Particle beam without spin into a magnetic field

  • Thread starter Telemachus
  • Start date
  • #1
833
30
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 [tex]V(x) = - \gamma m_z H_z[/tex] 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.
 

Answers and Replies

  • #2
833
30
I mean, something like this:
For x>0:
[tex]\psi=Ae^{i \omega x}+Be^{-i \omega x}[/tex]
[tex]\omega=\sqrt{\left (E+\frac{\gamma m_z H_z}{\hbar^2} \right ) } 2m[/tex]

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

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?
 
  • #3
TSny
Homework Helper
Gold Member
13,159
3,458
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 mz 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 mz 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.
 
  • #4
833
30
So I got
[tex]-1<m_z<1[/tex]
[tex]m_z=-1,0,1[/tex]
So I get the reflection coefficient for those values of mz and thats it.

Thank you Tsny.

Ok. I defined:
[tex]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[/tex]
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:
[tex]R=\left | \frac{D^1}{C^1} +\frac{D^0}{C^0}+\frac{D^{-1}}{C^{-1}} \right |^2[/tex]?
 
Last edited:

Related Threads on Particle beam without spin into a magnetic field

Replies
4
Views
5K
  • Last Post
Replies
1
Views
2K
Replies
1
Views
1K
Replies
2
Views
796
  • Last Post
Replies
0
Views
3K
Replies
2
Views
4K
  • Last Post
Replies
2
Views
889
Replies
7
Views
4K
  • Last Post
Replies
1
Views
3K
Replies
12
Views
7K
Top