Particle beam without spin into a magnetic field

Telemachus
Messages
820
Reaction score
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 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.
 
Physics news on Phys.org
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&#039; x}+De^{-i \omega&#039; x}
\omega&#039;=\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?
 
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.
 
So I got
-1&lt;m_z&lt;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?
 
Last edited:
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top