References: Shankar, R. (1994), *Principles of Quantum Mechanics*, Plenum Press. Exercises 1.9.1 – 1.9.3.

One of the most common ways to define a function of an operator is to consider the case where the function can be expressed as a power series. That is, given an operator , a function can be defined as

where the coefficients are, in general, complex scalars. This definition can still be difficult to deal with if is not diagonalizable since, in that case, powers of have no simple form, so it can be hard to tell if the series converges.

We can avoid this problem by restricting ourselves to hermitian operators, since such operators are always diagonalizable according to the spectral theorem and all eigenvalues of hermitian operators are real. Then powers of are easy to calculate, since if the th diagonal element of is , the th diagonal element of is . The problem of finding is then reduced to examining whether the series converges for each diagonal element.

Example 1Suppose we have the simplest such power series

If we look at this series in the eigenbasis (the basis of orthonormal eigenvectors that diagonalizes ), then we have

here is an matrix with eigenvalues , (it’s possible that some of the eigenvalues could be equal, if is degenerate, but that doesn’t affect the argument).

It’s known that the geometric series

converges as shown, provided that . Thus we see that converges provided all its eigenvalues satisfy . The function is then

To see what operator it converges to, we consider the function

Still working in the eigenbasis where is diagonal, the matrix is also diagonal with diagonal elements . The inverse of a diagonal matrix is another diagonal matrix with diagonal elements equal to the reciprocal of the elements in the original matrix, so has diagonal elements so from 5 we see that

provided all the eigenvalues of satisfy .

Example 2If is a hermitian operator, then is unitary. To see this, we again work in the eigenbasis of . By expressing as a power series and using the same argument as in the previous example, we see that

The adjoint of is found by looking at the power series:

where in the third line we used the hermitian property . Therefore

Thus and is unitary.

From 8 we can find the determinant of :

since the trace of a hermitian matrix is the sum of its eigenvalues.

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