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Why is the singlet configuration anti-symmetric?

  1. Apr 1, 2010 #1
    Obviously because there is a minus sign in |0 0> = 1/sqrt(2)*((up)(down)-(down)(up)) but I guess I am not completely sure why |0 0> can't equal 1/sqrt(2)*((up)(down)+(down)(up))...
    m still equals zero, and aren't we pretty much arbitrarily deciding that the spins are oppositely aligned in the singlet configuration? (I am thinking that that is what the minus sign has to represent said opposite alignment, but don't know how to justify that...)
    Sorry for such a basic question!
     
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  3. Apr 1, 2010 #2

    Matterwave

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    If you transform the one with a plus sign to another basis (say x-basis), you find that the spins are no longer anti-aligned. If you transform the singlet (with the minus sign) to ANY basis, you will find that the spins are STILL anti-aligned.
     
  4. Apr 2, 2010 #3
    I believe Matterwave is not correct here. The triplet state has no net angular momentum no matter which way you turn it in space. You can verify this by looking at the three p orbitals of hydrogen, which are also a spin-1 system. The middle state has no net spin no matter which way you turn it. (It happens that if you take the superposition of the +1 and the -1 states, you get something that looks like the middle state except it's rotated 90 degrees.)

    I asked this question about a year ago and got an awesomely good answer from peteratcam, who makes an occasional appearance in this forum. If nobody posts before tomorrow morning, I will put in my interpretation of peter's explanation.

    CRANK ALERT: I am a crank.
     
  5. Apr 2, 2010 #4
    I try the more pragmatic apporach, it is the way it is due to the way we have defined the algebra of quantum angular momentum operators
     
  6. Apr 2, 2010 #5

    Matterwave

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    I have done the transformation of basis myself, and seen it done by my professor. Transforming the triplet states yield spins which are not necessarily anti-aligned like transforming the singlet state.

    It's possible that this is not a sufficient answer; however, I don't think it's an incorrect answer.
     
  7. Apr 2, 2010 #6
    Yes, Matterwave, your answer is more correct than I gave it credit for. I understood you to be saying that there is a net spin in some direction other than the z axis; which of course, there isn't. What you get is something like what I described pictorially, a superposition of spin up and spin down along the new axis.

    CRANK ALERT: I am a crank.
     
  8. Apr 2, 2010 #7

    SpectraCat

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    I think Matterwave's explanation, while correct, is a symptom rather than a cause of this. The underlying cause is the requirement that the overall wave-function be anti-symmetric with respect to exchange of the electrons, since they are fermions. For the electronic ground state, the spatial part of the wavefunction is symmetric with respect to exchange, and since the total wavefunction is the product of the spatial and spin parts, this means that the spin contribution *must* be antisymmetric.

    The symmetric combination you posted contributes to the triplet excited state, which of course has an anti-symmetric spatial wavefunction, to preserve the required overall fermionic exchange symmetry.
     
  9. Apr 2, 2010 #8

    Physics Monkey

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    Hi iluvphysics,

    Here are some further heuristics that may help.

    You would agree that the state [tex] | \uparrow \uparrow \rangle [/tex] is definitely s = 1 m =1. To find the s = 1 m = 0 state you must act on the s = 1 m = 1 state with the lowering operator [tex] S_- = S_{1 -} + S_{2 -} [/tex]. This operator converts up to down but doesn't change the sign or the total spin giving you the state [tex] | \uparrow \downarrow \rangle + | \downarrow \uparrow \rangle [/tex]. The other state with m = 0 (the s = 0 m = 0 state) must be orthogonal to this one and hence has a minus sign.

    Here is another way to think about it. There must be three s = 1 states and only one s = 0 state. The hard part of the identification is the question of what combinations of (up down) and (down up) give s = 0 m = 0 and s = 1 m = 0. You know the s = 1 m = 0 must somehow share the characteristics of the other s = 1 states. Both the s = 1 m = 1 and s = 1 m = -1 states are relatively easy to identify and are symmetric. This suggests that the s = 1 states all share the property of being symmetric. In fact this symmetry property is preserved under rotations (this is a great thing to check yourself), so you can be sure that whatever the s = 1 m = 0 state is it must be symmetric since it only mixes with the other s = 1 states under rotations (like Matterwave said). Thus again the orthogonal antisymmetric state is the singlet s = 0 m = 0.

    Another good check is to do what Matterwave suggested and rewrite the supposed singlet in terms of the x basis. In other words, write [tex] | \uparrow \rangle = \frac{1}{\sqrt{2}} ( | + X \rangle + | - X \rangle ) [/tex] and [tex] | \downarrow \rangle = \frac{1}{\sqrt{2}} ( | + X \rangle - | - X \rangle ) [/tex] and check that with the minus sign the singlet is preserved in the x basis. This is the unique property of the singlet state. And as you said, you can see in this formulation that the minus sign is necessary to cancel the + X + X and - X - X terms.
     
    Last edited: Apr 2, 2010
  10. Apr 2, 2010 #9

    SpectraCat

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    It is unclear what you are trying to say here, and your analysis looks incorrect. First of all, it is just a semantic point, but the p-orbitals of hydrogen are *not* a "spin-1 system" ... they have an orbital angular momentum of 1, which is not quite the same thing. The spin of any individual electron in a p-orbital will always be 1/2.

    Second, the triplet state has a spin of 1 by definition, so there is always a "net angular momentum", according to the usual definition of that phrase. The particular symmetric combination the OP mentioned corresponds to a perpendicular orientation of the spin-vector with an applied magnetic field, and so has a zero-length projection (i.e. mS=0), which is analogous to the pz orbital you mentioned. The other two microstates that contribute to the triplet state have non-zero projections, and while they can be combined into linear combinations with as you say, these states are not eigenstates of the Sz operator. It is true that their <Sz> expectation values will be zero, but those are just averages .. any individual measurement will still give +1 or -1 for mS.
     
  11. Apr 2, 2010 #10

    Physics Monkey

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    Actually I don't quite agree with you here. It's relatively minor, but the question I think is why the singlet is antisymmetric and this question doesn't have to involve fermions. If you do consider fermions such that the spatial part of the wavefunction is symmetric then of course I agree that the spin must be antisymmetric. But the question then becomes why this antisymmetric combination of spins is the singlet s = 0 state? Why isn't it the s = 1 m = 0 state, for example?
     
    Last edited: Apr 2, 2010
  12. Apr 2, 2010 #11

    SpectraCat

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    I don't get your question at all .. the spins are *different*. Simply apply the S^2 operator to the state and note the result. For the singlet state you get s(s+1)=0, for the triplet state you get s(s+1)=2. In the vector picture, the lengths of the vectors are different. Either way, the states are clearly distinguishable in terms of their overall spin.

    I am intrigued by your opening statement as well. I assumed the OP was talking about fermions because I work with QM in the context of atomic and molecular physics (i.e. electronic wavefunctions). That is also the context in which this material is typically taught in QM courses, when discussing the Pauli-exclusion principle. Of course there is no a priori reason to prevent one from looking at the spin wavefunction for a pair of equivalent, say, spin-1 bosons, and I can even see why the [STRIKE]antisymmetric[/STRIKE] combination corresponding to ms=0 would logically be called a singlet. However, what would you call the degenerate symmetric state? [STRIKE]On the one hand, it has a degeneracy of 3, and thus the "triplet" designation seems logical.However, the term "triplet" usually refers to the[/STRIKE] The spin multiplicity, which is calculated as 2S+1, for the symmetric spin-combinations for a pair of spin-1 bosons, is 5.

    EDIT: Corrected mis-statements above concerning symmetric and anti-symmetric states of spin-1 boson pair. The anti-symmetric combinations, which I don't mention above, give rise to a triplet state, bringing the total number of microstates to 9, as expected. (Thanks to Physics Monkey for noting my mistake)

    Also, are there physical systems where such bosonic spin-wavefunctions are important? I can't recall encountering any, and I can't really think of one of the top of my head .. maybe symmetric nuclear fission (which is extremely rare I think)?
     
    Last edited: Apr 2, 2010
  13. Apr 2, 2010 #12

    Physics Monkey

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    Yeah, I'm not saying you said anything incorrect. I certainly agree that you can apply the S^2 operator to tell the states apart. My impression was simply that the OP was asking for some more heuristic reasons why the minus sign was correct. Or phrased differently, how might one guess that the minus sign is right without doing any calculation?

    If you had two spin one bosons then there are indeed 5 symmetric states, but they correspond to S = 2 and not S = 1. The counting is still correct. There is also an S = 1 set of states but they are not totally symmetric. Lastly, there is a singlet state with S = 0. You started with 3x3 = 9 states and ended up with 5 + 3 + 1 = 9 states. The singlet designation refers to the number of states in the multiplet and also to the fact that the singlet sector corresponds to the trivial representation of the group SU(2).

    As for physical systems, you can certainly have S = 1 bosons in some spin systems (for example certain kinds of d or f shell materials). You can also have effectively S = 1/2 bosons as in some atomic systems where bosonic atoms have hyperfine levels. The "spin" here is not associated with rotations, but the algebra may be identical so that all the mathematical machinery carries over. There are other examples I can mention later.
     
  14. Apr 2, 2010 #13

    SpectraCat

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    Indeed, and the vector-based explanation you gave is nicely intuitive in that regard. However, you still can't address the question of which is the *ground* state without considering whether the system under consideration is fermionic or bosonic.


    Ok .. thanks for clearing that up. One personal pet-peeve: I realize that you are probably not responsible for introducing it in the first place, but I would *really* prefer that the phrase "S=1/2 boson" not ever appear :biggrin:. I understand what you are saying, but as you point out, you are really just talking about another resultant angular momentum term, and not the true "spin" that is used (and IMHO should be reserved) for talking about fundamental particles. I find such duplications of terminology can lead to massive confusion, which would likely have been avoided if a different term were chosen.
     
  15. Apr 2, 2010 #14
    There have been a number of good answers to this question, especially compared to the crappy answers I got when I posted the same question last year. But I promised I would give peteratcam's ingenious analysis, which takes a bit of operator algebra but is quite different from the other answers so far.

    You identify the vector operator

    [tex]\hat{i}S_x + \hat{j}S_y + \hat{k}S_z [/tex]

    which operates on a spin-1/2 state to give you the spin projection. You can make two operators now, one for electron1 and one for electron2, and take their dot product, which is a measure of their tendency to act in the same direction. It looks like this:

    [tex]S_x_1S_x_2 + S_{y1}S_{y2} + S_z_1S_z_2[/tex]

    When you operate on the two states in question, symmetric and antisymmetric, you get different results. It's a bit of spinor algebra but it's kind of worth doing once in your life at least. The antisymmetric combination turns out to be strongly anticorrelated and the symmetric state is weakly correlated. You can use your imagination to interpret these results. It is as though the in antisymmetric state the two spinors are always pointing the opposite way.

    My own variation on this approach was to evaluate [tex] S^2 [/tex] for the two states in question. The appropriate operator seemed to be just the sum of the x, y, and z components:

    [tex] (S_x_1 + S_x_2)^2 + (S_y_1 + S_y_2)^2 + (S_z_1 + S_z_2)^2[/tex]

    When you multiply this out, you get

    [tex] S_x_1^2 + S_x_2^2 + S_y_1^2 + S_y_2^2 + S_z_1^2 + S_z_2^2
    +2(S_x_1S_x_2 + S_{y1}S_{y2} + S_z_1S_z_2)
    [/tex]

    This is just peteratcam's dot prudoct operator (the mixed terms) plus a bunch of terms in [tex]S^2[/tex]. The nice thing is really there's no extra work in evaluating the [tex]S^2[/tex] terms because operating twice with the spin operator returns the same state. They easily add up to 6/4. It's the dot product which is critical: for the symmetric combination you get an extra 2/4, and for the antisymmetric case, you get a -6/4 which reduces it to zero.

    CRANK ALERT: I am a crank
     
  16. Apr 2, 2010 #15

    SpectraCat

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    First of all, please post a link to the thread with peteratcam's analysis, so that interested parties can review it for themselves. From what I have seen of his posts, peteratcam's treatment is likely completely correct, whereas your analysis below has a couple of (minor) issues. Furthermore, you need to be careful attributing this to peteratcam ... from what you have posted, it just looks like the standard QM treatment of angular momentum, which has been known for decades. Perhaps it is better to say that you appreciate his clear explanation of it, rather than calling it "his ingenious analysis", which suggests he himself developed it.

    No, that is the total spin operator S. A spin projection operator is any one of the component operators, for example Sz, which yields the projection of the spin on the z-axis. You can also construct other spin-projection operators to find the projection on any arbitrary axis in 3D-space.

    You need to define what you mean by "strongly anticorrelated" and "weakly correlated" ... those are not standard descriptions.

    Ok, I understand that when you say "my own variation on this", you probably mean that you derived this for yourself, but you should also be aware that it is a standard approach that appears in basically every Q.M. textbook. FWIW, you did do the math correctly, and you obtained the correct results.
     
  17. Apr 2, 2010 #16
    But I clearly identified myself as a crank so there should be no danger of people blaming peteratcam for my misrepresentation or butchery of his treatment.

    CRANK ALERT: I am a crank.
     
  18. Jul 24, 2010 #17
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