D.49 The generalized variational principle

The purpose of this note is to verify directly that the variation of the expectation energy is zero at any energy eigenstate, not just the ground state.

Suppose that you are trying to find some energy eigenstate $\psi_n$ with eigenvalue $E_n$, and that you are close to it, but no cigar. Then the wave function can be written as

\varepsilon_1 \psi_1
+ \varepsilon_2 \psi_2
+ ...
...epsilon_n) \psi_n
+ \varepsilon_{n+1} \psi_{n+1}
+ \ldots

where $\psi_n$ is the one you want and the remaining terms together are the small error in wave function, written in terms of the eigenfunctions. Their coefficients $\varepsilon_1,\varepsilon_2,\ldots$ are small.

The normalization condition $\langle\psi\vert\psi\rangle$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 is, using orthonormality:

1 =
+ \varepsilon_2^2
+ \ldots
+ \...
+ (1+\varepsilon_n)^2
+ \varepsilon_{n+1}^2

The expectation energy is

\big\langle E\big\rangle =
\varepsilon_1^2 E_1
+ \vare...
...varepsilon_n)^2 E_n
+ \varepsilon_{n+1}^2 E_{n+1}

or plugging in the normalization condition to eliminate $(1+\varepsilon_n)^2$

\big\langle E\big\rangle & = &
\varepsilon_1^2 (E_1-E_n)
+ E_n
+ \varepsilon_{n+1}^2 (E_{n+1}-E_n)

Assuming that the energy eigenvalues are arranged in increasing order, the terms before $E_n$ in this sum are negative and the ones behind $E_n$ positive. So $E_n$ is neither a maximum nor a minimum; depending on conditions $\big\langle E\big\rangle $ can be greater or smaller than $E_n$.

Now, if you make small changes in the wave function, the values of $\varepsilon_1,\varepsilon_2,\ldots$ will slightly change, by small amounts that will be indicated by $\delta\varepsilon_1,\delta\varepsilon_2,\ldots$, and you get

\delta\big\langle E\big\rangle & = &

This is zero when $\varepsilon_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\varepsilon_2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\ldots$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, so when $\psi$ is the exact eigenfunction $\psi_n$. And it is nonzero as soon as any of $\varepsilon_1,\varepsilon_2,\ldots$ is nonzero; a change in that coefficient will produce a nonzero change in expectation energy. So the variational condition $\delta\langle{E}\rangle$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 is satisfied at the exact eigenfunction $\psi_n$, but not at any nearby different wave functions.

The bottom line is that if you locate the nearest wave function for which $\delta\langle{E}\rangle$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 for all acceptable small changes in that wave function, well, if you are in the vicinity of an energy eigenfunction, you are going to find that eigenfunction.

One final note. If you look at the expression above, it seems like none of the other eigenfunctions are eigenfunctions. For example, the ground state would be the case that $\varepsilon_1$ is one, and all the other coefficients zero. So a small change in $\varepsilon_1$ would seem to produce a change $\delta\langle{E}\rangle$ in expectation energy, and the expectation energy is supposed to be constant at eigenstates.

The problem is the normalization condition, whose differential form says that

0 =
2 \varepsilon_1\delta\varepsilon_1
+ 2 \varepsilon...
+ 2 \varepsilon_{n+1}\delta\varepsilon_{n+1}
+ \ldots

At $\varepsilon_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 and $\varepsilon_2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\ldots$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\varepsilon_{n-1}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $1+\varepsilon_n$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\varepsilon_{n+1}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\ldots$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, this implies that the change $\delta\varepsilon_1$ must be zero. And that means that the change in expectation energy is in fact zero. You see that you really need to eliminate $\varepsilon_1$ from the list of coefficients near $\psi_1$, rather than $\varepsilon_n$ as the analysis for $\psi_n$ did, for the mathematics not to blow up. A coefficient that is not allowed to change at a point in the vicinity of interest is a confusing coefficient to work with.