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Unformatted text preview: of with eigenvalues and , with . If is Hermitian then (59) since as shown above. Because we assumed , we must have , i.e. and are orthogonal. Thus we have shown that eigenfunctions of a Hermitian operator with different eigenvalues are orthogonal. In the case of degeneracy (more than one eigenfunction with the same eigenvalue), we can choose the eigenfunctions to be orthogonal. We can easily show this for the case of two eigenfunctions of with the same eigenvalue. Suppose we have (60) We now want to take linear combinations of and to form two new eigenfunctions and , where and . Now we want and to be orthogonal, so (61) Thus we merely need to choose (62) and we obtain orthogonal eigenfunctions. This Schmidt-orthogonalization procedure can be extended to the case of n-fold degeneracy, so we have shown that for a Hermitian operator, the eigenvectors can be made orthogonal....
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