Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Beginner's quantum mechanic homework

  1. Dec 14, 2005 #1
    [tex]\widehat{H} = \frac{p^2}{2m} + V(x)[/tex]

    if eigenvalue of H operator is [tex]E_n[/tex] and eigenvectors are [tex]u_n[/tex], show that

    [tex]\Sigma_m (E_m-E_n) |x_{mn}|^2 = \frac{\hbar^2}{2m}[/tex]

    is true. here, [tex]x_{mn} = (u_m, xu_n)[/tex] is a matrix element.
     
  2. jcsd
  3. Dec 14, 2005 #2

    Physics Monkey

    User Avatar
    Science Advisor
    Homework Helper

    Could you please show us what you've done so far? We can't help until you do. Thanks.
     
  4. Dec 14, 2005 #3
    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...
     
  5. Dec 14, 2005 #4

    StatusX

    User Avatar
    Homework Helper

    I don't understand your question. Is that sum supposed to be over n and m? And what exactly is xmn? Is it a vector? A complex number? What does x represent? Please be clearer with your notation.
     
  6. Dec 14, 2005 #5

    Physics Monkey

    User Avatar
    Science Advisor
    Homework Helper

    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 [tex] \sum_m (E_m - E_n)\langle n | x | m \rangle \langle m | x | n \rangle, [/tex] where all I have done is make everthing 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, [tex] x_{n m} = \langle n | x | m \rangle [/tex] is a matrix element (a complex number) of the position operator [tex] x [/tex].
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook