Parallel and antiparallel spins

Question

When describing a system of two spin 1/2 particles it is common to separate between a singlet state (with spin zero):

$$ |\psi_0\rangle = \frac{1}{\sqrt2} \left( |+-\rangle - |-+\rangle \right) $$

And a triplet state (with spin 1):

$$ |\psi_1 \rangle = \frac{1}{\sqrt2} \left( |+-\rangle + |-+\rangle \right)$$ $$ |\psi_2 \rangle= |++\rangle $$ $$ |\psi_3 \rangle = |--\rangle $$

My teacher of undergraduate nuclear physics refers to the singlet state as having antiparallel spins, and to the triplet state as having parallel spins. However, I don't understand why the state $|\psi_1\rangle$ is thought of as having parallel spins, since clearly it is a linear combination of states with different spins.

I do understand that the behaviour of the $|\psi_1\rangle$ and $|\psi_0\rangle$ is different under rotations in the spin state, so I guess that my confusion arises from the relative phase of the two states.

Another guess is that I do not understand correctly the addition of angular momentums (spins in this case). The fact that we can separate between a singlet and triplet states is because the Hilbert space is a direct sum of two Hilbert spaces (with dimensions 1 and 3, respectively). Is this only a mathematical fact or does it have a physical interpretation?

Thank you in advance and sorry for the long question

Answer 1

(Anti)Parallel spins is mostly a classical colloquialism for (anti)symmetric. There are several reasons the "parallel" analogy is misleading:

1. The particles may not be in definite spin states ($j_z=0$, for example), so you can't even talk about their spin as a spin eigenstate.

2. The magnitude of the spin is not $\frac 1 2$, it is $\sqrt{j(j+1)} = \sqrt 3/2$, so even in the states where they are "parallel", they are not exactly parallel.

Nevertheless, the parallel/antiparallel nomenclature isn't going anywhere, so always keep in mind what it really means (a habit that helps in other areas of quantum mechanics, too).

Regarding:

$$ \bf 2 \otimes \bf 2=\bf 3 \oplus \bf 1$$

Yes, it has physical meaning. It tells you the irreducible subspaces that are invariant under rotations, and it works for physical rotations, rotations in isospin space (flavor SU(2)), and so on. Moreover, the generalization to higher rank tensors is central to understanding their physical meaning.

Answer 2

Let me offer a different perspective to the other answers:

We have that $$\vert\psi_1\rangle \propto \vert 00\rangle -\tfrac12\vert ++\rangle - \tfrac12\vert --\rangle\ , $$ (where $\vert0\rangle = (\vert+\rangle + \vert - \rangle)/\sqrt{2}$) that is, the state $\vert\psi_1\rangle$ is spanned by states with parallel spins (just in different bases).

More generally, the triplet space (i.e. the span of the three $\vert\psi_i$, $i=1,2,3$) is equal to the span of all states $\vert\phi,\phi\rangle$, i.e. all product states with two equal spins (=parallel spins in some direction).

Of course, as you noted, this does not imply that when you measure in a specific basis, you obtain the same value on both spins for all triplet states.

Answer 3

While I don't know if I agree with your teacher's statement that the spins are "parallel" also in the state that you labeled $|\psi_1\rangle$, there is a fundamental difference here and a way to understand his comments.

When adding two 1/2-spins you have $4$ possible states, and indeed we tend to separate it into a singlet and a triplet. The point is that they correspond to different behavior upon rotations of the spins, which correspond to the total angular momentum. The state in the singlet really has total angular momentum of zero, which means that $\langle S_j^2 \rangle=0$ for any $j=x,y,z$, when $S_j = s_{1,j}+s_{2,j}$ the addition of angular momenta. To contrast, in the triplet $\langle S_j^2 \rangle > 0$ for all states.

So one can say that only in the singlet the spins are truly "anti-parallel" in the sense that they "cancel" each other for all projections and moments of angular momentum.

More mathematically, it is related to the separation of the group $SU(2)$ into its irreducible representations, which determine how the state will behave under rotations of the spins.

Answer 4

The state $|\psi_1\rangle$ has a total angular momentum of 1, but the z component of angular momentum is 0. Classically to obtain a total angular momentum of 1 from 2 spins of 1/2, the spins would have to be aligned parallel. Consequently we interpret the state $|\psi_1\rangle$ as having its spins parallel, but pointing somewhere in the xy-plane.