# Why can any pure bipartite state be written in the Schmidt decomposition as $|\Psi\rangle=\cos\theta|00\rangle+\sin\theta|11\rangle$?

The beginning of this paper (pg.no. 1) on generalised Schmidt decomposition of three qubit states mentions the following:

The Schmidt decomposition allows one to write any pure state of a bipartitie system as a linear combination of biorthogonal product states or, equivalently, of a non-superfluous set of product states built from local bases. For two quantum-bits (qubits) it reads $$\tag{1} |\Psi \rangle = \cos \theta|00\rangle + \sin \theta|11 \rangle \ , \ 0 \leq \theta \leq \pi/4.$$

I have been trying to prove this, But to no result. Need some help.

I think this phrasing is a bit misleading if isolated. If this was true as written for any state $$\psi$$, then the tensor product of two qubits would have dimension $$2$$ instead of $$4$$. But if you look at the next sentence it is clear what the author actually means

Here $$|ii〉 ≡ |i〉_A⊗|i〉_B$$, both local bases $$\{|i〉\}_{A,B}$$ depend on the state $$|\Psi\rangle$$

i.e. here $$|0\rangle$$ and $$|1\rangle$$ are two orthogonal states that depend on $$|\Psi\rangle$$. In general if $$H_A$$ and $$H_B$$ are two Hilbert spaces , and $$|v\rangle\in H_A\otimes H_B$$, you can find $$|\psi_1\rangle\dots|\psi_n\rangle$$ and $$|\phi_1\rangle\dots|\phi_n\rangle$$ bases of $$H_A$$ and $$H_B$$ such that

$$|v\rangle=\sum_k a_k |\psi_k\rangle|\phi_k\rangle$$

for some coefficients $$a_k$$. This is the Schmidt decomposition and is easy to prove by writing

$$|v\rangle=\sum_{ij} b_{ij} |\eta_i\rangle|\gamma_j\rangle$$ for some orthonormal bases $$|\eta_i\rangle$$ of $$H_A$$ and $$|\gamma_i\rangle$$ of $$H_B$$, and taking a singular value decomposition of $$b_{ij}$$. Notice though that the two bases $$|\psi_i\rangle$$ and $$|\phi_i\rangle$$ depend on $$|v\rangle$$. For two qubits, calling $$|\psi_i\rangle=|i\rangle_A$$ and $$|\phi_i\rangle=|i\rangle_B$$, we get

$$|v\rangle=a_0|0\rangle_A|0\rangle_B + a_1 |1\rangle_A|1\rangle_B$$

by normalization constraints we can write $$a_0=\cos\theta$$ and $$a_1=\sin\theta$$, and any relative phase can just be absorbed in the definition of $$|0\rangle$$ and $$|1\rangle$$.

• This makes a lot of sense! Thanks! – user07 Jun 3 '20 at 10:16