How do I determine if a given pure two-qubit state is separable?

Question

I'm trying to self-study some topics about quantum computing and I came across a topic of state separability. Talking about that, I wanted to determine separability on the following state (from Qiskit Textbook):

$$ \frac{1}{\sqrt{2}}\left|00\right> + \frac{i}{\sqrt{2}}\left|01\right> $$

I know, that it is separable, as it can be rewritten as

$$ \left| 0\right> \otimes \frac{\left|0 \right> + i\left| 1\right>}{\sqrt{2}}, $$

but what's the way to determine if the state is separable or entangled without guessing?

It's stated in Is there an algorithm for determining if a given vector is separable or entangled? , there are ways to do this and

I came also across Peres-Horodecki separability criterion, often called also PPT Criterion.

I must admit, that due to my little previous experience, I don't understand everything well, so I'd appreciate an explanation on the abovementioned state, preferably with algebraic steps written, as they're mostly missing in many explanations of this topic.

---

My attempt

As I understand it, the very first thing is to rewrite the state-vector as a density matrix $\rho$:

$$ \left| \psi\right> = \left| 0\right> \otimes \frac{\left|0 \right> + i\left| 1\right>}{\sqrt{2}} = \frac{1}{\sqrt{2}} \begin{bmatrix} 1\\ i\\ 0\\ 0\end{bmatrix} $$

$$ \rho = \left| \psi \right > \left< \psi \right| = \begin{bmatrix} \frac{1}{2} & -\frac{i}{2} & 0 & 0\\ \frac{i}{2} & \frac{1}{2} & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{bmatrix} $$

And now I should perform "partial transform" and find eigenvalues of the partially transformed matrix. But I'm really confused here about A and B "party", as I have only one state $\left| \psi \right>$. How am I supposed to proceed now?

---

My attempt 2

I also tried to utilize a reduced density matrix:

$$ \left| \psi \right> = \frac{1}{\sqrt{2}}\left|00\right> + \frac{i}{\sqrt{2}}\left|01\right> = \frac{1}{\sqrt{2}} \left( \left| 0\right>_A \otimes \left| 0\right>_B + i\left(\left| 0\right>_A \otimes \left| 1 \right>_B \right)\right) $$

$$ \rho_A = Tr_B \left( \left|\psi \right> \left<\psi \right| \right) = \frac{1}{2} \left( \left| 0\right> \left< 0\right| + i\left( \left|0\right>\left<1\right| \right) \right) = \frac{1}{2} \begin{bmatrix}1 & i\\ 0 & 0 \end{bmatrix} $$

$$ Tr(\rho_A^2) = \frac{1}{4} Tr\left( \begin{bmatrix} 1 & i\\ 0 & 0 \end{bmatrix} \right) = \frac{1}{4} $$

As we can see, my results seems to be $\frac{1}{4}$, while it should be 1 for a separable state. What am I doing wrong?

Answer 1

Your approach is correct, but you are taking the partial trace wrong:

$$\rho_A=\text{Tr}_B\big( \rho \big) = \sum_{i} \langle i_B | \rho |i_B \rangle = \langle 0_B | \rho |0_B \rangle + \langle 1_B | \rho |1_B \rangle$$ $$=\frac{1}{2}|0_A \rangle \langle 0_A | + \frac{1}{2}|0_A \rangle \langle 0_A | = |0_A \rangle \langle 0_A | $$

Clearly you can see now: $\rho_A^2=\rho_A$ and so the subsystem is pure, therefore the composite system is unentangled.

A really easy way to remember partial traces in the computational basis for 2-bit systems is the following:

This picture is not mine, it is from Prof. Michele Mosca's Lecture Notes at University of Waterloo; publicly available.