Functions of Operators

The spectral theorem does more than diagonalize an operator — it tells you how to take any function of it. Want the square root of an operator? Its exponential? Its logarithm? You apply the ordinary scalar function to the eigenvalues and leave the eigenvectors alone. This functional calculus is what gives meaning to expressions like the time-evolution operator $e^{-iHt/\hbar}$ .

The definition

Let $A$ be Hermitian (or, more generally, normal) with spectral decomposition $A = \sum_k \lambda_k |k\rangle\langle k|$ . For any scalar function $f$ defined on the eigenvalues, define

f(A) = \sum_k f(\lambda_k)\,|k\rangle\langle k|.

The eigenvectors $|k\rangle$ are unchanged; only the weights are transformed from $\lambda_k$ to $f(\lambda_k)$ . In the eigenbasis, where $A = \operatorname{diag}(\lambda_1, \dots, \lambda_n)$ , this is simply

f(A) = \operatorname{diag}\big(f(\lambda_1), \dots, f(\lambda_n)\big).

Apply the function entry by entry along the diagonal — that is the whole idea.

Why this is the right definition

For a polynomial $f(x) = \sum_m c_m x^m$ there is no choice: $f(A)$ must mean $\sum_m c_m A^m$ , where $A^m$ is repeated operator multiplication. But powers of $A$ are easy in the eigenbasis, $A^m|k\rangle = \lambda_k^m |k\rangle$ , so

f(A)|k\rangle = \sum_m c_m A^m |k\rangle = \Big(\sum_m c_m \lambda_k^m\Big)|k\rangle = f(\lambda_k)|k\rangle.

The spectral formula agrees with the polynomial definition. For more general $f$ (square roots, exponentials, logarithms) we define $f(A)$ by the same rule, and it matches the limit of polynomial approximations whenever the power series converges. Consistency requires only that $f$ be defined at each eigenvalue.

The operator exponential

The most important example is $f(x) = e^x$ . The exponential of an operator,

e^{A} = \sum_{m=0}^{\infty} \frac{A^m}{m!} = \sum_k e^{\lambda_k}\,|k\rangle\langle k|,

is what turns a Hermitian Hamiltonian into a unitary evolution. If $H$ is Hermitian with real eigenvalues $E_k$ , then

U(t) = e^{-iHt/\hbar} = \sum_k e^{-iE_k t/\hbar}\,|E_k\rangle\langle E_k|

has eigenvalues $e^{-iE_k t/\hbar}$ of modulus one — a unitary, exactly as promised in the unitary- operators lesson. Each energy eigenstate simply picks up a rotating phase.

A trace shortcut

Because the trace is basis-independent and equals the sum of the eigenvalues, the trace of $f(A)$ is just

\operatorname{Tr}\big(f(A)\big) = \sum_k f(\lambda_k).

You never have to build the matrix $f(A)$ to get its trace — sum $f$ over the spectrum. For instance, with $A = \begin{pmatrix} 3 & 1 \\ 1 & 3\end{pmatrix}$ (eigenvalues $4$ and $2$ ), the operator square root satisfies $\operatorname{Tr}(\sqrt{A}) = \sqrt4 + \sqrt2 = 2 + \sqrt2$ . The same logic gives the partition function $\operatorname{Tr}(e^{-\beta H}) = \sum_k e^{-\beta E_k}$ of statistical mechanics for free.

Try it

Using the spectral rule, compute $\sqrt{A}$ for $A = \begin{pmatrix} 3 & 1 \\ 1 & 3\end{pmatrix}$ and return its trace $\operatorname{Tr}(\sqrt{A})$ as a plain number.

Run your code to see the quantum state.