This problem is from a "passing remark" in this lecture notes. With the help of some colleagues I managed to find a way for this supposedly elementary fact, but I would like to see if there is an alternative the details of which I will explain at the end.

Let $V=\mathcal{M}_{d,d'}(\mathbb{C})$ be the space of $d\times d'$ complex matrices and upgrade this into a Hilbert space by using the inner product \begin{align} \langle A,B\rangle := \text{Tr}(PA^\dagger B) \end{align} where $P>0$ is any positive-definite $d' \times d'$ matrix. Now construct an orthonormal basis (ONB) of $V$ using the inner product so that we get $dd'$ operator basis elements.

The claim I would like to prove is this: the ONB $\{A_j\}$ defines a completely-positive (CP) map $\Phi$ such that \begin{equation} \Phi(\rho):= \sum_{k=1}^{dd'}A_k^\dagger\rho A_k = \text{Tr}(\rho)P^{-1}. \end{equation} Although not necessary for the task, I believe this CP map is by construction maximum Kraus rank since the maximum rank of the Choi matrix is $dd'$. If this were a quantum channel (i.e., if we impose trace-preserving property), it would be a replacement channel that outputs a constant state proportional to $P^{-1}$.

This statement can be proven by drawing pictures using tensor network type language (not exactly obvious but once drawn it is straightforward to see why). For this it is easy to use a different set of Kraus operator $\\{B_k=A_k\sqrt{P}\\}$ for which the inner product becomes Hilbert-Schmidt which makes it easier.

The problem is as follows. First, since this appears in Chapter 2 of the lecture notes and is just a statement in passing, I would like to see if there is (1) straightforward and/or (2) transparent proof of this statement. By "transparent" I mean I should not, hopefully, need to rely on obscure facts in linear algebra. Second, this statement seems to be connected to the theory of fixed points of quantum channels (which the lecture notes discuss in Chapter 6). I am still learning about it, but if there is an insight or natural proof using Perron-Frobenius theory (irreducible maps, etc.), I would love to be inspired/enlightened. Last but not least, I am hoping for some intuition: in the case of qubits, this is the case of a completely depolarizing channel, but I am not sure if this intuition generalizes to higher dimensions or arbitrary $P$ (which needs not be diagonal). I guess by absorbing $P$ into the Kraus operators it does look like completely depolarizing channel anyway (though here we don't have yet trace-preserving property), but since the Kraus operators are not guaranteed to be unitary I am not sure if this is valid.

  • $\begingroup$ What are "obscure facts in linear algebra"? $\endgroup$ Feb 22 at 19:03
  • $\begingroup$ ... and would it be sufficient to prove this for the $B_k$ with the HS inner product? And what is wrong with a diagrammatic proof -- those can usually be easily rewritten without resorting to diagrammatic techniques, unless there are dozens of indices. $\endgroup$ Feb 22 at 19:05
  • $\begingroup$ @NorbertSchuch Ok I guess that is a fair point. I am just thinking from the perspective of students taking the course who may not be acquainted with diagrammatic techniques. And yes, for $B_k$ with HS inner product I think that's how the diagrammatic proof would look simplest. I was just wondering if there are more direct method without, as you say, using thousands of indices of brute force computation (which would amount to taking the diagram and translate them literally). If there is none I am happy with it. $\endgroup$ Feb 23 at 10:44

2 Answers 2



The matrix $P$ is a distraction, because the $dd'$ matrices $A_k$ are an orthonormal basis with respect to the inner product $\langle A,B\rangle_P:=\mathrm{tr}(PA^\dagger B)$ if and only if the $dd'$ matrices $A_k\sqrt{P}$ are an orthonormal basis with respect to the Hilbert-Schmidt inner product $\langle A,B\rangle_{HS}=\mathrm{tr}(A^\dagger B)$. Thus, no generality is lost by assuming that $P=I$.

Proof from state-channel duality

I argue that the proof below does not employ any "obscure facts in linear algebra". Admittedly though, the linear algebra below, as elementary as it is, relies heavily on the concepts of basic quantum information theory.

Proof: Let $|\psi\rangle:=\sum_{i=1}^{d'}|i\rangle|i\rangle\in\mathbb{C}^{d'}\otimes\mathbb{C}^{d'}$ be the unnormalized maximally entangled state with zero relative phases and let $$ |A\rangle\!\rangle:=(A\otimes I)|\psi\rangle\in\mathbb{C}^d\otimes\mathbb{C}^{d'}\tag1 $$ denote the vectorization of the operator $A\in\mathcal{M}_{d,d'}(\mathbb{C})$. By sending $|\psi\rangle$ into $\Phi\otimes I$, we find that the Choi matrix $C(\Phi)$ of $\Phi$ is $$ \begin{align} C(\Phi):&=(\Phi\otimes I)(|\psi\rangle\langle\psi|)\tag2\\ &=\sum_k(A_k^\dagger\otimes I)|\psi\rangle\langle\psi|(A_k\otimes I)\tag3\\ &=\sum_k|A_k^\dagger\rangle\!\rangle\langle\!\langle A_k^\dagger|.\tag4 \end{align} $$ But $\mathrm{tr}(A^\dagger B)=\langle\!\langle A|B\rangle\!\rangle$, so orthogonality of $A_k$ with respect to Hilbert-Schmidt inner product in $\mathcal{M}_{d,d'}(\mathbb{C})$ gives us orthogonality of $|A_k\rangle\!\rangle$, and hence of $|A_k^\dagger\rangle\!\rangle$, with respect to the standard inner product in $\mathbb{C}^d\otimes\mathbb{C}^{d'}$. In light of this fact, $(4)$ implies$^{1,2}$ that $C(\Phi)=I$.

It is easy to check that the Choi matrix $C(\Delta)$ of the the completely depolarizing channel $$ \Delta(\rho):=\mathrm{tr}(\rho)I\tag5 $$ is also the identity. Finally, $C(\cdot)$ is injective, so $\Phi=\Delta$. $\square$

$^1$ Another way to think about this: equation $(4)$ is an eigendecomposition of $C(\Phi)$. Identity is the only Hermitian operator with one as its sole eigenvalue.

$^2$ Yet another way: apply the operator in $(4)$ to any linear combination of the basis vectors $|A_k^\dagger\rangle\!\rangle$.


This solution uses the spectral theorem and some elementary linear algebra computations. If you let $P = \sum_{i = 1}^{d'} \lambda_i \vert{\psi_i}\rangle\langle{\psi_i}\vert$ be the spectral decomposition of $P$ and let $\vert{j}\rangle$'s form an orthonormal basis of $\mathbb{C}^{d}$, then $\frac{1}{\sqrt{\lambda_i}}{\vert\psi_i\rangle\langle{j}\vert}$'s form an orthonormal basis (with respect to $\langle{,}\rangle$) of $V$. The $A_k$'s also form an orthonormal basis and two orthonormal bases are related by a unitary matrix. It follows then by unitary freedom of completely positive maps that the $\frac{1}{\sqrt{\lambda_i}}{\vert\psi_i\rangle\langle{j}\vert}$'s also give a Kraus representation of $\Phi$ and so $$\Phi(\rho) = \sum_{i,j} \frac{1}{\lambda_i}{\vert\psi_i\rangle\langle{j}\vert}\rho{\vert{j}\rangle\langle{\psi_i}\vert} = \left(\sum_{j = 1}^{d} \langle{j}\vert\rho\vert{j}\rangle\right)\sum_{i = 1}^{d'} \frac{1}{\lambda_i}{\vert\psi_i\rangle\langle{\psi_i}\vert} = \operatorname{tr}(\rho) P^{-1}.$$

  • $\begingroup$ $A_k$ aren't guaranteed to be rank one. Pauli basis is a counterexample. $\endgroup$ Feb 27 at 13:28
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    $\begingroup$ @AdamZalcman Unless I am missing something, I don't think this proof requires the $A_k$'s to be rank one. I am using unitary freedom of CP maps to show that $\Phi$ has a Kraus representation consisting of these rank one matrices. $\endgroup$ Feb 27 at 20:09
  • $\begingroup$ Ah, sorry! You're right of course. Apologies for my sloppy reading of your nice answer. Upvoted. $\endgroup$ Feb 28 at 3:51
  • $\begingroup$ @user2533488 I would like to acknowledge that I like this solution too, but there can only be one ticked answer. $\endgroup$ Feb 28 at 12:44

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