}^{L}(2m(E-U_{0}))^{½}dx = nπħ.

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

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

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

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

Higher precision solution of E/2 - U_{0}/4 + (E/2)(1 - U_{0}/E)^{½}= E_{0}we obtain

E_{1}= E_{0}+ 0.506 U_{0}.

(c) First order perturbation theory:

E_{n}= n^{2}π^{2}ħ^{2}/(2mL^{2}) + (2/L)∫_{L/2}^{L}sin^{2}(πx/l)U_{0})dx = n^{2}π^{2}ħ^{2}/(2mL^{2}) + 0.5 U_{0}.

E_{1}= E_{0}+ 0.5 U_{0}.

(d) The variational method:

We have to choose a trial wave function with an adjustable parameter which satisfies the boundary conditions.

Choose ψ(x) = A (2/L)^{½}(sin(πx/L) + bsin(2πx/L)) = A(ψ_{1}(x) + bψ_{2}(x)).

Normalize: A^{2}(1 + b^{2}) = 1, A = (1/(1 + b^{2}))^{½}.

<H> = <ψ|H|ψ> = -(ħ^{2}/(2m))∫_{0}^{L}ψ*(∂^{2}/∂x^{2})ψ dx + U_{0}∫_{L/2}^{L}ψ*ψ dx.

<H> = A^{2}E_{1(unperturbed)}+ A^{2}b^{2}E_{2(unperturbed)}+ U_{0}[A^{2}∫_{L/2}^{L}ψ_{1}*ψ_{1}dx + A^{2}b^{2}∫_{L/2}^{L}ψ_{2}*ψ_{2}dx + 2A^{2}b(2/L)∫_{L/2}^{L}sin(πx/L) sin(2πx/L) dx]

= E_{0}[(1 + 4b^{2})/(1 + b^{2}) + 0.1*[0.5(1 + b^{2})/(1 + b^{2}) - (8/(3π))b/(1 + b^{2})]

= E_{0}[(1 + 4b^{2})/(1 + b^{2}) + 0.05 - (8/(30π))b/(1 + b^{2})].

d<H>/db = 0. 6b - (8/(30π))(1 - b^{2}) = 0. b^{2}+ 22.5π b - 1 = 0.

b = -11.25 π + ((11.25 π)^{2}+ 1)^{½}= +0.01414

<H>_{min}= E_{0}*1.0494

E_{1}= E_{0}+ 0.494 U_{0}is the best estimate for the ground state energy using the variational method.

Let

(a) Find the eigenvalues and eigenvectors of
H_{0}.

(b) Now derive the expressions for E^{(1)}, the change in
energy to first order in H_{1}.
Also derive an equation which must be satisfied for the first
order correction |ψ>^{(1)}.
Use these results to find E^{(1)} and |ψ>^{(1)}
for each of the unperturbed states.

(c) Now solve (H_{0 }+ H_{1})|ψ> = E|ψ>
exactly, obtaining both eigenvalues and eigenvectors. Expand your results to first order in
ε and verify that they agree with the results from part (b).
[Note: Your exact answers for the eigenvectors should be
normalized.]

Solution:

- Concepts:

Two-state problem, first-order perturbation theory - Reasoning:

We are asked to find the exact eigenvalues of a Hamiltonian operator in a two-dimensional vector space and to find its approximate eigenvalues to first order. - Details of the calculation:

(a) We diagonalize the matrix of H_{0}.

.

E^{2}= 1, E_{±}= ±1. |ψ_{+}> = 2^{-½}(|1> - |2>), |ψ_{-}> = 2^{-½}(|1> + |2>).

(b) For the two state system we have

H_{0}|ψ_{1}^{0}> = E_{0}^{1}|ψ_{1}^{0}> = 1*|ψ_{+}>, H_{0}|ψ_{2}^{0}> = E_{0}^{2}|ψ_{2}^{0}> = -1*|ψ_{-}>.

Let H = H_{0 }+ λW. Let |ψ_{1}> and |ψ_{2}> be the eigenstates of H.

H|ψ_{1}> = E^{1}|ψ_{1}>, H|ψ_{2}> = E^{2}|ψ_{2}>.

Assume λW = H_{1}, and λ is small.

Since λW is small, we assume that E and |ψ> can be expanded as a power series in λ.

E^{p }= E_{0}^{p }+ λE_{1}^{p }+ λ^{2}E_{2}^{p }+ ... ,

|ψ_{p}> = |ψ_{p}^{0}> + λ|ψ_{p}^{1}> + λ^{2}|ψ_{p}^{2}> + ... .

(For the two-state system p can only take on the values 1 or 2.)

We may then write

(H_{0 }+ λW)(|ψ_{p}^{0}> + λ|ψ_{p}^{1}> + λ^{2}|ψ_{p}^{2}> + ...)

= (E_{0}^{p }+ λE_{1}^{p }+ λ^{2}E_{2}^{p }+ ...)(|ψ_{p}^{0}> + λ|ψ_{p}^{1}> + λ^{2}|ψ_{p}^{2}> + ...).

This equation is must be valid over a continuous range of λ. Therefore we equate coefficients of equal powers of λ on both sides to obtain a series of equations that represent successively higher orders of the perturbation.

(i) (H_{0 }- E_{0}^{p})|ψ_{p}^{0}> = 0

(ii) (H_{0 }- E_{0}^{p})|ψ_{p}^{1}> = (E_{1}^{p }- W)|ψ_{p}^{0}>

(iii) (H_{0 }- E_{0}^{p})|ψ_{p}^{2}> = (E_{1}^{p }- W)|ψ_{p}^{1}> + E_{2}^{p}|ψ_{p}^{0}>

We choose <ψ_{p}^{0}|ψ_{p}^{0}> = 1. |ψ_{p}^{s}> is not uniquely defined. We can add an arbitrary multiple of |ψ_{p}^{0}> to each |ψ_{p}^{s}> without affecting the left hand side of the above equations. Most often this multiple is chosen so that <ψ_{p}^{0}|ψ_{p}^{s}> = 0. The perturbed ket is then**not**normalized.

Multiplying (ii) by <ψ_{p}^{0}| (and then setting λ = 1) we have E_{1}^{p }= <ψ_{p}^{0}|H_{1}|ψ_{p}^{0}>.

E_{1}^{p }is the first order energy correction.

|ψ_{p}^{1}> may be written as |ψ_{p}^{1}> = ∑_{p'}b_{p'}|ψ_{p'}^{0}>. Here b_{p}= 0 since <ψ_{p}^{0}|ψ_{p}^{1}> = 0.

For our two-state system we therefore can write

|ψ_{1}^{1}> = b_{2}|ψ_{2}^{0}>, |ψ_{2}^{1}> = b_{1}|ψ_{1}^{0}>.

Using (ii) we write

(H_{0 }- E_{0}^{1})|ψ_{1}^{1}> = (E_{1}^{1 }- H_{1})|ψ_{1}^{0}>,

b_{2}(H_{0 }- E_{0}^{1})|ψ_{2}^{0}> = b_{2}(E_{0}^{2 }- E_{0}^{1})|ψ_{2}^{0}> = (E_{1}^{1 }- H_{1})|ψ_{1}^{0}>.

Multiplying by <ψ_{2}^{0}| we obtain

b_{2}= <ψ_{2}^{0}|H_{1}|ψ_{1}^{0}>/(E_{0}^{1 }- E_{0}^{2}).

Similarly,

b_{1}= <ψ_{1}^{0}|H_{1}|ψ_{2}^{0}>/(E_{0}^{2 }- E_{0}^{1}).

In this problem

<ψ_{+}|H_{1}|ψ_{+}> = ½[<1|H_{1}|1> + <2|H_{1}|2> - <1|H_{1}|2> - <2|H_{1}|1>) = 0.

<ψ_{-}|H_{1}|ψ_{-}> = ½[<1|H_{1}|1> + <2|H_{1}|2> + <1|H_{1}|2> + <2|H_{1}|1>) = 0.

<ψ_{+}|H_{1}|ψ_{-}> = ½[<1|H_{1}|1> - <2|H_{1}|2> + <1|H_{1}|2> - <2|H_{1}|1>) = ε.

<ψ_{-}|H_{1}|ψ_{+}> = ½[<1|H_{1}|1> - <2|H_{1}|2> - <1|H_{1}|2> + <2|H_{1}|1>) = ε.

Therefore E_{1}^{1}= E_{1}^{2}= 0, the first order energy corrections are zero.

E^{1}= E_{+}and E^{2}= E_{-}to first order.

To first order the eigenstates of H are

|ψ_{1}> = |ψ_{+}> + (ε/2)|ψ_{-}>, |ψ_{2}> = |ψ_{-}> - (ε/2)|ψ_{+}>.

|ψ_{1}> = 2^{-½}[(1 + (ε/2))|1> - (1 - (ε/2))|2>],

|ψ_{2}> = 2^{-½}[(1 - (ε/2))|1> + (1 + (ε/2))|2>].

(c) We diagonalize the matrix of H.

.

(E - ε)(E + ε) = 1, E^{2}= 1 + ε^{2}, E = ±(1 + ε^{2})^{½}~ ±(1 + ε^{2}/2).

E^{1}= E_{+}= 1 and E^{2}= E_{-}= -1 to first order in ε.

|ψ_{1}> = a|1> + b|2>, where a(ε - (1 + ε^{2})^{½}) - b = 0, a^{2}+ b^{2}= 1.

|ψ_{2}> = c|1> + d|2>, where c(ε + (1 + ε^{2})^{½}) - d = 0, c^{2}+ d^{2}= 1.

Expand and keep only terms up to first order in ε.

a(ε - 1) - b = 0, a^{2}+ a^{2}(ε - 1)^{2}= 1, a^{2}+ a^{2}(1 - 2ε) = 1,

a^{2}= 1/(2 - 2ε), b^{2}= (1 - 2ε)/(2 - 2ε) ~ (1 - ε)/2

a ~ 2^{-½}((1 + (ε/2)), b ~ -2^{-½}((1 - (ε/2)).

c(ε + 1) - d = 0, c^{2}+ c^{2}(ε + 1)^{2}= 1, c^{2}+ c^{2}(1 + 2ε) = 1,

c^{2}= 1/(2 + 2ε), d^{2}= (1 + 2ε)/(2 + 2ε) ~( 1 + ε)/2.

c ~ 2^{-½}((1 - (ε/2)), d ~ 2^{-½}((1 + (ε/2)).

If we keep only terms up to first order in ε we get the same solution as in part b.