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Consider a Kraus representation $\{A_a\}_a$ of the identity channel $\mathcal{I}$ that maps any state to itself. Of course, $\{A_a\}_a$ are not the simplest Kraus operators, which would just be $\{I\}$, and they need not to be orthogonal. Is there a constraint on such Kraus operators $\{A_a\}_a$ that represent $\mathcal{I}$?

(By the isometric equivalence of Kraus operators, I suppose the constraint should be something like $\sum_a c_aA_a=I$ for some square-normalized complex vector $c_a$)

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3 Answers 3

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I can think of two other methods for addressing this question.

First, taking any set of basis states $|m\rangle$ for the Hilbert space of the state, we must have $$\sum_i \langle m|A_i|n\rangle\langle n| A_i^\dagger|m\rangle=\delta_{mn}$$ for all $m$ and $n$ (i.e., the transformation $|n\rangle\langle n|\to|n\rangle\langle n|$). Since the sum adds together only positive numbers $|\langle m|A_i|n\rangle|^2$, we immediately find the requirement $$\langle m|A_i|n\rangle=0\quad \forall m\neq n.$$ This means that, expressed in any basis, each of the Kraus operators must be diagonal! The only matrix that is diagonal in every basis is the one proportional to the identity matrix, so we discover that $$A_i\propto \mathbb{I}\quad \forall i.$$ Normalization is then the only degree of freedom left: the proportionality constants can be changed for each $i$ so long as their squares all sum to unity.

Second, we can appeal to a larger Hilbert space to write the channel as $$\rho\to \mathrm{Tr}_b( U_{ab} \rho\otimes |0\rangle_b\langle 0| U_{ab}^\dagger)=\rho.$$ I don't have the next step of the proof, but it seems that, for this to hold for all $\rho$, one requires $$U_{ab} \rho\otimes |0\rangle_b\langle 0|U_{ab}^\dagger=\rho\otimes \sigma.$$ Then, one must consider only unitaries of the form $U_{ab}=\mathbb{I}\otimes U_b$, which immediately yields Kraus operators of the form $$A_i=\langle i|_b U_{ab}|0\rangle_b=\mathbb{I}_a \langle i|U_b|0\rangle,$$ such that each Kraus operator is again proportional to the identity matrix for our state. Since we already have shown this result using the first method, it must be true, but this second proof lacks the proof that $U_{ab} \rho\otimes |0\rangle_b\langle 0|U_{ab}^\dagger=\rho\otimes \sigma$ must hold.

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Short version

Another approach: observe that finding the Kraus operators for a channel $\Phi$ is equivalent to finding a decomposition for the Choi $J(\Phi)$ in terms of positive-semidefinite unit-rank operators. But the Choi of the identity channel is $J(\operatorname{Id})=d\, |m\rangle\!\langle m|$, $|m\rangle\equiv\frac{1}{\sqrt d}\sum_i |i,i\rangle$, which has unit rank, hence its only possible such decompositions involve scalar multiples of the operator $|m\rangle\!\langle m|$, which corresponds to saying that the only possible Kraus operators for $\operatorname{Id}$ are multiples of the identity operator.

Longer version

To see the above, remember that the Choi is defined as $J(\Phi)\equiv d \,(\Phi\otimes I)(|m\rangle\!\langle m| )$ where $|m\rangle\equiv \frac{1}{\sqrt d}\sum_{a=1}^d |a,a\rangle$, and then for any Kraus decomposition $\Phi(\rho)=\sum_a A_a \rho A_a^\dagger$ we have $$J(\Phi) = \sum_a \mathbb{P}(\operatorname{vec}(A_a)), \qquad \mathbb{P}(u)\equiv uu^\dagger.$$ where $\operatorname{vec}(A_a)$ is the vectorisation of the operator $A_a$, that is, the vector such that $(\operatorname{vec}(A_a) )_{ij} = (A_a)_{ij}$. More explicitly, if $A\equiv \sum_{ij} A_{ij} |i\rangle\!\langle j|$, then $\operatorname{vec}(A)=\sum_{ij} A_{ij} |i,j\rangle$, and vice versa. Note in particular that $\mathbb{P}(\operatorname{vec}(A_a) )$ are positive semidefinite operators (and more specifically, they are multiples of unit-rank projections).

On the other hand, if $J(\Phi)=\sum_a u_a u_a^\dagger$ for some collection of vectors $u_a\equiv \sum_{i\alpha} u^{(a)}_{i,\alpha}|i,\alpha\rangle$, then $$\Phi(\rho) =\sum_a \tilde u_a \rho \tilde u_a^\dagger, \qquad \tilde u_a\equiv \sum_{i\alpha} u^{(a)}_{i,\alpha} |i\rangle\!\langle\alpha|\equiv \operatorname{unvec}(u_a).$$

Now, consider the identity channel $\operatorname{Id}$. This has Choi $$J(\operatorname{Id}) =d \,\mathbb{P}(|m\rangle) \equiv \sum_{i,j} |ii\rangle\!\langle jj|.$$ This Choi has unit rank. But then, in how many ways can one decompose a unit-rank positive semidefinite operators in terms of unit-rank positive semidefinite operators? It is not hard to see that the only possible such decomposition uses a set of operators which are all multiples of each others. In other words, $$|a\rangle\!\langle a| = \sum_a P_a\implies P_a = \lambda_a \, |a\rangle\!\langle a|,$$ for some $\lambda_a\ge 0$ with $\sum_a \lambda_a=1$ (under the assumption that each $P_a$ is positive semidefinite).

So, applying this to our Choi, we see that any decomposition for $J(\operatorname{Id})$ uses operators multiple of $\sum_{i,j} |ii\rangle\!\langle jj|$, which corresponds to any Kraus operator for $\Phi$ being a multiple for the identity.

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  • $\begingroup$ Nice answer. This probably provides the glue for my second proof - that the Choi matrix must be rank one. $\endgroup$ Commented Apr 13, 2022 at 21:24
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The most direct argument is to use the well-known fact that any two Kraus representations of the same channel are related by a partial isometry $V$ (i.e. $V^\dagger V$ is a projector), by virtue of $$ K_i' = \sum V_{ij} K_j\ . $$ Since you have one Kraus representation with a single Kraus operator $K_1=I$, this will thus hold for any other Kraus representation $\{K_i'\}$. (But, of course, you can still have several Kraus operators, all proportional to $I$.)

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