# Does the 4x4 matrix $|00\rangle\!\langle00|+|11\rangle\!\langle11|$ have a decomposition?

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)?

• @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. Jan 27, 2022 at 16:15
• @M.Stern this matrix does not correspond to a pure state Jan 27, 2022 at 17:46
• I believe the user means |1><1|+|3><3| ? Jan 28, 2022 at 2:13
• @User101 you're right Jan 28, 2022 at 6:48

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.

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

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.