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# 04 - Physics 6210/Spring 2007/Lecture 4 Lecture 4 Relevant...

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Physics 6210/Spring 2007/Lecture 4 Lecture 4 Relevant sections in text: § 1.2, 1.3, 1.4 The spin operators Finally we can discuss the definition of the spin observables for a spin 1/2 system. We will do this by giving the expansion of the operators in a particular basis. We denote the basis vectors representing states in which the z component of spin is known by := | S z , ± , and define S x = ¯ h 2 | + -| + |- + | S y = i ¯ h 2 |- + | - | + -| S z = ¯ h 2 | + + | - |- -| . Note that we have picked a direction, called it z , and used the corresponding spin states for a basis. Of course, any other direction could be chosen as well. You can now check that, with the above definition of S z , are in fact the eigenvectors of S z with eigenvalues ± ¯ h/ 2. Labeling matrix elements in this basis as A ij = + | A | + + | A |- -| A | + -| A |- , you can also verify the following matrix representations in the basis : ( S x ) ij = ¯ h 2 0 1 1 0 ( S y ) ij = ¯ h 2 0 - i i 0 ( S z ) ij = ¯ h 2 1 0 0 - 1 . Finally, you should check that all three spin operators are self-adjoint. It will be quite a while before you get a deep understanding of why these particular operators are chosen. For now, let us just take them as given and see what we can do with them. As a good exercise you can verify that | S x , ± = 1 2 ( | + ± |- ) , | S y , ± = 1 2 ( | + ± i |- ) are eigenvectors of S x and S y , respectively, with eigenvalues ± ¯ h/ 2. Note that these eigen- vectors are normalized to have norm (“length”) unity. The fact that these eigenvectors are distinct from those of S z will be dealt with a little later. For now, just note that the three operators do not share any eigenvectors. Note also that the eigenvalues of the spin operators are all non-degenerate – exercise. 1

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Physics 6210/Spring 2007/Lecture 4 Spectral decomposition The spin operators have the general form A = ij A ij | i j | , which we discussed earlier. Note, though, that S z has an especially simple, diagonal form. This is because it is being represented by an expansion in a basis of its eigenvectors. It is not hard to see that this result is quite general. If | i is an ON basis of eigenvectors of A with eigenvalues a i : A | i = a i | i ,
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