Question about wavenumbers and determining ?

[philmccrevace]

Homework Statement

The normalized wavefunctions for a particle confined to move on a circle are w(o) = sqrt(1/2pi) e^-imo where m = 0, ±1, ±2, ±3... and o is between 0 and 2pi. Determine o

w = psi
o = the o with the vertical line in the middle

Homework Equations

Not sure what would be relevant in this case. I've tried e^-ix = cos x - i sin x

The Attempt at a Solution

I've tried to use the euler method, but I'm not sure how to solve everything to get the o on one side to solve. The answer according to the teacher is pi, but I have no idea how to get there. How do I do this?

 

[benorin]

We have

$$\psi_m (\phi ) = \frac1{\sqrt{2\pi}}e^{-im\phi},\qquad\mbox{ where } m=0,\pm 1,\pm 2,\ldots\mbox{ and }0\leq \phi \leq\2\pi$$​

There might be something missing from your question: specifically, we are to determine $$\phi$$ under what condition? It is not normalization, for the condition that wavefunctions $$\psi_m (\phi)$$ are normalized is already met. The following proves this:

In this case, the wavefunctions $$\psi_m (\phi)$$ are normalized if, and only if

$$\int_{0}^{2\pi}\left|\psi_m (\phi)\right|^2 d\phi =1.$$​

Indeed this is already so since

$$\left|\psi_m (\phi)\right|^2 = \psi_m (\phi)\psi_m^* (\phi) = \left(\frac1{\sqrt{2\pi}}e^{-im\phi}\right)\left(\frac1{\sqrt{2\pi}}e^{im\phi}\right) =\frac1{2\pi}$$​

where $$\psi_m^* (\phi)$$ denotes the complex conjugate of $$\psi_m (\phi)$$ and hence we see that

$$\int_{0}^{2\pi}\left|\psi_m (\phi)\right|^2 d\phi =\int_{0}^{2\pi}\frac1{2\pi} d\phi=1.$$​

Thus the condition of normaliztion is already met.

 

[philmccrevace]

I think it wants to know the expectation value for phi. The problem has phi like <phi>.

 

[dextercioby]

So how do you compute the expectation value of $\varphi$ ?

Daniel.