Does independence from the input state imply a tensor product structure for the unitary?

Question

Let $U$ be a unitary acting on Hilbert space $\mathcal H = \mathcal H_A \otimes \mathcal H_B$ such that $$\mathrm{Tr}_A(U \vert \psi \rangle \langle \psi \vert_A \otimes \vert 0 \rangle \langle 0 \vert_B U^\dagger)$$ is a fixed matrix which independent of the input state $\vert \psi \rangle$. I strongly suspect that this implies that $U$ must have a tensor product structure, i.e. $$U = U_A \otimes U_B,$$ since these unitaries don't 'mix' the two parts of the system. At least, in this case, it is easy to see that the resulting state is independent of $\vert \psi \rangle$, but I am not sure if this is necessarily the case. How would one prove that there is no other unitary for which the input state is irrelevant?

It can be seen that if $$U = \sum_k U_A^k \otimes U_B^k,$$ then \begin{align*} \mathrm{Tr}_A(U \vert \psi \rangle \langle \psi \vert_A \otimes \vert 0 \rangle \langle 0 \vert_B U^\dagger) &= \sum_{k,\ell} \mathrm{Tr}_A(U_A^k \otimes U_B^k \vert \psi \rangle \langle \psi \vert_A \otimes \vert 0 \rangle \langle 0 \vert_B U_A^{\ell\dagger}\otimes U_B^{\ell\dagger})\\ &= \sum_{k,\ell} \mathrm{Tr}_A(U_A^k \vert \psi \rangle \langle \psi \vert_A U_A^{\ell\dagger} \otimes U_B^k \vert 0 \rangle \langle 0 \vert_B U_B^{\ell\dagger})\\ &= \sum_{k,\ell} \langle \psi \vert U_A^{\ell\dagger}U_A^k \vert \psi \rangle \cdot U_B^k \vert 0 \rangle \langle 0 \vert_B U_B^{\ell\dagger}\\ &= \sum_{k,\ell} c_{k,\ell,\psi} U_B^k \vert 0 \rangle \langle 0 \vert_B U_B^{\ell\dagger} \end{align*} So, it seems that if the tensor product for $U$ has more than one term, then the resulting state is not independent of $\vert \psi \rangle$. But what about unitaries that don't have a tensor structure at all? It seems like these should only be worse still but I'm struggling to see this rigorously.

Answer 1

You can frame this as a question about complementary channels to characterise if and when the statement applies. For starters, let me recall a few basic facts:

1. Any channel $\Phi$ can be represented via an isometry $V$ as $\Phi(\rho)=\operatorname{tr}_2[V\rho V^\dagger]$. This is the same as representing a channel via a unitary as $\Phi(\rho)=\operatorname{tr}_2[U(\rho\otimes |0\rangle\!\langle 0|)U^\dagger]$, with $U$ and $V$ related as $V=U(I\otimes |0\rangle)$, which is shorthand for $V |\psi\rangle=U(|\psi\rangle\otimes|0\rangle)$ (or in other words, $V$ is the "first block of columns" of $U$). Thinking in terms of $V$ is generally easier because it allows to not deal with the action of $U$ on all ancillary states orthogonal to $|0\rangle$, which is not physically relevant to the channel.

2. A channel $\Phi$ is isometric, that is, equals $\Phi(\rho)=V\rho V^\dagger$ for some isometry $V$, if and only if its Choi $J(\Phi)$ has unit rank.

3. Given a channel $\Phi(\rho)=\operatorname{tr}_2[V\rho V^\dagger]$, its complementary channel is defined as $\Phi^c(\rho)=\operatorname{tr}_1[V\rho V^\dagger]$. In words, it's the action you get if you interpret the channels as a unitary action in an enlarged space, but instead of looking at the "system" you look at the "environment" (that is, you now consider as output the space you were previously tracing out).

4. A quantum map $\Phi$ has output independent from its input if and only if it is a "replacement map", that is, it has the form $\Phi(\rho)=\langle B,\rho\rangle C\equiv \operatorname{tr}(B^\dagger \rho)C$ for some operators $B,C$. More to the point, a quantum channel has output independent from its input if and only if it has the form $\Phi(\rho)=\langle I,\rho\rangle \sigma\equiv \operatorname{tr}(\rho)\sigma$ for some state $\sigma$.

5. Saying that the unitary $U$ factorises as $U=U_A\otimes U_B$ is kinda like asking whether the isometry $V$ has the form $V=U_A\otimes |\phi\rangle$ for some fixed state $|\phi\rangle$, and unitary $U_A$. Except not exactly, as I'll discuss below.

Given the above background, note that your question can be reframed (and slightly generalised) as: what does $\Phi^c(\rho)=\operatorname{tr}(\rho)\sigma$ imply for the structure of $\Phi$? And in particular, does it imply that its isometric representation looks like $V=U_A\otimes |\phi\rangle$?

To answer this, let's observe the following:

1. If a channel has Kraus decomposition $\Phi(\rho) = \sum_i A_i \rho A_i^\dagger$, then its complementary looks like $$\Phi^c(\rho) = \sum_{ij} \langle A_i^\dagger A_j,\rho\rangle E_{ij}\equiv \sum_{ij} \operatorname{tr}(A_j^\dagger A_i \rho) |i\rangle\!\langle j|. $$ The Choi of $\Phi^c$ can then be written as $$J(\Phi^c) = \sum_{ij} E_{ij} \otimes (A_i^\dagger A_j)^*,$$ with $(A_i^\dagger A_j)^*$ complex conjugate of $A_i^\dagger A_j$.

2. The (a) Kraus decomposition of a replacement channel $\Phi(\rho)=\operatorname{tr}(\rho)\sigma$ is $$\Phi(\rho) = \sum_{ij} A_{ij}\rho A_{ij}^\dagger, \quad A_{ij} \equiv \sqrt{\lambda_i^\sigma} |\lambda_i^\sigma\rangle\!\langle j|,$$ where $\sigma$ eigendecomposes as $\sigma = \sum_i \lambda_i^\sigma |\lambda_i^\sigma\rangle\!\langle\lambda_i^\sigma|$. It follows that the complementary channel has the form $$\Phi^c(\rho) = \sum_{ij,mn} \langle A_{ij}^\dagger A_{mn} ,\rho\rangle |ij\rangle\!\langle mn| = \sum_{ijn} \lambda_i^\sigma \langle E_{jn},\rho\rangle |ij\rangle\!\langle in| = \tilde\sigma\otimes \rho,$$ where $\tilde\sigma\equiv \sum_i \lambda_i^\sigma E_{ii}$, and $E_{ii}\equiv |i\rangle\!\langle i|$.

This is sufficient to reach our general conclusion. Imagine the general action of a channel as an isometric evolution with one input and two output spaces. If looking at the first output space you have no information about the input (i.e. $\Phi$ is a replacement channel), then looking at the second output you must have all of $\rho$ (i.e. $\Phi^c(\rho)=\tilde\sigma\otimes \rho$, which is like saying the second output contains $\rho$ untouched, plus possibly some other fixed ancillary state). This is of course what you should have expected intuitively from the start: if the overall evolution is unitary no information is lost, therefore if the information is not at one hand, it has to be at the other.

^{I should note that I made some simplifications here: in particular, the complementary of a channel is not unique, for the same reason the isometric representation of a channel is not unique. Nonetheless, one can show that all complementary channels of a given channel are related by an isometry, so we don't lose much doing things this way. It does mean however that complementary channels can have the more general structure $W(\tilde\sigma\otimes\rho)W^\dagger$ for some isometry $W$. Furthermore, the dimension of $\tilde\sigma$ equals the number of indices $i$ in these expressions. This means in particular that it can have "dimension one", *e.g.* when $\Phi$ is isometric in the first place. That just means we'd get $\Phi^c(\rho)=\rho$, so not big deal.}

In conclusion: if $\Phi$ has two output spaces, the second one of which knows nothing about the input, then its first output must contain all information about $\rho$, however it can nonetheless be correlated with the second output, as you can see from the fact that $\Phi^c$ contains $\sigma^c$, which obviously is tightly connected to $\sigma$. This means that we cannot conclude that the dilation isometry $V$ representing the channel has the form $V=\tilde V\otimes |\phi\rangle$ for some state $|\phi\rangle$ and isometry $\tilde V$. For example, we could instead have a structure like the following:

But even if we could conclude that $V=\tilde V\otimes |\phi\rangle$, or even better that $V=U_A\otimes |\phi\rangle$ for some unitary $U_A$, we still couldn't conclude that the overall unitary $U$, the one such that $V=U(I\otimes |0\rangle)$, factorises. The reason being that $U$ could for example decompose as $U=\sum_i U^A_i\otimes |i\rangle\!\langle i|$, meaning that $U$ could have a more complex "entangling" action, which you however cannot see by fixing the initial ancilla to $|0\rangle$.

Answer 2

If you apply a controlled-$V$, controlled off system $B$ and targeting system $A$, then because system $B$ is in $|0\rangle$, it will do absolutely nothing. Thus, the output after tracing will be independent of $|\psi\rangle$, while the unitary does not have a tensor product structure.