On the one hand, in this article, symmetric states are defined as:

The $M$ symmetric quantum states which should be distinguished by means of the quantum measurement are described by statistical operators, $$\begin{align*} \hat{\rho}_j&=\left|\psi_j\middle\rangle\!\middle\langle\psi_j\right|\\ &= \hat{V}^{j-1}|\psi\rangle\!\langle\psi|\hat{V}^{\dagger j-1} \end{align*}$$ where $j=1,2,\ldots,M$.

[...] The operator $\hat{V}$ is unitary and satisfies the relation $$\hat{V}^M=\hat{I}$$ where $\hat{I}$ stands for an identity operator.

On the other hand, in this article, geometrically uniform states are defined as:

We define a geometrically uniform (GU) state set as a collection of vectors $S = \left\{|\phi_i\rangle = U_i|\phi\rangle,U_i\in\mathcal{G}\right\}$, where $\mathcal{G}$ is a finite abelian (commutative) group of $m$ unitary matrices $U_i$, and $|\psi\rangle$ is an arbitrary state.

What is the difference between these two notions?


1 Answer 1


This is only a partial answer. Finding an actual counterexample would be cool!

A set of symmetric quantum states is geometrically uniform, but the converse is not necessarily true.

Note that $\left\{\hat{V}^{j-1}\middle|j=1,2,\ldots,M\right\}$ is a finite abelian group of $M$ unitary matrices. As such, the set $\left\{\hat{V}^{j-1}|\phi\rangle\middle|j=1,2,\ldots,M\right\}$ satisfies the definition of geometrically uniform states. This set is (taking bras instead of brakets), according to the definition, a set of symmetric quantum states.

However, it is not immediate to claim that the converse is true. It is quite easy to find a group $\mathcal{G}$ of unitaries that is both abelian and finite and that is not generated by a single element of this group. For instance, one might take $U$ to be a controlled-$X$ gate with the control qubit being the first one and the target qubit to be the second one and $V$ to be $I\otimes X$, that is an $X$ gate on the second qubit. You can check that $\mathcal{G}=\{I, U, V, UV\}$ is a finite abelian group and it's definitely not generated by a single element.

However, we're interested in applying these elements to a starting point $|\psi\rangle$, and it may just be that there is a unitary $R$ that generates the exact same states. For instance, with the example above, if we start from $|\psi\rangle=\frac{|00\rangle+|11\rangle}{\sqrt{2}}$, then $R=\begin{pmatrix}0 & 0 & 0 & 1\\0 & 0 & 1 & 0\\1 & 0 & 0 & 0\\0 & 1 & 0 & 0\\\end{pmatrix}$ generates the same states.

All in all, there might be a result stating that a set of geometrically uniform states is always a set of symmetric states, but if it's the case then we would need a formal proof. At first sight at least, these two are different objects.


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