Is the eigenvalue of an eigenstate the same as its (global) phase? I'm trying to understand Shor's Algorithm by reading this Qiskit textbook. At some part the following equation comes up:
\begin{equation}
U |u_0\rangle = \frac{1}{\sqrt{12}} \begin{pmatrix} |3\rangle + |9\rangle + |27\rangle ... |12\rangle + |1\rangle\end{pmatrix} = |u_0\rangle,
\end{equation}
followed up by this sentence: "this eigenstate [of $U$] has an eigenvalue of 1, which isn’t very interesting".
I thought I understood the meaning of eigenstates and eigenvalues, but now I'm not sure whether this means that the eigenvalue of $|u_0\rangle$ is basically the same as its (global) phase, since the phase in this example is 1 as well.
 A: It's hard to write a clear answer because it appears you are trying to understand a fairly sophisticated piece of quantum theory without a very secure grounding or general familiarity with quantum theory. To get you from where you are to where you want to be would take, I think, a bit of a reminder-course in quantum theory. Two thoughts which are relevant are:

*

*an eigenvalue is not to do with a state alone, it is to do with an state and an operator. It says something about the effect of the operator on the state.


*a state vector such as $| u_0 \rangle$ does not have a phase in and of itself. It only has a relative phase with other states. For example, if I define
$$| v_0 \rangle = e^{i \phi} | u_0 \rangle
$$
then if $U |u_0 \rangle = \lambda |u_0\rangle$ then also $U |v_0 \rangle = \lambda | v_0 \rangle$. In this example both $|u_0\rangle$ and $|v_0\rangle$ are eigenstates of $U$ with the same eigenvalue $\lambda$, and furthermore they represent the same physical state (as I will explain in a moment). In the present example you have $\lambda = 1$.
My purpose in introducing $|v_0\rangle$ here is mainly to show that the notion of a global phase has got pretty much nothing to do with eigenvalues in general.
Note also that in this example $|v_0 \rangle$ and $| u_0 \rangle$ would be said to be the 'same' state for many purposes in quantum theory. This is because a physical state is associated with a direction in Hilbert space, which is not quite the same as a state vector. But of course if a system were in a superposition such as
$$
\frac{1}{\sqrt{2}}\left( | u_1 \rangle + e^{i \phi} | u_0 \rangle \right)
$$
where $| u_1 \rangle$ is some other state, with $\langle u_1 | u_0 \rangle = 0$, then the physical state would depend on $\phi$.
