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$\newcommand{\ket}[1]{\left|#1\right>}$Consider two qubits, $\ket{1}$ and $e^{i \theta}\ket{\psi}$ (what the qubits are doesn't actually matter). We express these in a single system using their tensor product, which would be $\ket{1} \otimes (e^{i \theta} \ket{\psi})$. The constant is free floating in the tensor product, so it can be expressed as $(e^{i \theta}\ket{1}) \otimes \ket{\psi}$. And so, you could say that the phase $e^{i \theta}$ belongs to the $\ket{1}$ qubit.

This post helped me, but didn't really answer my question entirely, https://physics.stackexchange.com/questions/77702/coefficients-of-the-vectors-in-a-tensor-product.

My question is, where does the phase belong? Is that even a question that makes sense asking? Does it not even matter, as it's global phase?

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We can write the total state as a tensor product of two qubits, but there is only one wavefunction of the total system. In fact $\vert1\rangle \otimes (e^{i\theta}\vert\psi\rangle)$ is exactly the same state as $(e^{i\theta}\vert1\rangle)\otimes\mid\psi\rangle$. The phase does not belong to either the first or the second qubit, it a property of the total state, i.e. a global phase.

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What do you mean with "where does the phase belong"?

As you noted yourself, tensor products are defined so that $(\lambda v)\otimes w$ is the same element as $v\otimes(\lambda w)$, for any scalar $\lambda$. In other words, you can write the scalar however you like, it doesn't matter, the result element is the same. Which is why this would be generally written as $\lambda(v\otimes w)$, to point out that $\lambda$ is not "attached" to any particular vector.

And note that the above issue is distinct from it being a global phase or not, it holds in general for tensor products of vector spaces and the corresponding scalar fields. Then, as an added element, you have the fact that ket states are defined as vectors up to complex scalar multiplication. Which means that $\lambda |\psi\rangle$ and $|\psi\rangle$ represent the same physical state (and in a more careful treatment, they are completely identified). But this holds more generally, you don't need to tensor product structure for this statement.

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  • $\begingroup$ What I meant by "where does the phase belong?", I mean if you were to look at each qubit's state, which one's state would it be in? The second qubit existed by itself with that phase in its state, but when expressing as part of a larger system, it could be considered that the second qubit did not have that phase, and the first qubit's state was $\left| 1 \right>$ times the phase. Then, you could take the combined system apart, and apparently be left with two different states than what you started with. I hope that helps; I'm not sure exactly how to say what I find wrong with this. $\endgroup$
    – tpws
    May 30, 2022 at 17:35
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    $\begingroup$ it seems to me like you keep making the mistake of thinking that $|\psi\rangle$ and $e^{i\alpha}|\psi\rangle$ are different states. They are not. Think of these are just two equivalent ways to express mathematically the same physical entity. See also quantumcomputing.stackexchange.com/q/5125/55 $\endgroup$
    – glS
    May 30, 2022 at 22:03

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