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?

  • $\begingroup$ could you clarify what you find unclear about e.g. the first answer to the linked question? You compute the reduced density matrix and check whether it's pure. Or are you asking about how to perform one of these two steps? $\endgroup$
    – glS
    Commented Feb 5, 2021 at 20:10
  • $\begingroup$ @glS I'm mostly stuck on how to compute the reduced density matrix now. $\endgroup$
    – Eenoku
    Commented Feb 5, 2021 at 20:12
  • 1
    $\begingroup$ you might want to check out these related question on reduced density matrices: How to find the reduced density matrix of a four-qubit system? and What is the Reduced Density Matrix? $\endgroup$
    – glS
    Commented Feb 5, 2021 at 20:13
  • $\begingroup$ @glS Thank you, I've tried another time using a reduced density matrix - added into my question. $\endgroup$
    – Eenoku
    Commented Feb 5, 2021 at 20:39

1 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:

enter image description here

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.