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

In summary, the conversation was about proving the equation \langle p \rangle = \langle -\frac{\partial V}{\partial x} \rangle and its relation to the derivatives of the wavefunction \Psi. The suggestion was made that since \Psi is a normalizable wavefunction, it must approach 0 as x \rightarrow \pm\infty, and therefore its derivatives must also approach 0. However, there was uncertainty about whether this argument holds.
  • #1
broegger
257
0
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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  • #2
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.
 
  • #3
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: :tongue2:
 
  • #4
I see your point :) I mean the time derivative, of course.
 
  • #5
That's something totally different.It is simply CORRECT...


Daniel.
 
  • #6
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.
 

1. How do you define normalizability in mathematical terms?

In mathematics, normalizability refers to the property of a function or wavefunction to have a finite integral when squared over its entire domain. This means that the total probability or amplitude of the function is equal to 1, indicating that it is a valid representation of a physical system.

2. Why is proving normalizability important in scientific research?

Proving normalizability is important because it ensures that the mathematical model being used to describe a physical system is valid and physically meaningful. It also allows for accurate calculations and predictions to be made based on the model.

3. What are the key steps in proving $\Psi$ and $\Psi'$ normalizable?

The key steps in proving normalizability of a function $\Psi$ and its derivative $\Psi'$ are: 1) setting up the integral for the squared function, 2) using integration by parts to simplify the integral, 3) applying the boundary conditions to determine the limits of integration, and 4) showing that the resulting integral is finite.

4. Can you give an example of a function that is not normalizable?

One example of a function that is not normalizable is the Dirac delta function, which is commonly used in quantum mechanics. This function has an infinite amplitude at a specific point, making its integral over its entire domain infinite and thus not normalizable.

5. How does the proof of normalizability relate to the physical interpretation of a wavefunction?

The proof of normalizability is closely related to the physical interpretation of a wavefunction, as it ensures that the function is a valid representation of a physical system. It also allows for the calculation of probabilities and amplitudes, which are essential in understanding the behavior of the system.

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