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Background

Suppose I have a quantum channel $\Phi:B(\mathcal{H}_1)\rightarrow B(\mathcal{H}_1)\otimes B(\mathcal{H}_2)$, such that there is some small $\epsilon$ such that for any two input states $\rho$ and $\sigma$

$$ \Vert \rho - \sigma\Vert_1 (1-\epsilon) \leq \Vert\text{Tr}_2(\Phi(\rho)) - \text{Tr}_2(\Phi(\sigma))\Vert_1.\tag{1}$$

That is, the channel almost preserves distance even if we trace out the second system. This makes me think that the second system can't have much dependence on the first system, i.e., there is some channel $\Phi_1:B(\mathcal{H}_1)\rightarrow B(\mathcal{H}_1)$ such that $\Phi$ is close to $\Phi_1\otimes \rho_0$, i.e., a channel that just applies some channel to the first system and tacks on a fixed state to the second system.

Somehow this needs to use something like no-cloning, because an ill-defined map $\Psi:\rho\mapsto \rho\otimes \rho$ satisfies the above inequality, but is not a quantum channel.

Question

Is there any way to prove that $\Phi$ has this form of "close to just adding a constant state to the second system"?

To phrase this formally: For any $\delta>0$, is there an $\epsilon >0$ such that for any channel $\Phi$ satisfying equation (1) for all input states, then there exists a channel $\Phi_1:B(\mathcal{H}_1)\rightarrow B(\mathcal{H}_1)$ and a state $\rho_0\in B(\mathcal{H}_2)$ such that $\Phi$ is within distance $\delta$ of the channel $\tilde{\Phi}:\rho\mapsto \Phi_1(\rho)\otimes \rho_0$?

Additional context

I am imagining two unitarities $U_1$ and $U_2$ whose action differs only on 2 basis states. That is, $\Vert U_1 - U_2\Vert = 2$, but there is a subspace $V$ of almost the full space such that $U_1\vert_V = U_2\vert_V$. Now I only have noisy channels $\mathcal{\tilde{U}}_{i}$ that implement these, i.e. $\mathcal{\tilde{U}_i} = (1-p)U_i + p \mathcal{D}$ for some noise channel $\mathcal{D}$. Then considering the channel

$$ (I\otimes \mathcal{\tilde{U}}_i)\Phi \tag{2}$$

I want to argue that there is some trade-off between the fidelity of this channel and its ability to distinguish between $U_1$ and $U_2$. That is, if I have two input states $\rho$ and $\sigma$ that distinguish $U_1$ and $U_2$, then after I apply $\Phi$, if too much information about the input state is in the second system, then the system has an irrecoverable loss because $\mathcal{\tilde{U}}_i$ is noisy, but if not enough information about the input state is int he second system, then it can't distinguish $U_1$ from $U_2$.

The original question should solve this (if $\Phi$ is close to $\Phi_1\otimes\rho_0$, then it can't distinguish $U_1$ from $U_2$ very well) but maybe there are other approaches.

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  • $\begingroup$ I haven't fully digested the question so maybe I'm way off here, but my gut reaction is that this seems like it's maybe related to some sort of approximately "degradable" or "antidegradable" quantum channel. $\endgroup$
    – Condo
    Aug 26 at 13:38
  • $\begingroup$ That might help, though the second system is not an environmental system. I thought about Stinespring-dilating into an environmental system to get, e.g., entropy equalities, but didn't find anything useful. $\endgroup$
    – Sam Jaques
    Aug 26 at 13:42
  • $\begingroup$ "Given a quantum channel and an epsilon, is there a way to prove that the channel satisfies Eq. 1 for all possible pairs of input states?" Is that your question? $\endgroup$ Aug 29 at 0:38
  • $\begingroup$ No, the question is: given an epsilon and a channel $\Phi$ that satisfies Eq. 1, is there a state $\rho_0$ and channel $\Phi_1$ such that $\Phi$ is close to $\Phi_1\otimes \rho_0$? $\endgroup$
    – Sam Jaques
    Aug 29 at 10:28
  • $\begingroup$ Thanks for modifying the question. How is the distance between $\Phi$ and $\tilde{\Phi}$ being defined? Diamond norm? $\endgroup$ Aug 29 at 18:50

2 Answers 2

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I suppose you're asking the following: for any $\epsilon\ge0$ and $\Phi$ that satisfy conditions, is there $\delta_\epsilon \rightarrow 0$ when $\epsilon \rightarrow 0$, such that there exists a special channel in delta proximity?

I think yes. This is not a full proof but the idea is following.

Let $\Psi(\rho) = {\rm Tr}_2(Φ(\rho))$, where $\Psi:B(\mathcal{H}_1)\rightarrow B(\mathcal{H}_1)$ is also a channel.

If $\epsilon=0$, i.e. $\Psi$ is distance preserving, then $\Psi(\rho) = U\rho U^\dagger$ for some unitary $U$. See, e.g., Theorem 7 in On Partially Trace Distance Preserving Maps and Reversible Quantum Channels. The way to prove this theorem is to observe that if $p$ and $q$ have orthogonal supports then $\Vert p-q \Vert_1 =2$. Hence $\Psi(p)$ and $\Psi(q)$ also have orthogonal supports. It follows that for any basis $|b_i\rangle \in \mathcal{H}_1$ the supports of operators $\Psi(|b_i\rangle\langle b_i|)$ are orthogonal to each other. Thus they are rank-1 projectors (assuming finite dimensional case). This means $\Psi$ is rank preserving (on Hermitian operators). Distance preservation implies that fidelity is also preserved for rank-1 projectors. Which means that for any unit vectors $|a\rangle,|b\rangle$ $$ {\rm Tr}(\Psi(|a\rangle\langle a|)\Psi(|b\rangle\langle b|)) = |\langle a|b \rangle|^2. $$

Though, I don't see an easy way to complete the proof of Theorem 7 from here.

Anyway, once we know that $\Psi(\rho) = U\rho U^\dagger$ we can show that $\Phi = \Psi \otimes \rho_0$. This is indeed a version of no-cloning.

It's well known that if the reduced state $\rho_A$ of a bipartite pure state $\rho_{AB}$ is pure, then the state must be a tensor product of pure states, i.e. $\rho_{AB} = \rho_A \otimes \rho_B$. Hence, if for a mixed $\rho_{AB}$ the reduced state $\rho_A$ is pure, then $\rho_{AB} = \rho_A \otimes \rho_B$ as well due to linearity. For any pure $\rho$ we have that ${\rm Tr}_2(\Phi(\rho)) = U\rho U^\dagger$ is pure. Thus $\Phi(\rho) = U\rho U^\dagger \otimes {\rm Tr}_1(\Phi(\rho))$ for pure $\rho$.

Now assume that ${\rm Tr}_1(Φ(\rho_1)) \neq {\rm Tr}_1(Φ(\rho_2))$ for two different pure non-orthogonal states $\rho_1,\rho_2$. It's easy to see that given channels $Φ$ and $\Psi$ we can clone ${\rm Tr}_1(Φ(\rho))$ as much as we want given only a single copy of $\rho$. Therefore we can discriminate $\rho_1,\rho_2$ with the access to channels $Φ, \Psi$, which is known to be impossible in theory. Thus ${\rm Tr}_1(Φ(\rho))$ must be constant.

Now let $\epsilon > 0$. In this case $\Psi$ is almost distance preserving. Yet, it's possible to prove that $\Psi$ must be close to a distance preserving map if $\epsilon$ is close to $0$.

Again, we have that $\Vert \Psi(\rho_i) - \Psi(\rho_j) \Vert_1 \approx 2$ for a complete set of pure $\{\rho_i\}_i$ where $\rho_i \perp \rho_j$. Using the inequality $D(p,q) \le \sqrt{1-F(p,q)}$ between distance and fidelity you can show that $F(\Psi(\rho_i),\Psi(\rho_j)) \approx 0$, and thus ${\rm Tr}(\Psi(\rho_i),\Psi(\rho_j)) \approx 0$. So that $\Psi(\rho_i)$ are almost orthogonal to each other. It follows that they are almost rank-1 projections. Thus $\Psi$ is close to $\Psi'$ that preserves distances exactly. That is, $\Psi'(\rho) = U\rho U^\dagger$.

Consider the channel $\Pi(\rho) = (U^\dagger \otimes I)\Phi(\rho)(U \otimes I)$. The channel ${\rm Tr}_2(\Pi(\rho)) = U^\dagger\Psi(\rho)U$ must be close to identity. To prove that ${\rm Tr}_1(\Pi(\rho))$ is close to a constant we can use the same no-cloning argument. That is, we can clone ${\rm Tr}_1(\Pi(\rho))$ by iterative application of $\Pi$, up to some error dependent on $\epsilon$ and iteration step. Of course, $\epsilon$ has to be small enough for this to work.

Update

A bit more details. Let $d = \dim(\mathcal{H}_1)$, and for a set of pure $\{\rho_i\}_{i=1}^d$ where $\rho_i \perp \rho_j $ we have $$ {\rm Tr}(\Psi(\rho_i)\Psi(\rho_j)) < \epsilon_1 $$ for any $i\neq j$ and small $\epsilon_1>0$. Consider the sum $$ S = \frac{1}{d} \sum_{i=1}^d \Psi(\rho_i). $$ It's a state since $S\ge 0$ and ${\rm Tr}(S)=1$. For any state we have ${\rm Tr}(S^2) \ge \frac{1}{d}$ with the equality only if $S$ is maximally mixed. Let $$ m = \sum_i {\rm Tr}(\Psi(\rho_i)^2). $$ Then $$ \frac{1}{d} \le {\rm Tr}(S^2) = \frac{1}{d^2} (\sum_i {\rm Tr}(\Psi(\rho_i)^2) + \sum_{i \neq j}{\rm Tr}(\Psi(\rho_i)\Psi(\rho_j)) ) \le \frac{1}{d^2} (m + d(d-1)\epsilon_1). $$ Thus $$ \sum_i {\rm Tr}(\Psi(\rho_i)^2) = m \ge d - d(d-1)\epsilon_1 \approx d, $$ hence ${\rm Tr}(\Psi(\rho_i)^2) \approx 1$ for all $i$, so that each $\Psi(\rho_i)$ must be close to pure.

It's indeed not that trivial to show from this that $\Psi$ is close to a unitary channel, even though we have such a property in the exact case. But it looks very natural. I think there must be a proof of Theorem 7 that could be modified for this case.

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  • $\begingroup$ This looks like it does it, but I'm having a lot of trouble showing that if $\text{Tr}(\Psi(\rho_i),\Psi(\rho_j)) \geq 1-\epsilon $, that $\Psi$ must be close to unitary (or even that $\{\Psi(\rho_i)\}$ are close to mutually orthogonal rank-1 projections). The usual proofs that fidelity preservation implies unitarity look very exact. $\endgroup$
    – Sam Jaques
    Sep 1 at 10:37
  • $\begingroup$ Yeah, apparently it's not that trivial. I've updated the answer. $\endgroup$
    – Danylo Y
    Sep 1 at 12:00
  • $\begingroup$ Ah I think I can see an ugly way to do this from what you've written. If we can show that for any given set of orthogonal pure states, $\Psi$ maps it to close to a set of orthogonal pure states, then on those states, $\Psi$ is close to a unitary. Then we can go through all $2\times 2$ subsystems and pick a different pair of orthogonal pure states -- on which $\Psi$ is close to a different unitary -- and use a simple inner product argument to show that the two unitaries are close to each other. This extends to show that $\Psi$ is similar to one of the unitaries on all rank-1 matrices. $\endgroup$
    – Sam Jaques
    Sep 1 at 12:38
  • $\begingroup$ I think something like that should work. But at first we need to prove the theorem for $2 \times 2$ systems :) $\endgroup$
    – Danylo Y
    Sep 1 at 12:57
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I think that $\Phi$ is indeed close to a channel of the form $\Phi_1\otimes\rho_0$. My argument consists of two parts. First, I rigorously bound the diamond distance between the two channels in terms the trace distance between some operators. Then, I argue informally why the latter should be small.

Upper bound on diamond distance

Let $J(\Phi)$ denote the state corresponding to $\Phi$ under Choi-Jamiołkowski isomorophism, i.e. $J(\Phi)=\frac{1}{N_1}\sum_{ij}\Phi(|i\rangle\langle j|)\otimes|i\rangle\langle j|$ where $N_1=\dim\mathcal{H}_1$. Using one of the inequalities in this answer, we can bound the diamond distance in terms of the trace distance of the Choi matrices $$ \begin{align} \|\Phi-\Phi_1\otimes\rho_0\|_\diamond&\le N_1\|J(\Phi)-J(\Phi_1\otimes\rho_0)\|_1.\tag{a} \end{align} $$

A Choi matrix can be written in block form $$ \begin{align} J(\Phi)&=\frac{1}{N_1}\left\|\sum_{i,j=1}^{N_1}\Phi(|i\rangle\langle j|)\otimes|i\rangle\langle j|\right\|_1\\ &=\frac{1}{N_1}\begin{bmatrix}\Phi(E_{11})&\dots&\Phi(E_{1N_1})\\\vdots&&\vdots\\\Phi(E_{N_11})&\dots&\Phi(E_{N_1N_1})\end{bmatrix} \end{align}\tag{b} $$ where $E_{ij}=|i\rangle\langle j|$. We can bound the trace norm of $n\times n$ block matrix of this form in terms of its rank and the trace norm of its blocks $$ \begin{align} \left\|\sum_{ij}A_{ij}\otimes|i\rangle\langle j|\right\|_1&= \|\begin{bmatrix}A_{11}&\dots&A_{1n}\\\vdots&&\vdots\\A_{n1}&\dots&A_{nn}\end{bmatrix}\|_1 \le\sqrt{r}\,\|\begin{bmatrix}A_{11}&\dots&A_{1n}\\\vdots&&\vdots\\A_{n1}&\dots&A_{nn}\end{bmatrix}\|_F\\ &=\sqrt{r\sum_{i,j=1}^n\|A_{ij}\|_F^2}\le\sqrt{r\sum_{i,j=1}^n\|A_{ij}\|_1^2} \end{align}\tag{c} $$ where $r$ denotes the rank of the full matrix $A$ and $\|.\|_F$ denotes the Frobenius norm. Using $(a)$, $(b)$ and $(c)$, we get $$ \|\Phi-\Phi_1\otimes\rho_0\|_\diamond\le\sqrt{r_J\sum_{i,j=1}^{N_1}\|\Phi(|i\rangle\langle j|)-\Phi_1(|i\rangle\langle j|))\otimes\rho_0\|_1^2}\tag{d} $$ where $r_J$ denotes the rank of $J(\Phi-\Phi_1\otimes\rho_0)$.

Entanglement harms distinguishability on subsystems

Now I will informally argue that $(1)$ implies that $\|\Phi(X)-\Phi(X)_1\otimes\rho_0\|_1$ - and therefore also the right-hand side of $(d)$ - is small for any operator $X$. We set $\Phi_1=\mathrm{tr}_2\circ\Phi$ where $\circ$ denotes channel composition.

First, note that using Stinespring dilation, we can extend the codomain of $\Phi$ from $B(\mathcal{H}_1)\otimes B(\mathcal{H}_2)$ to $B(\mathcal{H}_1)\otimes B(\mathcal{H}_2)\otimes B(\mathcal{H}_3)$ in such a way that $\Phi(\rho)=\mathrm{tr}_3(\Phi'(\rho))$ and $\Phi'(\rho)=U(\rho\otimes|0\rangle_{2,3}\langle 0|)U^\dagger$ for a unitary $U$. Moreover, $\Phi'$ satisfies $(1)$ as long as we replace the partial trace over subsystem $2$ with the partial trace over subsystems $2$ and $3$. Now, choose $\rho$ and $\sigma$ to be pure orthogonal states and suppose that $U$ maps $\rho\otimes|0\rangle_{2,3}\langle 0|$ to a fairly entangled state. In this case, the left-hand side of $(1)$ is $2-2\epsilon$ which is nearly the maximum possible value $2$. On the other hand, $\mathrm{tr}_{2,3}(\Phi'(\rho))$ is fairly mixed. Thus, we should be able to choose $\sigma$ so that $\mathrm{tr}_{2,3}(\Phi'(\rho))$ and $\mathrm{tr}_{2,3}(\Phi'(\sigma))$ have non-negligible overlap. Therefore, the states are not easily distinguishable by observables acting on subsystem $2$ alone. Consequently, the right-hand side of $(1)$ will be much less than $2$. Thus, $U$ cannot map $\rho\otimes|0\rangle_{2,3}\langle 0|$ to states with significant amount of entanglement. Therefore, we can take $U$ to be nearly a product unitary $U\approx U_1\otimes U_{2,3}$. However, in this case $\|\Phi(X)-\mathrm{tr}_2(\Phi(X))\otimes\rho_0\|_1$ is very small and therefore our upper bound $(d)$ on the diamond distance $\|\Phi-\Phi_1\otimes\rho_0\|_\diamond$ is also very small.

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    $\begingroup$ I think for my end goal the 1-norm is good enough. Couldn't I have $\text{tr}_{2,3}(\Phi'(\rho))$ be highly mixed (because the larger system is entangled), but it's still constrained to a subspace that is orthogonal to $\text{tr}_{2,3}(\Phi'(\sigma)$? $\endgroup$
    – Sam Jaques
    Sep 1 at 10:47
  • $\begingroup$ Yes, $\mathrm{tr}_{2,3}(\Phi'(\sigma))$ could live in a subspace orthogonal to $\mathrm{tr}_{2,3}(\Phi'(\rho))$, which is why I wrote that we should be able to choose $\sigma$ so that the overlap is non-negligible. We're able to choose $\sigma$ this way because $\mathcal{H}_1$ is finite-dimensional. Essentially, while it is possible for any given $\mathrm{tr}_{2,3}(\Phi'(\sigma))$ to have small overlap with $\mathrm{tr}_{2,3}(\Phi'(\rho))$ it is not possible that $N_1-1$ mutually orthonormal candidate $\sigma$s do, unless $\mathrm{tr}_{2,3}(\Phi'(\rho))$ is itself nearly pure. $\endgroup$ Sep 1 at 16:51
  • $\begingroup$ Okay, thanks, I think I get the rough idea now, although it seems like quantifying the details of how much mixing implies how much entanglement implies how close to the other state might be very tricky. I think it will be easier to flesh out the details of Danylo Y's answer. $\endgroup$
    – Sam Jaques
    Sep 1 at 16:59
  • $\begingroup$ Fair enough! Danylo did write a great answer! Thanks for an interesting question :-) $\endgroup$ Sep 1 at 17:01
  • $\begingroup$ Actually I tried to work out the details here, and this led to showing that the Schmidt decomposition of a rank-1 projection must have one coefficient of something like $1-O(\epsilon)$. But this hits roughly the same sticking point as Danylo's answer: showing that mutually orthogonal rank-1 projections each map to close to some other set of mutually orthogonal rank-1 projections. $\endgroup$
    – Sam Jaques
    Sep 2 at 10:53

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