# Questions on Quantum Numbers (introductory physics)

#### nonequilibrium

Hello, I'm taking a course in introductory modern physics, and I have a few questions on the nature of quantum numbers:

1. Why is $$\Delta l = \pm 1$$? My book simply states this is due the angular momentum of photons, but we have that $$L_{photon} = \pm 1 \hbar$$ and $$L_{electron} = \sqrt{l(l+1)} \hbar$$. The fact that l must change with one doesn't seem compatible...?
2. (possibly vague question, skip if confusing) What is the meaning of $$m_l$$ (measure for the projection of $$\vec L$$ on the z-axis) if we don't define a z-axis? Or in other words: sure we can always define a z-axis, but this can be totally arbitrary (assuming no external B-field or something), suggesting that due to symmetry it's true for every axis, but that is not true, so what is up with that?
3. Why doesn't spin influence the spectral lines of certain atoms? I've read somewhere it's because sometimes the net spin of the atom is zero, but I don't see the relevance of that. For example: say we have an electron with spin up (and there's an external magnetic field) and it drops down from $$l=1,m_l=1$$ to $$l=0,m_l=0$$. Now wouldn't we always expect a different energy jump (and thus some other frequency of the emitted light) if the electron had spin down? (due to the external magnetic field)
4. Trying to understand the Stern & Gerlach experiment: looking up silver, we see that its configuration is $$(Kr) 4d^{10} 5s^{1}$$, so if I understand correctly the reason $$m_l$$ does not play a role in the experiment is because in the s-orbit, $$m_l = 0$$, and in the other orbits for every $$m_l$$ there's another one with the opposing sign? But there is a net spin, resulting in the famous result of the experiment. Okay, that sounds understandable. This, however, means that quantummechanically, without the knowledge of spin, we would expect one line, i.e. no deviation (because net magnetic orbital number is zero). However, the experiment was to measure the quantization of $$L$$: how were they planning to measure that? Or in other words: I don't understand what they (Stern & Gerlach) were expecting to see (which indicates I might be misunderstanding the whole experiment).

Thank you!

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#### DrClaude

Mentor
Why is $$\Delta l = \pm 1$$? My book simply states this is due the angular momentum of photons, but we have that $$L_{photon} = \pm 1 \hbar$$ and $$L_{electron} = \sqrt{l(l+1)} \hbar$$. The fact that l must change with one doesn't seem compatible...?
One has to be careful not to mix up a particle's angular momentum and the projection of that angular momentum. A photon is a spin-1 particle with projections $\pm \hbar$, but the magnitude of the spin is $\sqrt{s ( s+1)} = \sqrt{2} \hbar$, so there is no difference with how we consider angular momentum for the electron.

What is the meaning of $$m_l$$ (measure for the projection of $$\vec L$$ on the z-axis) if we don't define a z-axis? Or in other words: sure we can always define a z-axis, but this can be totally arbitrary (assuming no external B-field or something), suggesting that due to symmetry it's true for every axis, but that is not true, so what is up with that?
There is no $m$ quantum number without a defined axis, as one must specify which operator one is considering. If an atom is unpolarized (no preferred quantization axis), then it would be found in a superposition of $m$ states, whichever axis is chosen.

Why doesn't spin influence the spectral lines of certain atoms? I've read somewhere it's because sometimes the net spin of the atom is zero, but I don't see the relevance of that. For example: say we have an electron with spin up (and there's an external magnetic field) and it drops down from $$l=1,m_l=1$$ to $$l=0,m_l=0$$. Now wouldn't we always expect a different energy jump (and thus some other frequency of the emitted light) if the electron had spin down? (due to the external magnetic field)
At the level of theory used, the electromagnetic field responsible for the transition between levels does not couple with the spin, therefore transitions involving a change in total spin can be ignored. However, because of spin-orbit coupling, some transitions will be possible or not depending on whether the spin of the electrons can help conserve angular momentum.

Trying to understand the Stern & Gerlach experiment: looking up silver, we see that its configuration is $$(Kr) 4d^{10} 5s^{1}$$, so if I understand correctly the reason $$m_l$$ does not play a role in the experiment is because in the s-orbit, $$m_l = 0$$, and in the other orbits for every $$m_l$$ there's another one with the opposing sign? But there is a net spin, resulting in the famous result of the experiment. Okay, that sounds understandable. This, however, means that quantummechanically, without the knowledge of spin, we would expect one line, i.e. no deviation (because net magnetic orbital number is zero). However, the experiment was to measure the quantization of $$L$$: how were they planning to measure that? Or in other words: I don't understand what they (Stern & Gerlach) were expecting to see (which indicates I might be misunderstanding the whole experiment).
In a classical world, one would still find that electrons have spin (spin is actually a consequence of special relativity), but any orientation would be possible. The SG experiment would thus have resulted in an elongated splotch. The fact that two spots appeared meant that spin was quantized (Stern and Gerlach could also deduce the value of that spin and found it was related to Planck's constant).

"Questions on Quantum Numbers (introductory physics)"

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