Proof that two sets of quantum maps are equivalent only when they are related by a unitary transformation

Question

I am trying to show that the two different quantum maps $\rho'=\sum_{\alpha} K_{\alpha} \rho K_{\alpha}^{\dagger}$ and $\rho''=\sum_{\beta} L_{\beta} \rho L_{\beta}^{\dagger}$ are equivalent i.e. $\rho'=\rho''$ iff \begin{equation} K_{\alpha}=\sum_{\beta}U_{\alpha\beta}L_{\beta}\tag{1} \end{equation} with $U_{\alpha\beta}$ being unitary.

I am able to do the 'if' part of the proof where by assuming the existence of such an unitary $U$, we show that $\rho'$ equals $\rho''$. However, I am getting stuck when I try to show the other part. I start by assuming that \begin{equation} K_{\alpha}=\sum_{\beta}A_{\alpha\beta}L_{\beta} \tag{2} \end{equation} where $A$ is arbitrarily chosen. Then after setting $\rho'=\rho''$ and plugging in the expression for $K_{\alpha}$, I arrive at an expression \begin{equation} \sum_{\alpha}A^{*}_{\alpha\beta'}A_{\alpha\beta}=\delta_{\beta'\beta} \tag{3} \end{equation} which gives me the condition on $A$ as $A^{\dagger}A=I$.

I also need to show that $AA^{\dagger}=I$ in order to prove that $A$ must be unitary. From the equation $(3)$ however I am unable to derive it.

Any help regarding how to proceed would be most helpful.

Answer 1

Let's take as a given that $A^\dagger A=I$, and that $A$ is square (that just means being careful about indices in your derivation, and perhaps introducing some padding of 0 operators). What is $AA^\dagger$? Let's call it $X$. It is Hermitian, and hence has a spectral decomposition.

Now, $$ X^2=A(A^\dagger A)A^\dagger=X. $$ Thus, $X$ has eigenvalues 0,1. Moreover, $$ \text{Tr}(X)=\text{Tr}(A^\dagger A)=\text{Tr}(I), $$ and the trace is equal to the sum of the eigenvalues. Hence, it must be that $X$ is a Hermitian operator with all eigenvalues 1. It is the identity.

In other words, for a square matrix, you never have to show that both $AA^\dagger=I$ and $A^\dagger A=I$. If one is true, both are true.

Answer 2

The result is actually a bit more general than that, as the connecting coefficients are not necessarily elements of a unitary. To see this, observe the following:

1. Let $\Phi$ be an arbitrary quantum channel. This means its Choi representation, defined as $$J(\Phi)=\sum_{ij} \Phi(E_{ij})\otimes E_{ij}, \qquad E_{ij}\equiv|i\rangle\!\langle j|,$$ is positive semidefinite. The connection between Kraus operators and Choi representation is that $\Phi(X)=\sum_a A_a X A_a^\dagger$ iff $J(\Phi)=\sum_a \mathbb{P}(\operatorname{vec}(A_a))$, where here $\operatorname{vec}(A)\equiv \sum_{ij} A_{ij} |i,j\rangle$ is the vectorisation of the operator $A$, and $\mathbb{P}(v)\equiv vv^\dagger$.

2. Note that more generally, given any linear operator $A$, linear decompositions of the form $A=\sum_i \mathbf v_i\mathbf v_i^\dagger$ for some set of vectors $\mathbf v_i$ can be equivalently written as $A=VV^\dagger$ with $V\equiv \sum_i \mathbf v_i \mathbf e_i^\dagger$ the matrix whose $i$-th column is $\mathbf v_i$.

3. Given any Hermitian positive semidefinite matrix $A\ge0$, if $A=BB^\dagger=CC^\dagger$, then there must be an isometry $V$ such that $C=BV^\dagger$. I show this e.g. in the last paragraph of this other answer, but it is relatively straightforward to show reasoning in terms of SVD of $B$ and $C$.

4. Putting the above together, we see that the Choi of a quantum channel has the form $J(\Phi)=\tilde A\tilde A^\dagger$ for some operator $\tilde A$ (which is representable as the matrix whose columns are the vectorisations of the Kraus operators $A_a$). Furthermore, any set of Kraus operators corresponds to a different such decomposition. Therefore two sets of Kraus operators correspond to decompositions $J(\Phi)=\tilde A\tilde A^\dagger=\tilde B\tilde B^\dagger$, which as discussed above mean they are related as $\tilde A=\tilde BV^\dagger$ for some isometry $V$.

   The conclusion is now straightforward: the individual Kraus operators (or more precisely, their vectorisations) are recovered as the columns of these $\tilde A,\tilde B$, so that for example $\operatorname{vec}(A_a)=\tilde A |a\rangle$. Therefore $$\tilde A |a\rangle = \tilde B V^\dagger |a\rangle = \sum_b \tilde B |b\rangle \bar V_{ab} \iff \operatorname{vec}(A_a) = \sum_b \bar V_{ab} \operatorname{vec}(B_b) \iff A_a =\sum_b \bar V_{ab} B_b.$$

In conclusion, we showed that if $$\Phi(X) = \sum_a A_a X A_a^\dagger = \sum_b B_b X B_b^\dagger,$$ then there is some isometry $V$ such that $$A_a = \sum_b \bar V_{ab} B_b.$$ In the special cases where the number of operators $\{A_a\}_a$ is equal to the number of operators in $\{B_b\}_b$, the isometry $V$ is also unitary, and therefore so is $V^\dagger$, and we recover the original statement in the question.