Mixed method problems
Consider the 1D infinite well, U(x) = 0 for 0 < x < L, U(x) infinite
The eigenfunctions and eigenvalues of the Hamiltonian for a particle of
mass m trapped in this well are
ψn(x) = (2/L)½sin(nπx/L),
En = n2π2ħ2/(2mL2),
Now consider the perturbed 1D infinite well, U(x) = 0 for 0 < x < L/2,
U(x) = U0 for L/2 < x < L, U(x) infinite everywhere else.
Let E0 denote the ground state energy of the unperturbed
Assume U0 = E0/10 = 0.1*π2ħ2/(2mL2),
i.e. U0 = 0.1 times the ground state energy of the
(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
(c) Solve for the energy eigenvalues of the perturbed well using 1st order
(c) Solve for the lowest energy eigenvalues of the perturbed well using
the variational method.
Comparing approximation methods
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
ψ1(x) = A sin(k1x)
The most general solution of the time independent solution in region 2
satisfying the boundary condition at x = L is
ψ2(x) = B sin(k2(L-x)).
Here k12 =
(2m/ħ2)E and k22 =
(2m/ħ2)(E - U0).
The boundary conditions are that
ψ(x) and (∂/∂x)ψ(x)
are continuous at x = L/2.
A sin(k1L/2) = B sin(k2L/2).
k1A cos(k1L/2) = -k2B cos(k2L/2).
This yields tan(k1L/2)/(k1L/2) = -tan(k2L/2)/(k2L/2).
This equation can be solved numerically. We expect solutions for
values of k1L/2 = nπ/2 + ε1 and k2L/2 = nπ/2 -
ε2, where ε1 and ε2 are small positive
Using Excel to find the ground state energy of the perturbed well E1
we obtain E1 = 1.04936 E0.
E1 = π2ħ2/(2mL2) +
(b) The WKB approximation:
In one dimension, if U(x) finite everywhere, then the WKB approximation
requires ∫cylcepdx = ∫cylceħkdx
= (n + ½)h.
Here U(x) is infinite everywhere except in some finite region and ∫cylcepdx =
∫0L ħkdx = nh/2 = nπħ. ∫0L/2
(2mE)½dx + ∫L/2L (2m(E-U0))½dx =
(2mE)½L/2 + (2m(E-U0))½L/2 = nπħ.
E/4 + (E - U0)/4 + (E(E-U0))½/2 = n2π2ħ2/(2mL2)
E/2 - U0/4 + (E/2)(1 - U0/E)½ ~ E/2 - U0/4
+ E/2 - U0/4 = E - U0/2 = n2E0.
E = n2E0 + U0/2.
Higher precision solution of
E/2 - U0/4 + (E/2)(1 - U0/E)½
= E0: E1 = π2ħ2/(2mL2)
+ 0.506 U0.
(c) First order perturbation theory:
En = n2π2ħ2/(2mL2)
sin2(πx/l)U0)dx = n2π2ħ2/(2mL2)
+ 0.5 U0.
E1 = π2ħ2/(2mL2)
+ 0.5 U0.
(d) The variational method:
We have to choose a trial wave function with an adjustable parameter which
satisfies the boundary conditions.
ψ(x) = A (2/L)½(sin(πx/L) + bsin(2πx/L)) = A(ψ1(x)
Normalize: A2(1 + b2) = 1, A = (1/(1 + b2))½.
<H> = <ψ|H|ψ> = -(ħ2/(2m))∫0L
ψ*(∂2/∂x2)ψ dx + U0∫L/2L
<H> = A2E1(unperturbed) + A2b2E2(unperturbed) +
+ A2b2∫L/2Lψ1*ψ1 dx
+ 2A2b(2/L)∫L/2Lsin(πx/L) sin(2πx/L)
= E0[(1 + 4b2)/(1+ b2) + 0.1*[0.5(1 + b2)/(1
+ b2) - (8/(3π))b/(1 + b2)]
= E0[(1 + 4b2)/(1+ b2) + 0.05 - (8/(30π))b/(1
d<H>/db = 0. 6b - (8/(30π))(1 - b2) = 0. b2
+ 22.5π b - 1 = 0.
b = -11.25 π + ((11.25 π)2 + 1)½ = +0.01414
<H>min = E0*1.0494
E1 = π2ħ2/(2mL2) +
0.494 U0 is the best estimate for the ground state energy
using the variational method.
(a) Find the eigenvalues and eigenvectors of
(b) Now derive the expressions for E(1), the change in
energy to first order in H1.
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 (H0 + H1)|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
Two-state problem, first-order perturbation theory
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
- Details of the calculation:
(a) We diagonalize the matrix of H0.
E2 = 1, E± =
= 2-½(|1> - |2>), |ψ->
= 2-½(|1> + |2>).
(b) For the two state system we have
H0|ψ10> = E01|ψ10>
H0|ψ20> = E02|ψ20>
Let H = H0 + λW. Let |ψ1>
and |ψ2> be the eigenstates of H.
H|ψ1> = E1|ψ1>,
H|ψ2> = E2|ψ2>.
Assume λW = H1, and λ is small.
is small, we assume that E and |ψ> can be expanded
as a power series in λ.
Ep = E0p + λE1p
+ λ2E2p +
|ψp> = |ψp0>
+ ... .
(For the two-state system p can only take on the values 1 or 2.)
We may then write
= (E0p + λE1p
+ λ2E2p +
...)(|ψp0> + λ|ψp1> + λ2|ψp2> +
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.
(ii) (H0 - E0p)|ψp1>
= (E1p - W)|ψp0>
= (E1p - W)|ψp1>
We choose <ψp0|ψp0>
= 1. |ψps> is not uniquely
defined. We can add an arbitrary multiple of |ψp0>
to each |ψps> without
affecting the left hand side of the above equations. Most often this multiple
is chosen so that <ψp0|ψps>
= 0. The perturbed ket is then not normalized.
Multiplying (ii) by <ψp0| (and
then setting λ = 1) we have E1p
is the first order energy correction.
|ψp1> may be written as |ψp1>
Here bp = 0 since <ψp0|ψp1>
For our two-state system we therefore can write
|ψ11> = b2|ψ20>,
|ψ21> = b1|ψ10>.
Using (ii) we write
= (E11 - H1)|ψ10>,
= b2(E02 - E01)|ψ20>
= (E11 - H1)|ψ10>.
Multiplying by <ψ20| we
b2 = <ψ20|H1|ψ10>/(E01
b1 = <ψ10|H1|ψ20>/(E02
In this problem
+ <2|H1|2> - <1|H1|2>
- <2|H1|1>) = 0.
+ <2|H1|2> + <1|H1|2>
+ <2|H1|1>) = 0.
- <2|H1|2> + <1|H1|2>
- <2|H1|1>) =
- <2|H1|2> - <1|H1|2>
+ <2|H1|1>) =
Therefore E11 = E12 = 0, the first
order energy corrections are zero.
E1 = E+ and E2 = E- to first
To first order the eigenstates of H are
|ψ1> = |ψ+>
|ψ2> = |ψ->
|ψ1> = 2-½[(1 + (ε/2))|1>
- (1 - (ε/2))|2>],
|ψ2> = 2-½[(1 - (ε/2))|1>
+ (1 + (ε/2))|2>].
(c) We diagonalize the matrix of H.
(E - ε)(E + ε) = 1, E2
= 1 + ε2, E =
±(1 + ε2)½
~ ±(1 +
E1 = E+ = 1
and E2 = E- = -1 to first order in ε.
|ψ1> = a|1> + b|2>, where a(ε
- (1 + ε2)½)
- b = 0, a2 + b2 = 1.
|ψ2> = c|1> + d|2>, where c(ε
+ (1 + ε2)½)
- d = 0, c2 + d2 = 1.
Expand and keep only terms up to first order in ε.
- 1) - b = 0, a2
- 1)2 = 1,
a2 + a2(1 - 2ε)
a2 = 1/(2 - 2ε),
b2 = (1 - 2ε)/(2
(1 - ε)/2
2-½((1 + (ε/2)), b ~ -2-½((1 - (ε/2)).
+ 1) - d = 0, c2
+ 1)2 = 1,
c2 + c2(1 + 2ε)
c2 = 1/(2 + 2ε),
d2 = (1 + 2ε)/(2
1 + ε)/2.
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.