# Exponentials of operators – Hadamard’s lemma

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:

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

Here we’ll look at a special function of two operators of the form

$\displaystyle h\left(A,B\right)=e^{A}Be^{-A} \ \ \ \ \ (2)$

If ${\left[A,B\right]=0}$, we can cancel the two exponentials and get the result ${h\left(A,B\right)=B}$. However, if ${\left[A,B\right]\ne0}$ the two exponentials must remain separated by the middle ${B}$ operator. To get a simpler form for this function, we’ll consider the auxiliary function

$\displaystyle f\left(t\right)=e^{tA}Be^{-tA} \ \ \ \ \ (3)$

where ${t}$ is some parameter. We’ll need the first 3 derivatives at ${t=0}$:

 $\displaystyle f\left(0\right)$ $\displaystyle =$ $\displaystyle B\ \ \ \ \ (4)$ $\displaystyle f^{\prime}\left(t\right)$ $\displaystyle =$ $\displaystyle Ae^{tA}Be^{-tA}-e^{tA}Be^{-tA}A\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle e^{tA}\left[A,B\right]e^{-tA}\ \ \ \ \ (6)$ $\displaystyle f^{\prime}\left(0\right)$ $\displaystyle =$ $\displaystyle \left[A,B\right]\ \ \ \ \ (7)$ $\displaystyle f^{\prime\prime}\left(t\right)$ $\displaystyle =$ $\displaystyle Ae^{tA}\left[A,B\right]e^{-tA}-e^{tA}\left[A,B\right]e^{-tA}A\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle e^{tA}\left[A,\left[A,B\right]\right]e^{-tA}\ \ \ \ \ (9)$ $\displaystyle f^{\prime\prime}\left(0\right)$ $\displaystyle =$ $\displaystyle \left[A,\left[A,B\right]\right]\ \ \ \ \ (10)$ $\displaystyle f^{\prime\prime\prime}\left(t\right)$ $\displaystyle =$ $\displaystyle e^{tA}\left[A,\left[A,\left[A,B\right]\right]\right]e^{-tA}\ \ \ \ \ (11)$ $\displaystyle f^{\prime\prime\prime}\left(0\right)$ $\displaystyle =$ $\displaystyle \left[A,\left[A,\left[A,B\right]\right]\right] \ \ \ \ \ (12)$

We can now write a Taylor expansion of 3 around ${t=0}$:

 $\displaystyle e^{tA}Be^{-tA}$ $\displaystyle =$ $\displaystyle f\left(0\right)+tf^{\prime}\left(0\right)+\frac{t^{2}}{2}f^{\prime\prime}\left(0\right)+\frac{t^{3}}{6}f^{\prime\prime\prime}\left(0\right)+\ldots\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle B+\left[A,B\right]t+\left[A,\left[A,B\right]\right]\frac{t^{2}}{2!}+\left[A,\left[A,\left[A,B\right]\right]\right]\frac{t^{3}}{3!}+\ldots \ \ \ \ \ (14)$

Taking ${t=1}$ gives the required expansion

$\displaystyle e^{A}Be^{-A}=B+\left[A,B\right]+\frac{1}{2!}\left[A,\left[A,B\right]\right]+\frac{1}{3!}\left[A,\left[A,\left[A,B\right]\right]\right]+\ldots \ \ \ \ \ (15)$

This is known as Hadamard’s lemma.

If we introduce the notation

 $\displaystyle \mbox{ad}_{A}\left(B\right)$ $\displaystyle \equiv$ $\displaystyle \left[A,B\right]\ \ \ \ \ (16)$ $\displaystyle \mbox{ad}_{A}\mbox{ad}_{A}\left(B\right)$ $\displaystyle =$ $\displaystyle \left[A,\left[A,B\right]\right] \ \ \ \ \ (17)$

and in general ${\left(\mbox{ad}_{A}\right)^{n}\left(B\right)}$ is the ${n}$th order commutator of ${A}$ with ${B}$, then we can write Hadamard’s lemma as

 $\displaystyle e^{A}Be^{-A}$ $\displaystyle =$ $\displaystyle \sum_{n=0}^{\infty}\frac{1}{n!}\left(\mbox{ad}_{A}\right)^{n}\left(B\right)\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \exp\left(\mbox{ad}_{A}\right)\left(B\right) \ \ \ \ \ (19)$