# What does it mean for a channel to be independent of the input state?

A 2007 paper shows how to construct quantum channels on finite-dimensional Hilbert spaces $$\sigma=\Phi(\rho)=\sum_i K_i \rho K_i^\dagger,\qquad \sum_i K_i^\dagger K_i=\mathbb{I}$$ for which $$\Phi(\rho)=\sigma$$ is independent of $$\rho$$. This seems to be useful for preparing desired final states $$\sigma$$ without having to worry about the input state, acting as some sort of better-than-possible quantum filter (there is 100% probability of success even when the input state is orthogonal to the output state). The construction is as follows:

1. Suppose the output is always some fiducial pure state $$\Phi(\rho)=|0\rangle\langle 0|$$ and we can, without loss of generality, span the Hilbert space with the orthonormal basis $$\{|0\rangle,|1\rangle,\cdots|N\rangle\}$$. Then the set of Kraus operators $$K_i=|0\rangle\langle i|,\qquad i\in\left(0,N\right)$$ does the intended task, even for mixed input states. This is seen by defining $$\rho=\sum_{j,k=0}^N\rho_{j,k}|j\rangle\langle k|$$ such that \begin{align} \Phi(\rho)&=\sum_{i=0}^N |0\rangle\langle i| \left(\sum_{j,k=0}^N\rho_{j,k}|j\rangle\langle k|\right) |i\rangle\langle 0|\\ &= |0\rangle\mathrm{Tr}(\rho)\langle 0|=|0\rangle\langle 0|. \end{align} We seem to have a perfect purification channel!
2. Supposed the output is now some fiducial mixed state $$\Phi(\rho)=\sigma\equiv\sum_{i=0}^N p_i|\phi_i\rangle\langle \phi_i|$$, where some of the $$p_i$$ may be zero. Then the set of Kraus operators $$K_{i,j}=\sqrt{p_j}|\phi_j\rangle\langle i|,\qquad i,j\in\left(0,N\right)$$ does the intended task, regardless of input state. This is similarly seen through \begin{align} \Phi(\rho)&=\sum_{i,j=0}^N \sqrt{p_j}|\phi_j\rangle\langle i| \left(\sum_{k,l=0}^N\rho_{k,l}|k\rangle\langle l|\right) |i\rangle\langle \phi_j|\sqrt{p_j}\\ &= \sum_{j=0}^N p_j|\phi_j\rangle(\sum_{i=0}^N\rho_{i,i})\langle \phi_j|=\sigma. \end{align}

What does this channel imply physically? It is clearly different from unitary channels, which are reversible, because there is no way of determining the input state from the output state. Are such channels feasible in practice, or are they only theoretical constructs?

The given form of the Kraus operators, though not unique, tells us what is physically happening.

1. In the case of pure-state outputs, each Kraus operator takes one of the possible basis states and swaps it with the desired output state. All basis states thus become $$|0\rangle$$ through some sort of swapping mechanism.
2. In the case of mixed-state outputs, each Kraus operator takes one of the possible basis states to one of the desired output states with a probability corresponding to the probability of the output state. We thus have some sort of probabilistic swapping mechanism that swaps in a state $$|i\rangle$$ with probability $$p_i$$.

The key concept above is swapping: one can easily think of enabling this transformation as first constructing the desired output state $$\sigma$$ in some other place and then simply swapping it in place of the input state. In some larger Hilbert space, this would look like the unitary transformation $$\rho\otimes\sigma\to \mathrm{SWAP}(\rho\otimes\sigma)=\sigma\otimes\rho,$$ and we would find the quantum channel to have originated from tracing out this second system: $$\Phi(\rho)=\mathrm{Tr}_2[\mathrm{SWAP}(\rho\otimes\sigma)]=\sigma.$$ In something like a completely depolarizing channel $$\Phi(\rho)=\mathbb{I}$$, this means that we cannot distinguish between random operations acting on a state and the action of the state simply being replaced by a random state. However, this is unlikely to be a physically useful channel for state preparation, as it defers the problem of creating the final state $$\sigma$$ to an auxiliary system.

## Is there any other way of enacting the quantum channels $$\Phi(\rho)$$ without needing an auxiliary system to first create the desired final state $$\sigma$$?

No. Any other physical process leading to the same channel $$\Phi(\rho)$$ must have Kraus operators related to the original ones by a unitary transformation $$E_i=\sum_{j}u_{i,j} K_j,\qquad \sum_{j}u_{i,j}u_{k,j}^*=\delta_{i,k}.$$ This is physically equivalent to performing a local unitary $$V$$ on the auxiliary degree of freedom, as can be seen from the definition of the Kraus operators as being elements of the unitary operators $$U$$ acting on the joint Hilbert spaces: $$K_i=\,_2\langle i|U|0\rangle_2 ,\qquad E_i=\left(\sum_{j}u_{i,j}\,_2\langle j|\right)U|0\rangle_2\equiv \,_2\langle i|(\mathbb{I}\otimes V^\dagger)U|0\rangle_2.$$ We can also consider changing the fiducial state of the auxiliary freedom $$|0\rangle_2$$ in the same way without loss of generality using another unitary $$\mathbb{I}\otimes W$$. The overall operation thus looks like $$\Phi(\rho)= \mathrm{Tr}_2\left[(\mathbb{I}\otimes V^\dagger)U \left(\rho\otimes W|0\rangle\langle 0|W^\dagger\right) U^\dagger (\mathbb{I}\otimes V)\right],$$ from which it is clear that $$V$$ has no impact on the quantum channel and amounts to a local transformation of the part of the Hilbert space that we ultimately ignore.

When $$\sigma=|\psi\rangle\langle\psi|$$ is pure, the channel can be achieved with the local unitary defined through $$|\psi\rangle=W|0\rangle$$ and global unitary $$U$$ swapping the two systems. The only remaining degrees of freedom are the other elements of the unitary $$W$$, which have no effect here. This means that the only channel that can create the pure final state $$\sigma$$ for arbitrary input states $$\rho$$ must first create $$\sigma$$ in an auxiliary system before swapping it with the input state - the channel itself is not the physical miracle it seems to promise. We learn from this unitary description of the overall system that, if the auxiliary system is not prepared in the correct state, the output from the channel will not be the correct state. This reinstates the reversibility of unitary quantum mechanics and explains why the channel $$\Phi(\rho)=\sigma$$ is apparently irreversible: there was an auxiliary system that did not change with changing $$\rho$$.

When $$\sigma$$ is mixed, we need $$|0\rangle$$ to be defined on a larger Hilbert space so that the reduced dynamics seem to incorporate a mixed state. We take $$W|0\rangle$$ to be a purification of $$\sigma$$ and enforce that $$U$$ only act trivially on this auxiliary system so that the overall dynamics are again reproduced (think of splitting $$\mathrm{Tr}_2$$ into first tracing over the auxiliary system introduced for the purification and then providing the partial trace from the case of pure $$\sigma$$). By acting trivially we here mean that $$U$$ act unitarily on the additional system introduced for purification and that tracing out that additional system yields the usual swapping unitary.

## Summary

We learn that the most general transformation $$\Phi(\rho)=\sigma$$ must first create $$\sigma$$ in an auxiliary system, perform some trivial unitaries in auxiliary systems, and swap the created $$\sigma$$ with our input state $$\rho$$. There is no way of creating a magical channel that always outputs your desired state without secretly creating that state every time you use the channel.

• I mean... you could also just throw away the input state and generate $\sigma$. This type of channel could simply model what happens if you throw away a state and consider the state generated by some other completely independent experimental device. Though I suppose here you're wondering about the "larger unitary" operations implementing a channel that forgets its input? In which case the need for an auxiliary system makes sense, as the information would need to go somewhere
– glS
Sep 17 '21 at 21:59
• @glS yes, that's exactly the same thing. It turns out that's the only way to do it, which confirms my intuition that you could never make a channel that always gives you some result for free. The original paper seemed to show that this magic construction exists, whereas I now understand that it simply shifts the burden of constructing the state to somewhere else Sep 18 '21 at 20:22