Surely, these conditions are all well-defined and well-known (via the Choi, Kraus, and Stinespring presentations). Is the following 'definition' valid? Does it make sense?

"The map is CPTP if, when expressed as a matrix acting on a vectorized density operator -in an arbitrary basis - and then subjected to the superdecoherence (i.e., in the corresponding Choi state all the off-diagonal elements are washed out) yields a stochastic matrix"

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    $\begingroup$ Can you maybe define more precisely what you mean by superdecoherence? Where did you get this definition? For it to be a "valid" definition you need to prove that it holds iff one of the other conditions hold. $\endgroup$
    – Rammus
    Commented Oct 19, 2023 at 19:09
  • $\begingroup$ "Superdecoherence" is the following procedure: You transform the map into the corresponding Choi state. Then the Choi state is written down in some bi-basis. Finally, all elements of this Chou matrix are nullified - except the diagonal one. Last step: reshuffle the obtained diagonal state into a map. The resulting map will be a stochastic matrix plus a set of decoupled exponentially decaying modes. $\endgroup$
    – trurl
    Commented Oct 20, 2023 at 13:17
  • $\begingroup$ so you're saying to put to zero all off-diagonal elements of the Choi, and obtain a "stochastic matrix", which for a diagonal matrix means to equal the identity? That doesn't seem right. You'd be saying that the Choi needs to have 1 on the diagonal, which is not true $\endgroup$
    – glS
    Commented Oct 20, 2023 at 14:33
  • $\begingroup$ @glS I am saying the following: If you decohere the Choi state, i.e. 'put to zero all off-diagonal elements' (in any basis) and then reshuffle the so obtained diagonal Choi it into a map - you end up with a classical stochastic matrix 'coated' with N^2-N decoupled modes $\endgroup$
    – trurl
    Commented Oct 20, 2023 at 21:28
  • $\begingroup$ so in practice you're just saying: a map is a channel iff its Choi is positive semidefinite, ok. But only if you're saying you get a positive semidefinite diagonal matrix after decoherence in all possible bases (which maybe is what you meant, but it wasn't entirely clear to me) $\endgroup$
    – glS
    Commented Oct 20, 2023 at 23:01

2 Answers 2


First the short answer:

The procedure desribed in the original post & subsequent comments guarantees that the map is positive, trace preserving but does, surprisingly, not characterize complete positivity. A counterexample is given at the very end of this answer: The procedure turns the transposition map (not completely positive!) into a stochastic matrix, regardless of the chosen orthonormal bases on the individual factors.

As for the long answer: let me first make sure I understand the procedure & I can give an appropriate answer by summarizing the method as explained in the comments:

  1. Starting from a linear map $\Phi:\mathbb C^{n\times n}\to \mathbb C^{n\times n}$ "transform the map into the corresponding Choi state" $\mathsf C(\Phi)=\sum_{j ,k=1}^n|j\rangle\langle k|\otimes\Phi(|j\rangle\langle k|)$.
  2. "Then the Choi state is written down in some bi-basis", that is, given some orthonormal bases $G:=\{g_j\}_{j=1}^n$, $H:=\{h_j\}_{j=1}^n$ of $\mathbb C^n$ we consider the matrix \begin{align*} \mathsf C^{G,H}(\Phi):=&\;\sum_{a,b,c,d=1}^n|a\otimes c\rangle\langle g_a\otimes h_c|\mathsf C(\Phi)|g_b\otimes h_d\rangle\langle b\otimes d|\\ =&\; \sum_{a,b,c,d}\langle g_a\otimes h_c|\mathsf C(\Phi)|g_b\otimes h_d\rangle|a\rangle\langle b|\otimes|c\rangle\langle d| \end{align*} where $|j\rangle$ refers to the standard (computational) basis.
  3. "All elements of this Choi matrix are nullified - except the diagonal ones". Setting the off-diagonal elements to zero is of course equivalent to just extracting the diagonal and ignoring the rest, i.e. we look at the vector $v(\mathsf C^{G,H}(\Phi)):=\sum_{j ,k=1}^n\langle j\otimes k|\mathsf C^{G,H}(\Phi)|j\otimes k\rangle |j\otimes k\rangle$
  4. "Reshuffle the obtained diagonal state into a map. The resulting map will be a stochastic matrix [...]". This means to use the inverse of the vectorization procedure to turn this vector of length $n^2$ into an $n\times n$ stochastic matrix. Doing so results in $$ \Xi(G,H,\Phi):={\rm vec}^{-1}\big( v(\mathsf C^{G,H}(\Phi)) \big)=\sum_{j ,k=1}^n\langle j\otimes k|\mathsf C^{G,H}(\Phi)|j\otimes k\rangle |k\rangle\langle j| $$

The first thing we do is find a simple expression for the entries of $\Xi(G,H,\Phi)$:

Lemma. Given any orthonormal bases $G,H$ of $\mathbb C^n$ and any linear map $\Phi:\mathbb C^{n\times n}\to \mathbb C^{n\times n}$ it for all $j ,k=1,\ldots,n$ holds that $$ \big(\Xi(G,H,\Phi)\big)_{kj}=\langle g_j\otimes h_k|\mathsf C(\Phi)|g_j\otimes h_k\rangle=\langle h_k|\Phi(|\overline{g_j}\rangle\langle\overline{g_j}|)h_k\rangle $$ where $|\overline{x}\rangle:=\sum_{j=1}^n\langle x|j\rangle|j\rangle$ is the entrywise complex conjugate of $x$ in the standard basis.

Proof. This is a straightforward computation: \begin{align*} \big(\Xi(G,H,\Phi)\big)_{kj}&=\langle j\otimes k|\mathsf C^{G,H}(\Phi)|j\otimes k\rangle\\ &=\langle g_j\otimes h_k|\mathsf C(\Phi)|g_j\otimes h_k\rangle\\ &=\sum_{a,b=1}^n\langle g_j|a\rangle\langle b|g_j\rangle\langle h_k|\Phi(|a\rangle\langle b|)|h_k\rangle\\ &=\Big\langle h_k\Big|\Phi\Big(\sum_{a=1}^n\langle g_j|a\rangle|a\rangle\sum_{b=1}^n\langle b|\langle b|g_j\rangle\Big)\Big|h_k\Big\rangle\\ &=\langle h_k|\Phi(|\overline{g_j}\rangle\langle\overline{g_j}|)h_k\rangle\tag*{$\square$} \end{align*}

With this identity we can easily make sure that there is no fundamental flaw in this procedure, that is, if $\Phi$ is (completely) positive and trace preserving, then $\Xi(G,H,\Phi)$ is a stochastic matrix for all orthonormal bases $G,H$ of $\mathbb C^n$. Indeed, $\Phi$ being positive shows $\Phi(|\overline{g_j}\rangle\langle\overline{g_j}|)\geq 0$ so $(\Xi(G,H,\Phi))_{kj} \geq 0$ for all $j ,k=1,\ldots,n$ by the above lemma. Moreover, $\Phi$ being trace preserving shows \begin{align*} \sum_{k=1}^n(\Xi(G,H,\Phi))_{kj}&=\sum_{k=1}^n \langle h_k|\Phi(|\overline{g_j}\rangle\langle\overline{g_j}|)h_k\rangle\\ &={\rm tr}\big( \Phi(|\overline{g_j}\rangle\langle\overline{g_j}|) \big)\\ &={\rm tr}(|\overline{g_j}\rangle\langle\overline{g_j}|) =\|\overline{g_j}\|^2 \end{align*} for all $k=1,\ldots,n$. Indeed, $\|\overline{g_j}\|^2=\sum_{k=1}^n|\langle k|g_j\rangle|^2=\|g_j\|^2=1$ by the Parseval identity so, as desired, $\Xi(G,H,\Phi)$ is a stochastic matrix, regardless of the chosen orthonormal bases $G,H$.

This finally gets us to the main question:

Does the converse also hold? That is, if $\Xi(G,H,\Phi)$ is a stochastic matrix for all orthonormal bases $G,H$ of $\mathbb C^n$ is it true that $\Phi$ is completely positive?

At first glance this is not a bad guess: after all, on the level of states such a characterization does hold. More precisely, it is readily verified that a matrix $A\in\mathbb C^{n\times n}$ is a density operator if and only if $(\langle g_j|A|g_j\rangle)_{j=1}^n$ is a probability vector for all orthonormal bases $\{g_j\}_{j=1}^n$ of $\mathbb C^n$.

Interestingly, this characterization falls short when going from states to channels. The problem is that in order to "extract" stochastic matrices we are limited to product bases $\{g_j\otimes h_k\}_{j,k}$ on the larger space---but testing things on product vectors only guarantees positivity as for complete positivity we would need access to some entanglement. This was originally observed in A. Jamiołkowski, "Linear Transformations Which Preserve Trace and Positive Semidefiniteness of Operators", Reports on Mathematical Physics 3(4), p.275-278 (1972), Theorem 1 therein. Now to once again be precise let us prove the following:

Theorem. Let $\Phi:\mathbb C^{n\times n}\to \mathbb C^{n\times n}$ linear be given. The following statements are equivalent.

  1. $\Xi(G,H,\Phi)$ is a stochastic matrix for all orthonormal bases $G,H$ of $\mathbb C^n$
  2. $\Phi$ is positive and trace preserving

Proof. We already showed 2. $\Rightarrow$ 1. so we only need to prove the converse. First we show that $\Phi$ is trace preserving. For this it, by linearity, suffices to prove that ${\rm tr}(\Phi(A))={\rm tr}(A)$ for all Hermitian matrices $A$ (as every matrix can be written as a linear combination of two Hermitian matrices). Because $A$ is Hermitian we can spectrally decompose it into $A=\sum_{j=1}^n a_j|f_j\rangle\langle f_j|$. Defining the "conjugate" orthonormal basis $F:=\{\overline{f_j}\}_{j=1}^n$ and choosing any orthonormal basis $H$ of $\mathbb C^n$, using the above lemma we compute \begin{align*} {\rm tr}(\Phi(A))&=\sum_{j,k=1}^na_j\langle h_k|\Phi(|f_j\rangle\langle f_j|)|h_k\rangle \\ &=\sum_{j=1}^na_j\sum_{k=1}^n(\Xi(F,H,\Phi))_{kj}\\ &=\sum_{j=1}^na_j={\rm tr}(A)\,. \end{align*} As for positivity we use that $\Phi$ is positive if and only if $\langle y|\Phi(|x\rangle\langle x|)|y\rangle\geq 0$ for all $x,y\in\mathbb C^n$; this condition is, by definition, equivalent to $\Phi(|x\rangle\langle x|)\geq 0$ for all $x$ which---again by spectral decomposition and linearity---is equivalent to $\Phi(A)\geq 0$ for all $A\geq 0$. Now if either $x$ or $y$ are $0$, then this is trivially true. Else $x,\overline{y}\neq 0$ we can normalize them and complete them to respective orthonormal bases $X,Y$ (i.e. $x_1=\frac{x}{\|x\|}$, $y_1=\frac{\overline{y}}{\|y\|}$). Then by assumption $(\Xi(X,Y,\Phi))_{11}\geq 0$ which by the above lemma translates to \begin{align*} 0\leq \|x\|^2\|y\|^2 (\Xi(X,Y,\Phi))_{11}&=\|x\|^2\|y\|^2\langle y_1|\Phi(|x_1\rangle\langle x_1|)y_1\rangle\\ &=\langle y|\Phi(|x\rangle\langle x|)|y\rangle\,, \end{align*} as desired. $\square$

Finally, this characterization allows us to construct a counterexample by choosing $\Phi$ positive and trace preserving but not completely positive. The standard example of such a map is of course the transposition map $\mathsf T$. Indeed, for any orthonormal bases $G,H$ we find \begin{align*} \big(\Xi(G,H,\mathsf T)\big)_{kj}&=\langle h_k|(|\overline{g_j}\rangle\langle\overline{g_j}|)^T|h_k\rangle\\ &=\langle h_k|g_j\rangle\langle g_j|h_k\rangle=|\langle g_j|h_k\rangle|^2 \end{align*} so $\Xi(G,H,\mathsf T)$ is a stochastic matrix for all $G,H$ meaning that, as expected, $\Xi$ fails to detect that $\mathsf T$ is not completely positive.

  • $\begingroup$ Thank you for the detailed answer, I'll read it carefully. But just a quick question: Do you agree with that the question I asked is identical to the following one "Is a matrix, that has all diagonal elements non-negative - in any basis - is positive semidefinite"? Because this matrix is the Choi matrix in our case. $\endgroup$
    – trurl
    Commented Apr 15 at 17:11
  • $\begingroup$ It is related but not identical. After all, if you represent the Choi matrix in a basis which it not of product form $\{f_j\otimes g_k\}_{j,k}$, then the diagonal---while non-negative---will in general not correspond to a stochastic matrix for lack of trace-preservation. So if you just wanted to check complete positivity then yes, that's equivalent to the Choi matrix having a non-negative diagonal in every basis. But because you asked about characterizing CPTP the bases you can use in the first place is limited which is the reason why the equivalence doesn't transfer over. $\endgroup$ Commented Apr 15 at 17:23
  • $\begingroup$ Thank you for addressing the comment/question. This is the key point - it is not any bases (in the $d^2$ space) but has to be a product basis. But to have non-negative diagonal in any product basis is not enough to have it non-negative in any basis (and be, therefore, PSD). Is it a correct reproduction of your argument? $\endgroup$
    – trurl
    Commented Apr 15 at 17:32
  • $\begingroup$ That's exactly right. To be explicit, looking again at the transposition map $\rho\mapsto\rho^T$ (which I considered at the very end of my answer) the diagonal of its Choi matrix in any product basis corresponds to a stochastic matrix. However, the Choi matrix has a negative eigenvalue ($-1$) because transposition is not completely positive so if we choose the global diagonalizing unitary $U$ we can "produce" this $-1$ on the diagonal. $\endgroup$ Commented Apr 15 at 18:06
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    $\begingroup$ And, of course, the diagonal of the Choi matrix of a P-map has to be non-negative in any bi-basis - simply because the original map is P (it is clear in retrospection, though... thank you for explaining this) $\endgroup$
    – trurl
    Commented Apr 16 at 10:46

The answer is indeed 'yes' because it is a tautology: Just another way to say that the Choi state is non-negative and the original map is trace preserving. So, it is a CPTP map.

To unfold: A state, i. e., a density operator, is a positive semidefinite operator. It means that if it is expressed in the form of a density matrix - in any basis - the matrix diagonal consists of non-negative elements only (additionally normalized so they are probabilities but this is not important in the current context). Next, a map is completely positive iff its Choi operator is positive-semidefinite.

Another words, the Choi operator, when expressed in any bi-basis, has non-negative diagonal elements. Superdecoherence (which is basis specific) washes out all of-diagonal elements of the Choi matrix and keeps only diagonal elements. Note that this 'washed-out' Choi matrix still represents a state -- it is still positive-semidefinite. Now this Choi matrix is reshuffled into a map. This map is a stochastic matrix 'coated' with $N-1$ decoupled exponentially decaying - and $N$-time degenerated - modes. Therefore, for a CPTP map, superdecoherence will always result in a stochastic matrix - for any basis - simply becuase the corresponding Choi operator is a state (semidefinite positive).

  • $\begingroup$ Can you elaborate? $\endgroup$
    – FDGod
    Commented Oct 20, 2023 at 14:41
  • $\begingroup$ @FDGod Yes, sure. See above. $\endgroup$
    – trurl
    Commented Oct 20, 2023 at 21:23
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    $\begingroup$ You are right that "for a CPTP map, superdecoherence will always result in a stochastic matrix - for any basis" but---assuming I understood your procedure correctly---the converse does not hold so this is not a characterization: getting a stochastic matrix for any, as you call it "bi-basis", only guarantees that the map is positive and trace preserving, but not completely positive. See my answer for more detail. $\endgroup$ Commented Apr 14 at 19:14

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