# Coefficients of the vectors in a tensor product

The postulate says that if we want to build the compound state of two sub-systems, we just take the tensor product $\otimes$ of the respective state vectors. This means that if one of the vectors has an overall phase $e^{i \phi}$, that can be interpreted either as the phase of any of the two vectors in the subspaces, or of the compound one, and this can be easily seen "in the formulas" when using the bracket notation $e^{i\phi}(|\psi\rangle\otimes|\zeta\rangle)=|\psi\rangle\otimes(e^{i\phi}|\zeta\rangle)=(e^{i\phi}|\psi\rangle)\otimes|\zeta\rangle$, or even more directly if we use the matrix representation.

This seems a technicality, as usually we just elide the overall phase because it's not measurable. But, I came across this quantum computation problem.

Consider the "phase estimation" algorithm presented in Quantum Computation and Quantum Information, Nielsen & Chuang, II edition, page 221 (also Wikipedia). Consider only a part of the circuit described: a qubit register $|t\rangle$ controls (i.e., activates or inhibits) the application of an operator $U$ to a second qubit register, which is initially in the state $|u\rangle$. This $|u\rangle$ happens to be an eigenvector of $U$, so $U|u\rangle=e^{i\phi}|u\rangle$. Now, the algorithm starts by setting $|t\rangle = 0$ and then applying an Hadamard transform, $|0\rangle \rightarrow H|0\rangle = \frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)$, and then let it do the job on the second register. The output is $|tu\rangle_{final}=\frac{1}{\sqrt{2}}(|0\rangle+e^{i\phi}|1\rangle)|u\rangle$, for the aforementioned tensor product rule. At the end, we do some operations on $|t\rangle$ and we measure it getting some information on the phase $e^{i\phi}$, completely ignoring the state of $|u\rangle$ (which N&C regard as unchanged, since the phase somehow "passed" to the register $|t\rangle$).

I can't fit together these two facts: $|t\rangle$ is supposed to be "controlling" the transform on $|u\rangle$, so it should not change at all; yet, thanks to the tensor product and its "permeability" to numeric factors, I can just shuttle around the $e^{i\phi}$ phase, and decide to measure it (somehow) on the first register, instead of the second, where my intuition would tell me that it resides. Most importantly, as I said initially, putting the phase here and there seems natural, possible in the formulas and completely arbitrary, so I don't see why I shouldn't let the phase stay "attached" to $|u\rangle$ and then get... basically nothing useful for my quantum algorithm.

How can I get out of this apparent contradiction? Is this a clever trick set into the very core rules of quantum mechanics, to encode the fact that in two interacting systems there can't really be one spectator (or controller) and one affected (controlled)?

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In this case, the use of the terminology "control" and "target" derived from classical computer science is a little confusing. Your intuition here may be helped here by realising that there is no way in general to split the operation $U$ into parts that act on the two qubits separately$^{\ast}$. In general, $U$ acts non-trivially on the joint state of the qubit pair, so there is no reason to expect that the control qubit should be left unchanged.
If you write your output state like $$|0\rangle |u\rangle + e^{i\phi}|1\rangle|u\rangle,$$ it becomes clear that the phase is not "attached" to either of the individual qubits: control or target. Rather the phase is attached to one branch of the wavefunction, i.e. just one of the pair of two-qubit states that appears in the superposition. Namely, the phase appears in the branch where the control qubit was in the state $|1\rangle$, thus applying a phase shift to that part of the global wavefunction. The lesson here is that one cannot usually think of quantum states of many-particle systems as being associated to one particle or another. As Lev Vaidman niftily wrote: "the quantum wavefunction describes all systems together."
$^{\ast}$ However this situation, as described in the OP's question, could also arise from the unitary $U = V \otimes I$, where $I$ is the identity on the target qubit and $V$ acts on the control qubit as $$V|0\rangle = |0\rangle \qquad V|1\rangle = e^{i\phi}|1\rangle.$$ So actually $U$ must act non-trivially at least on the control qubit, perhaps surprisingly.