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.

2 thoughts on “Exponentials of operators – Baker-Campbell-Hausdorff formula

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