Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

This question developed out of conversation between myself and Joe Fitzsimons. Is there a succinct stabilizer representation for symmetric states, on systems of n spin-1/2 or (more generally) n higher spin particles?

By a "stabilizer representation", I mean that:

  • every symmetric state (or some notable, non-trivial family of them which contains more than just product states) is represented as the unique +1-eigenstate of some operator or the unique joint +1-eigenstate of a list of operators, where

  • each element of this set of stabilizing operators can be succinctly described, as an operator on the larger Hilbert space (i.e. not only as a transformation restricted to the symmetric subspace itself), and

  • where the stabilizing operators transform in a nice way in the Heisenberg picture under symmetric local unitaries (i.e. unitary transformations of the form U⊗n).

Ideally, one would be able to efficiently describe all sorts of transformations between various symmetric states; but one cannot have everything.

The constraint of being a unique +1-eigenstate of the list of stabilizing operators could also be made subject to the constraint of being a symmetric state. (For instance, many states on n spin-1/2 particles are stabilized by a σz operator on a single spin, but exactly one symmetric state is stabilized by that operator. Not that I would expect such an operator necessarily to arise in the formalism...)

Does a representation with the above properties (or one close to it) exist?

share|improve this question

1 Answer 1

up vote 7 down vote accepted

I believe there is such a representation, as follows:

You need only consider operators which can be written as $\Omega = \sum_k \alpha_k \gamma_k$, where $\gamma_k = \sum_\ell P_\ell \big( \sigma_{k_1} \otimes \cdots \otimes \sigma_{k_N} \big)P_\ell^\dagger$ with $P_\ell$ being the operator corresponding to the $\ell^{\text{th}}$ permutation of qubits. Note that $\gamma_k$ is defined on the whole space, and is essentially a generalization of the number operator. $\gamma_k$ can be uniquely identified by each of the number of each kind of Pauli matrix it contains, and so is polynomial in the number of qubits. The same works for the case of qudits, though the degree of the polynomial will scale with the dimensionality of the local systems. For qubits, we have 3 relevant numbers: 1) $N_X$ the number of sites where the operator acts as $\sigma_X$, 2) $N_Y$ the number of sites where the operator acts as $\sigma_Y$ and 3) $N_Z$, the number of sites where the operator acts as $\sigma_Z$, subject to the constraint that $N_X + N_Y + N_Z \leq N$. So instead, we can relabel the $\gamma$ matrices as $\gamma_{N_X,N_Y,N_Z}$ for the two dimensional case.

Notice that the set of possible $\Omega$ form a group under multiplication, and that each element of the group has a polynomial description (up to approximation of the complex coefficients).


  1. Since the $\gamma$ operators form a basis for Hermitian matrices which are invariant under permutation of the local Hilbert spaces, they satisfy your first criterion.

  2. Any stabilizing operator $\Omega$ can be described by a set of real numbers (or approximations there of), the number of which is polynomial in the number of subsystems and exponential in their local dimension, thus satisfying your second criterion.

  3. Symmetric local unitaries can also be written in terms of a sum of $\gamma$ matrices (albeit it with an additional restriction on the values of $\alpha_k$), and thus by the group structure, the outcome can still be represented within this framework. When a unitary is applied, the stabilizers transform as $U \Omega U^\dagger$, which is efficiently computable if $U$ is symmetric given the reduced basis with which both $U$ and $\Omega$ can be expressed, satisfying your last criterion.

share|improve this answer
We can certainly represent any density operator as a sum of $\gamma$ matrices, but they aren't actually symmetric states themselves. I suppose this is irrelevant, however, as for any pure symmetric state, we can compute the decomposition of the density operator into $\gamma$ operators (e.g. from the Majorana representation), which then is its own stabilizer group. It is not clear how to actually carry out the operator multiplication of operators expressed in the $\gamma$ basis, but I suppose that it is enough that this is possible. –  Niel de Beaudrap Oct 2 '11 at 17:43
Your description is really a way to restrict operators to the symmetric subspace implicitly, by symmetrizing the Pauli operators involved. The work is entirely in finding the polynomial-length decompositions in terms of an operator basis, and then multiplying them out. (This also happens to be true in the Pauli stabilizer case, but which is not the mechanism by which states are actually expressed, and in any case the multiplications are simpler.) But while this formalism does not actually use the fact that any operator stabilizes any state, it does meet the criteria which I first laid out. –  Niel de Beaudrap Oct 2 '11 at 17:54
I can this that this approach will probably work, but I don't quite see the multiplication rule (and hence also the time evolution) yet. That is, how to calculate multiplications in polynomial time. For example, $(X_{1}+X_{2})(Y_{1}+Y_{2})=X_{1}Y_{1}+X_{2}Y_{1}+X_{2}Y_{2}+X_{1}Y_{2}$ –  Earl Oct 3 '11 at 7:09
@Earl: Even with a look-up table you get efficient multiplication, since the table contains only a polynomial number of entries. –  Joe Fitzsimons Oct 3 '11 at 7:12
sorry... starting editing my comment and it should that I can only edit for 5 minutes: copy and pasted: I can this that this approach will probably work, but I don't quite see the multiplication rule (and hence also the time evolution) yet. That is, how to calculate multiplications in polynomial time. For example, $(X_{1}+X_{2})(Y_{1}+Y_{2})=X_{1}Y_{1}+X_{2}Y_{1}+X_{2}Y_{2}+X_{1}Y_{2}$, which looks like a some of canonical $\gamma$ terms, but not manifestly like one. –  Earl Oct 3 '11 at 7:27

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.