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I Where Does Larmor Precession Occur?

  1. Sep 5, 2016 #1

    Twigg

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    I have been doing a lot of exercises out of Griffiths lately, and one thing I spent a lot of time thinking about was Larmor precession. It's a fun thing to work out, but where does it actually happen? I'm having a hard time to see where it would apply.

    The first condition for Larmor precession to occur is that you have a spin-1/2 particle at rest. First off, the at rest part bugs me in part because it seems impossibly difficult to have a free electron "at rest" relative to your lab. In theory, I get that it works out. "At rest" in this context means that ##\langle \vec{p} \rangle = 0##, which is well and good, since it implies that ##\langle \vec{p} \cdot \vec{A} \rangle = 0##. This means there should be no canonical momentum funky business coming into the Pauli equation, and so the particle should look like a standing wavepacket, leaving only the spin-y stuff affecting the Hamiltonian. What bugs me is that it seems impossibly difficult to get a free electron at rest in the lab frame. For starters, you would need to do the experiment in a Faraday cage to keep out the air waves, but with the walls far from the electrons so they don't have a mirror charge reaction, which would make the electron move. Earnshaw's theorem would suggest that this experiment can't be done without something forcing the electron to stay at rest. You can't optically trap a free electron, because there is no restoring force, so your laser will always be blue detuned off resonance no matter what. I suppose the next best thing to a free electron is a conduction electron in a material, but common sense says that if you put a magnetic field in a conductor it heats up because of eddy currents: moving electrons. That empirical fact would seem to suggest that conduction band electrons don't satisfy ##\langle \vec{p} \rangle = 0##, I suspect due to thermal motion. Say you got it down to 1.8K with a helium cryo, assuming the conductor doesn't start superconducting, then maybe you could get the electrons to precess a little before they start gaining momentum. Still, I highly doubt that we care about Larmor precession for that application, if it even worked.

    I have a limited understanding of these things, but I suspect that the real reason we care about Larmor precession is that the behavior of an atom in a magnetic field can be decomposed into an atomic part describing atom-y stuff (Zeeman splitting of orbits, etc.) and a part describing the precession of the electron as it "orbits" the nucleus, via an interaction picture or some other fancy transformation of the Pauli equation Hamiltonian. This seems a lot more important an application from an atomic physics point of view. Am I getting warmer? If so, I would actually like to try this problem. Can I get out the Larmor precession Hamiltonian as an interaction term starting with the Pauli equation for a hydrogen atom in a magnetic field? If this is not the case, please let me know so I don't sink a bunch of hours into it.
     
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  3. Sep 5, 2016 #2

    Vanadium 50

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    But you're OK with frictionless planes, stretchless ropes, perfectly elastic springs, etc. ?
     
  4. Sep 5, 2016 #3

    Twigg

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    Touche! :DD

    Still, frictionless planes and stuff have their uses analytically because there are conditions under which the effect of friction is negligible. But for the free electron in a magnetic field, if the electron moves at all the system may not be stable. I can't honestly tell offhand if it would orbit like any old charged particle or if the spin affects the trajectory at all. More importantly, I can't tell if having the electron move changes its precession at all (if it still precesses). Maybe that's the issue. How would I determine this from the Pauli equation?
     
  5. Sep 5, 2016 #4
    @l Twigg.....you forget Larmor precession is not limited to 'electrons" . Nuclear protons (which are 'essentially' at 'rest') in a lattice are characterized by specific Larmor frequency, and used extensively in Nuclear magnetic resonance with great effect as imaging systems. And the being "at rest" is somewhat of a moot point because even in liquid environments the precession rate is not determined by the orientation of the spins with respect to the B field, the Larmor frequency being determined essentially only by the gyromagnetic ratio and the Magnetic field strength
     
    Last edited: Sep 5, 2016
  6. Sep 5, 2016 #5

    Twigg

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    @Pet Scan That makes a lot of sense. I was aware that you would get some sort of more complicated precession for higher spin fermions, but I did not think of nuclei. Glad to finally have a sense of what NMR is. I can definitely see the importance. I'm still confused whether the spin will precess at the same rate when the particle is moving with constant velocity. I'll have a go at the Pauli equation, and if the answer isn't already posted here, I'll give my results if they make sense.
     
  7. Sep 5, 2016 #6

    Vanadium 50

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    When doing any kind of calculation, you'll be making approximations and idealizations. If you want to get anywhere, you have to - and then, if necessary, perturb the solution to handle the imperfections of the modeling. There are few areas of space with exactly zero electric field,but we don't worry about the Stark effect when first learning the hydrogen atom. It obscures rather than clarifies - especially when you start working in parabolic coordinates.
     
  8. Sep 10, 2016 #7

    Twigg

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    So I did what I said and I looked at what happens to the Pauli equation for a constant B-field when the electron is moving at non-relativistic speeds. I didn't solve the equation, but I got pretty far simplifying the differential equations and made some educated guesses at what happens. As a caveat, I made conscious decisions to treat the B field and spin angular momentum of the electron as the same in the electron's rest frame as in the laboratory frame. The result I found after expanding the Pauli equation is that the difference between the eigenvalues of the Hamiltonian for the spin-up and spin-down states of the electron differ only by ##\hbar \omega_L## where ##\omega_L## is the Larmor frequency. Based on this, I concluded that the moving electron's spin angular momentum precesses at the same rate as the electron at rest (the Larmor frequency). In practice, though, the resonance with electromagnetic radiation would appear Doppler shifted. Does this sound about right? Thanks again all!

    Thanks for the advice. After thinking about what you said, I was able to limit the amount of work I had to do to simply looking at the eigenvalue equations obtained from the Pauli equation, to see what the precession frequency was, rather than solve the Pauli equation. All I wanted to know was whether the moving electron precesses at a different rate.
     
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