Let $f$ be a (complex) function with Taylor expansion \[f(z) = \sum_{n=0}^{\infty} a_n z^n,\] and radius of convergence $|z| \leq r$.

We can extend $f$ to be a function on linear operators by defining \[f(\hat{A}) = \sum_{n=0}^{\infty} a_n \hat{A}^n.\]

It is possible to prove that this series converges if \[\sup_{\{\ket{\psi}| \| \psi \| = 1 \}} \Abs{\sand{\psi}{\hat{A}}{\psi}} \leq r.\]

Since $[\hat{A}+\hat{B},\hat{C}] = [\hat{A},\hat{C}] + [\hat{B},\hat{C}]$ and $[\hat{A}^{n},\hat{B}] = 0$ whenever $[\hat{A},\hat{B}] = 0$, the fact that functions are defined in terms of power series means that

• If $[\hat{A},\hat{B}] = 0$ then $[f(\hat{A}),\hat{B}] = 0$ for any function $f$.
• In particular, since $[\hat{A},\hat{A}] = 0$, we have $[f(\hat{A}),g(\hat{A})] = 0$ for any functions $f$ and $g$.

Since $(\hat{A} + \hat{B})^{\dagger} = \hat{A}^{\dagger} + \hat{B}^{\dagger}$ and $\left ( \hat{A}^n \right )^{\dagger} = \left ( \hat{A}^{\dagger} \right )^n$, for a function \[f(\hat{A}) = \sum_{n=0}^{\infty} a_n \hat{A}^n,\] we have \[f(\hat{A})^{\dagger} = \sum_{n=0}^{\infty} a_n^* \hat{A^{\dagger}}^n.\]

Note, if $\hat{A}$ is a Hermitian operator then $f(\hat{A})$ is a Hermitian operator if and only if the expansion coefficients $a_n$ are real.

The exponential function is an operator function that is used a lot in quantum mechanics. It is defined via its power series as \[e^{a\hat{A}} = \sum_{n=0}^{\infty} \frac{a^n}{n!} \hat{A}^n = \hat{I} + a\hat{A} + \frac{a^2}{2}\hat{A}^2 + \cdots.\]

Just like its classical counterpart, you can prove via the binomial theorem that this is equivalent to \[e^{a\hat{A}} = \lim_{n\rightarrow \infty} \left ( \hat{I} + \frac{a}{n} \hat{A}\right )^n\]

If $\hat{A}$ and $\hat{B}$ commute then the usual rules for multiplying exponentials apply \[e^{a\hat{A}}e^{b\hat{B}} = e^{a\hat{A} + b\hat{B}},\] but in general this is not true. Instead \[e^{a\hat{A}}e^{b\hat{B}} = e^{a\hat{A} + b\hat{B}} e^{ab[\hat{A},\hat{B}]/2},\] which can be proved by a somewhat laborious process of multiplying together the power series and collecting terms.

As a special case of the rules for commutators, if $[\hat{A},\hat{B}] = 0$ then we will have $[e^{a\hat{A}},\hat{B}] = [\hat{A},e^{b\hat{B}}] = [e^{a\hat{A}},e^{b\hat{B}}] = 0$, but these will not be true if $\hat{A}$ and $\hat{B}$ do not commute.

The Hermitian adjoint of an exponential can be taken as follows. \[\left ( e^{a\hat{A}}\right )^{\dagger} = \sum_{n=0}^{\infty} \frac{\left (a^*\right )^n}{n!} \left ( \hat{A}^{\dagger}\right )^n = e^{a^*\hat{A}^{\dagger}},\] i.e. you just take the Hermitian adjoint of the term inside the exponential.

A special case of this is that if $\hat{A}$ is Hermitian then $\left ( e^{\hat{A}} \right )^{\dagger} = e^{\hat{A}}$, so $e^{\hat{A}}$ is Hermitian as well. However, for the operator $e^{i\hat{A}}$, which is the type of operator that describes time evolution in quantum mechanics, we have $\left ( e^{i\hat{A}} \right )^{\dagger} = e^{-i\hat{A}}$, so this is not a Hermitian operator.

If it exists, the inverse $\hat{A}^{-1}$ of an operator $\hat{A}$ is defined by \[\hat{A}^{-1}\hat{A} = \hat{A}\hat{A}^{-1} = \hat{I}.\]

Note that we do not use the notation $\frac{\hat{A}}{\hat{B}}$ for $\hat{A}\hat{B}^{-1}$ because, in general $\hat{A}\hat{B}^{-1} \neq \hat{B}^{-1}\hat{A}$, so $\frac{\hat{A}}{\hat{B}}$ would have no unique meaning.

Taking inverses reverses the order of products and commutes with taking powers, i.e. \begin{align*} \left ( \hat{A}\hat{B}\hat{C}\cdots\right )^{-1} & = \cdots\hat{C}^{-1}\hat{B}^{-1}\hat{A}^{-1}, & \left ( \hat{A}^{-1} \right )^{n} & = \left ( \hat{A}^n \right )^{-1}. \end{align*} You will prove the first of these in an in class activity.

An operator is unitary if $\hat{U}^{\dagger} = \hat{U}^{-1}$ or, equivalently $\hat{U}\hat{U}^{\dagger} = \hat{U}^{\dagger}\hat{U} = \hat{I}$.

If $\hat{U}$ and $\hat{V}$ are unitary then so is their product $\hat{U}\hat{V}$, since \[\left ( \hat{U}\hat{V} \right )^{\dagger} \left ( \hat{U}\hat{V} \right ) = \hat{V}^{\dagger} \hat{U}^{\dagger} \hat{U}\hat{V} = \hat{V}^{\dagger} \hat{I} \hat{V} = \hat{V}^{\dagger}\hat{V} = \hat{I}.\] More generally, if $\hat{A}, \hat{B}, \hat{C}, \cdots$ are unitary then so is the product $\hat{A}\hat{B}\hat{C}\cdots$.

If $a$ is real and $\hat{A}$ is Hermitian then $e^{ia\hat{A}}$ is unitary because \[\left ( e^{ia\hat{A}}\right )^{\dagger} e^{ia\hat{A}} = e^{-ia\hat{A}}e^{ia\hat{A}} = e^{i(-a\hat{A} + a\hat{A})} = e^{i0} = \hat{I}.\]

1. Prove that $\left ( \hat{A} \hat{B} \right )^{-1} = \hat{B}^{-1}\hat{A}^{-1}$