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In Sakurai section 2.1, he works through an example of spin precession where we come to the conclusion that, when working in the z-basis and applying a magnetic field in the z-direction, the spin precesses in the xy-plane. I am wondering how I can conceptually interpret this precession?

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  • $\begingroup$ I'm not sure what you're looking for. Do you know what precession is? If you do, then that's exactly what's going on. The spin vector, which creates its own little magnetic dipole moment, is precessing around the direction of the magnetic field. I'm not sure what you need beyond that exactly. Can you clarify? $\endgroup$
    – march
    Commented Feb 6, 2023 at 4:43
  • $\begingroup$ When you studied electromagnetism, did you learn that a magnetic field exerts a torque on a magnetic dipole? $\endgroup$
    – Ghoster
    Commented Feb 6, 2023 at 4:46
  • $\begingroup$ well, i have a follow up comment and question. my question is why does the spin create a dipole moment. my comment is that in my quantum mechanics I class, we learned that the precession of the angular momentum vector is a statement of uncertainty. if we know the angular momentum in z, it must precess in x and y to maintain some kind of uncertainty and satisfy commutation relations. i was wondering if something similar is going on here and what it means for the quantity that is spin to be precessing. [...] $\endgroup$ Commented Feb 6, 2023 at 4:57
  • $\begingroup$ [...] does it just mean this system's spin "wobbles" between having a definite z spin and some combination of x and y spin? i am not sure any of this is right. @march $\endgroup$ Commented Feb 6, 2023 at 4:57
  • $\begingroup$ why does the spin create a dipole moment The classical intuition for this comes from considering a rotating charged sphere. (It has angular momentum and it has a magnetic dipole moment, and they’re proportional.) The quantum derivation involves the Dirac equation. $\endgroup$
    – Ghoster
    Commented Feb 6, 2023 at 5:02

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It’s essentially the same as when rotating objects precess. For objects which are actually spinning, there is some axis about which they spin. This axis is likewise described as the direction of the angular momentum of that object where the sign denotes which direction they are spinning (right hand rule and all that jazz). Precessions are just when this axis is not stationary, but rather is also rotating. If you google “spinning top precessions” and look at the videos, there should be some good examples of what precession means physically in the context of objects actually spinning.

Now for elementary particles, their ‘spin’ is not quite the same as they are not actually physically spinning, but they still do have angular momentum which is what the spin is accounting for. Essentially by saying that a particle has spin, all we are saying is that it has some intrinsic angular momentum (but again, it isn’t actually spinning). As with before, this angular momentum is defined in some direction which classically would be the axis in which the thing is rotating about, but in the quantum world, we have other issues to worry about, mainly the Heisenberg Uncertainty Principle.

Because of Heisenberg, I am not allowed to know my particles momentum exactly, and this includes the angular momentum. As you likely saw in the calculations you are referring to, we can measure our angular momentum in one direction and have absolute precision in that measurement simply because angular momentum occurs in three dimensions, and with information from two directions still missing we are left not knowing the exact angular momentum of our particle and thus are safe. This is where the precession comes from. We know the magnitude of our angular momentum, and ‘how much of it’ is in the axis we choose to measure it in (the z axis in your case), but we don’t know what’s going on in the other two directions (x and y), so by Heisenberg, our particle is essentially allowed to distribute its “remaining” angular momentum anywhere it likes in the x and y directions respectively, thus it precesses about the z axis since the Heisenberg Uncertainty Principle tells us that we can’t know exactly how much angular momentum it has in those directions (note that I know the total angular momentum and the angular momentum in the z axis, so if I knew how much angular momentum was in either the x OR y direction, I would know the angular momentum completely since knowing x would give me y and vice versa, so I can’t know either).

The physical intuition for precessing is as I described it at first for classical spin, but for particles it’s more abstract since they aren’t actually spinning. Basically the precession comes from the fact that particles have intrinsic angular momentum (spin), but the Heisenberg Uncertainty Principle tells us that we can’t know exactly where that spin is, so if I measure it about some axis, it must precess about that axis in order to ensure that we don’t know where it is. In other words it just comes from our uncertainty about the spin, but this is an intrinsic uncertainty, not just an uncertainty so to bad measurements. Hopefully this long winded answer helps somewhat.

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  • $\begingroup$ hello, thank you for your answer. i am wondering if i have a misunderstanding about the initial conditions of this problem. by understanding is that we are in a state of only spin in the z direction, then we apply a magnetic field, and we see this precession. if we are in a state of definite z spin to begin with, this should violate the uncertainty principle. where am i going wrong here? $\endgroup$ Commented Feb 6, 2023 at 5:13
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    $\begingroup$ So the confusion here is that we are not in a state of only z spin to begin with. The point is that we don’t know what the spin of our particle is until we measure it, so it could be anywhere. When we apply a magnetic field along the z direction, we are essentially measuring the spin along that axis, at which point the precession starts for the reasons I mentioned, mainly that we now know how much is in the z axis so we cannot know well where it is in the other directions. $\endgroup$ Commented Feb 7, 2023 at 16:01

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