# Symmetric Under Particle Exchange?

The issue is easily resolved if you explicitly label your kets using particle numbers: $$ \vert\psi_\pm\rangle=\frac{1}{\sqrt{2}}\left(\vert 0\rangle_1\vert 1\rangle_2\pm \vert 1\rangle_1\vert 0\rangle_2\right) $$ so that the action of the permutation group is on the particle labels $1$ and $2$. Thus $$ P_{12}\vert\psi_\pm\rangle = \frac{1}{\sqrt{2}}\left(\vert 0\rangle_2\vert 1\rangle_1\pm \vert 1\rangle_2\vert 0\rangle_1\right) =\frac{1}{\sqrt{2}}\left(\vert 1\rangle_1\vert 0\rangle_2\pm \vert 0\rangle_1\vert 1\rangle_2\right) =\pm \vert\psi\rangle $$

In this fashion writing $$ \frac{1}{\sqrt{2}}\left(\vert 0\rangle_1\vert 0\rangle_2-\vert 1\rangle_1\vert 1\rangle_2 \right) $$ is clearly symmetric under interchange of $1$ and $2$.

(Note there is another action which interchanges *the states* $0$ and $1$, but the symmetry character of the state is normally defined under permutation not of the states but of particle labels.)

Alternatively one can understand $|\Psi\rangle=|ab\rangle$ as a wavefunction statement to the effect of $\Psi(x_1, x_2) = \psi_a(x_1)\psi_b(x_2).$ The particle-permutation operator can be written easily in the wavefunction picture as $P[\Psi](x_1, x_2) = \Psi(x_2, x_1)$ and therefore $\hat P |ab\rangle = |ba\rangle.$

It is linear, so $\hat P\big( |00\rangle - |11\rangle\big) = \hat P |00\rangle - \hat P |11 \rangle = |00\rangle - |11\rangle.$

I think it should be symmetric. When we write $\mid 10\rangle$, we mean to say that this ket is a tensor product of two individual kets $|1\rangle\in\mathcal{H}_1$ and $|0\rangle\in\mathcal{H}_2$, so that $|10\rangle=|1\rangle\otimes|0\rangle$, which is an element of the composite Hilbert space $\mathcal{H}=\mathcal{H}_1\otimes\mathcal{H}_2$. When you're exchanging particles, you're essentially taking the first ket to the second Hilbert space and vice versa. So under an exchange of particles, we get $|10\rangle\to|01\rangle$. Thus, $|00\rangle\to|00\rangle$ and $|11\rangle\to|11\rangle$, which means the state

$\mid\psi\rangle=\frac{1}{\sqrt{2}}(|00\rangle-|11\rangle)\to\frac{1}{\sqrt{2}}(|00\rangle-|11\rangle)$

is indeed symmetric under exchange of particles.