1.6: Example of matrix representation method and choice of basis

In practical quantum problems, we almost always describe the state of the system in terms of some basis set. Consider a simple spin 1/2 system, choosing as basis states $S_z = \pm \frac{1}{2}$. Consider this system in a magnetic field pointing in the $x$ direction, the operator corresponding to this is $\mu B \hat{S}_x$. We wish to find the eigenstates and eigenenergies.

Evaluating the required matrix elements such as $\langle S_z = \frac{1}{2} |\mu B\hat{S}_x | S_z = \frac{1}{2} \rangle$ (see QP3) gives a matrix:

$$\begin{pmatrix} 0 & \mu B/2 \\ \mu B/2 & 0 \end{pmatrix} \nonumber$$

The normalised eigenvectors of this matrix are $(\sqrt{\frac{1}{2}}, \sqrt{\frac{1}{2}})$ and $(\sqrt{\frac{1}{2}}, -\sqrt{\frac{1}{2}})$ with eigenvalues $(\mu B/2)$ and $(-\mu B/2)$. Of course these represent the eigenstates $|S_x = \pm \frac{1}{2} \rangle$ in the basis of $|S_z = \pm \frac{1}{2} \rangle$:

$$|S_x = \pm \frac{1}{2} \rangle = \left[|S_z = \frac{1}{2} \rangle \pm |S_z = -\frac{1}{2} \rangle \right] / \sqrt{2} \nonumber$$

Had we chosen $|S_y = \pm \frac{1}{2} \rangle$ as our basis set, then the matrix would have been:

$$\begin{pmatrix} 0 & -i\mu B/2 \\ i\mu B/2 & 0 \end{pmatrix} \nonumber$$

Once again, the eigenvalues of this matrix are $(\mu B/2)$ and $(-\mu B/2)$, as they must be since these are the measurable quantities. Coincidently, the eigenvectors in this basis set are also $(\sqrt{\frac{1}{2}}, \sqrt{\frac{1}{2}})$ and $(\sqrt{\frac{1}{2}}, -\sqrt{\frac{1}{2}})$.

Had we chosen $|S_x = \pm \frac{1}{2} \rangle$ as our basis set in the first place, the problem would have been much simplified. The matrix would then be:

$$\begin{pmatrix} \mu B/2 & 0 \\ 0 & -\mu B/2 \end{pmatrix} \nonumber$$

Once again, the eigenvalues of this matrix are $(\mu B/2)$ and $(-\mu B/2)$, and now the eigenvectors are (1,0) and (0,1): i.e. the eigenstates are simply the basis states.