# Entangled vectors in hilbert space

We consider a system of two particles of spin $\frac{1}{2}$, each described by the two-dimensional one-particle Hilbert space $\mathcal{H}$. Let $|\pm\rangle\in\mathcal{H}$ denote the eigenvectors of the spin operator $\hat{S}_3$.

I want to show, that the vectors \begin{align} \varphi^\pm&=\frac{1}{\sqrt{2}}(|+\rangle\otimes|+\rangle\pm|-\rangle\otimes|-\rangle)\\ \psi^\pm&=\frac{1}{\sqrt{2}}(|+\rangle\otimes|-\rangle\pm|-\rangle\otimes|+\rangle)\end{align} form an orthonormal basis of maximally entangled vectors (Bell basis) in $\mathcal{H}\otimes\mathcal{H}$.

Well, it is very easy to show, that the vectors are orthonormal and form a basis in the Hilbert space. I have problems to proof, that the vectors are maximally entangled.

I know that a pure state is called maximally entangled, if the eigenvalues of the reduced density matrix are equal. So should I calculate the density matrix $\varrho=|\varphi^\pm\rangle\langle\varphi^\pm|$. Can I really assume, that the vectors are pure states?

• What's a pure state, if not a single vector? Also, the two Bell vectors are not a basis of the full Hilbert space. – ACuriousMind Oct 29 '14 at 13:28
• @ACuriousMind Perhaps you missed the fact that the vectors are four! They are a orthonormal basis of ${\mathbb C}^2 \otimes {\mathbb C}^2$ since they are four, pairwise orthogonal and normalised. – Valter Moretti Oct 29 '14 at 13:32
• @ValterMoretti: Oh, you're right, of course. I'd still ask what the OP's definition of a pure state is, though. – ACuriousMind Oct 29 '14 at 13:33
• If he's expressing the state as vectors then they're pure states. – QuantumBrick Oct 29 '14 at 13:39

As @ACuriousMind pointed out in his comment, the definition of a pure state is one that is a vector of $\mathcal{H}$, in this case $\Bbb{C}^2\otimes\Bbb{C}^2$. If you take for example $\rho=|\varphi^+\rangle\langle\varphi^+|$, the reduced density matrix on the first space is \begin{align}\rho_1&=\operatorname{Tr}_2\rho\\ &=(I\otimes\langle+|)\rho(I\otimes|+\rangle)+(I\otimes\langle-|)\rho(I\otimes|-\rangle)\\ &=(I\otimes\langle+|)|\varphi^+\rangle\langle\varphi^+|(I\otimes|+\rangle)+(I\otimes\langle-|)|\varphi^+\rangle\langle\varphi^+|(I\otimes|-\rangle)\\ &=\frac{1}{2}|+\rangle\langle+|+\frac{1}{2}|-\rangle\langle-|\\ &=\frac{I}2\end{align} So, $\rho_1$ has equals eigenvalues $1/2$. If you do similar calculations for the other vectors or the second spce you will get the same result, which shows that these are maximally entangled.