# QM Variation Method

## Homework Statement

Show that variation principle (parameters ci) leads to equations
$$\sum\limits_{i = 1}^n {\left\langle i \right|H\left| j \right\rangle c_j = Ec_i {\rm{ where }}} \left\langle j \right|H\left| i \right\rangle = \int {d\textbf{r}^3 \chi _j^* \left( \textbf{r} \right)\left( {H\chi _i \left( \textbf{r} \right)} \right)}$$

## Homework Equations

I've got $$\psi \left( \textbf{r} \right) = \sum\limits_{i = 1}^n {c_i \chi _i \left( \textbf{r} \right)}$$
, but

## The Attempt at a Solution

I'm confused about what's being asked, and what the expected result means. Indices as kets and bras? If they're different vectors within the Hilbert space, won't they be orthonormal implying all i not equal j would be zero? I have a nagging feeling I'm either confused by notation or overlooking something basic.

## The Attempt at a Solution

The left hand side is just the matrix elements of the hamiltonian in some orthonormal basis. If they are eigenstates of the Hamiltonian, this matrix will be diagonal.

The expression they give you for the matrix elements is the position space representation

Beyond that, I don't know what you're asking.

Well, thanks for your response. First, let me say that the problem statement as shown was exactly as presented, and we were given no other information. The only thing I can think of is that the indices in the bras and kets are really intended to be $$\chi_{i}$$ and $$\chi_{j}$$, but this seems to be trivial, because in that case wouldn't $$\left\langle j \right|H\left| i \right\rangle = \int {d\textbf{r}^3 \chi _j^* \left( \textbf{r} \right)\left( {H\chi _i \left( \textbf{r} \right)} \right)}$$
imply $$\left\langle i \right|H\left| j \right\rangle = \int {d\textbf{r}^3 \chi _i^* \left( \textbf{r} \right)\left( {H\chi _j \left( \textbf{r} \right)} \right)}$$
? And if that's the case, doesn't that simply mean that the $$\chi$$'s are simply the eigenstates of the Hamiltonian?