Given a quantum channel (CPTP map) $\Phi:\mathcal X\to\mathcal Y$, its adjoint is the CPTP map $\Phi^\dagger:\mathcal Y\to\mathcal X$ such that, for all $X\in\mathcal X$ and $Y\in\mathcal Y$, $$\langle Y,\Phi(X)\rangle= \langle \Phi^\dagger(Y),X\rangle,$$ where $\newcommand{\tr}{\operatorname{tr}}\langle X,Y\rangle\equiv \tr(X^\dagger Y)$.

For example, if $\Phi$ is the trace map, $\Phi(X)=\tr(X)$, then $\Phi^\dagger(\alpha)=\alpha I$ for $\alpha\in\mathbb C$, as follows from $\langle \alpha,\Phi(Y)\rangle = \tr(Y) \alpha^* = \langle \Phi^\dagger(\alpha),Y\rangle$.

Another example is the partial trace map. If $\Phi(X)\equiv\tr_2(X)$, then $\Phi^\dagger(Y)=Y\otimes I$.

Is there any general physical interpretation for the adjoint channel?


The adjoint of a channel $\Phi$ represents how observables transform (in the Heisenburg picture), under the physical process for which $\Phi$ is the description of how states transform (in the Schrödinger picture). So, in particular, the expected value of a measurement of the observable $E$ on a state $\Phi(\rho)$ is equivalent to the expected value of the observable $\Phi^\dagger(E)$ on the state $\rho$.


The key is to utilize the Kraus decomposition along with the Hilbert-Schmidt inner product: Given a quantum channel, $\mathcal{N}$ with Kraus operators $\left\{V_{l}\right\}$, we have, $$ \begin{align}\langle Y, \mathcal{N}(X)\rangle &=\operatorname{Tr}\left\{Y^{\dagger} \sum_{l} V_{l} X V_{l}^{\dagger}\right\}=\operatorname{Tr}\left\{\sum_{l} V_{l}^{\dagger} Y^{\dagger} V_{l} X\right\} \\ &=\operatorname{Tr}\left\{\left(\sum_{l} V_{l}^{\dagger} Y V_{l}\right)^{\dagger} X\right\}=\left\langle\sum_{l} V_{l}^{\dagger} Y V_{l}, X\right\rangle \end{align} $$

Therefore, the adjoint of a quantum channel $\mathcal{N}$ is given by $$ \mathcal{N}^{\dagger}(Y)=\sum_{l} V_{l}^{\dagger} Y V_{l} $$

Notice that the adjoint channel is CP (since it admits a Kraus decomposition) and unital (from the trace-preserving property of the original channel). Now, here's a way to interpret the adjoint channel: Let $\{ \Lambda^{j} \}$ be a POVM, then the probability of getting outcome $j$ from a measurement on state $\rho$ is $$ p_{J}(j)=\operatorname{Tr}\left\{\Lambda^{j} \mathcal{N}\left(\rho\right)\right\}=\operatorname{Tr}\left\{\mathcal{N}^{\dagger}\left(\Lambda^{j}\right) \rho\right\} $$

The latter expression can be interpreted as the Heisenberg picture, where we evolve the ``observables'' instead of the state $\rho$ under the action of the channel $\mathcal{N}$.

You can find more details in these lecture notes by Mark Wilde.


This may be broader than what you're looking for, but it's clear from your question that you've read up on the QIT materials on the subject already. So I'll try to give a different perspective (more GR-ish) that I think is much more intuitive. The concepts are very portable, so hopefully it's helpful.

I usually think of adjoint operations in terms of pullbacks and their adjoint pushforwards. For a simple example, assume we have smooth maps $f: \mathcal{Y} \rightarrow \mathbb{R}$ and $g: \mathcal{X} \rightarrow \mathcal{Y}$, as shown below. In this case, the pullback of $f$ to $\mathcal{X}$ is simply the composition $\Phi = f \circ g$.


While it's straightforward to pull functions on $\mathcal{Y}$ back to $\mathcal{X}$, even if we had a function mapping $\mathcal{X} \rightarrow \mathbb{R}$ there would be no way to push that function forward to $\mathcal{Y}$. The maps we have available aren't sufficient to define that kind of transfer.

However we can define the pushforward of a vector from $\mathcal{X}$ to $\mathcal{Y}$, which is the adjoint to the pullback described above. This is possible because we can treat vectors as derivative operators that map functions to $\mathbb{R}$.

For a vector at point $x$ on $\mathcal{X}$, say $V(x)$, the pushforward vector $\Phi^\dagger V$ at point $g(x)$ on $\mathcal{Y}$ can be defined in terms of its action on functions of $\mathcal{Y}$: $$(\Phi^\dagger V)(f) = V(\Phi f).$$ So the action of $\Phi^\dagger V$ on a function is the action of $V$ on the pullback of that function.


From a practical standpoint, we can take a basis for vectors on $\mathcal{X}$ as ${\partial {}_\mu} = {\partial }/{\partial x^\mu}$ and the same for $\mathcal{Y}$, ${\partial {}_\nu} = {\partial }/{\partial y^\nu}$. To relate $V = V^\mu \partial {}_\mu$ to $(\Phi^\dagger V)=(\Phi^\dagger V)^\nu \partial {}_\nu$ we only need the chain rule: $$(\Phi^\dagger V)^\nu \partial {}_\nu f = V^\mu \partial {}_\mu(\Phi f) = V^\mu \partial {}_\mu(f \circ g) = V^\mu(\partial y^\nu / \partial x^\mu) \partial {}_\nu f.$$ This leads directly to the matrix $$(\Phi^\dagger)^\nu{}_\mu = \partial y^\nu / \partial x^\mu.$$ You can see after all this that the adjoint of the pullback, a vector pushforward, is essentially a generalization of a coordinate transformation.

This was a bit long winded, but still doesn't do the subject justice. If you think this approach to building intuition might be helpful, Sean Carroll has a phenomenal exposition on the subject in Appendix A, Maps between Manifolds, in Spacetime And Geometry.

  • $\begingroup$ Perhaps you should clarify how this provides a physical interpretation of the adjoint channel. $\endgroup$ – Niel de Beaudrap Jul 18 '20 at 19:56
  • $\begingroup$ @NieldeBeaudrap Physical interpretation is open to interpretation when talking about quantum mechanics. For my part, I think of GR and differential geometry as a much more natural and physical setting to get a deep understanding of what an adjoint operation is, and the concepts are easy to adapt to things like quantum channels once you understand their nature and purpose. Also, my past interactions with glS suggests he likes to dig into topics and understand them deeply. In my opinion GR has the best resources to do that (or at least to supplement) for this particular subject. $\endgroup$ – Jonathan Trousdale Jul 18 '20 at 23:06
  • $\begingroup$ Ultimately it's a very open ended question. I just answered it by trying to walk through the concept in a way that was, and still is, very helpful to me. Hopefully it's helpful to someone else. $\endgroup$ – Jonathan Trousdale Jul 18 '20 at 23:15

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