30 Matrix exponentials and operator exponentials
Matrix and operator exponentials occur throughout quantum mechanics and quantum chemistry. They appear in time evolution, unitary transformations, orbital rotations, coupled-cluster theory, and many other settings.
For a square matrix or linear operator \(A\), the exponential is defined by its power series
\[ e^A = I + A + \frac{1}{2!}A^2 + \frac{1}{3!}A^3 + \cdots = \sum_{n=0}^\infty \frac{A^n}{n!}. \]
This definition is the direct analogue of the exponential series for a number. The important difference is that matrices and operators generally do not commute. If \(A\) and \(B\) are operators, it is therefore generally true that \[ \text{In general:} \quad e^{A+B} \neq e^Ae^B. \]
If \(A\) and \(B\) commute, however, the familiar rule is recovered:
\[ [A,B]=AB-BA=0 \quad\Longrightarrow\quad e^{A+B}=e^Ae^B=e^Be^A. \]
The inverse of an exponential is particularly simple:
\[ (e^A)^{-1}=e^{-A}. \]
Consequently,
\[ e^Ae^{-A}=e^{-A}e^A=I. \]
This result does not require any additional commutation assumptions, since \(A\) always commutes with itself.
A central operation in quantum mechanics is a similarity transformation
\[ \widetilde B=e^A B e^{-A}. \]
To understand such expressions, define the commutator
\[ [A,B]=AB-BA. \]
Repeated or nested commutators are written, for example,
\[ [A,[A,B]], \qquad [A,[A,[A,B]]]. \]
The exponential similarity transformation can then be expanded as
\[ e^A B e^{-A} = B+[A,B]+\frac{1}{2!}[A,[A,B]] +\frac{1}{3!}[A,[A,[A,B]]]+\cdots. \]
This is sometimes called the Hadamard lemma. Physicists sometimes call it the Baker–Campbell–Hausdorff formula, although the two are not exactly the same.
The above expansion is especially useful when the nested commutators eventually vanish or follow a simple pattern, such as in coupled-cluster theory.
A closely related problem is to combine two exponentials into one. In general,
\[ e^Ae^B=e^C, \]
where \(C\) is not simply \(A+B\). The Baker–Campbell–Hausdorff (BCH) expansion begins
\[ C=A+B+\frac12[A,B]+\frac1{12}[A,[A,B]]+\frac1{12}[B,[B,A]]+\cdots. \]
Thus, the corrections to the familiar scalar rule are expressed entirely in terms of commutators. When \([A,B]=0\), all these corrections disappear and \(C=A+B\).
These formulas are useful not only for formal manipulations. They provide a systematic way of understanding how noncommuting transformations combine, and why operator ordering matters in quantum mechanics.
Exercises
Exercise 30.1 (Exponential of a diagonal matrix) Consider the matrix
\[ A=\begin{pmatrix}a&0\\0&b\end{pmatrix}. \]
Calculate \(A^2\).
Calculate \(A^3\).
Use the power-series definition \(e^A=I+A+A^2/2!+\cdots\) to show that
\[ e^A=\begin{pmatrix}e^a&0\\0&e^b\end{pmatrix}. \]
Find \(e^{-A}\).
Verify explicitly that \(e^Ae^{-A}=I\).
Exercise 30.2 (An exponential whose series terminates) In coupled-cluster theory, the cluster operator \(T\) is nilpotent. This means that some power of \(T\) vanishes. In this exercise, we explore a simple example of a nilpotent matrix and its exponential. Consider
\[ N=\begin{pmatrix}0&1\\0&0\end{pmatrix}. \]
Calculate \(N^2\).
Calculate \(N^3\).
Use the power-series definition of the exponential to calculate \(e^N\).
Calculate \(e^{-N}\).
Multiply your two matrices and verify that \(e^Ne^{-N}=I\).
Explain why the exponential series in this example contains only finitely many nonzero terms.
Exercise 30.3 (Exponentials of commuting matrices) Consider
\[ A=\begin{pmatrix}a&0\\0&b\end{pmatrix}, \qquad B=\begin{pmatrix}c&0\\0&d\end{pmatrix}. \]
Calculate \(AB\) and \(BA\).
Calculate the commutator \([A,B]\).
Calculate \(e^A\), \(e^B\), and \(e^{A+B}\).
Verify directly that
\[ e^Ae^B=e^{A+B}. \]
Which property of \(A\) and \(B\) makes the familiar exponential rule work in this example?
Exercise 30.4 (A matrix exponential and rotations) Consider
\[ J=\begin{pmatrix}0&-1\\1&0\end{pmatrix}. \]
Calculate \(J^2\), \(J^3\), and \(J^4\).
Use the power series of \(e^{\theta J}\) and separate terms containing even and odd powers of \(J\).
Recognize the Taylor series of sine and cosine and show that
\[ e^{\theta J}= \begin{pmatrix} \cos\theta&-\sin\theta\\ \sin\theta&\cos\theta \end{pmatrix}. \]
Show that \(e^{\theta J}e^{\phi J}=e^{(\theta+\phi)J}\).
Explain why the result in part d does not contradict the fact that \(e^Ae^B\ne e^{A+B}\) in general.
Exercise 30.5 (Anti-Hermitian matrices and unitary matrices) A matrix \(A\) is anti-Hermitian if
\[ A^\dagger=-A, \]
and a matrix \(U\) is unitary if
\[ U^\dagger U=UU^\dagger=I. \]
Show from the power-series definition of the matrix exponential that
\[ (e^A)^\dagger=e^{A^\dagger}. \]
Let \(A\) be anti-Hermitian and define \(U=e^A\). Show that
\[ U^\dagger=e^{-A}. \]
Use the result from part b to prove that \(U\) is unitary.
Show that if \(H\) is Hermitian and \(\alpha\) is real, then \(A=i\alpha H\) is anti-Hermitian. Conclude that
\[ U=e^{i\alpha H} \]
is unitary.
Consider the Hermitian matrix
\[ H=\begin{pmatrix}1&0\\0&-1\end{pmatrix}. \]
Calculate \(U=e^{i\theta H}\) explicitly and verify directly that \(U^\dagger U=I\).
The converse is also true: every unitary matrix can be written as the exponential of an anti-Hermitian matrix. Suppose that \(U\) has an orthonormal basis of eigenvectors with eigenvalues \(e^{i\theta_j}\). Explain how to construct an anti-Hermitian matrix \(A\) such that \(U=e^A\).
Exercise 30.6 (A terminating similarity transformation) Let
\[ A=\begin{pmatrix}0&a\\0&0\end{pmatrix}, \qquad B=\begin{pmatrix}b&0\\0&c\end{pmatrix}. \]
Calculate the commutator \([A,B]\).
Calculate the nested commutator \([A,[A,B]]\).
Use the expansion
\[ e^A B e^{-A}=B+[A,B]+\frac12[A,[A,B]]+\cdots \]
to calculate \(e^A B e^{-A}\).
Calculate \(e^A\) and \(e^{-A}\) directly and verify your result by matrix multiplication.
Why did the commutator expansion terminate after only a few terms?
Exercise 30.7 (A simple BCH expansion) Suppose that two operators \(A\) and \(B\) satisfy
\[ [A,B]=C, \]
and that \(C\) commutes with both \(A\) and \(B\):
\[ [A,C]=[B,C]=0. \]
Write the BCH expansion for \(\log(e^Ae^B)\) through the first nested commutators.
Show that all nested commutators appearing after \([A,B]\) vanish.
Conclude that
\[ e^Ae^B=e^{A+B+\frac12[A,B]}. \]
Similarly, find an expression for \(e^Be^A\) as a single exponential.
Use your answers to show that
\[ e^Ae^B=e^Be^A e^{[A,B]}. \]
What does this result reduce to when \([A,B]=0\)?
Exercise 30.8 (Similarity transformation of an eigenoperator) Suppose \(A\) and \(B\) satisfy
\[ [A,B]=\lambda B, \]
where \(\lambda\) is a scalar.
Show that
\[ [A,[A,B]]=\lambda^2B. \]
Show more generally that the \(n\)-fold nested commutator of \(B\) with \(A\) is
\[ \underbrace{[A,[A,\ldots,[A}_{n\text{ times}},B]\ldots]]=\lambda^nB. \]
Insert this result into the commutator expansion of \(e^ABe^{-A}\).
Sum the resulting series and show that
\[ e^ABe^{-A}=e^\lambda B. \]
In the harmonic oscillator, the number operator \(\hat N\) and creation operator \(a^\dagger\) satisfy
\[ [\hat N,a^\dagger]=a^\dagger. \]
Use the result above to evaluate
\[ e^{\alpha\hat N}a^\dagger e^{-\alpha\hat N}. \]
The annihilation operator satisfies \([\hat N,a]=-a\). Find \(e^{\alpha\hat N}ae^{-\alpha\hat N}\).
Exercise 30.9 (BCH expansion for noncommuting quantum operators) Let
\[ A=i\alpha\sigma_x, \qquad B=i\beta\sigma_y, \]
where \(\alpha\) and \(\beta\) are real and the Pauli matrices satisfy
\[ [\sigma_i,\sigma_j]=2i\epsilon_{ijk}\sigma_k. \]
Calculate \([A,B]\).
Calculate \([A,[A,B]]\).
Calculate \([B,[B,A]]\).
Use the BCH expansion
\[ \log(e^Ae^B)=A+B+\frac12[A,B]+\frac1{12}[A,[A,B]]+\frac1{12}[B,[B,A]]+\cdots \]
to obtain the exponent through terms of total degree three in \(\alpha\) and \(\beta\).
Explain why the BCH series does not terminate in this case.
Interpret the calculation in terms of successive rotations: why should two rotations about different axes not simply combine by adding their generators?