# Prove that there are infinitely many entanglement classes under LU

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.

• 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. Jan 5 at 18:08
• Okay thank you, I'll edit it for a moment Jan 5 at 18:10

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 $$U\otimes V$$.
• 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? Jan 6 at 1:39
• Because on $[0,\frac{\pi}{2}]$ cosine (and sine) is an invertible function. Jan 6 at 1:42
• 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. Jan 6 at 1:45
• 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) Jan 6 at 2:43
• 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 $a\ne b$. Jan 6 at 3:43