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

[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?

 

[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:

<br /> \langle L_z \rangle = \int \psi^* \hat{L}_z \psi~d\tau<br />

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

<br /> \hat{L}_z = i\hbar \frac{\partial }{\partial \varphi}<br />

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

<br /> \int_0^{2\pi} \psi^* \psi ~d\varphi = 1,<br />

so you will have to change your function little bit.

 

[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

 

[Jano L.]

How did you get m into denominator?