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.

  • $\begingroup$ Please do not use images for text and equations. Images cannot be searched and copied and often don't render in a way consistent with text. $\endgroup$ Jan 5, 2023 at 18:08
  • $\begingroup$ Okay thank you, I'll edit it for a moment $\endgroup$
    – Steve J.
    Jan 5, 2023 at 18:10

1 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 $U\otimes V$.

  • $\begingroup$ Why is the $\theta$ unique when we choose $\lambda = \cos \theta$ and $\kappa = \sin \theta$? Moreover, it's true that abritary state can be equivalent to $\cos \theta |00\rangle+\sin \theta |11\rangle+$, but how this form yields an interpretation that the number of equivalence classes is infinite under LOCC? $\endgroup$
    – Steve J.
    Jan 6, 2023 at 1:39
  • $\begingroup$ Because on $[0,\frac{\pi}{2}]$ cosine (and sine) is an invertible function. $\endgroup$ Jan 6, 2023 at 1:42
  • $\begingroup$ Assuming you mean "under LU" not "under LOCC". The proof shows that $\theta$ is preserved under LU. Therefore, there is a one-to-one correspondence between the numbers in $[0,\frac{\pi}{2}]$ and the equivalence classes under LU. $\endgroup$ Jan 6, 2023 at 1:45
  • $\begingroup$ Does it means that if there is two states, e.g. $\cos a |00\rangle + \sin a |00\rangle$ and $\cos b |00\rangle + \sin b |00\rangle$ such that $a,b \in [0,\pi/2] ; a \ne b$, these two states are already not equivalent under LU? If so, I want to prove it explicitly, is there any some clue or hint for how to proving it? (My apologies for my weird english) $\endgroup$
    – Steve J.
    Jan 6, 2023 at 2:43
  • 1
    $\begingroup$ To see the inequivalence of your states more explicitly, ignore the second qubit and let $\rho_1$ be the density matrix of the first qubit in the first state and $\rho_2$ the density matrix of the first qubit in the second state. Then $\det\rho_1=\frac14\sin^2(2a)$ and $\det\rho_2=\frac14\sin^2(2b)$. Now, if you act with any $U\otimes V$ on the two qubits then $\rho_1$ becomes $U\rho_1 U^\dagger$ which is a similarity transformation and hence cannot change the determinant. Thus, no product unitary can turn the first state into the second if $a\ne b$. $\endgroup$ Jan 6, 2023 at 3:43

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