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Some similar questions have been ask before, but I still don't really get the definition of seperable states in quantum mechanics.

Consider a bell state of a two qubit system. \begin{align} \left|\Psi\right> = \frac{1}{\sqrt{2}}(\left|0 \right>_A\otimes \left|0\right>_B + \left|1 \right>_A\otimes \left|1\right>_B) \end{align} where the indices A and B denote the one qubit subsystems. I see, that this is not a product state and is therefore called an entangled state. So this means that the corresponding density matrix can not be seperable. So let's check this. The density matrix corresponding to the Bell state above is \begin{align} \rho &= \left|\Psi \right>\left<\Psi\right| \\ &= \frac{1}{2}(\left|0\right>_A\left<0\right|_A\otimes \left|0\right>_B\left<0\right|_B + \left|0\right>_A\left<1\right|_A\otimes \left|0\right>_B\left<1\right|_B + \left|1\right>_A\left<0\right|_A\otimes \left|1\right>_B\left<0\right|_B + \left|1\right>_A\left<1\right|_A\otimes \left|1\right>_B\left<1\right|_B ) \end{align}

Now a density matrix is called seperable, if it can be written as a convex combination of tensor products of the subsystem \begin{align} \rho_{sep} = \sum_ip_i (\rho^i_A\otimes \rho^i_B) \end{align} But if we take all the $p_i=\frac{1}{2}$ and take \begin{align} \rho^1_A &= \left|0\right>_A\left<0\right|_A \\ \rho^1_B &= \left|0\right>_B\left<0\right|_B \\ \\ \rho^2_A &= \left|0\right>_A\left<1\right|_A \\ \rho_B^2 &= \left|0\right>_B\left<1\right|_B \end{align} and so on we see that the density matrix corresponding to the Bell state can be written as a convex combination and therefore is seperable.

There must be a conceptual misunderstanding and I would be really thankful if one could help me cure this.

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    $\begingroup$ Your $\rho$ (the Bell state) is pure, while you seem to suggest that you can write it as a non-trivial convex combination of density matrices--which is an inherent contradiction... Hint: Are $\rho^2_A$ or $\rho^2_B$ density matrices? $\endgroup$
    – Tobias Fünke
    Commented May 22 at 11:49
  • $\begingroup$ The statements you wrote are not consistent. The density matrix you gave for the bell state and the convex combination you gave do not yield the same result. (e.g. the latter never produces the $ |11 \rangle \langle 11|$ term. ) $\endgroup$
    – Thomas Tappeiner
    Commented May 22 at 11:50
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    $\begingroup$ your $\rho^2$s are not Hermitian and therefore not density matrices $\endgroup$
    – By Symmetry
    Commented May 22 at 11:51
  • $\begingroup$ @ThomasTappeiner i didnt write all density matrixes down. Following the same precedure the matrices $\rho^4_A$ and $\rho^4_B$ would produce this term $\endgroup$
    – Aralian
    Commented May 22 at 11:57
  • $\begingroup$ Instead of writing the edit, just write an answer yourself. You can also accept your answer if you want. This way, the question is "tagged" as solved, and potential future readers might benefit from it (the question and the answer/resolution). $\endgroup$
    – Tobias Fünke
    Commented May 22 at 12:05

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As the comments suggest, what I didn't see is that for example $\rho^2_A$ and $\rho^2_B$ are not hermitian and therefore are no density matrices. I've also noticed, that we also have the condition $\sum_i p_i=1$ since we need to take convex combinations. Also this condition is violated in my example.

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  • $\begingroup$ This should be an edit to the original post, not an answer. $\endgroup$
    – Matt Hanson
    Commented May 22 at 12:17
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    $\begingroup$ I had it as an edit but another user told me I should post it as an answer. $\endgroup$
    – Aralian
    Commented May 22 at 12:19
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    $\begingroup$ @MattHanson I suggested otherwise. Why an edit? It answers the question. Why should we leave the question open if the OP noticed (with a little help) what is wrong? A question is a question; an answer an answer. One shouldn't edit the question to include an answer. One should write an answer instead. $\endgroup$
    – Tobias Fünke
    Commented May 22 at 12:19
  • $\begingroup$ Fair enough. At a glance I didn’t realize this fully answered the question. $\endgroup$
    – Matt Hanson
    Commented May 22 at 19:55

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