# 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

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.
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).