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


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).