Normalisation Constant In Wavefunction

  • #1
laser1
147
23
TL;DR Summary
normalisation constant
When doing the standard procedure for normalisation a wavefunction, I get $$|A|=\sqrt \frac{2}{L}$$ where A is the normalisation constant. It is mathematically correct to say that $$A=\sqrt \frac{2}{L} e^{i\theta}$$ but would this be a valid answer for the normalisation constant? In my lecture notes and in books, the ##e^{i\theta}## term is not included. I remember reading something about this once online, but I can't find the source. Thank you
 
Physics news on Phys.org
  • #2
laser1 said:
but would this be a valid answer for the normalisation constant?

Yes.

laser1 said:
In my lecture notes and in books, the eiθ term is not included.

Because it's arbitrary, and can be set to 1. If you're a rebel you can use some other ##\theta\neq 0##. And sometimes people do, if it's convenient.
 
  • Like
Likes laser1
  • #3
weirdoguy said:
Yes.



Because it's arbitrary, and can be set to 1. If you're a rebel you can use some other ##\theta\neq 0##. And sometimes people do, if it's convenient.
What does the ##\theta## mean, physically? Or does it have no meaning other than mathematical.
 
  • #4
laser1 said:
What does the ##\theta## mean, physically? Or does it have no meaning other than mathematical.
Phase angles are not physically significant but the difference between two phase angles is.
 
  • #5
Nugatory said:
Phase angles are not physically significant but the difference between two phase angles is.
I don't follow.
 
  • #6
laser1 said:
TL;DR Summary: normalisation constant

When doing the standard procedure for normalisation a wavefunction, I get $$|A|=\sqrt \frac{2}{L}$$ where A is the normalisation constant. It is mathematically correct to say that $$A=\sqrt \frac{2}{L} e^{i\theta}$$ but would this be a valid answer for the normalisation constant? In my lecture notes and in books, the ##e^{i\theta}## term is not included. I remember reading something about this once online, but I can't find the source. Thank you
Is this from the infinite potential well? As others have said, yes you can choose any ##e^{i\theta}## but this has no physical consequences (think of the measurement postulate, it depends on the absolute squared value ##|\psi|^2=|e^{i\theta}\psi|^2##).

There is some useful relation that you can prove. When your Hamiltonian is time-reversal symmetric, you can choose the eigenfunctions to be real.
 
  • #7
laser1 said:
I don't follow.

##\theta## does not mean anything physically. But when you add two wavefunctions with different phase angles the difference of those becomes physically significant since: ##\vert e^{i\theta_1}\psi_1+ e^{i\theta_2}\psi_2\vert^2=\ldots## where dots represent something I wanted to write but I gave up, because I'm on my phone and writing lateX on your phone may kill you 💀 I leave that for you as an exercise. You'll se that there will be a term with difference of phase angles.
 
  • Like
Likes laser1 and Nugatory
  • #8
pines-demon said:
Is this from the infinite potential well?
Yes.

Take, for example, the below wavefunction. With method 1, I get an extra phase. With method 2, I do not. (In method 2 I admit I forgot the negative root, but even so, that's only 2 values of theta, namely, 0, and pi.) Why does method 2 not yield that e^i term? Also, I realise this is kind of a "homework" question now so I might be in the wrong thread.

1738699854957.png
 
  • #9
laser1 said:
. With method 2, I do not.

You do if you realise that A is (or may be) complex. So in your integral there should be A*, and that yields |A|^2, and so on. Btw, both of your 'methods' are exactly the same, since |z|^2=zz*. You just wrote out different steps in left and right columns.
 
  • Like
Likes pines-demon and laser1

Similar threads

Replies
10
Views
880
Replies
12
Views
710
Replies
2
Views
1K
Replies
5
Views
2K
Replies
6
Views
1K
Replies
1
Views
2K
Replies
4
Views
2K
Replies
5
Views
998
Replies
5
Views
1K
Back
Top