With density matrix $\rho=\sum_{a,b=0}^1\rho_{a,b}|a\rangle\langle b|$ and it's transpose $\rho^T=\sum_{a,b=0}^1\rho_{a,b}|b\rangle\langle a|$. How to confirm that $\rho^T$ is positive and trace preserving.
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3 Answers
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Transposing a matrix is trace preserving since for $\rho = \sum_{a,b} \rho_{a,b} | a \rangle \langle b |$:
$$\text{Tr}(\rho)= \sum_c \langle c| \big( \sum_{a,b} \rho_{a,b} | a \rangle \langle b | \big) | c \rangle = \sum_{a,b,c} \rho_{a,b} \delta_{a,c} \delta_{b,c} = \sum_c \rho_{c,c}$$
$$\text{Tr}(\rho^T)= \sum_c \langle c| \big( \sum_{a,b} \rho_{a,b} | b \rangle \langle a | \big) | c \rangle = \sum_{a,b,c} \rho_{a,b} \delta_{b,c} \delta_{a,c} = \sum_c \rho_{c,c}$$ Therefore: $$\text{Tr}(\rho^T) = \text{Tr}(\rho)$$
For positivity, assuming $\rho$ is nonnegative, which it should be since it is a density matrix, we know that for all $| \psi \rangle$, $\langle \psi| \rho |\psi \rangle \geq 0 \rightarrow \big( \langle \psi| \rho |\psi \rangle \big)^T \geq 0 \rightarrow \langle \psi| \rho^T |\psi \rangle \geq 0 \rightarrow \rho^T$ must be nonnegative.
answered Feb 8, 2021 at 20:14
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$\begingroup$ But not all density matrices are positive. So this assumption: "assuming ρ is positive, which it should be since it is a density matrix" is not correct. $\endgroup$ Commented Feb 8, 2021 at 20:19
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$\begingroup$ All density operators are however non negative; I will amend the solution. $\endgroup$ Commented Feb 8, 2021 at 20:21
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$\begingroup$ That's not true either. See the definition of positive and non-negative in my answer. $\endgroup$ Commented Feb 8, 2021 at 20:40
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2$\begingroup$ All density matrices are positive semidefinite - which is generally written $\rho \geq 0$ and means that all eigenvalues of $\rho$ are nonnegative. Your argument still holds though, because exactly for all positive semidefinite matrices we have $\langle \psi|\rho|\psi\rangle \geq 0$. $\endgroup$– JSdJCommented Feb 8, 2021 at 20:40
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$\begingroup$ What @JSdJ said. $\endgroup$ Commented Feb 8, 2021 at 20:41
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Trace preservation
The trace must be preserved in the transpose of a matrix, because the trace is the sum of the diagonal elements. When transposing a matrix, you do not change the diagonals at all! Only the off-diagonals change when you transpose a matrix.
"Positive" preservation
"A positive matrix is a matrix in which all the elements are strictly greater than zero."
- Neither the density matrix not its transpose has to be positive in general, but if you're talking about positive maps which are transformations that send positive elements to positive elements, then again if $\rho$ is positive, then $\rho^T$ must be too, because transposing does nothing to the sign of the elements it only rearranges them.
- Alternatively, if you're using the word "positive" as a synonym for "positive-semi-definite" as Adam Zalcman pointed out some people do then I would recommend not to do that in the future, because you can just write "positive semi-definite" and no one will get confused about what you mean. If this is what you mean by "positive", then you mean that the eigenvalues of the matrix are all more than or equal to zero. Since the transpose operation does not change the eigenvalues, it will also preserve positive semi-definiteness.
answered Feb 8, 2021 at 20:05
user1271772 No more free timeuser1271772 No more free time
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1$\begingroup$ @JSdJ But the OP user is clearly a beginner, and introducing terminology that is very foreign for the user, such as "quantum channel" and "completely positive" might be over-complicating things. The answer is simply that transposing the matrix doesn't change the sign of any of the elements. So a positive matrix remains positive. $\endgroup$ Commented Feb 8, 2021 at 20:47
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1$\begingroup$ I have misread your answer - I thought you were talking about quantum maps. After rereading more carefully, I now agree with you. I'll delete my comments $\endgroup$– JSdJCommented Feb 8, 2021 at 20:55
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1$\begingroup$ Unfortunately, the term "positive" is ambiguous. Here are three possible interpretations. First, as this answer says, when applied to a matrix $A$, the term might mean that all entries of $A$ are positive. $\endgroup$ Commented Feb 8, 2021 at 21:21
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1$\begingroup$ Second, also applied to a matrix $A$, it might mean that eigenvalues of $A$ are non-negative, i.e. it is a synonym to "positive semi-definite". The second use of the term originates in the study of $C^*$-algebras where some elements of an algebra are said to be "positive elements". If an algebra is a matrix algebra then positive elements are precisely the positive semi-definite matrices. $\endgroup$ Commented Feb 8, 2021 at 21:22
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2$\begingroup$ I think that in quantum information science, the term generally has the second meaning (when it is applied to a matrix) or the third (when it is applied to a map between matrix algebras, i.e. to a quantum channel). I think it is very rare to see the first meaning and the linked wikipedia article even warns that this meaning is implied only occasionally due to possible confusion with the more common second meaning (=positive semi-definite). $\endgroup$ Commented Feb 8, 2021 at 21:25
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Given an arbitrary matrix $A$, its trace equals the sum of its diagonal elements. Because the transpose leaves the diagonal elements invariant, we must have $\operatorname{Tr}(A)=\operatorname{Tr}(A^T)$.
The trace can also be seen to equal the sum of the eigenvalues of $A$, counted with their multiplicities. But again, the transposition doesn't affect the eigenvalues of a matrix, as seen e.g. from the fact that $\det(A-\lambda I)=\det(A^T-\lambda I)$. Therefore any property of $A$ which depends on the eigenvalues is unaffected by transposition.
In particular, a normal matrix $H$ is positive semi-definite, $H\ge0$, iff its eigenvalues are all real positive numbers. Because $H^T$ will have the same eigenvalues, we have $H^T\ge0$.