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

For two spin 1/2 particles I understand that the triplet states ($S = 1$) are: $\newcommand\ket[1]{\left|{#1}\right>} \newcommand\up\uparrow \newcommand\dn\downarrow \newcommand\lf\leftarrow \newcommand\rt\rightarrow $

\begin{align} \ket{1,1} &= \ket{\up\up} \\ \ket{1,0} &= \frac{\ket{\up\dn} + \ket{\dn\up}}{\sqrt2} \\ \ket{1,-1} &= \ket{\dn\dn} \end{align}

And that the singlet state ($S = 0$) is:

$$ \ket{0,0} = \frac{\ket{\up\dn} - \ket{\dn\up}}{\sqrt2} $$

What I'm not too sure about is why the singlet state cannot be $(\ket{↑↓} + \ket{↑↓})/\sqrt2$ while one of the triplet states can then be $(\ket{↑↓} - \ket{↑↓})/\sqrt2$. I know they must be orthogonal, but why are they defined the way they are?

Answer

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.