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Is there some reason behind the appearance of spin effects only under the application of variable magnetic field.

I know that we need variable magnetic field to act force on charge particle with some angular momentum, but I am comparing spin angular momentum with orbital angular momentum, i.e. we can detect orbital angular momentum without application of magnetic field, so why not spin angular momentum?

Do we consider this as just inherent property of spin, or do we have suitable reasons to justify it?

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    $\begingroup$ Since the energy is given by $E=-\mu \cdot B$ the atomic levels depend on the magnitude of the B-field. Hence, the transition energies depend on B and not on its gradient. Could you please try to specify your question/the system under consideration further. Currently your question is "incorrect". $\endgroup$ – Semoi Jul 12 at 12:27
  • $\begingroup$ I am confused between orbital angular momentum and spin angular momentum, I know that for a single electron, If I want to see the energy splitting due to spin, I need to apply spatially varying magnetic field(Stern Gerlach experiment), but for two different elections, If I want to see the energy splitting due to orbital angular momentum, do I need the same setup? @Semoi $\endgroup$ – Shine kk Jul 12 at 13:47
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    $\begingroup$ First of all I suggest that you rephrase your question. As I said before, in an atom the orbital angular momentum certainly gives rise to an effect due to a constant B field, which is measurable. In the case of two orbiting electrons I am not sure what physical system you consider: Two free electrons do not orbit each another. So unless you clarify the question it's hard to answer it. I also do not understand why we are discussing this, when you already accepted an answer. $\endgroup$ – Semoi Jul 12 at 14:23
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Quantization in a narrow sense refers to discreteness of energy levels. In the context of atomic levels such discreteness is usually observed via optical experiments -e.g., via absorption, which has resonance at the frequency corresponding to the level spacing: $\hbar\omega = E_2 - E_1$. Studying optical absorption spectrum necessarily implies the use electromagnetic radiation, i.e. of time-dependent electromagnetic field.

Orbital momentum designates the states of a charge particle (electron) in the electrostatic field of the nucleus. As such these states are coupled to the electromagnetic field via charge, i.e. they are coupled to the electric field. In order to observe absorption on needs both constant and time-dependent fields: the former one to cause level splitting (via the Stark effect) and the latter to case absorption.

Spin is a kind of magnetic moment, i.e. it is coupled to the magnetic field, and consequently a constant magnetic field is needed to split spin levels, and a time-dependent field is necessary to cause absorption.

Finally, an experimental setup my use additional time-dependent fields to facilitate more precise measurements -e.g., by causing the Rabi oscillations.

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Spin effects do not only occur in a variable magnetic field. It is certainly true that a Stern-Gerlach type experiment is a clear demonstration of the existence of spin-1/2 objects, but the existence of spin could have (eventually) been deduced without this result.

Important examples of where spin comes up explicitly are the Pauli exclusion principle, the computation of average values in fermionic systems, and various spin-orbit interactions affecting the energy levels at the atomic or nuclear level.

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  • $\begingroup$ If we want to see splitting in energy due to orbital angular momentum or spin angular momentum, do we have to use the same setup? the reason behind asking this is, I am confused between the two, In H-atom model, we incorporate the degeneracy due to orbital angular momentum and not due to spin angular momentum. why? $\endgroup$ – Shine kk Jul 12 at 13:41
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    $\begingroup$ One does not need an external magnetic field to see a spin-orbit splitting. See en.m.wikipedia.org/wiki/Spin%E2%80%93orbit_interaction. The spin-orbit term in hydrogen is treated as a perturbation although - because it depends on the derivative $dV/dr$ - the effect is much more significant in nuclei. $\endgroup$ – ZeroTheHero Jul 12 at 13:54

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