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.