Proving sum rule for position matrix elements in quantum mechanics

[gulsen]

$$\widehat{H} = \frac{p^2}{2m} + V(x)$$

if eigenvalue of H operator is $$E_n$$ and eigenvectors are $$u_n$$, show that

$$\Sigma_m (E_m-E_n) |x_{mn}|^2 = \frac{\hbar^2}{2m}$$

is true. here, $$x_{mn} = (u_m, xu_n)$$ is a matrix element.

 

[Physics Monkey]

Could you please show us what you've done so far? We can't help until you do. Thanks.

 

[gulsen]

well, we've read the question over and over again and, ummm, that's almost it i guess. expect we tried orthagonality relation -maybe it had something with the question- and realized we've all messed it up...

in short, we couldn't manage to get anything worth to mention...

 

[StatusX]

I don't understand your question. Is that sum supposed to be over n and m? And what exactly is x_mn? Is it a vector? A complex number? What does x represent? Please be clearer with your notation.

 

[Physics Monkey]

gulsen,

Here are some thoughts to help you get started. The first thing I would do is expand out the left hand side so you can see the structure. In other words, write it something like $$\sum_m (E_m - E_n)\langle n | x | m \rangle \langle m | x | n \rangle,$$ where all I have done is make everything very explicit. From this expression is should be clear that you can perform the m sum, so why don't you try again with this hint.

StatusX.

The sum is just over m, it just so happens that the result is independent of n. Also, $$x_{n m} = \langle n | x | m \rangle$$ is a matrix element (a complex number) of the position operator $$x$$.