2
$\begingroup$

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.

$\endgroup$

2 Answers 2

4
$\begingroup$

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,
$\textbf{v}=\left(\begin{matrix}\cos\alpha\\e^{i\phi}\sin\alpha\end{matrix}\right)$.

$\endgroup$
2
  • 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? $\endgroup$ 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
    Jul 25, 2022 at 17:26
3
$\begingroup$

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).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.