PHYS 424 Notes

Suppose that we have a Hamiltonian that can be written as the sum of two parts,

H = H₀ + lH'

If we had the first part alone we could solve for its eigenstates and eigenvalues,

H₀y_n ⁽⁰⁾ = E_n⁽⁰⁾ y_n⁽⁰⁾,

but with both parts included the Schroedinger Equation is not exactly soluble. For instance, in Griffiths Problem 6.1, a delta-function potential with a small coefficient in front is inserted into the center of an infinite well, so that

H₀= p²/(2m) + V(x)

      V(x) = 0 for |x|<a
      V(x) = ¥ for |x|>a

H' = a d(x - a/2)

In this example a is itself small, and the factor l in front of H' is not really needed. Similar thing happen in all examples, so l is not really needed but can simply be set equal to 1. We will eventually do just that, but for the derivation of formulas l is a convenient marker of which terms are smaller and which larger. It merely serves as a flag for the size of iindividual terms.

Assume that we can expand E_n and y_n as

E_n = E_n⁽⁰⁾ + lE_n⁽¹⁾ + l²E_n⁽²⁾ + . . .
and
y_n = y_n⁽⁰⁾ + ly_n⁽¹⁾ + l²y_n⁽²⁾ + . . .

and substitute into (H₀ + lH') y_n - E_n y_n = 0

to get

   l⁰ [ H₀ y_n⁽⁰⁾ - E_n⁽⁰⁾ y_n⁽⁰⁾ ]
+ l¹ [ H₀y_n⁽¹⁾ + H'y_n⁽⁰⁾ - E_n⁽⁰⁾y_n⁽¹⁾ - E_n⁽¹⁾y_n⁽⁰⁾ ]
+ l² [ H₀y_n⁽²⁾ + H'y_n⁽¹⁾ - E_n⁽⁰⁾y_n⁽²⁾ - E_n⁽¹⁾y_n⁽¹⁾ - E_n⁽²⁾y_n⁽⁰⁾ ]
+ ...
= 0.

If the series for E_n and y_n converged, we could set the coefficients of each power of l separately to zero. In practice, the best you can do is to establish that the two series are asymptotic, not convergent. However, if we can find expressions that make each coefficient vanish, we would (might) have a practical set of approximations to the solution of the Hamiltonian H. So we make the attempt:

From the coefficient of l⁰ we have

H₀ y_n⁽⁰⁾ = E_n⁽⁰⁾ y_n⁽⁰⁾

so that y_n⁽⁰⁾ and E_n⁽⁰⁾ are the eigenstates and eigenvalues of H₀. We can use the known solutions of H₀ as the simplest approximations to the solutions of H₀ + H'. This situation is physically sensible. The quantity E_n⁽⁰⁾ is known as the "unperturbed" energy; similarly y_n⁽⁰⁾ is called the "unperturbed" wave function.

Where convenient in the remainder of the derivation, we will write y_n⁽⁰⁾ as |n> and y_n^{(0) *} as <n|.

If we take the equation obtained by setting the coefficent of l¹ equal to zero, multiply by <n|, and integrate over all coordinates, we get

<n|H₀|y_n⁽¹⁾> + <n|H'|n> = E_n⁽⁰⁾<n|y_n⁽¹⁾> +E_n⁽¹⁾<n|n>

But <n|H₀ = E_n⁽⁰⁾<n| and <n|n> = 1, and we get

Hence the first correction to the energy is easily calculated. It is just the expectation value, in the unperturbed states, of the perturbing Hamiltonian.

The first correction to the wave function is a bit more complicated. There is enough information in the coefficient of l ¹ to get the correction, but a bit of work is required to extract the result. The eigenfunctions are presumably a complete set, so we can expand y_n⁽¹⁾ in terms of the |n> The resulting y_n⁽⁰⁾ + l y_n⁽¹⁾ must be normalized to order (l). It turns out that the normalization condition can and must be satisfied by omitting the term m=n from the expansion! As a result the expansion is

y_n⁽¹⁾ = S _{m ¹ n} a_m|m>

Substituting this expansion into the (vanishing) coefficient of l¹ gives

H₀ S _{m ¹ n} a_m |m> + H' |n> = E_n⁽⁰⁾ S _{m ¹ n} a_m|m> + E_n⁽¹⁾ |n>

If we multiply by <m'|, we get

S _{m ¹ n} [ a_m <m'|H₀|m> ] + <m'|H'|n> = E_n⁽⁰⁾ S _{m ¹ n} a_m<m'|m> + E_n⁽¹⁾ <m'|n>

If we choose m'¹n the last term in this expression will vanish using the orthonormality of the unperturbed wave functions. Then using H₀|n> = E_n⁽⁰⁾|n>, and the orthonrmality relation <m'|m> = d_{m' m} we get

a_m' (E_m'⁽⁰⁾ - E_n⁽⁰⁾) = - <m'|H'|n>.

a_m' = <m'|H'|n> / (E_n⁽⁰⁾ - E_m'⁽⁰⁾)

with m' different from n so that the denominator is OK. Now, dropping the unnecessary prime from m'

A similar analysis, taking into account the orthogonality of the zero-order and first-order wave functions, turns the coefficient of l² into

E_n⁽²⁾ = S_m¹n |<m|H'|n>|² / (E_n⁽⁰⁾ - E_m⁽⁰⁾)

Notice that in both E⁽¹⁾ and y⁽¹⁾ it is vital that there are no values of n and m for which E_n⁽⁰⁾ = E_m⁽⁰⁾. Hence this procedure only works for Hamiltonians with nondegenerate unperturbed energies, or at least for unperturbed states which are not degenerate with any other unperturbed state.

Now let's look at Problem 6.1, where we put a delta-function perturbation in the center of the infinite well:

H' = a d(x - a/2)

The unperturbed states and energies are

|n> = sqrt(2/a) sin (npx/a)       E_n = (n p hbar)² / (2ma²)

We need the matrix elements
<q|H'|n> = òdx (2/a) a sin (qpx/a) d(x - a/2) sin (npx/a)
              = (2/a) a sin (qp/2) sin (npa/2)
              = (2a/a) (-1)^(q+n-2)/2,       q and n odd
              = 0                                 q or n even

The first-order perturbations to the energy vanish for even n, since there is zero probability for the unperturbed particle to be inside the perturbing potential. For odd n, the energies including first-order perturbations are

E_n = (n p hbar)² / (2ma²) + 2a / a To see a change in the energies of the even-n states, we must go to second order. The result is

DE_n = (2a/a)² S_{q odd} [1/(E_q(⁰⁾ - E_n⁽⁰⁾)]
       = [8 m a² / (p ² hbar²)] S_{q odd} [1/(q² - n²)]