1
$\begingroup$

The usual way of solving a QM problem with a small perturbation operator $V$ is done in the following way: Of course I assume that the solutions (eigen-functions $\psi^0$ and eigenvalues $E^0$) of the unperturbed system are known and form a complete and discrete system: $H_{0}\psi^0_m = E^0_m \psi^0_m$.

We want to know the (approximate) solutions of the equation:

$$ H\psi = (H_0 +V)\psi =E\psi$$ (1)

As we know of the completeness of the functions $\psi^0$ we can expand the searched solution $\psi$ in $\psi^0_n$:

$$ \psi = \sum\limits_m c_m \psi^0_m$$

The plug this expansion into (1) and get:

$$ \sum\limits_m c_m(E^0_m +V)\psi^0_m = \sum\limits_m c_m E \psi^0_m$$

Both sides are multiplied by $\psi^{(0)\ast}_k$, integrate and use the orthogonality of the functions $\psi^0_m$:

$$ (E-E^0_k)c_k = \sum\limits_m V_{km} c_m$$ (2)

where the matrix elements $V_{km} $ are defined as

$$ V_{km} = \int \psi^{(0)\ast}_k V \psi^{(0)}_m d^3x $$.

So far so good. For the calculation of the 1. correction to the $k^{th}$ eigenvalue the following approximation is usually done:

$ c^0_k =1$ and $c^0_m =0$ with $m\neq k $

Plugged into equation (2) I get:

$E-E^0_k = V_{kk}$ respectively $E^{(1)} = V_{kk}$

This is standard approximate solution of the given perturbation problem and corresponding eigenfunction is $\psi^0_k$, actually the same as for the unperturbed solution assuming that the perturbation is rather small. The next order solution can be obtained in a rather similar way ( $c^1_k = V_{km} /(E^{(0)}_m - E_k^0)$ applying $k\neq m$).

However, I think, a better solution could be obtained. Actually equation (2) could be written like this ($\delta_{km}$ being the kronecker symbol):

$$\sum\limits_m ( V_{km}- \delta_{km}(E-E^0_k))c_m =0$$

and non-trivial solutions could be obtained by requiring (the secular equation):

$det(V_{km}- \delta_{km}(E-E^0_k))=0$

For each $k$ an eigenvalue $\lambda^k \equiv E^{k(1)}=(E-E^0_k)$ would be obtained and for each of these eigenvalues I would get coefficients $c^{k}_m$ which would provide a better solution of the eigenfunctions $\psi_k = \sum\limits_m c^k_m \psi^0_m$.
So my question is: Why this way is not considered ? Why the secular equation solution is only considered for degenerate energy eigenvalue systems ?

$\endgroup$
0
$\begingroup$

Actually, the standard strategy in perturbation theory as presented in the question is only valid if the inequality

$$ |V_{km}| << | E^0_k - E^0_m| $$

is fulfilled, in other words, the perturbation has to be small compared to the distance of the affected unperturbated energy levels.

On the other hand if the affected energy levels are close to each other compared to the perturbation, the solution of

$$(E-E^0_k)c_k = \sum\limits_m V_{km}c_m$$

by the secular equation can indeed be reasonable. It is demonstrated in the book Landau/Lifshitz III (QM) in the chapter on the two-atomic molecule. This particular case distinguish from the standard perturbation method by the use of a parameter which "tunes" the strength of the perturbation. For the two-atomic molecule it is actually the distance $r$ between the nuclei of both atoms. This parameter can make $\hat{V}$ "small" or "large", or even better the whole physical range of the parameter $r$ can be considered. In that case it is reasonable to solve the corresponding secular equation in order to find when for instance 2 originally separated energy levels cross (couple), if the latter happens the condition $ |V_{km}| << | E^0_k - E^0_m| $ is of course no longer fulfilled.

In particular upon solution of the secular equation of a 2-level system the following energy eigenvalues are found:

$$E=\frac{1}{2}(E_1+E_2+V_{11}+V_{22})\pm\sqrt{\frac{1}{4}(E_1-E_2+V_{11}-V_{22})^2 +|V_{12}|^2}$$

which yields for $|V_{12}|\approx 0$ (what corresponds to $ |V_{km}| << | E^0_k - E^0_m| $):

$$E=E_1 + V_{11}$$ and as second equation $$E = E_2 +V_{22}$$.

The results of the standard perturbation theory are reproduced.

I hope that this insight is also interesting for other users.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.