Symmetric Under Particle Exchange? Usually, in undergrad QM, we see states such as:
$$|\psi\rangle_{\pm}=\frac{1}{\sqrt 2}(|01\rangle\pm|10\rangle)$$
Where trivially the + state is symmetric, and the $-$ state is antisymmetric. However, I am a bit confused by the particle exchange procedure. I've mostly convinced myself it is swapping the contents within each ket, rather than swapping the kets themselves. I reason this because there may be a state $|\phi\rangle=|00\rangle$, which has to be symmetric, but there are no other terms to swap it with, although I'm not confident. 
I ask this question because I'm wondering about the symmetry of the state $$\frac{1}{\sqrt 2}(|00\rangle-|11\rangle).$$
If in fact (as I think), we flip the numbers within each ket, then this state is symmetric. However if we flip the kets themselves, then this state is antisymmetric. Can someone provide insight into this with a physical interpretation?
 A: 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 $\hat P\big( |00\rangle - |11\rangle\big) = \hat P |00\rangle - \hat P |11 \rangle = |00\rangle - |11\rangle.$
A: 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.
A: 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.)
