Normalising phi for the Hydrogen atom.

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Homework Help Overview

The discussion revolves around normalizing the wave function \(\Phi\) for the hydrogen atom, specifically focusing on the normalization constant \(A\) and the calculation of the integral involved in the process. Participants are examining the implications of the absolute square of the wave function and its role in probability density.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to compute the normalization constant \(A\) by evaluating an integral involving the wave function \(\Phi\). There are questions about the absolute square of \(\Phi\) and how it relates to the normalization process. Some participants are clarifying the definition of the absolute square and its computation.

Discussion Status

The discussion is active, with participants exploring different interpretations of the absolute square of the wave function. There is a mix of attempts to clarify the normalization process and some guidance on how to compute the integral correctly. No explicit consensus has been reached, but there are indications of progress in understanding the normalization concept.

Contextual Notes

Participants are working within the constraints of a homework assignment, which may limit the information they can share or the methods they can use. There is an emphasis on ensuring that the calculations adhere to the definitions and properties of wave functions in quantum mechanics.

mrausum
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This is most likely very simple, but I can't figure it out.

http://www.sussex.ac.uk/physics/teaching/btv/Lect02_2006.pdf

Step 5 they've got an equation for \Phi. They then normalise it to get A = \frac{1}{\sqrt{2\pi}}. Every time I do the integral I get:

A^2.^{2\pi}_{0}[ \frac{exp(2i\sqrt{\Lambda}\Phi)}{2i\sqrt{\Lambda}}] = 1

Which makes the integral go to zero when you rewrite the exp using Euler's and and take into account \sqrt{\Lambda} must be an integer?
 
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What is the absolute square of phi?
 
Edgardo said:
What is the absolute square of phi?

The probability density function. So?
 
No, I mean write down phi and then compute |phi|^2. What do you get for |phi|^2?
 
Edgardo said:
No, I mean write down phi and then compute |phi|^2. What do you get for |phi|^2?

A^2.exp(2i\sqrt{\Lambda}\Phi)
 
No, look again. It is the absolute square.
 
Edgardo said:
No, look again. It is the absolute square.

So I can't take the two inside the exp? I don't really see why not? So I'm guessing you're saying that I have to do the integral of phi multiplied by its conjugate?

Thanks anyway, that's solved it :)
 
Last edited:

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