I Zeeman effect after a Stern-Gerlach experiment

svletana
Messages
20
Reaction score
1
Let's say you take a beam of particles and pass it through a Stern Gerlach apparatus and you select one of the outgoing beams, therefore collapsing the wavefunction to certain values of orbital angular momentum and spin angular momentum.

If you performed the Zeeman experiment on that beam, I'm assuming you wouldn't see any splitting because the particles in this beam have very well defined orbital and spin angular momentum? Assuming they all have the same quantum number n.

For example in the anomalous Zeeman effect the splitting occurs from the quantum number m_j, where j is the total angular momentum, correct? So since we have a very well defined value for j, there shouldn't be any splitting in that case.

Is this reasoning correct?
 
Physics news on Phys.org
I would like to make an attempt to answer this question, but I don't know that this answer is necessarily correct. First of all, presumably a magnetic field is applied in the same direction as the inhomogeneous field of the Stern Gerlach apparatus to try to cause a splitting of the levels. One question I have is, can we assume the atoms from the Stern-Gerlach apparatus are in an excited state? Assuming they are, the different ## m_j ## or ## m_s ## will be separated by the Stern-Gerlach apparatus, so that one portion of the split beam consists of only one ## m_j ## or one ## m_s ##. It is somewhat of a puzzle how the Zeeman effect is going to be observed=at least for me=normally the Zeeman effect is observed by having an arc lamp placed in a magnetic field to create the splittings. The discharge (arc lamp) is necessary because that is the easiest way to get excited states and observed spectral lines. In the Stern Gerlach apparatus, I don't know that you have a sufficient number of excited states. Presuming that you do, I think there could be a splitting of the levels of the lower state, and some Zeeman effect could then be observed, but an experiment such as this would need to have sufficient excited states. Perhaps someone with more expertise in this area could help to fully answer this question. @vanhees71 Might you have an input?
 
  • Like
Likes svletana and dlgoff
svletana said:
For example in the anomalous Zeeman effect the splitting occurs from the quantum number mjm_j, where j is the total angular momentum, correct? So since we have a very well defined value for j, there shouldn't be any splitting in that case.
Charles Link said:
can we assume the atoms from the Stern-Gerlach apparatus are in an excited state?
That's a good point.
According to the abstract of this paper, Zeeman splitting can be observed in the fluorescence emission from a group of atoms/molecules by first exciting those atoms/molecules to some excited state manifold. Thus, if you want to run the atoms through a Stern-Gerlach separator first before observing the emission from one of the output channels, you need to ensure that the lifetime of the excited manifold is longer than the time the atoms take starting from excitation to the output channel.
I doubt though if one can do this experiment by observing the absorption from the ground state (hence eliminating the need of excitation) since no ground state is degenerate and thus won't split under whatever perturbation.
 
Last edited:
  • Like
Likes svletana, Charles Link, vanhees71 and 1 other person
blue_leaf77 said:
That's a good point.
According to the abstract of this paper, Zeeman splitting can be observed in the fluorescence emission from a group of atoms/molecules by first exciting those atoms/molecules to some excited state manifold. Thus, if you want to run the atoms through a Stern-Gerlach separator first before observing the emission from one of the output channels, you need to ensure that the lifetime of the excited manifold is longer than the time the atoms take starting from excitation to the output channel.
I doubt though if one can do this experiment by observing the absorption from the ground state (hence eliminating the need of excitation) since no ground state is degenerate and thus won't split under whatever perturbation.

Great Answer, thank you! The answers took a turn I did not expect :)
 
  • Like
Likes Charles Link
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA
If we release an electron around a positively charged sphere, the initial state of electron is a linear combination of Hydrogen-like states. According to quantum mechanics, evolution of time would not change this initial state because the potential is time independent. However, classically we expect the electron to collide with the sphere. So, it seems that the quantum and classics predict different behaviours!
Back
Top