I saw answers such as this and this which provide examples of tripartite system that don't take a Schmidt decomposition, but I wonder if there's an explicit condition that can tell whether a state is or is not Schmidt decomposable. Does anyone have an idea? I know a bipartite sytem always have one, so a tripartite's condition would be enough. If you can tell me about the multipartite system it'd be even better! Thank you!

Edit: After reading the comment, I understand that tripartite system also always have Schmidt decomposition, but for each specific biparition with different Schmidt coefficients. I think my question is that what is the condition that the entire state can be written as i.e. $\alpha|000\rangle+\beta|111\rangle$? Is it neccesarily true that if those Schmidt coefficients of the 3 possible bipartitions are the same, then the general state will admit a Schmidt decomposition like that? Is there a proof?

  • $\begingroup$ The given answer in the second question you link cites this paper and notes that it gives both necessary and sufficient conditions. $\endgroup$ – Rammus Feb 19 at 14:38
  • $\begingroup$ You got me there lol. The problem is that I don't really understand it, so I think I can ask it here so someone could give me an intuitive answer :D $\endgroup$ – Kim Dong Feb 19 at 14:46
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    $\begingroup$ I wouldn't say that those states "don't take a Schmidt decomposition". Any state admits a Schmidt decomposition, it's just that such decomposition is defined with respect to a specific bipartition, of which that are many for multipartite systems. Regardless, what is the exact requirement? In a schmidt decomposition, $|\psi\rangle=\sum_I\lambda_i|i_A\rangle|i_B\rangle$, you also require $\lambda_i\ge0$ and $\langle i_A|j_A\rangle=\langle i_B|j_B\rangle=\delta_{ij}$. Are you requiring the same here? $\endgroup$ – glS Feb 20 at 16:42
  • $\begingroup$ I understand. So for each bipartition we can have a Schmidt decomposition, but they may admit different coefficients. I think my question is that what is the condition that the entire state can be written as i.e. $\alpha |000> + \beta |111>$? Is it neccesarily true that if those Schmidt coefficients of the 3 possible bipartitions are the same, then the general state will admit a Schmidt decomposition like that? Is there a proof? $\endgroup$ – Kim Dong Feb 20 at 17:33
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    $\begingroup$ @KimDong well I guess it depends on how you choose to define the "Schmidt decomposition" in the multipartite case. For bipartites states it's essentially a singular value decomposition of the state, so I'd say getting real positive coefficients would be a defining property. Then again that's not necessarily what would be looking for so it depends. This paper arxiv.org/abs/quant-ph/0006125 discusses a multipartite version of Schmidt decomposition and seems to also require the positivity property $\endgroup$ – glS Feb 22 at 10:34

Generalization of Schmidt decomposition

The Schmidt decomposition $|\psi_{AB}\rangle = \sum_i\lambda_i|i_A\rangle|i_B\rangle$, with $\lambda_i$ positive real numbers and $|i_A\rangle$ and $|i_B\rangle$ (possibly incomplete) orthonormal bases, of a bipartite state $|\psi_{AB}\rangle$ can be thought of as a type of singular-value decomposition. Specifically, if the amplitudes $\psi_{ij} = \langle i|\langle j|\psi_{AB}\rangle$ are arranged into a $2\times 2$ matrix rather than the usual $4$-vector, then $\lambda_i$ are precisely the non-zero singular values of the matrix $\psi_{ij}$. This explains why $\lambda_i$ are positive real numbers. See the proof of theorem 2.7 on page 109 in Nielsen & Chuang for more details.

The decomposition can be generalized to $n$-partite state as

$$ |\psi_{A_1A_2 \dots A_n}\rangle = \sum_i \lambda_i |i_{A_1}\rangle |i_{A_2}\rangle\dots |i_{A_n}\rangle\tag1 $$

where as before $\lambda_i$ are positive real numbers and $|i_{A_k}\rangle$ are (possibly incomplete) orthonormal bases. If $n=2$, then decomposition $(1)$ exists for every pure state $|\psi_{A_1A_2}\rangle$. If $n>2$, then there are pure states without decomposition $(1)$.

Physical characterization

Physically, the $n$-partite states that admit decomposition $(1)$ can be characterized as the states that are free of $m$-partite entanglement for every $1<m<n$. More precisely, $|\psi_{A_1A_2 \dots A_n}\rangle$ admits decomposition $(1)$ if and only if for every $k = 1,\dots n-1$ and subsystems $A_{i_1}\dots A_{i_k}$, the state

$$ \rho_{A_{i_1}\dots A_{i_k}} = \mathrm{tr}_{A_{i_1}\dots A_{i_k}}\left(|\psi_{A_1A_2 \dots A_n}\rangle\langle\psi_{A_1A_2 \dots A_n}|\right) $$

is separable. One way of thinking about these states is that all (if any) entanglement that they have lives between all $n$ subsystems. Conversely, states that do not admit $(1)$ necessarily contain some entanglement between fewer than $n$ subsystems.

This characterization is not very useful to computationally check whether a given $n$-partite state admits decomposition $(1)$. Nevertheless, it helps to intuitively understand the class of Schmidt-decomposable states. In particular, it explains why a generic $n$-partite state for $n>2$ does not admit $(1)$. It also explains why all bipartite states admit $(1)$ $-$ it is impossible to entangle fewer than $2$ subsystems.

Necessary and sufficient conditions in the tripartite case

The necessary and sufficient conditions for a state to admit decomposition $(1)$ are given in the paper cited in a comment above by @Rammus. We reproduce the results of the paper in the case when the Schmidt decomposition

$$ |\psi_{ABC}\rangle = \sum_i \lambda_i |i_A\rangle|i_{BC}\rangle\tag2 $$

of a tripartite state $|\psi_{ABC}\rangle$ associated with the partitioning of $ABC$ into $A$ and $BC$ has distinct coefficients, i.e. $\lambda_i \ne \lambda_j$ for $i \ne j$. Define the matrices $\Omega_i$ as

$$ \Omega_{i,jk} = \langle j_B|\langle k_C|i_{BC}\rangle. $$

In other words, $\Omega_i$ is the matrix of amplitudes of the state $|i_{BC}\rangle$. The necessary conditions for $|\psi_{ABC}\rangle$ to admit decomposition $(1)$ is for all $\Omega_i$ to be rank one and $\Omega_i^\dagger\Omega_{i'} = 0$ and $\Omega_i\Omega_{i'}^\dagger = 0$ for $i\ne i'$.

If $\lambda_i$ are not all distinct, then the necessary and sufficient conditions become more complicated, because $(2)$ and thus $\Omega_i$ are no longer unique. See the paper for more details.

Example: $W$ state

The $W$ state is

$$ |W\rangle = \frac{|001\rangle + |010\rangle + |100\rangle}{\sqrt{3}} = \frac{\sqrt{2}}{\sqrt{3}}|0\rangle\otimes\frac{|01\rangle + |10\rangle}{\sqrt{2}} + \frac{1}{\sqrt{3}}|1\rangle\otimes |00\rangle. $$

Thus, $\lambda_0=\frac{\sqrt{2}}{\sqrt{3}}$, $\lambda_1 = \frac{1}{\sqrt{3}}$, $|0_{BC}\rangle = (|01\rangle+|10\rangle)/\sqrt{2}$ and $|1_{BC}\rangle = |00\rangle$ so

$$ \Omega_0 = \frac{1}{\sqrt{2}}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \quad \Omega_1 = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} $$

and we see that $\Omega_1$ is rank one, but $\Omega_0$ is not. Therefore, $W$ does not admit decomposition $(1)$.

  • $\begingroup$ is this approach analogous to what is done in the context of tensor networks, where (I think) you recursively do Schmidt decompositions on the different bipartitions? It seems the iff conditions you mention here is essentially saying that you can always reach the "full Schmidt decomposition" (when one exists) by starting from decomposing on any bipartition, which is interesting $\endgroup$ – glS Feb 23 at 9:17
  • $\begingroup$ There is a difference (at least comparing to the use of Schmidt decomposition in the context of tensor networks that I'm aware of). Specifically, in the context of TNs you start with a node $N_1$, use Schmidt/SVD to split it into $M_1D_1N_2$ where $D$ is a diagonal matrix, i.e. a two-legged node, often drawn as a diamond. Then you do this again to $N_2$ and you end up wit $M_1D_1M_2D_2N_3$, then again to $N_3$ and so on. You wind up with a chain of nodes connected by diagonal nodes where each diagonal node is a matrix, i.e. has two legs. $\endgroup$ – Adam Zalcman Feb 23 at 18:51
  • $\begingroup$ On the other hand, if we represent the generalized Schmidt decomposition above as a tensor network transformation, the first step would also start from $N_1$ and end as $M_1DN_2$, but then the second step exploits the special properties of $N_2$ (in particular, that it is rank one) to add a leg to the diagonal node $D$! So now we end up with a network that one may write as $D(M_1, M_2, N_3)$ where I used parenthesis to highlight the fact that the three nodes are connected to $D$ (and they are not connected to each other which signifies lack of $m$-partite entanglement for $m<n$). $\endgroup$ – Adam Zalcman Feb 23 at 18:53
  • $\begingroup$ Continuing, we end up in $D(M_1, M_2, \dots, M_n)$ which is a star, not a chain as before. Also, our sole diagonal node has $n$ legs whereas before we had $n-1$ diagonal nodes each with two legs. Two-legs diamond can always be extracted by Schmidt decomposition. On the other hand, it is not always possible to extract multi-legged diamonds because the generalized Schmidt decomposition is not always possible. $\endgroup$ – Adam Zalcman Feb 23 at 18:54

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