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I The ##1s^2## configuration of helium can't be triplet state?

  1. Oct 31, 2016 #1
    I thought this could be explained as follows: a singlet state is one with ##S=0##. Electrons have spin ##\frac{1}{2}## or ##\frac{-1}{2}##, and in the ##1s^2## state there are two electrons. For it to be a triplet state both electrons would have to be spin ##\frac{1}{2}## which isn't allowed, by the Pauli exclusion principle. So one electron must have spin ##\frac{1}{2}## and the other spin ##\frac{-1}{2}##. Then total spin is zero and the state is a singlet.

    When I gave this answer (not as coursework, as part of a discussion) I was told it was incomplete. Does that mean some part of it is wrong? Or if it's a correct explanation, what else is there to add?! Bit worried I've misunderstood or omitted some fundamental concept. Thanks for any help or suggestions!
     
  2. jcsd
  3. Oct 31, 2016 #2

    DrClaude

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    That's not strictly correct.

    Also, to be pedantic, I would consider
    to be incorrect. The spin of an electron is always 1/2. Its projection can be ±1/2.
     
  4. Oct 31, 2016 #3
    Why not?

    Second point accepted! An electron can have projected spin of ##\pm \frac{1}{2}## then?
     
  5. Oct 31, 2016 #4

    blue_leaf77

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    The more fundamental message about Pauli principle is actually about antisymmetrization. When the two electrons are in 1s2 state, the spatial wavefunction is necessarily symmetric and consequently the spin part must be antisymmetric. Singlet state has this required antisymmetry so this state must represent the spin part. All triplet states are symmetric so they cannot be chosen.
     
  6. Oct 31, 2016 #5

    DrClaude

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    Because opposite spin projections can also form a triplet state.

    Yes. Again, this was a pedantic point. I think it is the comment about "opposite spins forming a singlet" being the point of contention.
     
  7. Oct 31, 2016 #6
    As in with more than two electrons there could be some with one spin projection and some with the other and overall that would form a triplet state? Because with just two electrons if they have opposite spin projections how could they form a triplet?
     
  8. Oct 31, 2016 #7
    How do you know if the spatial part is symmetric or antisymmetric? And whether the spin part is symmetric or not? If it's a long calculation it's probably beyond the scope of my course.

    OK, found out how to do that! In next week's lecture notes!
     
    Last edited: Oct 31, 2016
  9. Oct 31, 2016 #8

    DrClaude

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    What are the three spin states that form the triplet state?
     
  10. Oct 31, 2016 #9
    With two electrons? A triplet state would be if they both have spin ##\frac{1}{2}## or both spin ##-\frac{1}{2}##, and I'm not sure about a third one.
     
  11. Oct 31, 2016 #10
    These projections of spin are denoted by ##m_s## I think. Looking at wikipedia it says that you can have a state of up down + down up (unless I'm interpreting it wrong) and that's a triplet but how could that happen in helium? Wouldn't you need two spin up electrons and two spin down ones i.e. four in total? And anyway how do the spins not cancel to zero in that situation?!
     
    Last edited: Oct 31, 2016
  12. Oct 31, 2016 #11

    blue_leaf77

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    It's apparent that you still have a lot to learn about addition of angular momentum operators. Please refer to your prof's textbook on this subject.
     
  13. Oct 31, 2016 #12

    Nugatory

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    That's a quantum superposition: Two electrons whose state vector is the sum of the vectors "first one up; second one down" and "first one down; second one up".

    If you are not familiar with the concept of quantum superposition you'll need to back up some, work through the early chapters of an intro QM textbook. What's going on here is analogous to the famous double-slit experiment, where we write the state of a particle as the vector sum of "particle went through left slit" and "particle went through right slit" but that doesn't mean that there were two particles.

    The projections of the spins on the z-axis do cancel to zero in that state. If the total spin quantum number is 1, then the projection of the spin on the z axis can take on the values 1, 0, and -1. If the projection on the z axis is up for both, the total projection on the z-axis is 1; if it down for both, the total projection is -1; and here we have the case where the projection on the z-axis is up for one and down for the other so they cancel to zero. Of course you couldn't produce this configuration with classical spinning balls, but electrons and other quantum particles aren't classical spinning balls.
     
  14. Nov 1, 2016 #13
    I am familiar with the concept of superposition but I most certainly don't understand it. I've even done the double slit experiment and seen a superposition, not that that clarified anything. Although having read my textbook I do understand why that's a triplet state, which begs the question: why can't ground state helium be in this state? Since the question was why can't ground state helium be in a triplet state?
     
  15. Nov 1, 2016 #14

    Nugatory

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    In the ground state the two electrons are indistinguishable, so the overall wave function must change sign if they are exchanged. However, the spatial part of the wave function doesn't change sign under that exchange (##\psi_r(\vec{r_1},\vec{r_2})=\psi_r(\vec{r_2},\vec{r_1})##) so the spin part of the wave function must change sign. The triplet states don't change sign under exchange and the singlet state does; therefore the ground state spin must be the singlet state.
     
  16. Nov 1, 2016 #15
    Wow, I think I actually understand that. I suppose it took long enough! That's why my explanation was incomplete, I hadn't said why the superposition state wasn't possible.

    A huge thank you to everyone for your patience and your help, I really appreciate it! :)
     
  17. Nov 1, 2016 #16

    DrClaude

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    To make things clear, the spin singlet is
    $$
    \frac{1}{\sqrt{2}} \left[ | \uparrow \downarrow \rangle - | \downarrow \uparrow \rangle \right]
    $$
    and the three triplet states are
    $$
    | \uparrow \uparrow \rangle \\
    \frac{1}{\sqrt{2}} \left[ | \uparrow \downarrow \rangle + | \downarrow \uparrow \rangle \right] \\
    | \downarrow \downarrow \rangle
    $$
    So there are two ways to make linear combinations of the states with opposite spins such that there is an overall symmetry. One of them is symmetric with respect to particle interchange, and it is part of the triplet (it correponds to a state with total spin ##S = 1##).
     
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