Verify that this kinetic energy operator is Hermitian

[sa1988]

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:

The Attempt at a Solution

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



Can anyone see the gaping error in my working?

Thanks

 

[DrClaude]

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

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.

 

[sa1988]

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 :)

 

[Louis419]

Hi, may I have the full set solution of this question? Thank you

 

[PeroK]

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.

 

[DrClaude]

Hi, may I have the full set solution of this question? Thank you

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