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A density matrix is defined in general as the operator

$$ \hat{\rho} = \sum_i p_i |\psi_i\rangle \langle\psi_i| $$

for the most general mixture of states $\{p_i \longrightarrow |\psi_i\rangle\}_{i=1, ..., N}$.

Every density matrix has the following properties:

  1. Hermiticity: $\hat{\rho}^{\dagger} = \hat{\rho}$
  2. Normalisation: $\mathrm{Tr}\,(\hat{\rho}) = 1$
  3. Positiveness: $\langle\phi|\hat{\rho}|\phi\rangle\geq 0 \;\;\; \forall|\phi\rangle$
  4. If it describes a pure state, then it is also a projector

When describing a quantum channel, we are asking that the output of the channel is still a legitimate quantum state, therefore that is is described by a density matrix, so we ask that the map is completely positive (namely it must preserve positiveness, even if we only look at a sub-system of an entangled state) and trace preserving (CPTP).

My questions

  1. What about Hermiticity? Is this automatically assured by the two properties (CP) and (TP)?

  2. Are all these properties automatically guaranteed when implementing a quantum channel using Kraus operators?

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  • $\begingroup$ My guess is that since your eigenvalues are real (and positive on top of that) then your matrix is hermitian $\endgroup$ May 29, 2022 at 14:53

2 Answers 2

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I am not an expert regarding quantum channels etc., but I think this is a pure math question. Let's assume that we're working on a finite-dimensional complex Hilbert space.

In finite dimensions, the defining properties of a density matrix $\rho$ are positive semi-definiteness $\rho \geq 0$ and the trace-normalization $\mathrm{Tr}\,\rho=1$. Properties $1$ and $4$ are consequences of this definition. In particular, a positive semi-definite operator is hermitian$^\dagger$.

To answer your first question: Let $\phi$ denote a map which maps positive semi-definite operators to positive semi-definite operators. Following the considerations above, we see that if $\rho$ is positive semi-definite, then $\phi(\rho)$ is hermitian.

I don't know if I correctly understand your second question, but note that Kraus operators preserve the normalization and positive semi-definiteness. To see this, define for a density operator $\rho$ the following map: $$\varphi(\rho):= \sum\limits_k B_k \,\rho \, B_k^\dagger \quad , $$

with $\displaystyle \sum\limits_k B_k^\dagger\, B_k = \mathbb I$, where $\mathbb I$ denotes the identity operator. We find $$ \mathrm{Tr}\,\varphi(\rho) = \mathrm{Tr}\, \sum\limits_k B_k\, \rho\,B_k^\dagger = \mathrm{Tr}\, \sum\limits_k B_k^\dagger B_k\, \rho = \mathrm{Tr}\, \mathbb I\,\rho = \mathrm{Tr}\,\rho = 1 \quad , $$

where we've used the linearity and the cyclic property of the trace. Moreover:

$$\langle \psi|\varphi(\rho)|\psi\rangle = \sum\limits_k \langle \psi|B_k\, \rho \,B_k^\dagger|\psi\rangle = \sum\limits_k \langle \underbrace{\psi_k|\rho|\psi_k\rangle}_{\geq 0} \geq 0 \quad , $$

with $|\psi_k\rangle := B_k^\dagger|\psi\rangle$.

So $\varphi(\rho)$ is a density matrix again and therefore pure if and only if it is a projector. However, note that Kraus operators can change the purity of a density operator, cf. this quantum computing question.


$^\dagger$ This is easy to see: For a positive semi-definite $\rho$, define the following hermitian operators: \begin{align} \rho_1 &:= \frac{1}{2} (\rho+ \rho^\dagger) \\ \rho_2 &:= \frac{1}{2i} (\rho -\rho^\dagger) \quad . \end{align} Then obviously $\rho = \rho_1 + i\,\rho_2$ and since $\langle \psi |\rho|\psi\rangle , \langle \psi |\rho_1|\psi\rangle , \langle \psi |\rho_2|\psi\rangle \in \mathbb R$, it follows that $\langle \psi|\rho_2|\psi\rangle \overset{!}{=}0$ for all $|\psi\rangle$. Thus $\rho_2=0$ and $\rho=\rho_1$, i.e. $\rho$ is hermitian.

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    $\begingroup$ I agree with almost everything but I was just puzzled at the following thought: matrix $A=-\mathbb{I}$ is Hermitian but it doesn't seem positive definite as $\langle e_i|A|e_i\rangle=-1$, where is a basis element, i.e. $e_i=[0,..,1,...,0]$. Im surely making a silly mistake! $\endgroup$ May 29, 2022 at 17:20
  • $\begingroup$ The reverse direction of the statement in the answer is incorrect, as you've noticed correctly. In other words: From positive semi-definite, hermiticity follows. The reverse is wrong (you've provided a counter example). $\endgroup$ May 29, 2022 at 17:20
  • $\begingroup$ Of course, I see it now :). Silly of me... 😅 $\endgroup$ May 29, 2022 at 17:21
  • $\begingroup$ One further question that came to mind. I guess that given a $\rho$ there are many in general not positive/trace preserving maps that happen to preserve the positivity and trace of the particular given $\rho$. However, without any knowledge about $\rho$, other than it's trace and positivity, is a Krauss map the most general such map? I guess my question is a convoluted way of saying: is a Krauss map the unique general solution to a quantum map? $\endgroup$ Oct 7, 2022 at 21:11
  • $\begingroup$ Dear @FriendlyLagrangian sorry, I really don't know. Perhaps it would be the best to open a separate, new question. $\endgroup$ Oct 8, 2022 at 8:20
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Let $\Phi$ be a quantum map, that is, a linear function between linear operators. Say $\Phi:\operatorname{Lin}(\mathcal H_A)\to\operatorname{Lin}(\mathcal H_B)$.

The first property you discuss seems to be what is generally referred to as the map being "Hermitian-preserving". In other words, the property that $\Phi(X^\dagger)=\Phi(X)^\dagger$ for all $X\in\mathrm{Lin}(\mathcal H_A)$. This amounts to $\Phi(\rho)$ being Hermitian for any Hermitian $\rho$.

Every positive (and thus in particular every completely positive) quantum map is Hermitian preserving, but the opposite isn't true. For example, $\Phi(\rho)\equiv \operatorname{Tr}(\rho)\sigma_x$, with $\sigma_x$ Pauli $X$ matrix, is Hermitian-preserving but not positive.

One way to characterise this is to observe that a map is Hermitian preserving iff its Choi opreator, $J(\Phi)\equiv\sum_{ij} \Phi(|i\rangle\!\langle j|)\otimes|i\rangle\!\langle j|$, is Hermitian. On the other hand, a map is completely positive iff its Choi is (Hermitian and) positive semidefinite. For completeness, let me also mention that the map is positive iff its Choi is (Hermitian and) positive on product states (meaning $\langle u,v|J(\Phi)|u,v\rangle\ge0$ for all vectors $u,v$).

Regarding the Kraus decompositions: a map that can be written as $\Phi(X)=\sum_a A_a X A_a^\dagger$ is automatically completely positive (and furthermore a channel if also $\sum_a A_a^\dagger A_a=I$). If a map is just Hermitian preserving, you can still kinda write a "Kraus-like" decomposition, because any such map can be decomposed as $$\Phi(X)=\sum_a \epsilon_a A_aX A_a^\dagger, \,\,\epsilon_a\in\{-1,1\}.$$ This again follows from the tight relation between eigendecompositions of the Choi operator (or more generally its decompositions in terms of unit-rank positive operators), and Kraus decompositions of the corresponding channel. See also this answer on qc.SE about it, and links therein.

You'll find a lot more info aobut these topics e.g. in chapter 2 of Watrous' book.

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