# Expectation value of z component of angular momentum for a particle on a ring

1. Oct 14, 2012

### rmjmu507

I have to find the expectation value of the z component of the angular momentum for a particle on a ring and the expectation value of the z component of the angular momentum squared for a particle on a ring.

The wavefunction is e^((± imx))

I've determined that the expectation value for the z component is -$\hbar$/m and that the expectation value for the square of the z component is $\hbar$ squared over m squared.

This would mean that the uncertainty in the z component of the angular momentum for a particle on a ring is 0.

Is this correct?

2. Oct 15, 2012

### Jano L.

No, the angular momentum should come out as proportional to m. You have to calculate the average value of L_z in this way:

$$\langle L_z \rangle = \int \psi^* \hat{L}_z \psi~d\tau$$

where $\tau$ stands for the coordinates of the particle. On a circle, the position of particle is given by just one coordinate, usually angle. Let us denote it by $\varphi$. It takes values from 0 to 2$\pi$. In this coordinate, the operator of angular momentum is given by

$\hat{L}_z = i\hbar \frac{\partial }{\partial \varphi}$

Tha last thing you need is the wave function. You gave $e^{im\varphi}$, but this is not correct wave function because it is not normalized.

The correct function has to satisfy

$$\int_0^{2\pi} \psi^* \psi ~d\varphi = 1,$$

so you will have to change your function little bit.

3. Oct 15, 2012

### rmjmu507

I get

$\frac{1}{2π}$$\int$(-i$\hbar$/im) d$\varphi$ which, integrated over 0 to 2π yields -$\hbar$/m

following the same procedure, I find the expectation of L$_{z}^{2}$ is -$\hbar^{2}$/m$^{2}$

Still the same result

4. Oct 16, 2012

### Jano L.

How did you get m into denominator?