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

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 $$$$K_{\alpha}=\sum_{\beta}U_{\alpha\beta}L_{\beta}\tag{1}$$$$ with $$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 $$$$K_{\alpha}=\sum_{\beta}A_{\alpha\beta}L_{\beta} \tag{2}$$$$ where $$A$$ is arbitrarily chosen. Then after setting $$\rho'=\rho''$$ and plugging in the expression for $$K_{\alpha}$$, I arrive at an expression $$$$\sum_{\alpha}A^{*}_{\alpha\beta'}A_{\alpha\beta}=\delta_{\beta'\beta} \tag{3}$$$$ which gives me the condition on $$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.

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.

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.