Particle beam without spin into a magnetic field

[Telemachus]

Hi there. I have to solve this problem. The statement says as follows:

A particle beam with l=1, without spin and with momentum p is incident on the plane x= 0. In the region x<0 the particles move freely and for x>0 are under the action of a constant magnetic field H in the z direction $$V(x) = - \gamma m_z H_z$$ Assuming that the beam is not polarized what proportion of particles is reflected?

I thought of solving the schrödinger equation for the given potential in both regions of space. But I don't see how the angular momentum gets into the mud.

 

[Telemachus]

I mean, something like this:
For x>0:
$$\psi=Ae^{i \omega x}+Be^{-i \omega x}$$
$$\omega=\sqrt{\left (E+\frac{\gamma m_z H_z}{\hbar^2} \right ) } 2m$$

For x<0:
$$\psi=Ce^{i \omega' x}+De^{-i \omega' x}$$
$$\omega'=\sqrt{2mE}$$

Then I fit the continuity conditions for the wave function in x=0, and I can get the reflection coefficient from that. But I don't see how to get l=1 into the mud.

Anyone?

 

[TSny]

Hello, telemachus.

Your second post threw some light on your question for me. I'm really not sure but maybe it's something like this. Since l = 1 you know what the possible values of m_z are. For each of those values, can you find the reflection coefficient? Since the beam is unpolarized, I guess that would mean that each value of m_z is equally likely. Then you should be able to find the total fraction of particles in the beam that are reflected.

Hope I'm not misleading you. If we're lucky, others will chime in.

 

[Telemachus]

So I got
$$-1<m_z<1$$
$$m_z=-1,0,1$$
So I get the reflection coefficient for those values of mz and that's it.

Thank you Tsny.

Ok. I defined:
$$R=R^1+R^0+R^{-1}=\left | \frac{D^1}{C^1} \right |^2+\left | \frac{D^0}{C^0} \right |^2+\left | \frac{D^{-1}}{C^{-1}} \right |^2$$
But then I started to think if this wouldn't give an R bigger than 1, which would be an absurd. I thought of it as the superposition of the reflection for every wave corresponding to the random distribution of angular momenta. But perhaps I didn't do it well, what you say?

Should it be:
$$R=\left | \frac{D^1}{C^1} +\frac{D^0}{C^0}+\frac{D^{-1}}{C^{-1}} \right |^2$$?