Can any "control-something unitary" be written with control and target spaces flipped?

Question

Consider a simple two-qubit gate such as the CNOT. The typical presentation of this gate is $$\text{CNOT} = |0\rangle\!\langle0|\otimes I + |1\rangle\!\langle1|\otimes X,$$ with $X$ the Pauli $X$ gate. This gate can also be written reversing the role of control and target, as $$\text{CNOT} = I\otimes |+\rangle\!\langle+| + Z\otimes |-\rangle\!\langle-|.$$ Is this sort of trick a general feature of unitary gates which have a "control-target structure" (that is, unitaries acting on a bipartite space which are block-diagonal in some basis)? More precisely, consider a generic unitary $\mathcal U:\mathcal H_A\otimes\mathcal H_B\to\mathcal H_A\otimes\mathcal H_B$ which has the form $$\mathcal U = \sum_i |u_i\rangle\!\langle u_i|\otimes U_i,$$ for some orthonormal set $\{|u_i\rangle\}_i\in\mathcal H_A$ and collection of unitaries $\{U_i\}_i$.

Is there always a set of orthonormal vectors $\{|v_j\rangle\}_j\subset\mathcal H_B$ and collection of unitaries $\{V_j\}_j$ such that $$\mathcal U = \sum_j V_j\otimes |v_j\rangle\!\langle v_j|$$ also holds?

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This question is closely related to General approach for switching control and target qubit, albeit here I'm focusing on a more general case.

Answer 1

Something I realised while writing the question:

This is possible whenever the gates $U_i$ are simultaneously diagonalisable. Indeed, suppose this is the case, and we can thus write the eigendecompositions of the $U_i$ as $$U_i = \sum_j \lambda_{ij} |x_j\rangle\!\langle x_j|,$$ for some basis $\{|x_j\rangle\}_j$ (that is equal for all $i$), and with $\{\lambda_{ij}\}_j$ the eigenvalues of $U_i$. These must therefore all be such that $|\lambda_{ij}|=1$.

We then have $$\sum_i |u_i\rangle\!\langle u_i| \otimes U_i = \sum_{ij} |u_i\rangle\!\langle u_i| \otimes \lambda_{ij} |x_j\rangle\!\langle x_j| = \sum_j \left(\sum_i \lambda_{ij} |u_i\rangle\!\langle u_i|\right) \otimes |x_j\rangle\!\langle x_j|.$$ We can therefore define $V_j\equiv \sum_i \lambda_{ij} |u_i\rangle\!\langle u_i|$, and obtain a new set of unitaries which are "controlled", with the (orthonormal) states $|x_j\rangle$ acting as controls.

I'm not sure whether this condition is also necessary though.

Answer 2

Yes, you can always do this. E.g. see this blog post: https://algassert.com/post/1706

Suppose you have two simultaneously diagonalizable unitary operations $A = \sum_k a_k |\lambda_k\rangle\langle \lambda_k|$ and $B = \sum_k b_k |\lambda_k\rangle\langle \lambda_k|$. For example, if they operate on different qubits then this property is met. Let the "control product" of the two operations be the result of multiplying them in log space (this corresponds to pairwise multiplying the angles of the two operations' eigenvalues, rescaled so that 180 degrees is the identity angle):

$\text{Control}(A, B) = \exp\left(\frac{\ln(A) \ln(B)}{i \pi}\right)$

When you set $A$ to a Pauli $Z$ operation, this corresponds to controlling $B$ by $A$'s target. E.g. note that $\text{Control}(Z_1, X_2) = \text{CNOT}_{1,2}$ and $\text{Control}(Z_1, Z_2) = \text{CZ}_{1,2}$. So the control product actually generalizes the notion of controlling to things beyond $Z$ controls, since you can set $A$ to things besides $Z$.

Anyways, note that the definition is completely symmetric. $\text{Control}(A, B) = \text{Control}(B, A)$. Therefore there's no fundamental distinction between the control and the target. This is one of the ways to see why phase kickback occurs: the phase kickback from applying $\text{Control}(Z, U)$ is due to the Z "being controlled by the U".