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)

2 thoughts on “Exponentials of operators – Hadamard’s lemma

  1. Pingback: Exponentials of operators – Baker-Campbell-Hausdorff formula | Physics pages

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