Prove that $a|00\rangle+b|01\rangle+c|10\rangle+d|11\rangle$ is separable iff $ad-bc=0$

Question

I am having trouble rigorously proving that a two qubit state $|\psi\rangle=a|00\rangle+b|01\rangle+c|10\rangle+d|11\rangle$ is unentangled/separable if $ad-bc=0$

I currently have fairly little exposure to quantum information theory. I've heard 'Schmidt decomposition' mentioned as a possible solution. Alternately, this is problem 16b in this document which also gives the hint to apply a unitary matrix $$e^{i\phi}\begin{bmatrix} c& -a \\ a^* & c^* \end{bmatrix}$$ to one of the qubits. Any solution is appreciated but the preference is for one that uses less machinery, and in particular I'd be interested to see how to use the above hint to prove the result.

EDIT

To be clear, I know that separable $\implies ad-bc=0$ - this is not difficult to show. It is the other implication ($ad-bc=0\implies$ separable) that I am having trouble with.

Answer 1

Maybe this will help.

Suppose we have the state $|\psi \rangle = a|00\rangle+b|01\rangle+c|10\rangle+d|11\rangle = \begin{pmatrix} a \\ b \\ c \\d \end{pmatrix} $, then by definition, $|\psi \rangle$ is separable if and only if $|\psi\rangle = |\phi_1\rangle \otimes |\phi_2\rangle$ where $|\phi_1\rangle, |\phi_2\rangle \in \mathbb{C}^2$.

Now let $|\phi_1 \rangle = \begin{pmatrix} x \\ y \end{pmatrix} $ and $|\phi_2 \rangle = \begin{pmatrix} u \\ v \end{pmatrix} $ then we have that $$|\phi_1 \rangle \otimes |\phi_2 \rangle = \begin{pmatrix} xu\\ xv\\ yu\\yv \end{pmatrix} $$

this implies that if $|\psi \rangle = |\phi_1\rangle \otimes |\phi_2\rangle$ then $\begin{pmatrix} a \\ b \\ c \\d \end{pmatrix} = \begin{pmatrix} xu\\ xv\\ yu\\yv \end{pmatrix} $ then after a few algebra steps you should arrive at $ad = bc$ must be the neccesary condition for this to be true...

Answer 2

Here is a complete and elementary answer - the discussion in the comments was basically there.

Let $|\psi\rangle=a|00\rangle + b|01\rangle + c|10\rangle + d|11\rangle$ be given such that $ad-bc=0$. Additionally, we have $\| \psi\|^2=|a|^2+|b|^2+|c|^2+|d|^2=1$.

Now the latter condition implies that $a,b,c,d$ are not all zero. So WLOG assume $a\neq 0$.

Now we have a decomposition $$|\psi\rangle = |v\rangle\otimes |w\rangle= \left(a|0\rangle + c|1\rangle\right)\otimes\left(|0\rangle +\frac{b}{a}|1\rangle\right)=a|00\rangle+b|01\rangle+c|10\rangle+\frac{cb}{a}|11\rangle$$ where $\frac{cb}{a}=d$.

In this decomposition, $\|v\|^2=|a|^2+|c|^2$, $\|w\|^2=1+\left|\frac{b}{a} \right|^2$. So the vectors $v,w$ are not normalised in general. Hence introduce a rescaling parameter, say $\lambda\in \mathbb{R}_{>0}$, such that $$|\psi\rangle=\lambda |v\rangle \otimes \frac{1}{\lambda}|w\rangle =\left(\lambda a|0\rangle + \lambda c|1\rangle\right)\otimes\left(\frac{1}{\lambda}|0\rangle +\frac{b}{\lambda a}|1\rangle\right) .$$

Now note that \begin{align*} \|\lambda v\|^2 \|\frac{1}{\lambda }w\|^2 &=\|v\|^2\|w\|^2\\ &=(|a|^2+|c|^2)\left(1+\left|\frac{b}{a} \right|^2\right)\\ &=|a|^2+|b|^2+|c|^2+\left|\frac{cb}{a}\right|^2\\ &=|a|^2+|b|^2+|c|^2+|d|^2=1. \end{align*} So if we pick $\lambda=\frac{1}{\|v\|}=\frac{1}{\sqrt{|a|^2+|c|^2}}$, then $\|\lambda v\|=1$, and the above also gives $\|\frac{1}{\lambda}w\|=1$. This now gives a valid decomposition and shows $|\psi\rangle$ is separable.

(At a higher level, what we used in this last renormalisation step is actually the fact that for tensor products of Hilbert spaces, we have $\langle v_1\otimes w_1| v_2\otimes w_2\rangle_{H\otimes K} =\langle v_1|v_2\rangle_H \cdot \langle w_1|w_2\rangle_K$, where $v_1,v_2 \in H$, $w_1,w_2\in K$. In our case this gave $\|\psi\|^2=\|v\|^2\|w\|^2$. )

Analogous decompositions hold in the case that we assume $b,c$ or $d$ nonzero instead.

Answer 3

An easy way to see it is to notice that $ad-bc$ is the determinant of the matrix of coefficients of the state. This matrix is singular, and thus has rank $1$, if and only if it has zero determinant.

But the rank of this matrix is also equal to the number of nonzero coefficients in the Schmidt decomposition of the state.

It follows that the state has the form $|\psi\rangle=|u\rangle|v\rangle$ for some states $|u\rangle$ and $|v\rangle$ if and only if $ad-bc=0$. A pure two-qubit state is separable if and only it is a product state, hence the conclusion.