It is relatively simple to derive the Schmidt decomposition of a pure state $|{\psi}\rangle \in H_A \otimes H_B$ with the SVD decomposition theorem. There are plenty of examples (lecture notes, books, videos, etc.) on the subject.

The question I have is clearly stated in the title. Is it possible to derive a Schmidt decomposition for a mixed state? Such a state is expressed as a set of pure states $|{\psi_i}\rangle$ with probability distribution $p_i$, for $1 \leq i \leq t$.

  • $\begingroup$ Can you explain your reasoning as to why you are wondering that Schmidt decomposition for mixed state might not be possible? $\endgroup$
    – FDGod
    Commented Feb 27 at 3:32
  • $\begingroup$ What do you mean by "Schmidt decomposition" in the case of a mixed state -- which property would have to be satisfied, what would be the structure of the decomposition, etc.? It is entirely non-obvious to me why there would be an obvious way to do this. $\endgroup$ Commented Feb 27 at 22:13

1 Answer 1


Usually, the Schmidt rank of a mixed state $\rho$ is defined as the minimum of the maximal Schmidt rank of an element in any decomposition of $\rho$ into a mixture of pure states (there are many different decompositions if $\rho$ is not pure), see B. Terhal and P. Horodecki, A Schmidt number for density matrices.

In general, it's very hard to compute it. The separability problem is equivalent to asking if $\rho$ has Schmidt rank 1 or higher. And it's known that this problem is NP-hard.


A second meaning when we talk about Schmidt decomposition of mixed states is the following. We can consider a mixed state $\rho$ as a vector in the space of matrices $\mathcal{L}(H_A) \otimes \mathcal{L}(H_B)$. And thus we can write down a so-called operator-Schmidt decomposition $$ \rho = \sum_i \gamma_i A_i \otimes B_i, $$ where $\gamma_i \ge 0$ and $\{A_i\}$, $\{B_i\}$ are ONBs in $\mathcal{L}(H_A)$, $\mathcal{L}(H_B)$ respectively.

In general, matrices $\{A_i\}$ and $\{B_i\}$ won't be quantum states. Nevertheless, this decomposition can be useful in recognizing the separability of $\rho$ (see N. Johnston, What the Operator-Schmidt Decomposition Tells Us About Entanglement), as was pointed out by John Watrous in the comments.

Note that this decomposition also defines what is called an operator-Schmidt rank of $\rho$. This is a different number than Schmidt rank (or Schmidt number) of $\rho$ defined earlier.

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    $\begingroup$ I knew this fact, but I forgot about it. But I did not answer my question. Any NP-complete problem has a brute force algorithm to solve it. Since the density matrix of a bipartite mixed state in $H_A \otimes H_B$ can be expressed as $\sum_i p_i | \psi_i \rangle \langle \psi_i |$, if the $|{\psi_i}\rangle\langle{\psi_i}|$ forms a vector space of dimension $(\dim A \times \dim B)^2$, it should be possible to use the SVD theorem in this case to find a Schmidt decomposition. I continue my investigation. Thanks. $\endgroup$
    – JMark
    Commented Feb 27 at 12:53
  • $\begingroup$ State $\rho$ is itself a vector in the space of matrices $L(H_A) \otimes L(H_B)$, where $L(H_A)$ is the space of matrices on the first subsystem. You can write down a Schmidt decomposition $\rho = \sum_i c_i A_i \otimes B_i$ where $\{A_i\}$ form a ONB in $L(H_A)$ and similarly for $\{B_i\}$. The problem is $A_i$ and $B_i$ will be general matrices, not quantum states. Otherwise $\rho$ would be separable, which is not always true. $\endgroup$
    – Danylo Y
    Commented Feb 27 at 13:57
  • $\begingroup$ Interesting. Doesn't feel like a very well known concept (though indeed it seems a natural construction in the same spirit as entanglement of formation etc., given that the Schmidt rank is an entanglement monotone for pure states). Yet, I don't think people would call this a "Schmidt decomposition", as in the question. $\endgroup$ Commented Feb 28 at 10:39
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    $\begingroup$ This is called the operator Schmidt decomposition. Here's an interesting blog post about it by Nathaniel Johnston: njohnston.ca/2014/06/…. Those that are interested may be able to pick up the trail from there. It's also been considered for unitary operators (Michael Nielsen refers to it as the Schmidt number of a unitary in his thesis for instance). $\endgroup$ Commented Feb 28 at 14:31
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    $\begingroup$ I think I did not express myself clearly: My point was that if you have two different answers to the same question, it would make more sense to post them as two separate answers, and not in one single answer. $\endgroup$ Commented Feb 28 at 19:33

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