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Can the diagonal matrix $$\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 0& 0 \\0&0&0&0 \\ 0&0&0&1 \end{pmatrix}$$ be written as a tensor product $A\otimes B$ of two $2\times 2$ matrices $A$ and $B$ (or possibly some other decomposition)?

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    $\begingroup$ @M.Stern, $|00>+|11>$ has a density matrix similar to this, but with additional 1's at (1,4)th and (4,1)th positions, as well. $\endgroup$
    – User101
    Commented Jan 27, 2022 at 16:15
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    $\begingroup$ @M.Stern this matrix does not correspond to a pure state $\endgroup$ Commented Jan 27, 2022 at 17:46
  • $\begingroup$ I believe the user means |1><1|+|3><3| ? $\endgroup$
    – LeWoody
    Commented Jan 28, 2022 at 2:13
  • $\begingroup$ @User101 you're right $\endgroup$
    – M. Stern
    Commented Jan 28, 2022 at 6:48

2 Answers 2

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In short, no. If we label your density matrix by its elements $\rho_{ij}$ with $i$ and $j$ ranging from 1 to 4, we have $$\rho_{11}=\rho_{44}=1$$ and $\rho_{ij}=0$ otherwise. Using a similar set of labels for $A$ and $B$, the tensor product rule tells us that $$\rho_{11}=A_{11} B_{11},\quad \rho_{22}=A_{11} B_{22},\quad \rho_{33}=A_{22} B_{11},\quad \rho_{44}=A_{22} B_{22}.$$ For $\rho_{22}$ to vanish, we must either have $A_{11}$ or $B_{22}$ vanish, which would make either $\rho_{11}$ or $\rho_{44}$ vanish, but they do not. (One could similarly inspect $\rho_{33}$.) It is therefore impossible to construct matrices $A$ and $B$ that simultaneously satisfy all the requirements for $\rho=A\otimes B$.

In terms of states, this looks like $$\rho=\begin{pmatrix}1&0\\0&0\end{pmatrix}\otimes \begin{pmatrix}1&0\\0&0\end{pmatrix}+\begin{pmatrix}0&0\\0&1\end{pmatrix}\otimes \begin{pmatrix}0&0\\0&1\end{pmatrix}.$$ This means that we can write the density matrix as a convex combination $$\rho=|00\rangle\langle 00|+|11\rangle\langle 11|$$ in the computational basis. This definitely cannot be written in the form of $A\otimes B$, but will still be considered "separable" in the language of quantum entanglement theory because it is the convex combination of two separable states.

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  • $\begingroup$ I think it should be $\rho = |\Phi^+><\Phi^+| + |\Phi^-><\Phi^-| $, where $\Phi^{\pm}= |00> \pm |11>$ (Bell states). $\endgroup$
    – User101
    Commented Jan 27, 2022 at 18:28
  • $\begingroup$ @User101 that's the same thing; write it out! $\endgroup$ Commented Jan 28, 2022 at 1:31
  • $\begingroup$ thanks. I have this question regarding the name of the slice state. Could you take a look? quantumcomputing.stackexchange.com/questions/23836/… $\endgroup$
    – User101
    Commented Jan 28, 2022 at 11:55
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Let

$$ A=\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix}\quad B=\begin{bmatrix}b_{11}&b_{12}\\b_{21}&b_{22}\end{bmatrix}\tag1 $$

and suppose that

$$ A\otimes B=\begin{bmatrix} 1&0&0&0\\ 0&0&0&0\\ 0&0&0&0\\ 0&0&0&1 \end{bmatrix}.\tag2 $$

Comparing diagonal elements on both sides of $(2)$, we have

$$ \begin{align} a_{11}b_{11}&=1 \\ a_{11}b_{22}&=0 \\ a_{22}b_{11}&=0 \\ a_{22}b_{22}&=1. \end{align}\tag3 $$

However, the first and last equations in $(3)$ mean that none of the diagonal elements of $A$ and $B$ are zero which contradicts the other two equations in $(3)$. The contradiction means that no such $A$ and $B$ exist.

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