1
$\begingroup$

Almost everywhere in scientific literature, Larmor precession is introduced via an analogy with a spinning top. I understand that in a quantum framework, precession can be explained considering the average value of the spin components orthogonal to the external magnetic field. What I find obscure in the classical analogy with a spinning top could be summarized in the following questions:

  • A spinning top in a gravitational field, with the rotation axis exactly vertical (frictionless approximation), should not undergo precession. It seems to me that this is not true for an elementary particle with spin$\neq$0. Can this property be explained only with quantum mechanical arguments? Or am I missing something also in the classical theory of precession?
  • Does the angle that the magnetic moment forms with the applied magnetic field depend on the initial direction of the spin, or is it always the same regardless of initial conditions (i.e., when the field is switched on)? If the angle is always the same (as I understand from the literature), is it just a consequence of the quantization of angular momentum, or can this property be explained also in a classical framework?
Improve this question
$\endgroup$
5
  • $\begingroup$ > It seems to me that this is not true for an elementary particle with spin $\neq 0$ -- Why? Usual half-integer spin like in NMR does not precess if aligned with external field. $\endgroup$
    – Ján Lalinský
    Commented Jun 12 at 14:08
  • $\begingroup$ As far as I understant there will always be precession with Larmor frequency. Also in case of perfect alignment with the field. $\endgroup$
    – Popbatman
    Commented Jun 12 at 18:32
  • $\begingroup$ How did you come to such understanding? What does precession mean, when the two vectors align? $\endgroup$
    – Ján Lalinský
    Commented Jun 12 at 19:03
  • $\begingroup$ If you perform a measure of the $S_x$ or $S_y$ component of spin for a purely $S_z$ state in a magnetic field you obtain a time varying result that can be interpreted as precession. Is not that true? $\endgroup$
    – Popbatman
    Commented Jun 13 at 5:45
  • $\begingroup$ First you wrote "there will always be", now you write "if you perform a measure". So which is it? And why would we obtain a time varying result? Measurement in quantum theory obtains one result, we can write it down. The result is fixed, it does not change in time. $\endgroup$
    – Ján Lalinský
    Commented Jun 13 at 8:02

1 Answer 1

Reset to default
1
$\begingroup$

A spinning top in a gravitational field, with the rotation axis exactly vertical (frictionless approximation), should not undergo precession. It seems to me that this is not true for an elementary particle with spin≠0. Can this property be explained only with quantum mechanical arguments? Or am I missing something also in the classical theory of precession?

The "classical analogy" is not just an analogy, it is a limit. Meaning this is not just an analogy that inspires the derivation, but it is actually the result of the correct quantum mechanical equations for the classical limit. Specifically, the expectation value of the behavior of a large number of spins results in the classical precession formula. So, indeed, in the limit of a large number of spins whose spin expectation is exactly vertical the expectation does not precess.

A single spin does not have a state that corresponds to the classical state of "rotation axis exactly vertical". If the spin is in a definite spin-up (eigenstate) condition on the z axis then the spin on x and y will be uncertain.

Does the angle that the magnetic moment forms with the applied magnetic field depend on the initial direction of the spin, or is it always the same regardless of initial conditions (i.e., when the field is switched on)? If the angle is always the same (as I understand from the literature), is it just a consequence of the quantization of angular momentum, or can this property be explained also in a classical framework?

A single spin does not have a definite measurable "angle that the magnetic moment forms with the applied magnetic field". For a large number of spins where such an angle is defined then that angle does depend strongly on the initial conditions. So this is about the classical limit.

Your impression that the angle is always the same probably comes from reading many papers that discuss the precession starting from the fully-relaxed state. In that case it isn't that the angle doesn't depend on the initial condition, but rather that the initial condition is well known and easily accessible.

If you start from an already excited state then you will indeed get a dependency of the final state on the initial state.

Improve this answer
$\endgroup$
5
  • $\begingroup$ > A single spin does not have a definite "angle that the magnetic moment forms with the applied magnetic field" Actually, quantum state vector for spin 1/2 defines a definite direction in space, and this is the direction of expected value of magnetic moment vector for that state. $\endgroup$
    – Ján Lalinský
    Commented Jun 13 at 13:45
  • $\begingroup$ Only on one axis. A single spin state with a definite spin up in z does not have a definite spin in x or y. $\endgroup$
    – Dale
    Commented Jun 13 at 14:46
  • $\begingroup$ My objection is to "does not have a definite angle". The direction and the angle in any quantum state is definite. I agree this definite direction and angle does not determine the result of SG-type measurement in non-aligned directions. But that is a different thing - has direction and angle before measurement, vs. gets found by measurement. SG-type measurement can't find what the original quantum state and its direction in space was. $\endgroup$
    – Ján Lalinský
    Commented Jun 13 at 19:12
  • $\begingroup$ Fair enough. I added the qualifier “measurable”. I am not sure that it is actually necessary, but it doesn’t hurt. $\endgroup$
    – Dale
    Commented Jun 13 at 19:35
  • $\begingroup$ I agree the direction and angle is not measurable, but it is a mathematical property of the spin states, and it is preparable, so it is not irrelevant to the question. $\endgroup$
    – Ján Lalinský
    Commented Jun 14 at 13:23

Not the answer you're looking for? Browse other questions tagged or ask your own question.