# Why schmidt decomposition only holds for two component composite systems?

According to schmidt decomposition any pure state belonging to a composite system $AB$ can be written as $|\psi\rangle = \sum_i \lambda_i |i_A\rangle |i_B\rangle$ where $\lambda_i$ are non negative real numbers and $|i_A\rangle |$ and $|i_B\rangle$ are orthonormal basis for system $A$ and $B$ respectively. But in an exercise ( exercise 2.77 ) of Nielsen and Chuang it asks to show an example for a composite system $ABC$ where the pure state belonging to it cannot be written as $|\psi\rangle = \sum_i \lambda_i |i_A\rangle |i_B\rangle |i_C\rangle$. If I am not wrong then $\frac{1}{\sqrt{2}}(|000\rangle+|011\rangle)$ is one such example. But is there any physical significance behind it that schmidt decomposition holds for two component composite system only or is it just a mathematical result ? And is the absence of schmidt decomposition for higher component composite systems related to concept of entanglement ?

• The state $|0\rangle|0\rangle|+\rangle$ is patently in that form. You also presumably mean Nielsen instead of Neilson or Neilsen. – Emilio Pisanty May 9 '15 at 12:43
• @EmilioPisanty yeah. can't think of an example that fits it. will find one and edit the answer to include it. – sashas May 9 '15 at 12:45
• @EmilioPisanty I corrected the typo and included a different example I hope it is correct . – sashas May 9 '15 at 12:48
• Note the typo is still present (it's Nielsen, not Neilsen). Your new example is hardly new - it's exactly the same up to relabelling of the qubits. – Emilio Pisanty May 9 '15 at 12:50
• @EmilioPisanty sorry about that I corrected the typo will think and put up an example that satisfies the constraints. – sashas May 9 '15 at 12:54

## 1 Answer

This is a mathematical result. The Schmidt decomposition tells you that there are bases for two parties $A$ and $B$ such that

$$\sum_{ij} \lambda_{ij} |i_{A}\rangle |j_{B}\rangle = \sum_k \nu_k |\tilde{k}_A\rangle |\tilde{k}_B\rangle$$

with some orthonormal bases $|i_A\rangle,|\tilde{i}_A\rangle, |i_B\rangle, |\tilde{i}_B\rangle$. If you compare the two sides and consider the fact that orthonormal bases are related by a unitary matrix, this will lead you to the singular value decomposition. This means that the Schmidt decomposition is a (rather trivial) corollary to the singular value decomposition.

Now, there is a mathematical result that tells you that this is not possible in higher dimensions (you can remedy this to some degree, as noted in the comments). Sadly, I don't know a nice and intuitive argument of why this is not the case (you could work with Lagrangian multiplies and see that it is not possible, though).

What are the physical consequences? Well, in a sense, this will have consequences almost everywhere where we use the Schmidt decomposition. One very striking example is pure state LOCC-interconvertibility. In other words: Let $|\psi\rangle$ and $|\phi\rangle$ be two bipartite pure states. Can we find a transformations with local operations and classical communication from $|\psi\rangle$ to $|\phi\rangle$? In the bipartite case, we can if and only if the Schmidt coefficients of $|\phi\rangle$ majorize those of $|\psi\rangle$. This was already proven in the last millenium (arXiv).

Having a perfunctory look at the proof, it seems to me that if we had a Schmidt decomposition for arbitrary multipartite systems, essentially the same proof should hold (feel free to confirm this suspicion). This would in particular imply that starting out from one state the "maximally entangled state", we could reach all others, which is known to be wrong for multipartite systems. In any case, if there are physical consequences, this is where I would expect them to be at the very least: State interconversion with LOCC.