Particle beam without spin into a magnetic field

In summary, the statement says that 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? Your first post gave some insight into your question and I was able to find the reflection coefficient for each wave corresponding to the distribution of angular momenta.
  • #1
Telemachus
835
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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.
 
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  • #2
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
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
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 that's 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:
  • #5


Hello,

Thank you for reaching out regarding your problem. I would approach this problem by first examining the physical principles involved. In this case, we have a particle beam with a specific angular momentum (l=1) and no spin, incident on a magnetic field.

The key here is to understand how the magnetic field will affect the particles. In this case, the particles will experience a Lorentz force due to the magnetic field, which will cause them to change direction. This force is proportional to the particle's charge, momentum, and the strength of the magnetic field. Since the particles have no spin, we can assume that their charge remains constant.

To solve for the proportion of particles that are reflected, we need to consider the conservation of energy and momentum. The incident particles have a specific momentum (p) and energy (E), and after interacting with the magnetic field, they will have a different momentum and energy. By solving the Schrödinger equation for both regions of space, we can determine the change in momentum and energy of the particles.

From there, we can use the conservation laws to determine the proportion of particles that are reflected. This will depend on the strength of the magnetic field and the initial momentum of the particles.

I hope this helps guide you in your problem-solving process. If you have any further questions, please let me know.

Best,
 

1. What is a particle beam without spin?

A particle beam without spin refers to a beam of particles that do not have any intrinsic angular momentum, or spin. This means that the particles are not spinning on their own axis, unlike other types of particles such as electrons or protons.

2. What happens when a particle beam without spin enters a magnetic field?

When a particle beam without spin enters a magnetic field, it will experience a force known as the Lorentz force. This force will cause the particles to move in a circular path, with the radius of the circle depending on the strength of the magnetic field and the velocity of the particles.

3. What are the applications of particle beams without spin in a magnetic field?

Particle beams without spin in a magnetic field have various applications in scientific research, particularly in particle accelerators. They can also be used in medical treatments, such as proton therapy, which uses a beam of protons without spin to target and destroy cancer cells.

4. Can particle beams without spin be manipulated or controlled in a magnetic field?

Yes, particle beams without spin can be manipulated and controlled in a magnetic field through the use of specialized devices such as magnetic lenses and deflectors. These devices can alter the trajectory of the particles and steer them towards a desired destination.

5. What are the limitations of using particle beams without spin in a magnetic field?

One limitation of using particle beams without spin in a magnetic field is that the particles must have a high velocity in order to experience a significant force from the magnetic field. This can be difficult to achieve, especially with larger particles. Additionally, the particles may also experience a loss of energy due to collisions with other particles or interactions with the magnetic field itself.

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