Eigenvectors of $S_x$ for Spin ${1\over 2}$

First the quick solution. Since there is no difference between x and z, we know the eigenvalues of $S_x$ must be $\pm{\hbar\over 2}$. So, factoring out the constant, we have

$$\begin{eqnarray*} & \pmatrix{0&1\cr 1&0}\pmatrix{a\cr b}=\pm \pmatrix{a\cr b} \\... ...hi^{(x)}_{-}=\pmatrix{{1\over\sqrt{2}}\cr {-1\over\sqrt{2}}} \\ \end{eqnarray*}$$

These are the eigenvectors of $S_x$. We see that if we are in an eigenstate of $S_x$ the spin measured in the z direction is equally likely to be up and down since the absolute square of either amplitude is ${1\over 2}$.

The remainder of this section goes into more detail on this calculation but is currently notationally challenged.

Recall the standard method of finding eigenvectors and eigenvalues:

$$\bgroup\color{black}A\psi=\alpha\psi\egroup$$

$$\bgroup\color{black}\left( A - \alpha \right)\psi=0\egroup$$

For spin $1\over2$ system we have, in matrix notation,

$$\bgroup\color{black}\left(\matrix{a_1&a_2 \cr a_3&a_4}\right)... ...cr 0&1}\right) \left(\matrix{\chi_1 \cr \chi_2}\right )\egroup$$

$$\bgroup\color{black}\Rightarrow \qquad\left(\matrix{a_1-\alph... ...pha}\right) \left(\matrix{\chi_1 \cr \chi_2}\right ) =0\egroup$$

For a matrix times a nonzero vector to give zero, the determinant of the matrix must be zero. This gives the ``characteristic equation'' which for spin $1\over2$ systems will be a quadratic equation in the eigenvalue $\alpha$:

$$\bgroup\color{black}\left\vert\matrix{a_1-\alpha &a_2 \cr a_3&a_4-\alpha}\right\vert=(a_1-\alpha)(a_4-\alpha)-a_2 a_3=0\egroup$$

$$\bgroup\color{black}\alpha^2-(a_1+a_4)\alpha + (a_1a_4-a_2a_3)=0\egroup$$

whose solution is

$$\bgroup\color{black}\alpha_\pm={(a_1+a_4)\over 4}\pm \sqrt{{(a_1+a_4)\over 2}^2-(a_1a_4-a_2a_3)}\egroup$$

To find the eigenvectors, we simply replace (one at a time) each of the eigenvalues above into the equation

$$\bgroup\color{black}\left(\matrix{a_1-\alpha &a_2 \cr a_3&a_4... ...pha}\right) \left(\matrix{\chi_1 \cr \chi_2}\right ) =0\egroup$$

and solve for $\chi_1$ and $\chi_2$.

Now specifically, for the operator $A=S_x={\hbar\over 2}\left(\matrix{0&1 \cr 1&0}\right)$, the eigenvalue equation $(S_x-\alpha)\chi =0$ becomes, in matrix notation,

$$\bgroup\color{black}{\hbar\over 2}\left(\matrix{0&1 \cr 1&0}\... ...pha}\right) \left(\matrix{\chi_1 \cr \chi_2}\right ) = 0\egroup$$

$$\bgroup\color{black}\Rightarrow\qquad\left(\matrix{-\alpha&{\... ...lpha}\right) \left(\matrix{\chi_1 \cr \chi_2}\right )=0 \egroup$$

The characteristic equation is $det\vert S_x-\alpha\vert=0$, or

$$\bgroup\color{black}\alpha^2-{\hbar^2\over 4}=0\qquad\Rightarrow\qquad\alpha=\pm{\hbar\over 2}\egroup$$

These are the two eigenvalues (we knew this, of course). Now, substituting $\alpha_+$ back into the eigenvalue equation, we obtain

$$\bgroup\color{black}\left(\matrix{-\alpha_+&\hbar/ 2 \cr \hb... ...&-1}\right) \left(\matrix{\chi_1 \cr \chi_2}\right ) =0 \egroup$$

The last equality is satisfied only if $\chi_1=\chi_2$ (just write out the two component equations to see this). Hence the normalized eigenvector corresponding to the eigenvalue $\alpha=+\hbar/2$ is

$$\bgroup\color{black}\chi^{(x)}_{+}={1\over\sqrt{2}}\left(\matrix{1\cr 1}\right).\egroup$$

Similarly, we find for the eigenvalue $\alpha=-\hbar/2$,

$$\bgroup\color{black}\chi^{(x)}_{-}={1\over\sqrt{2}}\left(\matrix{1\cr -1}\right).\egroup$$