Proof that d/dx is anti-hermitian by integration by parts

  • Thread starter HotMintea
  • Start date
  • #1
43
0
The attempt at a solution

[tex]
\begin{equation*}
\begin{split}
\ -\ i\int\psi^* \frac{\partial{\psi}}{\partial x}= \\ -i(\psi^*\psi\ - \int\psi \frac{\partial\psi^*}{\partial x})\space\ (?)
\end{split}
\end{equation*}
[/tex]

I thought [itex] \psi^*\psi\ = \ constant\neq\ 0.[/itex] However, it vanishes in the correct proof.
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
25,832
251
Hi HotMintea! :smile:

Not sure what this is :redface:, but doesn't [constant]ab = 0 ?
 
  • #3
Matterwave
Science Advisor
Gold Member
3,965
326
As tiny-tim pointed out, you need limits of integration.

However, psi*psi is not a constant. It is only a constant when integrated over all space, but it is not integrated in this case.

The real reason the first term vanishes is that for limits at plus or minus infinity, the wave-function must decay to 0 to be normalizable. For periodic boundary conditions, the two limits are the same point so they must subtract to 0.
 
  • #4
43
0
Hi tiny-tim and Matterwave! :smile:
As tiny-tim pointed out, you need limits of integration.
Not sure what this is :redface:, but doesn't [constant]ab = 0 ?
For a decaying wave function, is the following correct?:
[tex]
\begin{equation*}
\begin{split}
\ -\ i\int_{-\infty}^{\infty} \psi^*\frac{\partial \psi}{\partial x}\ =\\-\ i\ (\psi^*\psi\ |_{-\infty}^{\infty} - \int_{-\infty}^{\infty} \psi\frac{\partial \psi^*}{\partial x})\ = \\ i\int_{-\infty}^{\infty}\psi\frac{\partial\psi^*}{\partial x}
\end{split}
\end{equation*}
[/tex]
The real reason the first term vanishes is that for limits at plus or minus infinity, the wave-function must decay to 0 to be normalizable.
Thus, [itex] \psi\ = \ 0 \ at \ x\ = -\infty\ \ and \ \infty [/itex]?
For periodic boundary conditions, the two limits are the same point so they must subtract to 0.
For periodic boundary conditions, [itex]\ -\ i\int_{a}^{a} \psi^*\frac{\partial \psi}{\partial x}\ = \ i\int_{a}^{a}\psi\frac{\partial\psi^*}{\partial x}\ = \ 0 [/itex] ?
However, psi*psi is not a constant. It is only a constant when integrated over all space, but it is not integrated in this case.
For [itex] \psi\ = \ Ae^{i(k x-\omega t)} [/itex] where A is an arbitrary constant, [itex] \psi^* \psi\ = \ A^2 \ = \ constant. [/itex] Is [itex] \psi^* \psi\ = \ constant [/itex] not true in general? (Example taken from: http://en.wikipedia.org/wiki/Normalizable_wave_function#Example_of_normalization)
 
Last edited:
  • #5
tiny-tim
Science Advisor
Homework Helper
25,832
251
Hi HotMintea! :smile:
Thus, [itex] \psi\ = \ 0 \ at \ x\ = -\infty\ \ and \ \infty [/itex]?
Yes.
For periodic boundary conditions, [itex]\ -\ i\int_{a}^{a} \psi^*\frac{\partial \psi}{\partial x}\ = \ i\int_{a}^{a}\psi\frac{\partial\psi^*}{\partial x}\ = \ 0 [/itex] ?
No, [itex] \psi[/itex] is the same (at each boundary), so the [] is zero, not the ∫.
For [itex] \psi\ = \ Ae^{i(k x-\omega t)} [/itex] where A is an arbitrary constant, [itex] \psi^* \psi\ = \ A^2 \ = \ constant. [/itex] Is [itex] \psi^* \psi\ = \ constant [/itex] not true in general?
That's a featureless plane wave, so of course [itex] \psi^* \psi[/itex] is constant.

Not all waves are so uninteresting. :wink:
 
  • #6
43
0
Not all waves are so uninteresting. :wink:
:wink:!
No, [itex] \psi[/itex] is the same (at each boundary), so the [] is zero, not the ∫.
For the periodic boundary conditions, is the following correct?:
[tex]
\begin{equation*}
\begin{split}
\ -\ i\int_{a}^{b} \psi^*\frac{\partial \psi}{\partial x}\ = -\ i\ (\psi^*\psi\ |_{a}^{b} - \int_{a}^{b}\psi\frac{\partial\psi^*}{\partial x})\
\end{split}
\end{equation*}
[/tex]
However, [itex] \psi(a)\ = \psi(b).[/itex] Moreover, conjugates [itex] \psi^*(a)\ = \psi^*(b). [/itex] Thus, [itex] \psi^*\psi\ |_{a}^{b}\ = \ 0.[/itex]

Therefore,
[tex]\begin{equation*}\begin{split} -\ i\int_{a}^{b} \psi^*\frac{\partial \psi}{\partial x}\ = \ i\int_{a}^{b}\psi\frac{\partial\psi^*}{\partial x} \end{split}\end{equation*}[/tex]
 
Last edited:
  • #7
tiny-tim
Science Advisor
Homework Helper
25,832
251
Yes! :smile:
 
  • #8
43
0
Thanks for your help! :smile:
 

Related Threads on Proof that d/dx is anti-hermitian by integration by parts

  • Last Post
Replies
3
Views
10K
Replies
1
Views
1K
Replies
7
Views
1K
Replies
2
Views
555
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
10
Views
8K
  • Last Post
Replies
20
Views
2K
  • Last Post
Replies
2
Views
853
Replies
2
Views
13K
Replies
1
Views
1K
Top