Proving $\Psi$ and $\Psi'$ Normalizable: A Mathematical Proof

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Discussion Overview

The discussion revolves around proving the normalizability of wavefunctions, specifically focusing on the mathematical expressions related to the expectation value of momentum and its relation to potential energy. Participants explore the conditions under which certain boundary terms vanish, which is critical for establishing the validity of the equations presented.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests that if the wavefunction $\Psi$ is normalizable, it must approach 0 as $x \rightarrow \pm\infty$, which would imply that certain boundary terms vanish.
  • Another participant questions the validity of the initial claims regarding the expectation value of momentum and its relation to the potential energy gradient.
  • A later reply humorously challenges the correctness of the proof, indicating a potential disagreement about the underlying assumptions or conclusions.
  • There is a clarification regarding the time derivative of the expectation value of momentum, indicating a possible miscommunication about the original statements.

Areas of Agreement / Disagreement

Participants express differing views on the correctness of the initial claims and the implications of normalizability. The discussion remains unresolved, with multiple competing interpretations of the mathematical relationships presented.

Contextual Notes

Participants do not fully agree on the implications of normalizability and the correctness of the mathematical expressions. There are unresolved assumptions regarding the behavior of wavefunctions and their derivatives at infinity.

broegger
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I am trying to prove that

[tex]\frac{d\langle p \rangle}{dt} = \langle -\frac{\partial V}{\partial x} \rangle[/tex]

I am done if I can just prove that

[tex]\left[ \Psi^*\frac{\partial^2 \Psi}{\partial x^2} \right]_{-\infty}^{\infty} = 0[/tex]

[tex]\left[ \frac{\partial \Psi}{\partial x} \frac{\partial \Psi^*}{\partial x} \right]_{-\infty}^{\infty} = 0[/tex]

My suggestion is that since [tex]\Psi[/tex] is a wavefunction, it is normalizable and must approach 0 as [tex]x \rightarrow \pm\infty[/tex], and so must its derivatives. I don't know if this argument holds?
 
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broegger said:
I am trying to prove that

[tex]\langle p \rangle = \langle -\frac{\partial V}{\partial x} \rangle[/tex]

I am done if I can just prove that

[tex]\left[ \Psi^*\frac{\partial^2 \Psi}{\partial x^2} \right]_{-\infty}^{\infty} = 0[/tex]

[tex]\left[ \frac{\partial \Psi}{\partial x} \frac{\partial \Psi^*}{\partial x} \right]_{-\infty}^{\infty} = 0[/tex]

My suggestion is that since [tex]\Psi[/tex] is a wavefunction, it is normalizable and must approach 0 as [tex]x \rightarrow \pm\infty[/tex], and so must its derivatives. I don't know if this argument holds?


I would think that's a valid reason, no ?

cheers,
Patrick.
 
broegger said:
I am trying to prove that

[tex]\langle p \rangle = \langle -\frac{\partial V}{\partial x} \rangle[/tex]

Good luck proving something incorrect! :smile: :biggrin: :-p
 
I see your point :) I mean the time derivative, of course.
 
That's something totally different.It is simply CORRECT...


Daniel.
 
dextercioby said:
That's something totally different.It is simply CORRECT...

No, it also excaped me, but of course what is correct is:

d/dt <p> = <- dV/dx >

cheers,
Patrick.
 

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