2
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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

4
$\begingroup$

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 $U\otimes V$.

$\endgroup$
8
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.