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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.

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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 $|1\rangle$.

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

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