$\newcommand{\bra}[1]{\langle#1\rvert} % Bra \newcommand{\ket}[1]{\lvert#1\rangle} % Ket \newcommand{\qprod}[2]{ \langle #1 | #2 \rangle} %Inner Product \newcommand{\braopket}[3]{\langle #1 | #2 | #3\rangle} % Matrix Element \newcommand{\expect}[1]{ \langle #1 \rangle} % Expectation value$ I am working through the book Quantum Computing: A Gentle Introduction and I was working on problem 3.2. There are no solutions in the back of the book, so I wanted to double-check this one because I was unsure if I was correct or not. (I probably could do this for all of these questions, but I don't want to spam the board) The problem is:

Show by example that a linear combination of entangled states is not necessarily entangled.

I read this and thought that the only way a linear combination of entangled states would be not entangled is if they could be measured from a different basis. My thought was to find a linear combination of Bell states where the outcome is of the form $\ket{v} = a\ket{00} + 0\ket{11}$ which would be the standard basis state $\ket{00}$. That equation looks like this:

$\ket{\phi^+} + \ket{\phi^-} = \frac{2}{\sqrt{2}}\ket{00} + 0\ket{11}$

Is my thought process correct? If it's not, can you explain why it's wrong and what a correct answer to this question would look like?

  • $\begingroup$ Note that your final state is not normalized. The superposition should be $\frac{1}{\sqrt{2}}(|\phi^+\rangle+|\phi^-\rangle)$. $\endgroup$ Dec 25, 2021 at 9:30

2 Answers 2


Yes, that's correct, and your example is exactly what I had in mind after reading the question title. More generally, the four Bell states form a basis in the space of all 2-qubit states, so you can express any state using a linear combination of them, including all unentangled states.


That's an appropriate example, yes. The underlying reason this happens is that the set of entangled states is not convex, which roughly speaking means you sum entangled states and obtain (upon renormalisation) separable ones.

As a slight generalisation of the example you already provide, you can consider any pair of entangled states with Schmidt decompositions $$|\psi\rangle=\sum_{k=0}^s \sqrt{p_k}|u_k\rangle|v_k\rangle, \qquad |\phi\rangle=\sqrt{p_0}|u_0\rangle|v_0\rangle-\sum_{k=1}^s \sqrt{p_k}|u_k\rangle|v_k\rangle.$$ Then, $|\psi\rangle+|\phi\rangle$ will clearly always be a product state, even though both $|\psi\rangle$ and $|\phi\rangle$ are entangled (provided the Schmidt decompositions are nontrivial).

  • 1
    $\begingroup$ I appreciate the in-depth answer, but I'm not far enough in my studies to understand the generalization nor what it means for a set of entangled states to be not convex. D: The answer from Mariia makes sense to me as a beginner, so I will mark that as the answer in case another beginner stumbles across this post. $\endgroup$
    – Wadwamille
    Dec 24, 2021 at 19:25
  • $\begingroup$ Ah thanks this was the answer I was looking for. I'm also a beginner but I'm not reading that book, and just faced this question without being primed to think of doing it by adding Bell states. I was just thinking about it in terms of mathematical seperability, so this contrivance directly constructs unseperable things that add up to a seperable thing. $\endgroup$ Aug 19, 2022 at 16:17

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