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Is there a Majorana-like representation for singlet states?

+ 14 like - 0 dislike

I mean the Majorana representation of symmetric states, i.e., states of $n$ qubits invariant under a permutation of the qudits. See, for example, D. Markham, "Entanglement and symmetry in permutation symmetric states", arXiv:1001.0343v2.

By Majorana representation I mean decomposition of a state $$|\psi\rangle = \text{normalization} \times \sum_{perm} |\eta_1\rangle |\eta_2\rangle \cdots |\eta_n\rangle,$$ where $|\eta_k\rangle$ are uniquely determined (up to a global phase in each and the permutation) qubit states.

This post has been migrated from (A51.SE)
asked Sep 15, 2011 in Theoretical Physics by Joshua Herman (100 points) [ no revision ]
retagged Mar 7, 2014 by dimension10
I think the question should be expanded a little bit.

This post has been migrated from (A51.SE)
@UGPhysics: I don't think his question is specific to Markham's paper, but rather he gives it as an example. The Majorana representation was already in existence prior to that paper.

This post has been migrated from (A51.SE)

3 Answers

+ 13 like - 0 dislike

Well, there is certainly not a Majorana representation, since any decomposition will have two terms which differ by a phase of -1, you can't find Majorana points. The singlet state is anti-symmetric, so there is no way as writing it in the form $\frac{e^{i\alpha}}{\sqrt{2}} \sum_{j=0}^1 | \phi_{1\oplus j} \rangle \otimes | \phi_{0\oplus j} \rangle$ since it is always in the state $\frac{e^{i\alpha}}{\sqrt{2}} \sum_{j=1}^2 (-1)^j| \phi_{1\oplus j} \rangle \otimes | \phi_{0\oplus j} \rangle$. Here I have taken $\{\phi_{0}, \phi_{1}\}$ as a basis for the single qubit Hilbert space.

However, if you want something kinda-sorta like the Majorana representation, you can do the following. The Majorana representation is effectively treating the subsystems like bosons, and hence we are stuck working in the symmetric subspace. However, you can do the exact same thing treating the subsystems as fermions, which will the restrict you to the antisymmetric state for a Hilbert space of that dimension.

Another route would simply be to consider states which are LU equivalent to Majorana states, but I have no idea whether this is useful to you (you haven't explained exactly what you want or why you want it). If you just care about entanglement (which is a very common usage) then LU equivalence should be fine.

This post has been migrated from (A51.SE)
answered Sep 15, 2011 by Joe Fitzsimons (3,555 points) [ no revision ]
+ 6 like - 0 dislike

In the general - the answer is no.

Majorana representation's key point is to express a composite state of $n$ qubits as $n$ points is such way, that action of a collective rotation (i.e. $|\psi \rangle \mapsto U^{\otimes n} |\psi \rangle$ for $U\in \text{SU}(2)$) rotates each points in the same way (i.e. $| \eta_k\rangle \mapsto U | \eta_k\rangle $). In other words, $| \eta_k\rangle$ are covariant.

Singlet states are, by definition, invariant under the application of the same unitary operation (i.e. $U^{\otimes n}|\psi \rangle = |\psi \rangle$). So its is not possible to create a representation by covariant states.

However, if you are asking about a general way to tackle singlet states, there is for example a paper introducing a (non-orthohonal) basis (but with nice properties) for qubit singlet states (D. Lyons, S. Walck, PRA 2008, arXiv:0808.2989). Their reasoning can be generalized to qudit singlet states (I'm writing a paper on it, feel free to ask more).

This post has been migrated from (A51.SE)
answered Sep 18, 2011 by Piotr Migdal (1,255 points) [ no revision ]
+ 4 like - 0 dislike

It seems that you are asking if the fundamental theorem of algebra is true. A symmetric $n$ qubit state can be seen as a homogeneous polynomial of degree $n$ in two variables. The factorization corresponds to the roots of the dehomogenized version of this polynomial.

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answered Oct 13, 2011 by Abdelmalek Abdessela (430 points) [ no revision ]

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