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

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.

• Have you tried applying the unitary? What state do you get out afterwards? Jan 28, 2021 at 16:17
• I don't know if I'm applying it correctly, but for instance $(U\otimes I) \begin{bmatrix} a\\b\\c\\d\end{bmatrix}$ gives $(0,b c e^{i \phi }-a d e^{i \phi },a e^{i \phi } a^*+c e^{i \phi } c^*,b e^{i \phi } a^*+d e^{i \phi } c^*)$ where $U$ is the matrix. Of course the 2nd entry in this vector = 0 by assumption. Jan 28, 2021 at 16:56
• In general I don't think that matrix is unitary for the $a, b, c, d$ since to be unitary $|a|^2 + |c|^2=1$ which would imply that in the state $d=b=0$, the solution presented below is valid regardless; but the premise of your question is flawed (unless we are talking about a different $a \text{and} c$ which appears to be so looking at the exercise sheet) Jan 28, 2021 at 17:14
• Yes, I noticed that problem too - hence why I said I wasn't sure if I was applying the hint correctly Jan 29, 2021 at 3:10
• As a continuation for the analogous question of whether a state in $\mathbb{C}^m \otimes \mathbb{C}^n$ is separable, it is also determinant-like conditions. Like here your answer is cut out by the polynomial equation $ad-bc=0$ which is a single 2 by 2 determinant. You can see a general proof that all these Segre varieties are determinantal and what equations replace $ad-bc=0$. Feb 3, 2021 at 20:02

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...

• Downvoter left no comment, so I was wondering what their reason may have been. The answer is a nice simple argument that avoids the use of machinery like e.g. the determinant (which is what I'd use). I think their issue may have been with direction of implication. The question asks for a proof that $|\psi\rangle$ is separable if $ad=bc$, but the answer proves that $ad=bc$ if $|\psi\rangle$ is separable. Anyway, +1 to compensate since downvote is inappropriate here. Jan 29, 2021 at 2:51
• hi, thanks for the response. The trouble I am having is not showing that $ad=bc$ is a necessary condition for separability, but rather that it is a sufficient one. Am I correct in thinking your answer shows that it is necessary rather than sufficient? Jan 29, 2021 at 3:07
• $ad = bc$ is sufficient to allow you to rewrite the vector $|\psi \rangle = |\phi_1\rangle \otimes |\phi_2 \rangle$ since you can just consider $\begin{pmatrix}a \\ b \end{pmatrix} \otimes \begin{pmatrix} 1 \\ c/a \end{pmatrix}$ or $\begin{pmatrix} a \\ b \end{pmatrix} \otimes \begin{pmatrix} 1 \\ d/b \end{pmatrix}$ or $\begin{pmatrix} c \\ d \end{pmatrix} \otimes \begin{pmatrix} a /c\\ 1 \end{pmatrix}$ or $\begin{pmatrix} c \\ d \end{pmatrix} \otimes \begin{pmatrix} b /c\\ 1 \end{pmatrix}$ . Note that since $|a|^2 + |b|^2 + |c|^2 + |d|^2 = 1$ this implies that not all of them can be 0. Jan 29, 2021 at 3:34
• Very nice! I think there is a typo in the first coordinate of the last vector: $b/c$ should be $b/d$ :-) Jan 29, 2021 at 3:38
• @AdamZalcman Yes. It should be $b/d$. Thanks! Can't seem to edit comment though. Jan 29, 2021 at 3:41

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.

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.