Consider a linear map between linear maps of a Hilbert space, $\mathcal{E}: \mathcal{L}(\mathcal{H})\to\mathcal{L}(\mathcal{H})$. The standard definition I have encountered is that $\mathcal{E}$ is positive if $$\mathcal{E}(A) \geq 0 \quad \forall A\geq 0.$$

An element $A$ in $\mathcal{L}(\mathcal{H})$ is in turn said to be positive if $$\left<{\psi, A(\psi)}\right>\geq 0 \quad \forall \psi \in \mathcal{H}.$$

(Obs: I am thinking finite dimensions for now, because it is a very basic question, but feel free to discuss infinite dim too.)

Then, I wondered, since $\mathcal{L}(\mathcal{H})$ is also a Hilbert space, an analogous to the second definition also exists for linear operators in $\mathcal{L}(\mathcal{L}(\mathcal{H}))$ if we use the Hilbert-Schmidt inner product $\left<A,B\right>= \operatorname{Tr}(A^* B)$ in $\mathcal{L}(\mathcal{H})$, right?

With this, we would have that a map in $\mathcal{L}(\mathcal{L}(\mathcal{H}))$ is positive if $$\operatorname{Tr}(B^*\mathcal{E}(B))\geq 0 \quad \forall B \in \mathcal{L}(\mathcal{H}).$$ My question is: is this equivalent to taking positive maps to positive maps? I do not know how to prove the statement or even if is true. I also haven’t found that on the internet, and I realize that the first definition and some others equivalent to it are always used, with no mention to this. Is the equivalence valid at least in the simplest case?

  • $\begingroup$ Do you mind adding a sketch of the proof of what you have proven already? $\endgroup$ Commented Dec 30, 2022 at 20:14
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    $\begingroup$ Actually, I just realized I made a mistake in my proof, so I don’t know if any direction holds 🥲 $\endgroup$
    – Sahdo
    Commented Dec 30, 2022 at 22:51
  • $\begingroup$ For conjectures of such kind, first try some canonical examples of channels before trying to find a proof. It will save you a lot of time. (On this occasion, let me blatantly self-advertise my list of canonical examples of quantum channels: physics.stackexchange.com/questions/291810/…) $\endgroup$ Commented Dec 31, 2022 at 15:27
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    $\begingroup$ It should be noted that these two notions of positivity are not meant to be related. The first notion -- or more precisely, its stabilized version, complete positivity -- is the natural condition which one must imposed on maps in order to be able to implement them physically (at least with some probability). The fact that the same is being used can indeed be seen as unfortunate; but then again, this happens again and again, and there is a conflict between concise and precise terminology. $\endgroup$ Commented Dec 31, 2022 at 15:34
  • $\begingroup$ Good to know. I’ve been studying basics of quantum operations and networks for a while and yes, it is natural to ask maps to preserve positivity. So I guessed they just didn’t use the original definition because this “characterization” was more intuitive. Had never put much thought on whether they were different things until I affirmed the statement on a text and realized I did not know how to prove it. Probably there are more students out there confused about this too. Thanks for all the helpful answers and comments! $\endgroup$
    – Sahdo
    Commented Jan 2, 2023 at 0:16

4 Answers 4


I don't think that the converse statement holds true. Here is a counter example:

For $B, U\in\mathcal L(\mathcal H)$, let $\mathcal E(B):=U BU^*$. This defines a linear operator and it holds that if $B\geq 0$, then $\mathcal E(B) \geq 0$ on $\mathcal H$.

Now choose $B:=|\phi\rangle \langle \psi|$ for $\phi, \psi \in \mathcal H$ orthonormal (which means that $B\,\varphi:=\langle \psi,\varphi\rangle\, \phi$ for $\varphi \in \mathcal H$) and further pick $U:= -|\phi\rangle \langle \phi| + |\psi\rangle \langle \psi|$. With these choices, we find $$\mathcal E(B) \,\varphi = -\langle \psi,\varphi\rangle \phi $$ and hence

$$\mathrm{Tr}\,B^* \mathcal E(B) = \mathrm{Tr}\, |\psi\rangle\langle\phi|\, \mathcal E(B) = \langle \phi,\mathcal E(B)\,\psi\rangle = -1 \quad .$$

Thus, we have shown that there exists a linear map $\mathcal E$ on $\mathcal L(\mathcal H)$ which preserves positivity but does not fulfill the last inequality of the question.

As the other answer points out, the other direction does not hold either. However, as the OP suggested, we have that

$$ \mathrm{Tr}\, A\,\mathcal E(B) \geq 0 \quad \forall A,B\geq 0 \Longleftrightarrow \mathcal E(B)\geq 0 \quad \forall B\geq 0\quad .$$

To see this, note that if $\mathcal E$ preserves positivity, i.e. if $\mathcal E(B)\geq 0$ for $B\geq 0$, then for all $A\geq 0$ it holds that $\sqrt A\, \mathcal E(B)\, \sqrt A \geq 0$ and thus $ 0 \leq \mathrm{Tr}\,\sqrt A\, \mathcal E(B)\, \sqrt A = \mathrm{Tr}\, A\, \mathcal E(B)$ for all $B\geq 0$. The converse direction follows by noticing that the inequality in particular holds for all positive $A$ of the form $A_\psi:=|\psi\rangle\langle \psi|$, which yields $\langle \psi,\mathcal E(B)\,\psi\rangle \geq 0$ and hence $\mathcal E(B)\geq 0$, for all positive $B$.

  • $\begingroup$ Good! Thanks for the concise counterexample. About the other direction, it is potentially not true either. It seems like the condition for positive supermaps regarding trace is $\operatorname{Tr}⁡(A^∗\mathcal{E}(B))\geq0$ for all positive A and B. $\endgroup$
    – Sahdo
    Commented Dec 31, 2022 at 6:06
  • $\begingroup$ @Sahdo Glad I could help. You're right, and this is more or less trivial. I've added a proof in the answer. $\endgroup$ Commented Dec 31, 2022 at 8:09
  • $\begingroup$ while this is true, I think one can make the case that the statement is true if you restrict the action of the channel to the subspace of Hermitian operators. Any positive map is a sum of maps of the form $\rho\mapsto A\rho A^\dagger$, and for these you have ${\rm Tr}(P_\psi AP_\psi A^\dagger)\ge0$ for all $P_\psi\equiv|\psi\rangle\!\langle\psi|$. Rank-1 vectors span the Hermitian ops, and thus the statement holds in one direction. Not entirely sure about the other direction, but I feel like it should work assuming eg Hermitian-preserving maps, which always have Kraus-like decompositions $\endgroup$
    – glS
    Commented Feb 27, 2023 at 15:28
  • $\begingroup$ Dear @glS , thanks for your comment. Can you elaborate what exactly you mean? I don't know what you are referring to: do you mean the general statement in the question/last part of my answer? In other words: Do you mind explicitly state the equivalence you want to prove in mathematical terms? $\endgroup$ Commented Feb 27, 2023 at 15:43

In the fields of mathematics and quantum information science, the first concept is referred to as a positive map, although I believe that this terminology is misleading and we should instead refer to it as a positive preserving map (as I shall use in the following).

Indeed, the two concepts are completely distinct: a map can be positive preserving but not positive, positive preserving and positive, not positive preserving but positive, or neither positive preserving nor positive.

A simple example of a positive preserving but not a positive map has already been given by @Tobias Fünke.

To see the converse, first it is easy to see that a linear map ${\mathcal E}$ defined by ${\mathcal E}(X) := \mathrm{Tr}(A^\ast X) A$ with any $A \in {\mathcal L}({\mathcal H})$ is positive, as for all $B \in {\mathcal L}({\mathcal H})$, we have $\mathrm{Tr}(B^\ast {\mathcal E}(B)) = |\mathrm{Tr}(A^\ast B)|^2 \ge 0$. However, it is not necessary for ${\mathcal E}$ to be positive preserving unless $A$ is positive. For instance, if we let $A = |\phi_0 \rangle \langle \phi_0| - |\phi_1 \rangle \langle \phi_1|$ where $\phi_0, \phi_1 \in {\mathcal H}$ is an orthonormal system, then ${\mathcal E}(|\phi_0 \rangle \langle \phi_0 |) = \langle \phi_0, A \phi_0 \rangle A = A$, which is not positive as $A$ is not a positive operator (having an eigenvalue $-1$). Hence, this gives an example of positive but not positive preserving.

On the other hand, if we choose $A$ to be positive, the example gives a map that satisfies both definitions simultaneously. In fact, the example even gives a completely positive (preserving) map for any positive $A$. To see this, note that the Choi-Jamiolkowski isomorphism $\sum_{i,j}{\mathcal E}(|i \rangle \langle j|) \otimes |i \rangle \langle j|$ (with a fixed orthonormal basis ${ |i \rangle }$) is given by $A \otimes \overline{A}$ where $\overline{A}$ is the complex conjugate of $A$ (defined as $\overline{A} = \sum_{i,j} \langle i, A j\rangle |i\rangle \langle j|$). It is a well-known fact that complete positivity is equivalent to the positivity of the Choi-Jamiolkowski matrix. Moreover, if $A$ is positive, then $\overline{A}$ is also positive and $A \otimes \overline{A}$ is positive.

  • $\begingroup$ Great answer! I've added a link to the Wikipedia entry of "completely positive maps", as I can imagine that some readers may not know this concept. Feel free to undo this or replace it with a better link. $\endgroup$ Commented Dec 31, 2022 at 8:00
  • $\begingroup$ Thank you for your support. This was my first post, so I don't quite understand the system yet. Please forgive me if I have been rude ... $\endgroup$
    – Gen KIMURA
    Commented Dec 31, 2022 at 12:16
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    $\begingroup$ No worries, you were by no means rude. In any case, here is a link to the tour and a link to the help center, in case you have questions or anything. Keep up the good work. $\endgroup$ Commented Dec 31, 2022 at 12:55

The relation you speculate about is wrong not only for specific counterexamples, but it is wrong generically.

Namely, positive matrices have a positive spectrum, while a general completely positive superoperator (which is a special case of a positive superoperator) has a complex spectrum (with some constraints). Thus, any random positive map, or completely positive map, will be a counterexample.

Conversely, a positive spectrum of a superoperator is not enough to correspond by a positive matrix: You will need in addition that it is hermitian. Then, indeed, positivity of the corresponding matrix follows.

  • $\begingroup$ Hi @Norbert Schuch, do you know of any references to the spectrum of CP maps? $\endgroup$
    – Gen KIMURA
    Commented Jan 5, 2023 at 21:42
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    $\begingroup$ @GenKIMURA Michael Wolf's lecture notes are rather comprehensive: www-m5.ma.tum.de/foswiki/pub/M5/Allgemeines/MichaelWolf/… $\endgroup$ Commented Jan 6, 2023 at 22:28
  • $\begingroup$ Thank you for the information! I am sorry for the delay in replying. I forgot the information on this account and haven't been able to get in for a while... $\endgroup$
    – Gen KIMURA
    Commented Feb 18, 2023 at 2:57

Usually one endows $\mathcal{L}(\mathcal{H})$ with the operator-norm, i.e. $$ \| T \|_{\operatorname{op}} := \sup_{\| x \| = 1} \| Tx \|.$$ Then $\| \cdot \|_{\operatorname{op}}$ does not fulfill the parallelogram identity, so it is not induced by an inner product. However, since in finite dimensions all norms are equivalent, its topology is the same as the topology of a Hilbert space. The most natural inner product is then given by $\langle A, B \rangle := \operatorname{tr}(A^*B)$.

However in infinite dimensions not all norms are equivalent, and $\mathcal{L}(\mathcal{H})$ will not be isomorphic to a Hilbert space. Also, you cannot define the trace for all bounded linear maps on $\mathcal{H}$, so the above definition of an inner product will also not carry over.

  • $\begingroup$ What does this have to do with the question? $\endgroup$ Commented Dec 31, 2022 at 11:27
  • $\begingroup$ Yes, L(H) is not Hilbert in general. This inner product, however, generates the Hilbert space of Hilbert-Schmidt operators. So, I imagined that in infinite dimensions, positivity with the trace should make sense at least for operators inside that space. Then, it could be that positivity implied positive preserving, a condition that is less general but it would make positivity preservation look like a generalization of positivity defined as $\left<\psi, A\psi\right> \geq0$. That is not the case either, as @Gen Kimura pointed out. $\endgroup$
    – Sahdo
    Commented Dec 31, 2022 at 12:42

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