Perturbation Theory

Stationary perturbation theory is concerned with finding the changes in the discrete energy levels and the changes in the corresponding energy eigenfunctions of a system, when the Hamiltonian of a system is changed by a small amount. Let

H = H0 + H’ = H0 + λW

H0 is the unperturbed Hamiltonian whose eigenvalues E0p and eigenstates |φip> are known. Let {|φip>} denote an orthonormal eigenbasis of H0

H0 | φip> = E0p | φip >.

Here i denotes the degeneracy. Assume that the matrix elements of H’ in the eigenbasis of H0 are small compared to the matrix elements of H0.

<φip|H ' |φip>   <<   <φip | H0 | φip > = E0p.

We write

H’ = λW

with λ << 1 and <φip|W |φip> ≈ E0p.


Example:

H = H0 + H’ is the Hamiltonian of a perturbed one-dimensional harmonic oscillator.

We are looking for the eigenvalues E(λ) and the eigenstates |ψ(λ) > of H(λ) = H0 + λW. H(λ)|ψp> = Ep(λ)|ψp> or, keeping the dependence on λ in mind without specifically writing it down

H |ψp> = Ep |ψp>.

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

Ep=E0p+λE1p+λ2E2p+...,
|ψp>=|ψp0>+λ|ψp1>+λ2|ψp2>+....

We may then write

(H0 + λW)(|ψp0>+λ|ψp1> + λ2|ψp2> + ...)

=(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.
(H0-E0p)|ψp0>=0
(H0-E0p)|ψp1>=(E1p-W)|ψp0>
(H0-E0p)|ψp2>=(E1p-W)|ψp1>+E2p|ψp0>
(H0-E0p)|ψp3>=(E1p-W)|ψp2>+E2p|ψp1>+E3p|ψp0>
...

(H0-E0p)|ψp0> = 0 implies that |ψp0> is a linear combination of unperturbed eigenfuctionns |φip> with the corresponding eigenvalue E0.  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. We then have

0                              =   

To calculate the energy to sth order, we only need to know the state vector to order s-1.

First-order perturbation theory for non-degenerate levels

Consider a particular non-degenerate eigenvalue E0n of H0. H0| φn> = E0n| φn>.  The other eigenvalues of H may or may not be degenerate.  We have | ψn0> = | φn> and E1n = <φn |W| φn>. The first-order energy correction therefore is λE1n = < φn |H’| φn>. We have

En=E0n+<φn|H’|φn>+O(&lambda2)

Example:



En = E0n + O(λ2). To first order, the eigenvalues of H are equal to the eigenvalues of the unperturbed Hamiltonian H0.

First-order eigenvector corrections:

(H0-E0p)| ψp1> = (E1p-W)| ψp0 > = (E1p-W)| φp >.

|ψp0> is an eigenstate of the unperturbed Hamiltonian.  We may expand

in terms of the basis vectors |φip' >.  In the expansion bp = 0 because <ψp0| ψpi > = 0.

Multiply from the left by <φip'' |.

Therefore


Example:

Second-order perturbation theory for non-degenerate levels

Second-order energy corrections:

Since we have found the expression for the state vector to first order, we can now find the expression for the energy to second order.



Example:

Let H = H0 + H’ = H0 + λW.  In practice, after having derived the perturbation expansion, we often set λ = 1 and let H’ = W be small.

Problems:

Calculate the first-order shift in the ground state of the hydrogen atom caused by the finite size of the proton. Assume the proton is a uniformly charged sphere of radius r = 10-13cm. The ground state wave function of the hydrogen atom is and the Bohr constant is a0=0.53⋅10-10m .

Solution:

The ground state is not degenerate.  The potential energy of an electron outside a uniformly charged sphere of radius r0 and total charge qe is 

Inside the sphere the potential energy is 

From Gauss’ law we know that (SI units).

Let 

since r0 << a0,

=4⋅10-9eV.

A particle of mass m is in an infinite potential well perturbed as shown in the figure.

(a)   Calculate the first-order energy shift of the nth eigenvalue due to the perturbation.

(b)  Write out the first three non vanishing terms for the first-order perturbation expansion of the ground state in terms of the unperturbed eigenfunctions of the infinite well.

(c)  Calculate the second-order energy shift for the ground state.

Image4736.gif (2294 bytes)
Solution:

H = H0 + H’. 

The eigenvalues of H0 are not degenerate.





(a) 


(b)  First-order perturbation theory yields

  if n is odd.


(c) 

second-order energy shift for the ground state.

The second-order energy shift of the ground state is always negative.

Auf diesem Webangebot gilt die Datenschutzerklärung der TU Braunschweig mit Ausnahme der Abschnitte VI, VII und VIII.