Question

Show that Hermitian operators have real eigenvalues. Show that eigenvectors of a

Hermitian operator with unique eigenvalues are orthogonal. Use Dirac notation

for this problem.

Answer 1

Hermitian Operators

As mentioned previously, the expectation value of an operator is given by

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and all physical observables are represented by such expectation values. Obviously, the value of a physical observable such as energy or density must be real, so we require to be real. This means that we must have , or

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Operators which satisfy this condition are called Hermitian. One can also show that for a Hermitian operator,

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for any two states and .

An important property of Hermitian operators is that their eigenvalues are real. We can see this as follows: if we have an eigenfunction of with eigenvalue , i.e. , then for a Hermitian operator

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Since is never negative, we must have either or . Since is not an acceptable wavefunction, , so is real.

Another important property of Hermitian operators is that their eigenvectors are orthogonal (or can be chosen to be so). Suppose that and are eigenfunctions 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

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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

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Thus we merely need to choose

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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.