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

In summary, the attempted solution for the problem of integrating psi*psi over all space fails because the first term vanishes when limits are reached.
  • #1
HotMintea
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.
 
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  • #2
Hi HotMintea! :smile:

Not sure what this is :redface:, but doesn't [constant]ab = 0 ?
 
  • #3
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
Hi tiny-tim and Matterwave! :smile:
Matterwave said:
As tiny-tim pointed out, you need limits of integration.

tiny-tim said:
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]
Matterwave said:
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]?
Matterwave said:
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] ?
Matterwave said:
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
Hi HotMintea! :smile:
HotMintea said:
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
tiny-tim said:
Not all waves are so uninteresting. :wink:
:wink:!
tiny-tim said:
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
Yes! :smile:
 
  • #8
Thanks for your help! :smile:
 

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

1. What is the concept of anti-hermitian in mathematics?

In mathematics, the term anti-hermitian refers to a property of a mathematical operator or matrix. An operator is considered anti-hermitian if it is equal to its own negative transpose.

2. How does integration by parts prove that d/dx is anti-hermitian?

Integration by parts is a mathematical technique used to integrate the product of two functions. When this technique is applied to the derivative operator d/dx, it can be shown that the result is equal to the negative of the original operator, proving that d/dx is anti-hermitian.

3. What is the relationship between anti-hermitian operators and quantum mechanics?

In quantum mechanics, anti-hermitian operators play a crucial role in representing physical observables such as energy and momentum. The eigenvalues of these operators correspond to the possible outcomes of a measurement in quantum mechanics.

4. Are there any real-life applications of anti-hermitian operators?

Yes, anti-hermitian operators have various applications in fields such as physics, engineering, and signal processing. They are used to model physical systems, analyze data, and solve differential equations.

5. How does the proof of d/dx being anti-hermitian relate to other mathematical concepts?

The proof of d/dx being anti-hermitian is closely related to other mathematical concepts such as integration, differentiation, and linear algebra. It also has connections to other areas of mathematics, such as complex analysis and functional analysis.

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