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Verify kinetic energy operator is Hermitian.

  1. Oct 19, 2016 #1
    1. The problem statement, all variables and given/known data

    Not actually a homework question but is an exercise in my lecture notes.

    2. Relevant equations

    I'm following this which demonstrates that the momentum operator is Hermitian:

    103s7eg.png

    3. The attempt at a solution


    $$KE_{mn} = (\frac{-\hbar^2}{2m}) \int\Psi_{m}^{*} \Psi_{n}^{''} dx $$
    $$ by parts: \int uv' = uv - \int vu' $$
    $$ KE_{mn} = (\frac{-\hbar^2}{2m}) \Big( \Psi_{m}^* \Psi_{n}^{'} - \int \Psi_{n}^{'} \Psi_{m}^{'*} dx \Big) $$
    $$ KE_{mn} = (\frac{-\hbar^2}{2m}) \Big( \Psi_{m}^* \Psi_{n}^{'} - (\Psi_{n}^{'} \Psi_{m}^{*} - \int \Psi_{m}^{*}\Psi_{n}^{''} dx) \Big) $$
    $$ KE_{mn} = (\frac{-\hbar^2}{2m}) \int \Psi_{m}^{*}\Psi_{n}^{''} dx $$
    $$KE_{mn} = KE_{mn}$$

    :oldconfused::oldconfused::oldconfused:

    Can anyone see the gaping error in my working?

    Thanks :oldsmile:
     
  2. jcsd
  3. Oct 19, 2016 #2

    DrClaude

    User Avatar

    Staff: Mentor

    The second time you are doing the integration by parts, you are making the wrong choice for ##u## and ##v'##.

    Also, don't forget that this is a definite integral.
     
  4. Oct 19, 2016 #3
    Thanks for this. Relieving to see it was a fairly simple arithmetic error, but worrying that my eyes continually didn't pick up on it...

    Pretty late in the evening now and I don't have any pen or paper handy for going over it all, but I'll check it out tomorrow.

    Cheers :)
     
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