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

$$\newcommand{\bra}[1]{\langle#1\rvert} % Bra \newcommand{\ket}[1]{\lvert#1\rangle} % Ket \newcommand{\qprod}[2]{ \langle #1 | #2 \rangle} %Inner Product \newcommand{\braopket}[3]{\langle #1 | #2 | #3\rangle} % Matrix Element \newcommand{\expect}[1]{ \langle #1 \rangle} % Expectation value$$ 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?

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

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).