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I know that protons and electrons do not actually spin, although they have the property of spin.

I was learning about MRI. I was introduced to the idea that you have the spin of the proton in the external magnetic field. Since they are not perfectly aligned, it will cause precession just like in a spin top. I understand how precession occurs in spin tops and bike wheels, but that's dependent on the object actually spinning and having the velocity to change.

If protons don't spin, do they actually precess too? I would imagine not, because that would imply that they're actually spinning. If not actually precessing, then what is this precession?

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  • $\begingroup$ They do. You can even use that for measurements in destruction - free material analysis and medicine : MRT what the protons spin is excited and the decay of the aligned spin precession acid is used $\endgroup$ Dec 14, 2020 at 8:50
  • $\begingroup$ There is precession, but you don't need to invoke spinning for that. Weinberg, in Lectures On Quantum Mechanics (2012) treats this using Wigner-Eckart theorem, as far as I remember (sec 5.3) $\endgroup$
    – Cryo
    Dec 14, 2020 at 9:16
  • $\begingroup$ @planetmaker so if I had a super microscope, I could SEE them precessing like little spin tops?? $\endgroup$
    – John Hon
    Dec 14, 2020 at 9:31
  • $\begingroup$ @JohnHon no. protons are described quantum mechanically, which means all measurable quantities depend on a probability distribution. The probability distribution of many protons in the same boundary conditions will show the precession. $\endgroup$
    – anna v
    Dec 14, 2020 at 11:54
  • $\begingroup$ You will get more access to MRI if - only for your imagination - think about the spin together with the electrons magnetic dipole moment. Any external magnetic field will align a bit the electrons dipole and during the relaxation after switched-off external field photons get emitted.MRI is a lot of mathematics to reconstruct from this radiation the 3D picture of the inner structure of a body. $\endgroup$ Dec 20, 2020 at 17:51

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Existing answers seem a bit unclear to me, so here goes...

Yes, they do precess. The easiest way to see this is in the average properties (also called expectation value). You can define a vector $\bf \bar{S}$ with components as follows: $$ \bar{S}_x := \langle \psi(t)|\hat{S}_x |\psi(t)\rangle, \qquad \bar{S}_y := \langle \psi(t)|\hat{S}_y |\psi(t)\rangle,\qquad \bar{S}_z := \langle \psi(t)|\hat{S}_z |\psi(t)\rangle $$ where $|\psi(t)\rangle$ is the state of the spin as a function of time. The above is the standard notation in quantum mechanics for an average value. The quantities $\hat{S}_i$ are the operators representing the components of the spin angular momentum. Both the quantum state $| \psi(t) \rangle$ and the operators $\hat{S}_i$ are subtle beasts, but the average (also called expectation value) is a much more simple idea. All three of its components can be measured with perfect precision, in principle, all at the same time. Together they form a vector $$ {\bf \bar{S}} = ( \bar{S}_x,\;\bar{S}_y,\;\bar{S}_z ) $$ and this vector behaves just like a classical vector. For a proton or an electron in a magnetic field it precesses in exactly the same way that classical physics would propose for a system combining magnetic dipole moment and angular momentum.

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Proton and electron spin are like angular momentum in every aspect. It is just that we do not have a mechanical model of a rotating charge to explain it. Electrons and quarks, which give protons their spin, are point particles for all we know donut is hard to think of them as rotating charges.

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Protons and spin of particles of similar dimensions have to be described by quantum mechanical probability distributions which can be predicted using the solutions of the appropriate quantum mechanical equations.

Spin is part of angular momentum at the quantum frame, but neither can be directly observed as one would observe the trajectory of a spinning top. The (x,y,z) of particles in the quantum mechanical frame are described by orbitals , probability loci where a measurement will give the non-sequential (x,y,z) of a particle.

If one could measure a large number of same boundary condition protons, the accumulation of the loci for those protons would show the necessary precession due to angular momentum in the given state, to be checked with quantum mechanical calculations.

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These particles have half-integer spin, so their angular momentum can't be explained as if they're physical objects spinning rigidly. That would produce only integer values of angular momentum.

They do "really" precess in the following sense. Suppose you prepare an electron in state A, with its spin pointing along the positive x axis. Then you subject it to a magnetic field along the z axis and let it act for an amount of time that would, classically, cause a precession by some angle. For example, let's say it's 90 degrees counterclockwise about the z axis. Now it's in some state B. If you then measure the component of the electron's spin along the y axis, you will find what you expected, with 100% probability: the spin is aligned with the y axis.

But the results of other measurements on B will not give the results you would expect classically. For example, if you measure the component of the spin along the x axis, you will not get zero. (That's not a possible value for a spin-1/2 particle.) You will get either +1/2 or -1/2, with equal probability.

The answer by anna v is wrong. She says, "The (x,y,z) of particles in the quantum mechanical frame are described by orbitals , probability loci where a measurement will give the non-sequential (x,y,z) of a particle." The spin-1/2 of an electron has nothing to do with the x, y, and z degrees of freedom.

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A particle has an intrinsic property known as spin. Spin S is an associated angular momentum carried by the particle. Now a charge particle with spin S has a magnetic moment. Accordingly, the magnetic moment of a charge particle is given by, \begin{equation} \mu = \frac{g q}{2m} S , \end{equation}

where q is the charge of the particle, g is the g-factor, m is the mass and S is the spin of the particle.

Now if particle with magnetic moment interact with the magnetic field or in general EM field, then the magnetic moment will precess onto the direction of the magnetic field with frequency $w = -\gamma B $. This is what they called Larmor Precision. Now example of the charge particles that has a spin is protons and electrons. Hence these particles will precess too in the presence of magnetic field.

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