Take the 2-minute tour ×

There are two pure symmetric states $|\psi\rangle$ and $|\phi\rangle$ of $n$ qudits. Is there any known set of invariants $\{I_i:i\in\{1,\ldots,k\}\}$ which is equal for both states iff $|\phi\rangle=U^{\otimes n}|\psi\rangle$ for a $U\in\text{SU}(d)$?

There is a theorem by Hilbert saying that for a compact group acting on a linear space there is a finite number of polynomial invariants characterizing orbits. However, (as far as I know) it does not provide and explicit construction of the invariants.

The problem is easy when $n=2$ (Schmidt decomposition) and $d=2$ (Majorana representation). Partial solutions, and solutions with changed assumptions (e.g. linear operators instead of unitary operations) are welcome as well.

share|improve this question

5 Answers 5

up vote 10 down vote accepted

This is an algorithm for the computation of the homogeneous polynomial invariants in the general case, however without any attempt to reduce the algorithm's complexity. The basic needed ingredient of the algorithm is the ability to average over the Haar measure of the group of the local transformations. In the qubit case it is just an integration over copies of SU(2). But even in the more general cases of compact Lie groups, this task is possible but cumbersome, please, see for example the following parameterization of SU(N) by: Bertini, Cacciatori, Cerchiai that can be used for the integration.

The procedure is as follows:

First one computes the Molien function of the group of local transformations (The Wikipedia page describes the finite group case which can be generalized to compact groups):

$M(t) = \int \frac{d_H(g)}{det(1-tg)}$

The coefficient of $t^n$ in the Taylor expansion of $M(t)$ is the number of linearly independent homogeneous invariants of degree n.

Since the integrand is a class function the integration can be performed on the maximal torus by the Weyl integration formula.

Now, for each degree, one constructs all possible combinations of lower degree inavriants and if this number doesn't exhaust the required number from the Molien series, then the additional invariants are computed by averaging monomials of the required degree. This operation involves the integration overthe Haar measure which cannot be reduced to an integration on the maximal torus, which is the most complex step in the algorithm. This operation is repeated until a sufficient number of linearly independent polynomials is produced, then one passes to the next degree. This process is continued until the total number of invariants reaches the difference between the dimension of the vector space and the local symmetry group.

More practical variations of this procedure were used to construct polynomial invariants for special cases of entanglement problems, please, see for example the following work by: Grassl, Rotteler and Beth.

share|improve this answer
Thanks David, it looks nice. And thanks for the references. I will see if it is practical numerically. However, as far as I understand, the procedure is longer than the straightforward optimization of $|\langle \phi | U^{\otimes} |\psi \rangle|^2$ (unless a low degree invariant suffices to distinguish two states). –  Piotr Migdal Oct 6 '11 at 17:31
Take a look at: P. Migdał, J. Rodríguez-Laguna, M. Oszmaniec, M. Lewenstein, Which multiphoton states are related via linear optics?, arXiv:1403.3069 (where I acknowledge you and this thread). Thanks once again! –  Piotr Migdal Mar 13 at 11:12
@Piotr Migdal Thank you very much –  David Bar Moshe Mar 13 at 14:05
If you have any critical remarks, I would be grateful (email or scirate.com/arxiv/1403.3069). –  Piotr Migdal Mar 13 at 14:31

It seems to me that the stabilizer formalism provides an answer to your question (see section 3.1 of quant-ph/0603226 for an introduction to the formalism). Given the two states, you simply take the stabilizer group for that 2-dimensional subspace of the total Hilbert space, and they will give you a such a set of invariants. However, this of course has nothing to do with the globally symmetric operations you consider, and could be done with any pair of states. However, given that you are kept within the symmetric subspace, this insures that these stabilizers will always have a description at worst polynomial in the number of local systems.

If you wish to go further, and try to find some set of invariants which will uniquely identify states which are equivalent up to this form of globally symmetric local operation, then you are out of luck. This is because the set of product states which is globally symmetric spans the space of symmetric states, and hence any globally symmetric state can be written as a superposition of globally symmetric separable states.

That is to say, without fixing the states, it is impossible to produce an observable which is invariant between the two states but varies across the space of symmetric states. Thus the only invariants which exists depend only on the symmetry of the states.

UPDATE: I notice Norbert and I have interpreted the question somewhat differently. I have focused on the existence of observables which have the same value for the LU equivalent states. This is effectively answering the question of whether there is a measurement which distinguishes these states from other symmetric states. The paper Norbert links to is about the mathematical structure of the states, and is not testable with a single copy of the pair of states. I have no idea which setting Piotr had in mind (I had originally thought it was this one, but Norbert's answer has caused me to rethink that position).

share|improve this answer
Of course, it's important that not only does the symmetric subspace have polynomial dimension, but it has a basis which is easy to describe (in fact, it has a basis of product states); you can then compute the restriction of operators to the symmetric subspace efficiently. –  Niel de Beaudrap Oct 1 '11 at 14:29
@NieldeBeaudrap: Yes. In fact the basis of product states is exactly the reason why there cannot be some observable which has a different expectation for some entangled state on this space if it's expectation value is constant across all product states. –  Joe Fitzsimons Oct 1 '11 at 14:33
@JoeFitzsimons: I have in mind the mathematical structure of states, and invariants that can be computed numerically. For this question I'm not that interested if invariants are directly related to any measurements. –  Piotr Migdal Oct 3 '11 at 12:10
@Piotr: I posted a new answer which hopefully addresses your question as I now understand it. I'll leave this answer, as I think it is interesting in its own right, but is entirely different from the new answer. –  Joe Fitzsimons Oct 3 '11 at 16:22

It seems that this question is addressed in this paper:


(Edit: I just noticed this is seems to be restricted to qubits, so it probably does not answer your question ... )

share|improve this answer
Welcome to TP.SE! I wasn't aware of that paper, but it seems an interesting observation, and looks like it might generalize beyond qubits. –  Joe Fitzsimons Oct 1 '11 at 7:10

It seems from Piotr's comments on my other answer, that he is looking for an invariant of the mathematical representation of the state, rather than an observable that remains unchanged. In this case the answer is very different, and hence I am posting a new answer, rather than replacing the old one (since it would mean entirely rewriting it, and the current version may be of interest to some people).

Any density matrix can be written as $\rho = \sum_k \alpha_k \sigma_k$, where $\sigma_k=\bigotimes_i \sigma_{k_i}$ and $\{\sigma_{k_i}\}$ form an orthonormal basis for Hermitian matrices corresponding to the subsystem dimensionality and includes the identity matrix. When you apply local unitaries you obtain $\rho' = \big(\bigotimes_i U_i \big) \rho \big(\bigotimes_i U_i^\dagger \big)$. Now if you consider what happens term by term, you will notice that each operator $\sigma_k$ is mapped only to operators of the same weight (i.e. operating non-trivially on the same number of subsystems). I'll take $w_k = w(\sigma_k)$ to be the weight function for each operator $\sigma_k$. So, we have $w(\sigma_k) = w\big(\big(\bigotimes_i U_i \big) \sigma_k \big(\bigotimes_i U_i^\dagger \big)\big)$. This is trivially true, since for every subsystem where $\sigma_k$ in $\rho$ acts as the identity (i.e. for every $i$ such that $\sigma_{k_i}=\mathbb{I}$) $U_i$ and $U_i^\dagger$ cancel, and so the transformed operator also acts as the identity on that subsystem. Conversely, if $\sigma_{k_i}\neq \mathbb{I}$ then $U_i\sigma_{k_i}U_i^{\dagger}\neq \mathbb{I}$. There is a pretty intuitive reason for this: local operations should not create non-local correlations.

Now, from this, it should be clear that $\{\beta_w=\sum_{i:w(\sigma_i)=w} |\alpha_i |^2\}_{w=0}^n$ is invariant, since $U \sigma_k U^\dagger = \sum_j \gamma_{jk} \sigma_j$ such that $\sum_j |\gamma_{jk}|^2 = 1$.

This is a conserved quantity independent of the symmetry, and depends only on the fact that all applied unitaries are local, however I believe this is the kind of thing you want. Once you impose the criteria that the states and operations are symmetric, you have the additional criteria that $\alpha_i = \alpha_j$ if $\sigma_i$ can be obtained from $\sigma_j$ by permutation of the qudits, and hence $|\alpha_i-\alpha_j|$ is also invariant for all such pairs.

share|improve this answer
Thanks Joe - actually I was looking at something similar. Or rather - with more details (e.g. diagonalization of the matrix for $w=(\sigma_i)2$ (for symmetric states there are only 9 entries)). And do you have any idea which states have the same $\beta_w$s? –  Piotr Migdal Oct 6 '11 at 17:12

This answer is incomplete, but it should provide an answer for almost all symmetric states (i.e. it suffices for all but a set of symmetric states having measure zero).

The symmetric subspace is spanned by product states. We may then consider different ways in which a particular symmetric state decomposes into symmetric products; in particular, if any choice of decomposition naturally gives rise to an invariant.

A greedy way to go about decomposing a symmetric $\def\ket#1{|#1\rangle}\ket\psi$ state into symmetric products would be to simply look for the symmetric product $\ket\phi\ket\phi\cdots\ket\phi$ with which $\ket\psi$ has the overlap of the largest magnitude. Let $\ket{\phi_0}$ be the single-spin state satisfying this, and $$\def\bra#1{\langle#1|}\alpha_0 = \Bigl[\bra{\phi_0}\cdots\bra{\phi_0}\Bigr]\ket\psi$$ which without loss of generality is positive. With probability 1, the state $\ket{\phi_0}$ is unique (in that the set of symmetric states for which it is not unique has measure zero). Let $\ket{\psi_1}$ be the projection of $\ket{\psi}$ into the orthocomplement of $\ket{\phi_0}^{\otimes n}$: this is another symmetric state. So, we define $\ket{\phi_1}$ to be the (again with probability 1, unique) single spin state such that $\ket{\phi_1}^{\otimes n}$ has maximal overlap with $\ket{\psi_1}$; we let $\alpha_1$ be that overlap; and we define $\ket{\psi_2}$ to be the projection of $\ket{\psi}$ onto the orthocomplement of $\mathrm{span}\{\ket{\phi_0}^{\otimes n}\!\!,\;\; \ket{\phi_1}^{\otimes n}\}$. And so on.

By continually projecting $\ket{\psi}$ onto the orthocomplement of spans of larger and larger sets of symmetric products, we ensure that the resulting projections $\ket{\psi_j}$ will not have maximal overlap with any product state that came before, or more generally which can be spanned by the preceding symmetric products. So at each iteration we obtain a single-spin state $\ket{\phi_j}$ such that $\mathrm{span}\{\ket{\phi_0}^{\otimes n}\!\!,\;\ldots\,,\; \ket{\phi_j}^{\otimes n}\}$ has dimension one larger than in the previous iteration. In the end, we will obtain a collection of symmetric products which, if they don't span the symmetric subspace, at least contain $\ket{\psi}$ in their span. So we obtain a decomposition $$ \ket{\psi} \;=\; \sum_{j = 0}^\ell \alpha_j \ket{\phi_j}^{\otimes n} $$ where the sequence $\alpha_0, \alpha_1, \ldots$ is strictly decreasing. Let us call this the symmetric product decomposition of $\ket{\psi}$. (It wouldn't take too much to generalize this representation to one where the states $\ket{\phi_0}$, $\ket{\phi_1}$, etc. are non-unique; but for what comes next uniqueness will be important.) Given the symmetric product decomposition of $\ket{\psi}$, it is trivial to describe the corresponding representation for $U^{\otimes n} \ket{\psi}$: just multiply each of the $\ket{\phi_j}$ by $U$. And in fact, if you computed $U\ket{\psi}$ and then determined its symmetric product decomposition, the decomposition $$ U^{\otimes n}\ket{\psi} = \sum_{j=0}^\ell \alpha_j \Bigl[ U\ket{\phi_j} \Bigr]^{\otimes n} $$ is exactly what you would find: $\ket{\phi_0}^{\otimes n}$ has maximal overlap with $\ket{\psi}$ if and only if $[ U \ket{\phi_0} ]^{\otimes n}$ has maximal overlap with $U^{\otimes n} \ket\psi$, and so on. So to show that two symmetric states are LU-equivalent, it suffices to show that the sequence of amplitudes $\alpha_j$ are the same, and that the sequence of single-spin states $\ket{\phi_j}$ are related by a common unitary.

The last part can be most easily done by finding a normal form for the states which are equal, for sequences of single-spin states which are related by a common single-spin unitary. We can do this by finding a unitary $T$ which

  1. maps $\ket{\phi_0}$ to $\ket{0}$,
  2. maps $\ket{\phi_1}$ to a state $\ket{\beta_1}$ in the span of $\ket{0}$ and $\ket{1}$, with $\bra{1}\beta_1\rangle \geqslant 0$,
  3. and for each subsequent $j > 1$, maps $\ket{\phi_j}$ to some state $\ket{\beta_j}$ which is in the span of standard basis states $\ket{0}, \ldots, \ket{b_j}$ for $b_j$ as small as possible, with $\bra{b_j}\beta_j\rangle \geqslant 0$ if possible. (For any state $\ket{\phi_j}$ which is in the span of preceding states $\ket{\phi_h}$, the state $\ket{\beta_j}$ will similarly be determined by the states $\ket{\beta_h}$ for $h < j$, in which case we do not have a choice of the value of $\bra{b_j}\beta_j\rangle$.)

We then have a unitary such that $T \ket{\phi_j} = \ket{\beta_j}$; and for any two sequences of states $\ket{\phi_j}$ and $U\ket{\phi_j}$, we should obtain the same sequence of states $\ket{\beta_j}$. You can then determine that two symmetric states are equivalent if they give rise to the same sequence of amplitudes $\alpha_j$ and the same "normal form" single-spin states $\ket{\beta_j}$.

In the case that there is not a unique state $\ket{\phi_j}^{\otimes n}$ which has maximal overlap with $\ket{\psi_j}$ in the construction of the symmetric product decomposition, the problem is then in defining the normal form states $\ket{\beta_j}$. However, so long as the states $\ket{\phi_j}$ are unique, which happens with measure 1, you should have a polynomial-size invariant (up to precision limitations) for determining if two symmetric states are the same.

share|improve this answer
Thanks Niel. I like the approach, but (unfortunately) it has one drawback - interesting states tend to have a lot of symmetries and are often in the zero-measure set of all permutation symmetric states. E.g. for a Bell state $(|00\rangle+| 00\rangle)/\sqrt{2}$ the maximum is achieved on entire equator (i.e. separable states with $|\psi\rangle = \cos(\alpha)|0\rangle+\sin(\alpha)|1\rangle$) so your decomposition is not unique. –  Piotr Migdal Oct 6 '11 at 17:21
@PiotrMigdal: Fair enough; I'm too aware of the bitter-sweet irony that the nicest structures to study tend to be examples from a set of measure zero. Of course, if you have enough symmetry between a (discrete) set of maximizing products, such as with a Bell state, this would simplify the task of obtaining a normal form; perhaps one could formulate particular symmetry conditions for which a useful normal form is possible. –  Niel de Beaudrap Oct 6 '11 at 17:51

Your Answer


