Skip to main content

Section 6.10 Matrix Exponentials

How do you exponentiate matrices?

Recall the power series

\begin{equation} e^{i\theta} = 1 + i\theta - \frac{\theta^2}{2!} - i\frac{\theta^3}{3!} + ..\text{.}\tag{6.10.1} \end{equation}

which can famously be used to prove Euler's formula, namely

\begin{equation} e^{i\theta} = \cos\theta + i\sin\theta\text{.}\tag{6.10.2} \end{equation}

We can use this same strategy to exponentiate matrices, by using the corresponding power series. We therefore define

\begin{equation} e^{iM\theta} = I + iM\theta - \frac{M^2\theta^2}{2!} - i\frac{M^3\theta^3}{3!} + ..\text{.}\tag{6.10.3} \end{equation}

Let's consider some special cases. First consider a matrix \(M\) satisfying \(M^2=I\text{,}\) such as the Pauli matrices. For any such matrix, we have

\begin{align} e^{iM\theta} \amp = I + iM\theta - \frac{\theta^2}{2!} - i\frac{M\theta^3}{3!} + ...\notag\\ \amp = I\,\cos\theta + iM\,\sin\theta\tag{6.10.4} \end{align}

For example,

\begin{equation} e^{i\sigma_x\theta} = I\,\cos\theta + i\sigma_x\sin\theta = \begin{pmatrix} \cos\theta\amp i\sin\theta\\ i\sin\theta\amp \cos\theta \end{pmatrix} = U_x\text{,}\tag{6.10.5} \end{equation}

and similarly

\begin{align} e^{i\sigma_y\theta} \amp = U_y ,\tag{6.10.6}\\ e^{i\sigma_z\theta} \amp = U_z\text{.}\tag{6.10.7} \end{align}

In fact, exponentiating a Hermitian matrix in this way always yields a unitary matrix, since

\begin{equation} \left(e^{iM\theta}\right)^\dagger = e^{-iM^\dagger\theta}\text{,}\tag{6.10.8} \end{equation}

as can be verified by working term-by-term with the power series. The converse is also true; any unitary matrix \(U\) can be written as \(e^{iM\theta}\) for some Hermitian matrix \(M\text{.}\)

Remarkably, the eigenvector decompositions derived in the last section behave very nicely under exponentiation, as we now show.

Let \(M\) be Hermitian, and recall that we can expand \(M\) in terms of its eigenvectors and eigenvalues as

\begin{equation} M = \lambda |v \rangle \langle v| + \mu |w \rangle \langle w| + ... ~\text{.}\tag{6.10.9} \end{equation}

Notice that

\begin{equation} M^2 = \lambda^2 |v \rangle \langle v| + \mu^2 |w \rangle \langle w| + ... ~\text{.}\tag{6.10.10} \end{equation}

since all cross terms cancel, and since each projection matrix squares to itself. Furthermore, recall that the projections add up to the identity matrix, that is,

\begin{equation} I = |v \rangle \langle v| + |w \rangle \langle w| + ... ~\text{,}\tag{6.10.11} \end{equation}

Putting this all together, we see that

\begin{equation} e^{iM\theta} = e^{i\lambda\theta} |v \rangle \langle v| + e^{i\mu\theta} |w \rangle \langle w| + ... ~\text{,}\tag{6.10.12} \end{equation}

which shows explicitly how to relate the decompositions of Hermitian matrices and their corresponding unitary matrices. This remarkable result is much less surprising when expanded in terms of the given orthonormal basis, in which case \(M\) is diagonal, so that exponentiating the matrix is just exponentiating each of the eigenvalues.