Normalisation Constant In Wavefunction

[laser1]

TL;DR

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

 

[weirdoguy]

but would this be a valid answer for the normalisation constant?

Yes.

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.

 

[laser1]

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.

 

[Nugatory]

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.

 

[laser1]

Phase angles are not physically significant but the difference between two phase angles is.

I don't follow.

 

[pines-demon]

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.

 

[weirdoguy]

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.

 

[laser1]

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.

 

[weirdoguy]

. 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.