Suppose we have two spin-1/2 particles with no orbital angular momentum. We choose to work with the eigenbasis of total angular momentum $S^2$ and $S_z$, which gives us the triplet and the singlet states:

$$ \begin{align} (s=1, \rm triplet, \rm symmetric) \begin{cases} &|11\rangle=|\uparrow \uparrow\rangle \\ &|10\rangle=\frac{1}{\sqrt{2}}\left(|\uparrow \downarrow\rangle + |\downarrow \uparrow \rangle\right) \\ &|1-1\rangle=|\downarrow \downarrow\rangle \end{cases} \\ \\ (s=0, \rm singlet, \rm antisymmetric) \begin{cases} &|00\rangle=\frac{1}{\sqrt{2}}\left(|\uparrow \downarrow\rangle - |\downarrow \uparrow \rangle\right) \end{cases} \end{align} $$

  1. Both the triplet and the singlet states have integer-valued total spins. This suggests the composite system of two spin-1/2 particles behaves bosonically. Although the triplet state respects this by being totally symmetric, the singlet state is totally antisymmetric. Given there are no other parts to the wavefunction for us to antisymmetraize, we are stuck with a totally antisymmetric state describing $s=0$, which is for bosons. What am I missing here that gives rise to this contradiction?

  2. Are the four listed states above always allowed? Or, does it depend on whether the two spin-1/2 particles are identical or distinguishable?

    I'm thinking that if they are distinguishable, then all four states are allowed, keeping in mind my confusion described in Question 1 (that is, I think the system should behave bosonically, but the singlet state is antisymmetric).

    If the particles are identical, then I cannot tell them apart, and as far as I can tell, I have a composite system of two fermions, and I know the composite state must be antisymmetric. Therefore, only the singlet state would be allowed.


2 Answers 2

  1. There is no contradiction. A particle's spin is not its only attribute. A two-fermion state must be antisymmetric with respect to the exchange of all of their attributes, not just spin. If the state is symmetric with respect to the exchange of their spins, then it is antisymmetric with respect to the excahge of their other attributes (such as location or momentum, not shown in the OP), and conversely.

  2. The four states shown are allowed whether the particles are distinguishable or "identical" (same species). For the identical-particle case, consider the two electrons in orthohelium and parahelium (https://en.wikipedia.org/wiki/Helium_atom). For the non-identical-particle case, consider the various possible states of a hydrogen atom, taking into account the parallel or antiparallel configuration of the electron/nucleon spin (http://www.feynmanlectures.caltech.edu/III_12.html).

  • $\begingroup$ Ok, this is what I initially thought as well, but then I wasn't able to reason through Sakurai question 7.3. However, if you adopt the perspective presented in OP, then you get the right answer for problem 7.3. I didn't mention this in OP because I'm not asking a specific question about how to solve that problem, but I'm more interested in learning the concepts behind it. For instance, in the solution to 7.3, we claim that only the spin states can be made symmetric/antisymmetric, and so for that problem, the antisymmetric subset must be disallowed. Do you have access to that problem? $\endgroup$
    – Ptheguy
    Oct 24, 2019 at 4:26
  • $\begingroup$ I don't have access to Sakurai's book right now, so I'm not sure what the loophole is. Is question 7.3 considering a case in which the fermions are in the same spatial state, as described in @AndrewSteane's answer? $\endgroup$ Oct 24, 2019 at 12:49
  • $\begingroup$ Sakurai 7.3 concerns the restrictions we have on the spin states for the case of two spin-1 particles that are identical. Going off of your point in #1, I thought there would be no restrictions on the spin states because the spatial states could always be chosen to ensure an overall symmetric state. But according to the solution, we must disallow all antisymmetric states and only allow for the symmetric ones to exist. $\endgroup$
    – Ptheguy
    Oct 24, 2019 at 17:23
  • $\begingroup$ @Ptheguy For two spin-1 particles, the state must be symmetric with respect to the exchange of all of their attributes, assuming that the spin-1 particles are bosons as usual. I worded the answer for fermions, because the question specified two spin-1/2 particles (spin-statistics theorem), and this matched the equations written in the OP. For two spin-1 bosons, the overall state must be symmetric rather than antisymmetric, so if it's already symmetric in all the non-spin attributes, then it must also be symmetric in the spin. $\endgroup$ Oct 25, 2019 at 1:20
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    $\begingroup$ @Ptheguy I wonder if the author is using the words "the particles are identical" to mean that they're in the same state (except possibly for their spins) instead of merely being the same species. Then we would have enough information to determine the symmetry/antisymmetry of the spin-state. Whenever I read something in a reputable physics book that I don't understand, I ask myself: "Am I missing a key concept, or is this just a case of ambiguous language?" Often it turns out to be the latter, but that can be hard to diagnose. And I'm pretty sure I'm not the only one who struggles with this. $\endgroup$ Oct 25, 2019 at 14:39

First on your point 1. There is no contradiction between a two-spin system having an antisymmetric state with respect to exchange of its component parts, yet being symmetric with respect to exchange of that pair with some other pair.

The two-spin system is, overall, bosonic, as you say. In particular, the singlet state has the correct properties to be a $S=0$ state. For example, it does not change under rotations of the coordinate system.

One can take any number of pairs of fermions in such a state, and push them all into the same spatial state. If these fermions are all mutually indistinguishable then the state will be symmetric with respect to exchange of one pair with any other pair, and antisymmetric with respect to exchange of one fermion with any other fermion.

Now point 2. A pair of fermions has both spin and spatial properties. Their joint state can sometimes be factorised into a spatial part in a tensor product with a spin part. This does not always happen. Whether or not the state can be so factorized, it has to be antisymmetric with respect to exchange of those two fermions if the fermions are a pair of the same type of particle (e.g. two electrons). If the state can be factorized then its overall antisymmetry is achieved if:

either the spin state is the singlet and the spatial state is symmetric

or the spin state is the triplet and the spatial state is antisymmetric

Therefore if either kind of spatial state is available to the system, then so is either kind of spin state. If the two particles are in the same spatial state, then the overall spatial state cannot help but be symmetric, so in that case the spin state must be the singlet.

  • $\begingroup$ (1) In your point #2, it seems that the system of two electrons must have an overall antisymmetrical state. But, didn't we conclude that the system of two spin-1/2 particles behaves bosonically? I believe you approved of this conclusion in your post above. (2) Additionally, am I correctly understanding that when you have a system of identical particles and the spatial states are available, then those states are symmetrical; and if the system has non-identical particles, then we can form antisymmetrical spatial states? $\endgroup$
    – Ptheguy
    Oct 24, 2019 at 17:38
  • $\begingroup$ Follow-up: Regarding (1) in my previous comment, I believe my confusion came from overlooking your emphasis on exchanging a pair of fermions vs a single fermion. I think with that in mind, I'm good on this front. Also, I think the answer to (2) is no. The spatial part can be made into symmetric or antisymmetric just like any other states. $\endgroup$
    – Ptheguy
    Oct 24, 2019 at 19:26

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