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In the case of a cyclotron, with a constant magnetic field $B$ in the vertical direction, a moving electron circles in a horizontal orbit.

The cyclotron frequency is $\omega = eB/m$.

At the same time, the spin precesses around the magnetic field with the Larmor frequency. (I assume $g=2$ here, so neglect the anomalous magnetic moment.) For $g=2$, the Larmor frequency is the same as the cyclotron frequency, if I am not wrong. Therefore the spin points always away from the axis of rotation (which is the magnetic field direction). I understand that such experiments are regularly made with many electrons circling in ring accelerators.

Now here are my simple and my hard questions:

(1 - simple) (A) Is it correct to say that the electron(s), when looking in the direction of the field, orbit(s) with/along the clock, due to the negative charge? (B) And that the spin precesses in the same direction/sense?

(2 - simple) Is it correct to say that in spin up state, therefore the spin points always away from the axis above the rotation plane, and for spin down it always points away below the rotation plane? (Assuming "above" is where the magnetic field B points to.)

(3 - hard) Electrons have spin 1/2. This means that their wave function comes back to itself after a rotation by $4 \pi$. In the cyclotron motion, what happens to the wave function phase $\delta$ after one full orbit around $B$ in physical space: did the phase rotate by $2 \pi$ or by $4 \pi$, or by another value?

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(3 - hard) Electrons have spin 1/2. This means that their wave function comes back to itself after a rotation by 4π. In the cyclotron motion, what happens to the wave function phase after one full orbit around B in physical space: did the phase rotate by 2π or by 4π, or by another value?

Attempt of an answer: Along a path in a magnetic field with vector potential $A$, the phase $\delta$ acquired by the wave function of a charged particle is $(e/\hbar) \int A dx$. After one complete orbit, the phase change is therefore given by $\delta= e \Phi / h = e B \pi r^2 / h$, where $r$ is the radius of the orbit and where $h= 2 \pi \hbar$. Using the cyclotron radius $r=m v / eB$ we appear to get $$\delta = \pi (m v)^2/(e B h)$$

I hope this is correct! This result of the phase $\delta$ acquired by an orbiting electron thus depends on its speed and on the magnetic field. In any case, the relation for the orbital phase change is not a simple, fixed multiple of $\pi$, as we could have wished for.

(2 - simple) Is it correct to say that in spin up state, therefore the spin points always away from the axis above the rotation plane, and for spin down it always points away below the rotation plane? (Assuming "above" is where the magnetic field B points to.)

This question should be answered "yes" by a graph from hyperphysics:

spin 1/2 up and down

In the spin up state, the spin points always away from the axis and above the rotation plane, and for spin down it always points below the rotation plane and outside.

(1 - simple) (A) Is it correct to say that the electron(s), when looking in the direction of the field, orbit(s) with/along the clock, due to the negative charge? (B) And that the spin precesses in the same direction/sense?

(A) This should be answered "yes" by this graph from hyperphysics on the cyclotron, which is for positively charged particles though:

Indeed, the electron(s), when looking in the direction of the field, orbit(s) clockwise.

(B) The question on precession for negatively charged particles should be answered "yes" by this picture from wikipedia, where the large red arrow is the magnetic field:

The sense of precession is the same as the sense of rotation.

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  • $\begingroup$ Giulia, that’s interesting but hard to read. Could you please extend your answer? $\endgroup$ – HolgerFiedler Oct 15 '17 at 7:44
  • $\begingroup$ I tried to make it clearer. $\endgroup$ – Giulia Tozzi Oct 15 '17 at 10:53
  • $\begingroup$ Why do you think that the electron wavefunction phase will be that value? $\endgroup$ – anna v Oct 18 '17 at 16:46
  • $\begingroup$ I explained it in more detail. $\endgroup$ – Giulia Tozzi Oct 20 '17 at 17:48
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(3 - hard) Electrons have spin 1/2. This means that their wave function comes back to itself after a rotation by 4π. In the cyclotron motion, what happens to the phase after one full orbit around B in physical space: did the phase rotate by 2π or by 4π?

The rotation of the synchrotron is around a macroscopic axis, the wavefunction's 2 or 4π rotation and consequent phase is around an axis through the center of mass of the electron. The macroscopic rotation is not relevant.

If one wants to go into details there will be phases that could be modeled quantum mechanically, but no longer simple one electron states. Have a look to get at the complexity here.

Edit after comment by OP:

But in the cyclotron, the rotation of an electron around its own axis and the motion around the orbit are coupled. So the integral over the vector potential A can be taken. What is the result? That is what I want to know.

The electron is a quantum entity, and the phase is a phase within the Ψ wavefunction of the electron, which is a function of (x,y,z,t) and the energy momentum of the fourvector of the electron. In the probability distribution which is the complex conjugate squared of the wavefunction there are no phases. Phases have a measurable effect only in interference terms, otherwise they are irrelevant specialized experiments have to be designed to "measure" the phase.

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  • $\begingroup$ But in the cyclotron, the rotation of an electron around its own axis and the motion around the orbit are coupled. So the integral over the vector potential $A$ can be taken. What is the result? That is what I want to know. $\endgroup$ – Giulia Tozzi Oct 15 '17 at 10:49
  • $\begingroup$ I will elaborate $\endgroup$ – anna v Oct 15 '17 at 13:14
  • $\begingroup$ You are right, in nature, absolute quantum phases are not important, but phase differences are. I would like to know the phase difference between an electron at the start of the orbit and an electron at the end of the orbit! $\endgroup$ – Giulia Tozzi Oct 15 '17 at 14:46
  • $\begingroup$ It is not that they are not important. they are not measurable. How do you propose to measure the phase at the beginning and end of orbit of the same electron without an interaction? $\endgroup$ – anna v Oct 15 '17 at 14:52
  • $\begingroup$ I understand that you do not want to answer the question. I do not force you! But it can be answered, and I am just looking for the answer. A magnetic field produces a phase difference along a path of a moving electron (for closed and even for open paths), and I just want to know the value. Indeed, the Aharonov-Bohm is a tellign example. $\endgroup$ – Giulia Tozzi Oct 15 '17 at 16:38
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... in spin up state ... the spin points always away from the axis above the rotation plane, and for spin down it always points away below the rotation plane? (Assuming "above" is where the magnetic field B points to.)

The direction of the spin up and down has two meanings, which is unfortunately sometimes not distinguished.

  1. In atoms two electrons couldn’t be identical in all quantum numbers. From Wikipedia:

    if two electrons reside in the same orbital, and if their n, ℓ, and mℓ values are the same, then their $m_s$ (spin quantum number) must be different, and thus the electrons must have opposite half-integer spins of 1/2 and −1/2.

    This statement is misleading. The spin of the electron is always 1/2. But the orientation of the magnetic dipole moments of two electrons with identical n, ℓ, and mℓ must be different. The two electrons are anti-aligned and this is expressed as up and down.

  2. Spin and magnetic dipol moment of a subatomic particle are clearly coupled. Let as assume that for the electron - by arbitrary definition - the spin points into the direction of the north pole of its magnetic dipole moment and we may call this spin up or 1/2 spin. Then for the positron as well as for the proton the spin points in the opposite direction to the north pole and call then this spin down or to distinguish it -1/2 spin.

The magnetic dipole moment and the related spin of the subatomic particles are intrinsic (existing under all circumstances) properties and this is the reason that Larmor precession and Lorentz force are observable and the magnetic dipole moment is asssociated with a spin.

Comming back to your statement above, the magnetic dipole moment of all the electrons gets aligned by the external magnetic field and the spin of all the electrons always points in the same direction to the external magnetic field. (For the positron, their magnetic dipole moment gets aligned in the same manner as for electrons, but the spin points in the opposite to the electrons direction. The manifestation for this is the Lorentz force, which directions are opposite for electron and positron.)

Electrons have spin 1/2. This means that their wave function comes back to itself after a rotation by 4π. In the cyclotron motion, what happens to the wave function phase after one full orbit around B in physical space: did the phase rotate by 2π or by 4π, or by another value?

Due to Wikipedia a spin 1/2 particle is imaginable as follows:

A single point in space can spin continuously without becoming tangled. Notice that after a 360-degree rotation, the spiral flips between clockwise and counterclockwise orientations. It returns to its original configuration after spinning a full 720 degrees.

enter image description here

What has the imagination, of how a half spin could be visualized, to do with the behavior of a moving electron in a magnetic field? Does the trajectory of the electron changes after one or two revolutions in the cyclotron? The answer has to be clearly no. So the phase stays unremained.

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