In this paper: ArXiv and PRL links, the authors give the definition of two $d\times d$ qudits to be permutationally invariant states if $\varrho$ is invariant under exchanging the particles. This can be formalized by using the flip operator $$F=\sum_{i j}|i j\rangle\langle j i|$$ as states satisfying $$F \varrho F=\varrho.$$ A state to be symmetric if it acts on the symmetric subspace only. This space is spanned by the basis vectors $$\left|\phi_{k l}^{+}\right\rangle:=(|k\rangle|l\rangle+|l\rangle|k\rangle) / \sqrt{2}$$ for $k \neq l$ and $\left|\psi_{k}\right\rangle:=|k\rangle|k\rangle$.

Question: I can't see what's the difference of those two kinds of state, but the mathematical definition is different. Are there some examples that a state is permutationally invariant but not symmetric?


2 Answers 2


As they have mentioned, calling the first set of all states by $\mathcal{I}$ and the second set of states by $\mathcal{S}$ one has that $\mathcal{S} \subset \mathcal{I}$ meaning that the two sets are indeed not so unrelated. Nevertheless, note that $\mathcal{I}$ has elements that must not be in $\mathcal{S}$ so a simple example is the following.

Let $\vert \psi^- \rangle := \frac{1}{\sqrt{2}}(\vert 01 \rangle - \vert 10 \rangle)$, this state $\vert \psi^- \rangle \langle \psi^- \vert$ is not in $\mathcal{S}$, but is in $\mathcal{I}$.

$$F = \vert 00 \rangle \langle 00 \vert+\vert 01 \rangle \langle 10 \vert+\vert 10 \rangle \langle 01 \vert+\vert 11 \rangle \langle 11 \vert$$

$$F\vert \psi^- \rangle =\frac{1}{\sqrt{2}}(\vert 10 \rangle - \vert 01 \rangle ) = -\vert \psi^- \rangle$$

$$\langle \psi^-\vert F = \frac{1}{\sqrt{2}}(\langle 10 \vert- \langle 01 \vert) = - \langle \psi^- \vert$$

therefore we have that $F\vert \psi^- \rangle\langle \psi^- \vert F = (-1)(-1)\vert \psi^- \rangle\langle \psi^- \vert = \vert \psi^- \rangle\langle \psi^- \vert$ which implies that $\vert \psi^- \rangle\langle \psi^- \vert \in \mathcal{I}$ but $\vert \psi^- \rangle\langle \psi^- \vert \notin \mathcal{S}$ by definition.

So I guess this is the most important difference: one is a strictly larger set of states then the other. But this example helps understanding a bit the difference in terms of intuition as well, since it was crucial that although nonsymmetric, flipping twice preserved the invariance.

  • $\begingroup$ So it seems that they only differ with a relative phase? $\endgroup$
    – Sherlock
    Commented Mar 13, 2022 at 12:34

A state is permutationally invariant if it commutes with the swap operator (your $F$; this is equivalent to your condition $F\varrho F=\varrho$), while a symmetric state is one whose support is contained in the eigenspace of $F$ associated with the eigenvalue $+1$.

This relies on the observation that $F$ has eigenvalues $\pm1$, and the associated eigenspaces are symmetric and skew-symmetric subspaces.

Note that a pure state, if permutationally invariant, needs to be either symmetric or skew-symmetric, but this needs not be the case for general states. For example, $$\rho = \frac12 \left(\mathbb{P}(|00\rangle) + \frac12\mathbb{P}(|01\rangle-|10\rangle) \right), \qquad \mathbb{P}(|u\rangle)\equiv |u\rangle\!\langle u|,$$ is permutationally invariant, but neither symmetric nor skew-symmetric.

See e.g. Example 6.10 in Watrous' book for more details.


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