1.6: Example of matrix representation method and choice of basis
( \newcommand{\kernel}{\mathrm{null}\,}\)
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 Sz=±12. Consider this system in a magnetic field pointing in the x direction, the operator corresponding to this is μBˆSx. We wish to find the eigenstates and eigenenergies.
Evaluating the required matrix elements such as ⟨Sz=12|μBˆSx|Sz=12⟩ (see QP3) gives a matrix:
(0μB/2μB/20)
The normalised eigenvectors of this matrix are (√12,√12) and (√12,−√12) with eigenvalues (μB/2) and (−μB/2). Of course these represent the eigenstates |Sx=±12⟩ in the basis of |Sz=±12⟩:
|Sx=±12⟩=[|Sz=12⟩±|Sz=−12⟩]/√2
Had we chosen |Sy=±12⟩ as our basis set, then the matrix would have been:
(0−iμB/2iμB/20)
Once again, the eigenvalues of this matrix are (μB/2) and (−μB/2), as they must be since these are the measurable quantities. Coincidently, the eigenvectors in this basis set are also (√12,√12) and (√12,−√12).
Had we chosen |Sx=±12⟩ as our basis set in the first place, the problem would have been much simplified. The matrix would then be:
(μB/200−μB/2)
Once again, the eigenvalues of this matrix are (μB/2) and (−μB/2), and now the eigenvectors are (1,0) and (0,1): i.e. the eigenstates are simply the basis states.