For a spin of $\frac{1}{2}$ the angular momentum operator can be written as $\vec{S} = \frac{\hbar}{2} \vec{\sigma}$ in matrix form. Find the eigenvalues and eigenvectors of $S_x$ where $\sigma_x = \begin{pmatrix} 0 & 1 \\ 1 & 0 \\\end{pmatrix}$


  • Finding the eigenvalues:

$|S_x - \lambda I|=det\begin{bmatrix} -\lambda & \hbar/2 \\ \hbar/2 & -\lambda\\\end{bmatrix} = 0 \Rightarrow \lambda = \pm\hbar/2$

As far as I know that is correct so far.

  • Finding the eigenvectors:

Using $\lambda = -\hbar/2$ first: (I don't get a correct answer for this one, but I do for the other eigenvalue)

$\begin{bmatrix} \hbar/2 &\hbar/2 \\ \hbar/2 & \hbar/2\\\end{bmatrix}\begin{bmatrix} \alpha \\ \beta \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \end{bmatrix} $

$\Rightarrow (\hbar/2)\alpha + (\hbar/2)\beta = 0$

$\Rightarrow (\hbar/2)\alpha + (\hbar/2)\beta = 0$

Therefore the corresponding eigenvector is $c \begin{bmatrix} -1 \\ -1 \end{bmatrix} $ where $c$ is some constant. I know that the answer should be $c\begin{bmatrix} 1 \\ -1 \end{bmatrix} $ where after normalisation $c$ is found to be $1/\sqrt{2}$.

Where have I gone wrong? The eigenvalues are correct so it can't be that.


closed as off-topic by Brandon Enright, Emilio Pisanty, tpg2114, John Rennie, Qmechanic Nov 21 '13 at 8:59

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  • 1
    $\begingroup$ An eigenvector $(a,b)$ represents a state that has a probability of $|a|^2$ being spin up, and a probability of $|b|^2$ being spin down. For the probabilities to sum to one, the eigenvector must be normalized. $\endgroup$ – leongz Nov 20 '13 at 22:53
  • $\begingroup$ So I need to normalise my eigenvector. Aha... I see. That accounts for the $1/\sqrt{2}$. Great! $\endgroup$ – turnip Nov 20 '13 at 23:01
  • $\begingroup$ @leongz I get that now, but do you mind explaining the updated question? $\endgroup$ – turnip Nov 20 '13 at 23:19
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    $\begingroup$ @PPG: Well $(-1,-1)$ is not a solution of your equation, but $(1,-1)$ is... $\endgroup$ – Adam Nov 20 '13 at 23:36
  • $\begingroup$ @Adam Do you mind explaining/showing me why? $\endgroup$ – turnip Nov 20 '13 at 23:46

@PPG: Well (−1,−1) is not a solution of your equation, but (1,−1) is... – Adam

What he said was:

  • Finding the eigenvectors

You did right... but: $$(\hbar/2)\alpha+(\hbar/2)\beta=0\implies\alpha+\beta=0\implies\beta=-\alpha $$ Then, the corresponding eigenvector is: $ c\left[\begin{array}{c} 1 \\ -1 \end{array}\right]$.

  • $\begingroup$ Oh dear, my maths teacher never explained eigenvectors properly. I thought you have to find them a different way. That looks much easier (and correct) and I now feel stupid -_- $\endgroup$ – turnip Nov 21 '13 at 0:52
  • $\begingroup$ You're welcome!. Lol!. Don't fell bad! Just remember that for you next eigenvalue/eigenvector problem :P. $\endgroup$ – Wikitys Nov 21 '13 at 1:24
  • $\begingroup$ However, could I ask what would have happened if I had done a = -b. That would lead to c(-1 1) and isn't that a completely different answer? $\endgroup$ – turnip Nov 21 '13 at 1:33
  • $\begingroup$ Mmmm... No because $(1,-1) \wedge (-1,1) \in <\{(1,-1)\}>$ and, in quantum mechanics you are looking for any vector belonging to that space, because they differ in a "global" complex phase ($(1,-1)=-1(-1,1)=e^{i\pi}(-1,1)$). Am I making myself clear? $\endgroup$ – Wikitys Nov 21 '13 at 1:47
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    $\begingroup$ Yes, physically it's the same answer. That it's called the Gauge symmetry Wikipedia. $\endgroup$ – Wikitys Nov 21 '13 at 2:02

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