Eigenvectors of \bgroup\color{black}$S_x$\egroup for Spin \bgroup\color{black}${1\over 2}$\egroup

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

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


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

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:

\begin{displaymath}\bgroup\color{black}A\psi=\alpha\psi\egroup\end{displaymath}


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

For spin \bgroup\color{black}$1\over2$\egroup system we have, in matrix notation,

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


\begin{displaymath}\bgroup\color{black}\Rightarrow \qquad\left(\matrix{a_1-\alph...
...a}\right)
\left(\matrix{\chi_1 \cr \chi_2}\right )
=0\egroup\end{displaymath}

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 \bgroup\color{black}$1\over2$\egroup systems will be a quadratic equation in the eigenvalue \bgroup\color{black}$\alpha$\egroup:

\begin{displaymath}\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\end{displaymath}


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

whose solution is

\begin{displaymath}\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\end{displaymath}

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

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

and solve for \bgroup\color{black}$\chi_1$\egroup and \bgroup\color{black}$\chi_2$\egroup.

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

\begin{displaymath}\bgroup\color{black}{\hbar\over 2}\left(\matrix{0&1 \cr 1&0}\...
...a}\right)
\left(\matrix{\chi_1 \cr \chi_2}\right )
= 0\egroup\end{displaymath}


\begin{displaymath}\bgroup\color{black}\Rightarrow\qquad\left(\matrix{-\alpha&{\...
...pha}\right)
\left(\matrix{\chi_1 \cr \chi_2}\right )=0 \egroup\end{displaymath}

The characteristic equation is \bgroup\color{black}$det\vert S_x-\alpha\vert=0$\egroup, or

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

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

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

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

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

Similarly, we find for the eigenvalue \bgroup\color{black}$\alpha=-\hbar/2$\egroup,

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

Jim Branson 2013-04-22