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?