I have two states, $|\psi\rangle$ and $|\phi\rangle$. I have in mind that they live on a length $L$ spin chain with finite local Hilbert space dimension.

I know that for every Schmidt decomposition bipartitioning the system into a region $A$ and a region $B$, the following is true:

$$|\psi\rangle = \sum_{i=1}^n \lambda_i|a_i\rangle|b_i\rangle $$ $$|\phi\rangle = \sum_{i=1}^n \lambda_i|a'_i\rangle|b'_i\rangle $$

That is, while the states are not necessarily equal, their Schmidt coefficients are equal across every cut in physical space. The Schmidt coefficients may depend on the choice of cut. I imagine that all of my cuts are in physical space, but I do allow regions $A$ and $B$ to contain sites that are not contiguous; for example, $A$ could contain all even sites and $B$ can contain all odd sites.

Given this, am I guaranteed that there exists a unitary that is a tensor product of single-site unitaries, $U = \otimes_{i=1}^L U_i$, such that $ U|\psi\rangle = |\phi\rangle$?

Here are my thoughts. If I had the weaker statement that the Schmidt coefficients were equal $$|\psi\rangle = \sum_{i=1}^n \lambda_i|a_i\rangle|b_i\rangle $$ $$|\phi\rangle = \sum_{i=1}^n \lambda_i|a'_i\rangle|b'_i\rangle $$ for some specific regions $A$ and $B$, then I know I can make a unitary $U = U_A \otimes U_B$ with $U_A = \sum_i |a'_i\rangle \langle a_i|$ and $U_B = \sum_i |b'_i\rangle \langle b_i|$ that takes $|\psi\rangle$ to $|\phi\rangle$.

However, it's less clear to me how to use the information from all of the bipartite cuts together. I was thinking I could consider all $L$ cuts where $A$ contains a single site, and then attempt to argue that $U$ can be written as a product of single-site operators, but I'm not sure that will work. In particular, I'm getting suspicious that perhaps all of the bipartite cuts aren't enough, and that I'll need to know things about multipartite decompositions of the state.


1 Answer 1


Your claim already fails in the simplest non-trivial case, namely for three qubits.

This follows from a (the?) classic result in the theory of multipartite entanglement -- Three qubits can be entangled in two inequivalent ways by Dür, Vidal, and Cirac.

They show that for three qubits, there are two classes of genuine tripartite entangled states, the W class and the GHZ class. Given states $\lvert \phi_\mathrm{W}\rangle$ and $\lvert \phi_\mathrm{GHZ}\rangle$, from the two classes, it is impossible to convert between them using SLOCC, i.e., it is impossible to write $$ \lvert\phi_\mathrm{W}\rangle\propto (A\otimes B\otimes C)\lvert\phi_\mathrm{GHZ}\rangle $$ (or vice versa) for any $A$, $B$, and $C$. In particular, this rules out the possibility of unitaries doing the job, as you are asking.

What remains is to show that there are states $\lvert\phi_\mathrm{W}\rangle$ and $\lvert \phi_\mathrm{GHZ}\rangle$ with identical Schmidt coefficients in every bipartition. This is equivalent to demanding that their single-qubit reduced states have the same spectra.

Using the results of the paper of Dür, Vidal, and Cirac, it is now easy to find such states (and indeed, they really should exist -- it would be rather disappointing if the two classes were distinguished by the spectra of their reduced density matrices). For instance, you can choose $$ \lvert \phi_\mathrm{W}\rangle = \sqrt{\gamma}\lvert 001\rangle + \sqrt{\gamma}\lvert 010\rangle + \sqrt{\gamma}\lvert 100\rangle + \sqrt{1-3\gamma}\lvert 000\rangle $$ (cf. Eq. (20) in the paper), and $$ \lvert \phi_\mathrm{GHZ}\rangle \propto \lvert 0\rangle \lvert 0\rangle \lvert 0\rangle + \lvert \theta\rangle\lvert \theta\rangle\lvert \theta\rangle $$ (cf. Eq. (15)), with $\lvert\theta\rangle=\sqrt{\mu}\lvert 0\rangle + \sqrt{1-\mu}\lvert 1\rangle$. You can now easily check that for the spectra $(\lambda,1-\lambda)$ ($\lambda\le1/2$) of the single-qubit reduced states (which are all equal by symmetry)

  • for $\lvert\phi_\mathrm{W}\rangle$, all values $0<\lambda\le1/3$ can be obtained by varying $0<\gamma\le1/3$
  • for $\lvert\phi_\mathrm{GHZ}\rangle$, all values $0<\lambda\le1/2$ can be obtained by varying $0\le\mu< 1$.

Thus, you can easily find values $\mu$ and $\gamma$ where the reduced states have identical spectra, and thus the Schmidt spectra in all bipartitions are equal, yet the states cannnot be converted into each other by local unitaries (or even SLOCC).

  • $\begingroup$ I've accepted this because this perfectly answers my question, and the two different kinds of tripartite entanglement are very good to know! By any chance, is there any Hermitian $H$ such that the Schmidt coefficients of $e^{i H t} |\phi_W\rangle$ across any cut are independent of time, but such that there's a special time $T$ such that $e^{i H T} |\phi_W\rangle = |\phi_{GHZ}\rangle$, or is there an obstruction to that? I can also ask this as a separate question. $\endgroup$
    – user196574
    Jan 10 at 21:57
  • $\begingroup$ I'm not sure how to tackle this, but I'd try to look at the orbit of your evolution, and use that there is only a finite number of equivalence classes of tripartite states under LOCC (W, GHZ, two vs one, and product) following Dür et al.. The orbit of your evolution would always have to lie in one of these classes, probably this can be used to learn something about what is possible and impossible. Or one could try to think about the leading-order expansion of the evolution and interpret this as some SLOCC operation. $\endgroup$ Jan 10 at 22:02
  • $\begingroup$ ... another option would be to first characterize Hamiltonians whose evolution does not change the Schmidt coefficients when acting either on the W or the GHZ representative; this might be rather restrictive. In the end, these are just 8x8 matrices. $\endgroup$ Jan 10 at 22:05
  • $\begingroup$ Thanks! The evolution above was for general Hermitian $H$, so it might be possible that its expansion can't be viewed as LOCC. As a related aside, I'm trying to understand a bit better families of states that are equivalent under local unitaries, and if there might be a way to evolve strictly within a set of states under a time-independent Hamiltonian. I've written this as a question physics.stackexchange.com/questions/796857/… for a set of states satisfying a couple special properties. $\endgroup$
    – user196574
    Jan 11 at 0:50
  • $\begingroup$ On a second thought, since only W and GHZ have nontrivial Schmidt decomp. in all cuts, your orbit must always be in the W/GHZ family. This might help to understand if/how this is possible. $\endgroup$ Jan 11 at 8:33

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