Verify that this kinetic energy operator is Hermitian

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  • #1
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Homework Statement



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

Homework Equations



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

103s7eg.png


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:
 

Answers and Replies

  • #2
DrClaude
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$$ KE_{mn} = (\frac{-\hbar^2}{2m}) \Big( \Psi_{m}^* \Psi_{n}^{'} - \int \Psi_{n}^{'} \Psi_{m}^{'*} dx \Big) $$
$$ KE_{mn} = (\frac{-\hbar^2}{2m}) \Big( \Psi_{m}^* \Psi_{n}^{'} - (\Psi_{n}^{'} \Psi_{m}^{*} - \int \Psi_{m}^{*}\Psi_{n}^{''} dx) \Big) $$
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.
 
  • #3
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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.

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 :)
 
  • #4
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Hi, may I have the full set solution of this question? Thank you
 
  • #5
PeroK
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Hi, may I have the full set solution of this question? Thank you

Okay, but maybe you are two years too late!

Personally, I would have used the Hermitian property of ##p## to show the Hermitian property of ##p^2## directly.
 
  • #6
DrClaude
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Hi, may I have the full set solution of this question? Thank you
We don't give full solutions at PhysicsForums. Try it yourself and create a thread with your question if you are having some problems. We will help you get to the solution.
 

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