1
$\begingroup$

Suppose I have a matrix that is unknown whether it is a density matrix and assume that finding the eigenvalues of it is difficult because the matrix is expressed generally.

However, suppose that this matrix can be shown to be a tensor product of two matrix and hence separable. Does this guarantee that the matrix is a density matrix?

Edit: removed ambiguity in wording.

$\endgroup$
2
  • $\begingroup$ When you say the matrix can be shown to be a tensor product of two states. Do you mean $|\psi\rangle \otimes |\phi\rangle$ or $|\psi\rangle\langle \psi | \otimes |\phi\rangle\langle \phi |$? The former wouldn't be a density matrix as its the wrong shape. $\endgroup$
    – Callum
    Oct 24, 2023 at 13:20
  • 1
    $\begingroup$ @Callum I meant the latter. To your response, it wouldn't even be a matrix if the former. Edited the OP to remove ambiguity. $\endgroup$
    – Physkid
    Oct 24, 2023 at 13:21

2 Answers 2

2
$\begingroup$

No, that doesn't guarantee anything. A density matrix must have non-negative eigenvalues and unit trace. Say you have a matrix $D$ that has been factorized into two matrices: $$ D = A \otimes B $$ Then for $D$ to be a density operator, you have to satisfy two conditions:

  1. $\text{Tr}(D) = \text{Tr}(A) \text{Tr}(B) = 1$
  2. $D\geq 0$ (or $A \otimes B \geq 0$)

where $D\geq 0$ is shorthand for "$D$ is a positive semidefinite operator". Notice you don't necessarily need $A\geq 0$ and $\text{Tr}(A) = 1$, for instance $A = -2 \mathbb{I}_2$ and $B = -\mathbb{I}_2/8$ would produce a valid density operator if $A$ and $B$ are both qubit operators.


You can slightly refine the above conditions on $A$ and $B$, which might be useful for numerical work. In our setting, $D\geq 0$ implies $D^\dagger = D$, which further implies $A=A^\dagger$ and $B=B^\dagger$. This means $A$ and $B$ are diagonalizable with real eigenvalues. Now, suppose either $A$ or $B$ possess negative eigenvalues. Then the only way to get $D \geq 0$ is either:

  • $A\geq 0$ and $B \geq 0$
  • $A \leq 0$ and $B \leq 0$

So, conversely, $D$ is not a density matrix if any of the following conditions is true:

  • $\text{Tr}(A) \text{Tr}(B) \neq 1$
  • Exactly one of the following holds: $A \leq 0$ or $B\leq 0$ (but not both)

Unfortunately, you can't escape the task of testing the negativity of eigenvalues for the component matrices, but these rejection criteria means that sometimes you'll only have to check the smallest eigenvalue of just $A$, instead of $D$.

$\endgroup$
2
$\begingroup$

No, it doesn't. Consider the following counter example: $$ M=\begin{bmatrix} i & 0 \\ 0 & 0 \end{bmatrix}\otimes\begin{bmatrix} 1& 0 \\ 0 & 1\end{bmatrix}. $$ The matrix is clearly separable by design. But it is clearly not Hermitian, and not positive semi-definite (it has eigenvalues $i,i,0,0$), and does not have trace 1. So it certainly is not a density matrix.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.