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 |y(λ)> of H(λ)=H0+λW.

H(λ)|yp>=Ep(λ)|yp> or, keeping the dependence on λ in mind without specifically writing it down, H|yp>=Ep|yp>.

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

Ep=E0p+λE1p+λ2E2p+... ,
|yp>=|yp0>+λ|yp1>+λ2|yp2>+... .

We may then write

(H0+λW)(|yp0>+λ|yp1>+λ2|yp2>+...)

= (E0p+λE1p+λ2E2p+...)(|yp0>+λ|yp1>+λ2|yp2>+...).

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)|yp0>=0
(H0-E0p)|yp1>=(E1p-W)|yp0>
(H0-E0p)|yp2>=(E1p-W)|yp1>+E2p|yp0>
(H0-E0p)|yp3>=(E1p-W)|yp2>+E2p|yp1>+E3p|yp0>
...

(H0-E0p)|yp0>=0 implies that |yp0> is a linear combination of unperturbed eigenfuctionns |φip> with the corresponding eigenvalue E0.  We choose <yp0|yp0>=1. |yps> is not uniquely defined.  We can add an arbitrary multiple of |yp0> to each |yps> without affecting the left hand side of the above equations.  Most often this multiple is chosen so that <yp0|yps>=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 |yn0>=|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(l2).


Example:

En=E0n +O(l2). To first order, the eigenvalues of H are equal to the eigenvalues of the unperturbed Hamiltonian H0.

First-order eigenvector corrections:

(H0-E0p)|yp1> =( E1p-W)|yp0> = (E1p-W)|φp>. |yp0> is an eigenstate of the unperturbed Hamiltonian.  We may expand

in terms of the basis vectors |φip'>.  In the expansion bp=0 because <yp0|ypi>=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.