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Physics LibreTexts

1.6: Example of matrix representation method and choice of basis

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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:

(0iμB/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/200μB/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.


This page titled 1.6: Example of matrix representation method and choice of basis is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Graeme Ackland via source content that was edited to the style and standards of the LibreTexts platform.

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