This question is very homeworky, but I could not find an answer to this question anywhere.

The general question is: How do you recognise when phase factors should be included in an eigenvector?

Here is a question when this problem came up:

If a spin has been prepared $+1$ in the $z$-axis and then rotated an angle $\theta$ and $\phi$, as shown in the picture. What is the expected spin? The method is shown below:

Using the operator for spin values:$$\begin{bmatrix}n_z & n_x+in_y\\n_x-in_y & -n_z\end{bmatrix}$$ Substituting the values of $n_x$, $n_y$ and $n_z$ you get the matrix: $$\begin{bmatrix}cos\theta & \sin \theta \cos\phi − i \sin \theta\sin\phi\\ \sin \theta\cos\phi + i \sin \theta \sin\phi& -\cos\theta\end{bmatrix}$$ which can be simplified to: $$\begin{bmatrix}cos\theta &e^{-i\phi} \sin \theta \\e^{i\phi}\sin \theta & -\cos\theta\end{bmatrix}$$ Then using the trace and determinant to obtain the expected eigenvalues of $+1$ and $-1$.

To calculate the eigenvectors you then let:

$$|\lambda\rangle = \begin{bmatrix}\cos\alpha \\ \sin\alpha \end{bmatrix}$$

However, this is impossible to solve for $\alpha$ unless you add a phase factor making:

$$|\lambda\rangle = \begin{bmatrix}\cos\alpha \\ e^{i\phi} \sin\alpha \end{bmatrix}$$

How do you notice that:

  1. A phase factor needs to be added to make the equation solvable.
  2. Where the phase factor needs to be added.

Edit: When solving for eigenvectors, should they be defined as: $$|\lambda\rangle = \begin{bmatrix}e^{y \phi}\cos\alpha \\ e^{x\phi} \sin\alpha \end{bmatrix}$$

where $x$ and $y$ are integers.


2 Answers 2


You need to distinguish between global and relative phases of the 2 vector components. Of course, having an eigenvector $\lvert\lambda\rangle$ you can add any phase factor you want ($\rightarrow e^{i\phi}\lvert\lambda\rangle$) and it's still an eigenvector.

Shifting the relative phase between the components of your vector however does make a difference. Essentially your vector space ($\mathbb{C}^2$) has 4 degrees of freedom. We can parameterize it with 4 real (or 2 complex) variables by $\textbf{v}=\left(\begin{matrix}a_1+ib_1\\a_2+ib_2\end{matrix}\right)=\left(\begin{matrix}r_1e^{i\phi_1}\\r_2e^{i\phi_2}\end{matrix}\right)=e^{i\phi_1}\left(\begin{matrix}r_1\\r_2e^{i(\phi_2-\phi_1)}\end{matrix}\right)$.
Requiring a normalized vector eliminates 1 degree of freedom, because $\lvert r_1\rvert^2+\lvert r_2\rvert^2\overset{!}{=}1\implies(r_1,r_2)=(\cos\alpha,\sin\alpha)$ and we can ignore the global phase, so $\phi_1=0$.

This leaves you with the proper Ansatz for the eigenvector,

  • 2
    $\begingroup$ When calculating eigenvectors do we then always leave it in the form \begin{matrix}\cos\alpha\\e^{i\phi}\sin\alpha\end{matrix} as this allows a relative phase between the two vector components? $\endgroup$ Commented Jul 25, 2022 at 16:16
  • $\begingroup$ @explodingkitten500 Not sure if I understand the question correctly, but $\alpha$ and $\phi$ will be determined by the eigenvalue problem, so after the computation there is no "choice" to make. If the Eigenvalues are non degenerate, then this Ansatz gives you (up to a sign) one solution. $\endgroup$
    – go_science
    Commented Jul 25, 2022 at 17:26

Unless there is a specific reason to disregard it, a phase factor always needs to be considered. $[\cos(\alpha),\sin(\alpha)]^\top$ would be fine if we were only looking for real eigenvectors, but here we need to consider complex eigenvectors too.

Since only relative phase (i.e. between the x and y components here) matters and quantum states differing by an overall phase are physically indistinguishable, the quantum state $[e^{yi\phi} \cos(\alpha),e^{xi\phi}\sin(\alpha)]^\top$ is equivalent to $[\cos(\alpha),e^{(x-y)i\phi}\sin(\alpha)]^\top$. $[\cos(\alpha), e^{i\phi}sin(\alpha)]^\top$ is then a perfectly general way of considering a phase factor (redefining $(x-y)\phi \rightarrow \phi$ for convenience).


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