# Why is the singlet state for two spin 1/2 particles anti-symmetric?

Let's temporarily forget that the two $m=0$ states exist, and consider just the two completely aligned triplet states, $\newcommand\ket[1]{\left|{#1}\right>} \newcommand\up\uparrow \newcommand\dn\downarrow \newcommand\lf\leftarrow \newcommand\rt\rightarrow $ $\ket{\up\up}$ and $\ket{\dn\dn}$. There's not any physical difference between these: you can "transform" your state from one to the other by changing your coordinate system, or by standing on your head. So any physical observable between them must also be the same.

Either of the single-particle states are eigenstates of the spin operator on the $z$-axis, $$\sigma_z = \frac\hbar 2\left(\begin{array}{cc}1&\\&-1\end{array}\right),$$ and "standing on your head," or reversing the $z$-axis, is just the same as disagreeing about the sign of this operator.

But let's suppose that, on your way to reversing the $z$-axis, you get interrupted midway. Now I have a system which I think has two spins along the $z$-axis, but *you* are lying on your side and think that my spins are aligned along the $x$-axis. The $x$-axis spin operator is usually
$$\sigma_x = \frac\hbar 2\left(\begin{array}{cc}&1\\1\end{array}\right).$$

Where I see my single-particle spins are the eigenstates of $\sigma_z$,
$$\ket\up = {1\choose0} \quad\text{and}\quad \ket\dn = {0\choose1},$$ *you* see those single-particle states as eigenstates of $\sigma_x$,
\begin{align}
\ket\rt &= \frac1{\sqrt2}{1\choose1}
= \frac{\ket\up + \ket\dn}{\sqrt2}
\\
\ket\lf &= \frac1{\sqrt2}{1\choose-1}
\end{align}

If you are a $z$-axis chauvinist and insist on analyzing my carefully prepared $\ket{\rt\rt}$ state in your $\up\dn$ basis, you'll find this mess:

\begin{align} \ket{\rt\rt} = \ket\rt \otimes \ket\rt &= \frac{\ket\up + \ket\dn}{\sqrt2} \otimes \frac{\ket\up + \ket\dn}{\sqrt2} \\ &= \frac{\ket{\up\up}}2 + \frac{\ket{\up\dn} + \ket{\dn\up}}2 + \frac{\ket{\dn\dn}}2 \end{align}

This state, which has a clearly defined $m=1$ in my coordinate system, does *not* have a well-defined $m$ in your coordinate system: by turning your head and disagreeing about which way is up, you've introduced both $\ket{\up\up}$ and $\ket{\dn\dn}$ into your model. You've also introduced the symmetric combination $\ket{\up\dn} + \ket{\dn\up}$.

And this is where the symmetry argument comes in. The triplet and singlet states are distinguishable because they have different energies.
If you propose that the symmetric combination $\ket{\up\dn} + \ket{\dn\up}$ is the singlet state, then you and I will predict different energies for the system based only on how we have chosen to tilt our heads. Any model that says the energy of a system should depend on how I tilt my head when I look at it is *wrong*. So the $m=0$ projection of the triplet state must be symmetric, in order to have the same symmetry under exchange as the $m=\pm1$ projections.

There are at least 2 approaches. One is just show that it is symmetric by applying the lower operator for total spin to the maximal $S_z$ state which should satisfy ($\hbar=1$):

$$ S_-|1,1\rangle=\sqrt 2 |1, 0\rangle$$

so

$$|1, 0\rangle =\frac{1}{\sqrt 2}S_-|1,1\rangle=\frac{1}{\sqrt 2}(S_{1-}+S_{2-})\uparrow_1\uparrow_2$$

$$\frac{1}{\sqrt 2}[(S_{1-}\uparrow_1)\uparrow_2+\uparrow_1(S_{2-}\uparrow_2)] $$

$$ =\frac 1 {\sqrt 2}(\uparrow_1\downarrow_2 + \downarrow_1\uparrow_2)$$

That gives a nice demonstration of how to work with the ladder operators, but there is a much deeper reason it *must* be symmetric.

To find the rotationally invariant subspaces of a tensor product of $N$ states with dimension $d$ you do the following (this just a sketch of the procedure):

Find $N$. It is $N=2$, now we partition $2$ in every way possible:

$$ 2 = 2 $$

and

$$ 2 = 1 + 1 $$

For each of these partitions we draw the Young diagrams and connect those with irreducible representations of the permutation group on $N=2$ letters. This is called the Robinson-Schensted Correspondence.

Take the $2=2$ diagram an make a normal Young Tableau and then compute the Young symmetrizer. In this case, you get the purely symmetric operator: $S=(1 + e_{2,1})/\sqrt 2$

For $2=1+1$, you do the same an get the antisymmetric operator: $A=(1 - e_{2,1})/\sqrt 2$.

Schur-Weyl Duality tells us that applying these to the indices (here, particle labels), will tell us the rotationally invariant subspaces of this tensor product space; moreover, the remarkable Hook-Length Formula tells us the dimensions of the subspace, and the result for $d=2$ is the symmetric one has ${\bf 3}$ dimensions, and antisymmetric is ${\bf 1}$.

This is written:

$$ {\bf 2} \otimes {\bf 2} = {\bf 3}_S + {\bf 1}_A $$

So it simply has to be that all the states in the triplet have the same exchange symmetry.

Note that can add another spin $\frac 1 2$, and the whole procedure will show you that:

$$ {\bf 2} \otimes {\bf 2} \otimes {\bf 2}= {\bf 4}_S + {\bf 2}_M + {\bf 2}_M$$

which means the four $S=\frac 3 2$ states are symmetric and there are two doublet $S=\frac 1 2$ states with mixed symmetry, corresponding to the partitions:

$$ 3 = 3$$

and

$$ 3 = 2 + 1 $$

Note that the hook length formula for:

$$ 3 = 1 + 1 + 1 $$

yields a subspace of dimension zero: there is no antisymmetric combination of 3 spins.