Prove that there are infinitely many entanglement classes under LU

Question

Dur, 2000 states that

(...)But even in the simplest systems, $|\psi\rangle$ and $|\phi\rangle$ are typically not related by LU, and continuous parameters are needed to label all equivalence classes.

I've found some similar explanation in Ritz,2018

(...)The Schmidt decomposition of a two-qubit state has only one free parameter $$|\psi\rangle = \sqrt{\lambda_{0}}|00\rangle+\sqrt{\lambda_{1}}|11\rangle \quad ;\lambda_{0}+\lambda_{1}=1 \qquad\tag{2.85}$$ Thus, we can rewrite eq. (2.85) in terms of new parameter $\theta$ as $$|\psi\rangle = \cos \theta |00\rangle+\sin \theta|11\rangle \qquad \qquad \qquad \qquad \quad\tag{2.86}$$ Therefore, any two-qubit state can, under LU, be transformed to Eq. (2.86). Obviously there is still one continuous parameter, i.e. $θ$, left. Hence, even for the lowest possible dimension and particles, the number of equivalence classes under LU is infinite.

I couldn't understand, why any abritary two-qubit states under LU can be transformed to eq. (2.86)? If it's LU transformation, I think we should start from the general form of Unitary matrices itself, but it seems that we choose convinient transformation such that $\lambda_0 = \cos^2 \theta$ and $\lambda_1 = \sin^2 \theta$. If that so, why does continuous parameter make the number of classes infinte? I feel kinda clueless here.

Answer 1

TL;DR: This is an application of Schmidt decomposition followed by local basis change on each qubit.

By Schmidt decomposition any two-qubit state $|\psi\rangle$ may be written in the form $$ |\psi\rangle = \lambda|s\rangle|u\rangle + \kappa|t\rangle|v\rangle\tag1 $$ where $\lambda$ and $\kappa$ are non-negative real numbers such that $\lambda^2+\kappa^2=1$, the states $|s\rangle, |t\rangle$ are an orthonormal basis for the first qubit and the states $|u\rangle, |v\rangle$ are an orthonormal basis for the second qubit. This follows from the Singular Value Decomposition. See for example section $2.5$ on page $109$ in Nielsen & Chuang for more details.

Now, define single-qubit unitaries that send the basis $|s\rangle, |t\rangle$ (respectively, $|u\rangle, |v\rangle$) to the computational basis $$ \begin{align} U &= |0\rangle\langle s|+|1\rangle\langle t|\\ V &= |0\rangle\langle u|+|1\rangle\langle v| \end{align}\tag2 $$ and calculate $$ \begin{align} (U\otimes V)|\psi\rangle &= \lambda U|s\rangle V|u\rangle + \kappa U|t\rangle V|v\rangle\\ &= \lambda|00\rangle+\kappa|11\rangle.\tag3 \end{align} $$ Finally, by trigonometry, there is a unique $\theta\in[0,\frac{\pi}{2}]$ such that $$ \lambda=\cos\theta\quad \kappa=\sin\theta.\tag4 $$ Putting it all together we see that $|\psi\rangle$ is equivalent to $|\psi'\rangle:=\cos\theta\,|00\rangle+\sin\theta\,|11\rangle$ under local unitaries $U\otimes V$.