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):

$$ \frac{1}{\sqrt{2}}|00\rangle + \frac{i}{\sqrt{2}}|01\rangle $$

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

$$ | 0\rangle \otimes \frac{|0 \rangle + i| 1\rangle }{\sqrt{2}}, $$

but what's the way to determine if the state is separable or entangled without guessing? It's stated in this question that 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.

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My attempt

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

$$ | \psi\rangle = | 0\rangle \otimes \frac{|0 \rangle + i| 1\rangle }{\sqrt{2}} = \frac{1}{\sqrt{2}} \begin{bmatrix} 1\\ i\\ 0\\ 0\end{bmatrix} $$ and thus $$ \rho = | \psi \rangle \langle \psi | = \begin{bmatrix} \frac{1}{2} & -\frac{i}{2} & 0 & 0\\ \frac{i}{2} & \frac{1}{2} & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{bmatrix} $$

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

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My attempt 2

I also tried to utilize a reduced density matrix:

$$ | \psi \rangle = \frac{1}{\sqrt{2}}|00\rangle + \frac{i}{\sqrt{2}}|01\rangle = \frac{1}{\sqrt{2}} ( | 0\rangle _A \otimes | 0\rangle _B + i(| 0\rangle _A \otimes | 1 \rangle _B )) $$ and thus \begin{align*} \rho_A &= {\rm tr}_B( |\psi \rangle \langle \psi | ) = \frac{1}{2} ( | 0 \rangle \langle 0| + i |0 \rangle \langle 1| ) = \frac{1}{2} \begin{bmatrix}1 & i\\ 0 & 0 \end{bmatrix}\\ {\rm tr}(\rho_A^2) &= \frac{1}{4} {\rm tr} \begin{bmatrix} 1 & i\\ 0 & 0 \end{bmatrix} = \frac{1}{4} \end{align*}

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.