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# Stern Gerlach confusion watch

1. In this experiment, when hydrogen atoms are projected through the inhomogeneous magnetic field we get two distinct 'traces' on the screen because of the spin-up/spin-down possibilties.

But I'm struggling to extend this concept to atoms with more optically active electrons. Say we have helium atoms (2 o.a. electrons) being shot through the magnetic field - how do I determine the number of traces then? I understand that for a configuration with a spin S, there are 2S + 1 possibilties. So do I then consider the total spin - which can either be 0 or 1 in this case - and assume that we can have 2*1 + 1 = 3 possibilties = 3 traces for helium?

Thanks for any help.
2. (Original post by trm90)
In this experiment, when hydrogen atoms are projected through the inhomogeneous magnetic field we get two distinct 'traces' on the screen because of the spin-up/spin-down possibilties.

But I'm struggling to extend this concept to atoms with more optically active electrons. Say we have helium atoms (2 o.a. electrons) being shot through the magnetic field - how do I determine the number of traces then? I understand that for a configuration with a spin S, there are 2S + 1 possibilties. So do I then consider the total spin - which can either be 0 or 1 in this case - and assume that we can have 2*1 + 1 = 3 possibilties = 3 traces for helium?

Thanks for any help.
I would agree with you and say you consider the total electron spin, as effectively that is what you're doing in the case of hydrogen as well (total spin S = , hence there are 2 possible states)

That makes sense for helium as well as you could have the electron states (u being spin up, d being spin down):

u u (S = 1, Sz = 1) [bends 'upwards' through Stern-Gerlach equipment]
u d (S = 0, Sz = 0) [Straight through, no deviation]
d u (S = 0, Sz = 0) [Straight through, no deviation]
d d (S = 1, Sz = -1) [bends 'downwards' through Stern-Gerlach equipment]

i.e. three distinct paths (as the u d spin state is indistinguishable from the d u spin state)
3. Helium in the ground state is S=0 iirc so it should only have one trace. For excited states there should be either 1 or 3 traces depending on whether para or orthohelium is used.
I would agree with you and say you consider the total electron spin, as effectively that is what you're doing in the case of hydrogen as well (total spin S = , hence there are 2 possible states)

That makes sense for helium as well as you could have the electron states (u being spin up, d being spin down):

u u (S = 1, Sz = 1) [bends 'upwards' through Stern-Gerlach equipment]
u d (S = 0, Sz = 0) [Straight through, no deviation]
d u (S = 0, Sz = 0) [Straight through, no deviation]
d d (S = 1, Sz = -1) [bends 'downwards' through Stern-Gerlach equipment]

i.e. three distinct paths (as the u d spin state is indistinguishable from the d u spin state)
I'm pretty sure ud and du aren't physical states. You need linear combinations of those states ie ud+du and ud-du. So you end up with three S=1, and one S = 0 possible states depending on the symmetry of the spatial wavefunction.
5. (Original post by suneilr)
Helium in the ground state is S=0 iirc so it should only have one trace. For excited states there should be either 1 or 3 traces depending on whether para or orthohelium is used.
Ah right, so we assume it's parahelium... thanks

One more question. Now say I want to extend this trace problem to Boron, which has three optically active electrons. I'm told, as a hint, that Boron has two L-S coupling states. The electrons are in the second shell so that n = 2 and l = 1. I'm assuming the two L-S coupled states are basically either J = L + S or J = |L - S|, where L and S are the total angular momenta and spins, respectively.

But I'm still lost. No idea if I'm missing out on some 'rule' that governs how many traces there will be
6. (Original post by trm90)
Ah right, so we assume it's parahelium... thanks

One more question. Now say I want to extend this trace problem to Boron, which has three optically active electrons. I'm told, as a hint, that Boron has two L-S coupling states. The electrons are in the second shell so that n = 2 and l = 1. I'm assuming the two L-S coupled states are basically either J = L + S or J = |L - S|, where L and S are the total angular momenta and spins, respectively.

But I'm still lost. No idea if I'm missing out on some 'rule' that governs how many traces there will be
I'm pretty sure that the number of traces is given by the number of m_j values. Therefore the number of traces will depend on which LS coupling state the boron is in.
7. (Original post by suneilr)
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Right, so as I vaguely remember there are 2J + 1 mJ values as . That means J = L + S = 3 + 3/2 = 9/2, or J = L - S = 3 - 3/2 = 3/2. Which gives 2(9/2) + 1 = 10 traces or 2(3/2) + 1 = 4 traces.

So a total of 14 traces then?
8. (Original post by trm90)
Right, so as I vaguely remember there are 2J + 1 mJ values as . That means J = L + S = 3 + 3/2 = 9/2, or J = L - S = 3 - 3/2 = 3/2. Which gives 2(9/2) + 1 = 10 traces or 2(3/2) + 1 = 4 traces.

So a total of 14 traces then?
I think you've got your J values wrong. You've got electrons in n=2, l= 0 so these don't contribute to L. They also have to have opposite spin and and so S=0. So only the n=2, l=1 electron contributes, and since theres only one electron you can take L=1, and then the spin can be either aligned or antialigned with L, so you have J = L +S = 1 + 1/2 = 3/2 or J=L-S = 1-1/2 = 1/2 so you end up with either 2 or 4 traces depending on the LS state.
9. (Original post by suneilr)
I think you've got your J values wrong. You've got electrons in n=2, l= 0 so these don't contribute to L. They also have to have opposite spin and and so S=0. So only the n=2, l=1 electron contributes, and since theres only one electron you can take L=1, and then the spin can be either aligned or antialigned with L, so you have J = L =S = 1 + 1/2 = 3/2 or J=L-S = 1-1/2 = 1/2 so you end up with either 2 or 4 traces depending on the LS state.
Ohhhhhhhhhhhhhhhhhhhhhhhhh.

Damn, this is so embarrassing. I've been counting my electrons wrong: for some reason I thought I had to put 2 electrons in the first shell, then up to 8 in the subsequent shell and just didn't count it properly at all. I understand now that there is only one L-S coupled electron in the n = 2 l = 1 state.

Thanks again so much suneil, you're a star. And sorry for my denser-than-usual posts these days >_<
10. (Original post by trm90)
Ohhhhhhhhhhhhhhhhhhhhhhhhh.

Damn, this is so embarrassing. I've been counting my electrons wrong: for some reason I thought I had to put 2 electrons in the first shell, then up to 8 in the subsequent shell and just didn't count it properly at all. I understand now that there is only one L-S coupled electron in the n = 2 l = 1 state.

Thanks again so much suneil, you're a star. And sorry for my denser-than-usual posts these days >_<
Haha no worries. It's a good distraction from the 10000 words I'm supposed to be writing

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