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when the energy levels are about to be degenerate

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(1)

Second order perturbation: To calculate correction in second order, we will make use of λ2 equation (11),

( ˆH0−En(0)(2)n = −( ˆH0−En(1)(1)n +En(2)ψ(0)n (19) and go straight for expansion of ψ(2)n andψn(1) in terms of unperturbed wavefunction ψn(0),

ψ(1)n = X

m

c(1)m ψm(0) and ψn(2) = X

m

c(2)m ψ(0)m . (20) Substituting the above expression forψ(2)n and ψn(1) in (19), we obtain,

( ˆH0−En(0))X

m

c(2)m ψm(0) = −( ˆH0−En(1))X

m

c(1)m ψm(0)+En(2)ψn(0) X

m

(Em(0)−En(0))c(2)m ψm(0) = −X

m

( ˆH0−En(1))c(1)m ψm(0)+En(2)ψn(0). As before, we take inner product with the unperturbed wavefunctionψk(0),

X

m

(Em(0)−En(0))c(2)m

ψ(0)k , ψm(0)

=

−X

m

ψk(0),Hˆ0ψm(0)

c(1)m +X

m

En(1)c(1)m

ψ(0)k , ψ(0)m

+En(2)

ψk(0), ψ(0)n

or, X

m

(E(0)m −E(0)n )c(2)m δkm=−X

m

c(1)mkm0 +X

m

En(1)c(1)m δkm+En(2)δkn or, (Ek(0)−En(0))c(2)k =−X

m

c(1)mkm0 +En(1)c(1)k +En(2)δkn. (21) Fork=nwe get the second order correction to energy, keeping in mind that c(1)n = 0,

En(2) = X

m6=n

c(1)mnm0 = X

m6=n

mn0nm0 En(0)−Em(0)

= X

m6=n

|Hˆmn0 |2 E(0)n −E(0)m

. (22)

Fork6=n, we will getc(2)k knowingEn(1) (13) andc(1)m (17), c(2)k (Ek(0)−En(0)) = −X

m

c(1)mkm0 +c(1)knn0 or, c(2)k (Ek(0)−En(0)) = −X

m6=n

mn0km0 En(0)−Em(0)

+

kn0nn0 En(0)−Ek(0) or, c(2)k = X

m6=n

km0mn0

(En(0)−Em(0))(En(0)−Ek(0))

− Hˆkn0nn0 (En(0)−Ek(0))2

. (23)

Therefore, the energy En and wavefunctionψn of the full Hamiltonian (1) to second order in perturbation theory are,

En = En(0)+ ˆHnn0 + X

m6=n

|Hˆmn0 |2 En(0)−Em(0)

+· · · (24)

ψn = ψn(0)+ X

m6=n

"

mn0 En(0)−Em(0)

# ψm(0)

+X

m6=n

 X

k6=n

mk0kn0

(En(0)−Ek(0))(En(0)−Em(0))

− Hˆmn0nn0 (En(0)−Em(0))2

ψm(0)+· · · (25) 1

(2)

For perturbation theory to work, the corrections it produces must be small (not wildly different fromEn(0)). But onward second order corrections in energy (24) and first order in wavefunction (25) contain the term that must be small,

mn0 En(0)−Em(0)

1, n6=m

otherwise it has potential to grow large if En(0) ≈ E(0)m , i.e. when the energy levels are about to be degenerate. Therefore, degenerate energy levels have to be treated differently in perturbation theory.

Examples

1. Using first order perturbation theory, calculate the energy of then-th state for a particle of mass m moving in an infinite potential well of length 2L with wall at x = 0 and x = 2L, which is modified at the bottom by the perturbations: (i) λV0sin(πx/2L) and (ii) λV0δ(x−L), where λ1.

2. Calculate the energy of the n-th excited state to first order perturbation for a 1-dim infinite potential well of length 2L, with walls at x=−L and x =L, which is modified at the bottom by the following perturbations with V0 1,

0 =

−V0 −L≤x≤L

0 elsewhere Hˆ0 =

−V0 −L/2≤x≤L/2 0 elsewhere

0 =

−V0 −L/2≤x≤0

0 elsewhere Hˆ0 =

V0 0≤x≤L/2 0 elsewhere

3. For a 1-dim harmonic oscillator, the spring constant changes from k tok(1 +), where is small. Calculate the first order perturbation in the energy.

4. Calculate the first order perturbation in the energy for n-th state of a 1-dim harmonic oscillator subjected to perturbationβ x4,β is a constant.

5. Consider a quantum charged 1-dim harmonic oscillator, of chargeq, placed in an electric field E~ = Ex. Find the exact expression for the energy and then use perturbation theoryˆ to calculate the same.

6. For the following set of Hamiltonians, withλ1,

(i) E0

1 λ λ 3

(ii) E0

1 +λ 0 0 0

0 8 0 0

0 0 3 −2λ

0 0 −2λ 7

(iii) E0

−5 3λ 0 0

3λ 5 0 0

0 0 8 −λ

0 0 −λ −8

(iv) E0

3 2λ 0 0

2λ −3 0 0

0 0 −7 √

0 0 √

2λ 7

 ,

(a) find the eigenvalues and eigenvectors of the unperturbed Hamiltonian, (b) diagonalize the full Hamiltonian to find the exact eigenvalues and expand each eigenvalue to the second power ofλand (c) using the first and second order perturbation theory find the approximate energy eigenvalues and eigenstates of the full Hamiltonian.

2

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