# Exponentials of operators – Baker-Campbell-Hausdorff formula

References: Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 1.9.

Although the result in this post isn’t covered in Shankar’s book, it’s a result that is frequently used in quantum theory, so it’s worth including at this point.

We’ve seen how to define a function of an operator if that function can be expanded in a power series. A common operator function is the exponential:

$\displaystyle f\left(\Omega\right)=e^{i\Omega} \ \ \ \ \ (1)$

If ${\Omega}$ is hermitian, the exponential ${e^{i\Omega}}$ is unitary. If we try to calculate the exponential of two operators such as ${e^{A+B}}$, the result isn’t as simple as we might hope if ${A}$ and ${B}$ don’t commute. To see the problem, we can write this out as a power series

 $\displaystyle e^{A+B}$ $\displaystyle =$ $\displaystyle \sum_{n=0}^{\infty}\frac{\left(A+B\right)^{n}}{n!}\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle I+A+B+\frac{1}{2}\left(A+B\right)\left(A+B\right)+\ldots\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle I+A+B+\frac{1}{2}\left(A^{2}+AB+BA+B^{2}\right)+\ldots \ \ \ \ \ (4)$

The problem appears first in the fourth term in the series, since we can’t condense the ${AB+BA}$ sum into ${2AB}$ if ${\left[A,B\right]\ne0}$. In fact, the expansion of ${e^{A}e^{B}}$ can be written entirely in terms of the commutators of ${A}$ and ${B}$ with each other, nested to increasingly higher levels. This formula is known as the Baker-Campbell-Hausdorff formula. Up to the fourth order commutator, the BCH formula gives

$\displaystyle e^{A}e^{B}=\exp\left[A+B+\frac{1}{2}\left[A,B\right]+\frac{1}{12}\left(\left[A,\left[A,B\right]\right]+\left[B,\left[B,A\right]\right]\right)-\frac{1}{24}\left[B,\left[A,\left[A,B\right]\right]\right]+\ldots\right] \ \ \ \ \ (5)$

There is no known closed form expression for this result. However, an important special case that occurs frequently in quantum theory is the case where ${\left[A,B\right]=cI}$, where ${c}$ is a complex scalar and ${I}$ is the usual identity matrix. Since ${cI}$ commutes with all operators, all terms from the third order upwards are zero, and we have

$\displaystyle e^{A}e^{B}=e^{A+B+\frac{1}{2}\left[A,B\right]} \ \ \ \ \ (6)$

We can prove this result as follows. Start with the operator function

$\displaystyle G\left(t\right)\equiv e^{t\left(A+B\right)}e^{-tA} \ \ \ \ \ (7)$

where ${t}$ is a scalar parameter (not necessarily time!).

From its definition,

$\displaystyle G\left(0\right)=I \ \ \ \ \ (8)$

The inverse is

$\displaystyle G^{-1}\left(t\right)=e^{tA}e^{-t\left(A+B\right)} \ \ \ \ \ (9)$

and the derivative is

 $\displaystyle \frac{dG\left(t\right)}{dt}$ $\displaystyle =$ $\displaystyle \left(A+B\right)e^{t\left(A+B\right)}e^{-tA}-e^{t\left(A+B\right)}e^{-tA}A \ \ \ \ \ (10)$

Note that we have to keep the ${\left(A+B\right)}$ factor to the left of the ${A}$ factor because ${\left[A,B\right]\ne0}$. Now we multiply:

 $\displaystyle G^{-1}\frac{dG}{dt}$ $\displaystyle =$ $\displaystyle e^{tA}e^{-t\left(A+B\right)}\left[\left(A+B\right)e^{t\left(A+B\right)}e^{-tA}-e^{t\left(A+B\right)}e^{-tA}A\right]\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle e^{tA}\left(A+B\right)e^{-tA}-A\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle e^{tA}Ae^{-tA}+e^{tA}Be^{-tA}-A\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle e^{tA}Be^{-tA}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle B+t\left[A,B\right]\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle B+ctI \ \ \ \ \ (16)$

We used Hadamard’s lemma in the penultimate line, which in this case reduces to

$\displaystyle e^{tA}Be^{-tA}=B+t\left[A,B\right] \ \ \ \ \ (17)$

because ${\left[A,B\right]=cI}$ so all higher order commutators are zero.

We end up with an expression in which ${A}$ has disappeared. This gives the differential equation for ${G}$:

$\displaystyle G^{-1}\frac{dG}{dt}=B+ctI \ \ \ \ \ (18)$

We try a solution of the form (this apparently appears from divine inspiration):

$\displaystyle G\left(t\right)=e^{\alpha tB}e^{\beta ct^{2}} \ \ \ \ \ (19)$

From which we get

 $\displaystyle G^{-1}$ $\displaystyle =$ $\displaystyle e^{-\alpha tB}e^{-\beta ct^{2}}\ \ \ \ \ (20)$ $\displaystyle \frac{dG}{dt}$ $\displaystyle =$ $\displaystyle \left(\alpha B+2\beta ct\right)e^{\alpha tB}e^{\beta ct^{2}}\ \ \ \ \ (21)$ $\displaystyle G^{-1}\frac{dG}{dt}$ $\displaystyle =$ $\displaystyle \alpha B+2\beta ct \ \ \ \ \ (22)$

Comparing this to 18, we have

 $\displaystyle \alpha$ $\displaystyle =$ $\displaystyle 1\ \ \ \ \ (23)$ $\displaystyle \beta$ $\displaystyle =$ $\displaystyle \frac{1}{2}\ \ \ \ \ (24)$ $\displaystyle G\left(t\right)$ $\displaystyle =$ $\displaystyle e^{tB}e^{\frac{1}{2}ct^{2}} \ \ \ \ \ (25)$

Setting this equal to the original definition of ${G}$ in 7 and then taking ${t=1}$ we have

 $\displaystyle e^{A+B}e^{-A}$ $\displaystyle =$ $\displaystyle e^{B}e^{c/2}\ \ \ \ \ (26)$ $\displaystyle e^{A+B}$ $\displaystyle =$ $\displaystyle e^{B}e^{A}e^{\frac{1}{2}c}\ \ \ \ \ (27)$ $\displaystyle$ $\displaystyle =$ $\displaystyle e^{B}e^{A}e^{\frac{1}{2}\left[A,B\right]} \ \ \ \ \ (28)$

If we swap ${A}$ with ${B}$ and use the fact that ${A+B=B+A}$, and also ${\left[A,B\right]=-\left[B,A\right]}$, we have

$\displaystyle e^{A+B}=e^{A}e^{B}e^{-\frac{1}{2}\left[A,B\right]} \ \ \ \ \ (29)$

This is the restricted form of the BCH formula for the case where ${\left[A,B\right]}$ is a scalar.