# Exponential operator expansion

In my lectures, the professor discussed that for exponential linear operators it is $$\exp(\lambda A + \lambda B) \neq \exp(\lambda A)\exp(\lambda B)$$ for $$AB\neq BA$$.

Now I know that the exponential operator has a similar definition as the exponential function. So there must be also the notion of a Taylor series for operators.

Let us assume that $$\lambda \ll 1$$ such that we can neglect $$O(\lambda^3)$$ but we have to keep to second order in the real number $$\lambda$$. If the exponent of the exponential operator consists of only one small operator with the form $$\exp(A+\lambda B)$$ can we then expand to second order in $$\lambda$$ such that $$\exp(A+\lambda B)\approx \exp(A) + \exp(A)\lambda B + \frac{1}{2}\exp(A) \lambda^2 B^2 + .....?$$ I believe that this is not the case because we also have $$\exp(A)\exp(\lambda B)\approx \exp(A) (I+\lambda B + \frac{1}{2} \lambda^2 B^2 + .....)$$ which is the same as above and that contradicts $$AB\neq BA$$.

One can also apply the definition of the exponential operator directly by $$\exp(A+\lambda B) = \sum_{n=0}^\infty \frac{(A+\lambda B)^n}{n!}$$ but I am not sure how to decompose the "$$A$$" part (assuming we can neglect $$O(\lambda^3)$$) and also how the Taylor expansion would work applied to such an operator without bumping into that contradiction.

Thanks in advance!

## 2 Answers

The standard way to expand $$e^{A+\lambda B}$$ is the Dyson series https://en.wikipedia.org/wiki/Dyson_series.

I will give the derivation I know, though I am sure there must be a good online reference I just couldn't seem to track it down:

Consider $$e^{A+\lambda B}=e^{N((A+\lambda B)/N)}= \prod_{k=1}^N e^{(A+\lambda B)/N}$$

Thus $${d\over d\lambda}e^{A+\lambda B}=\sum_k e^{(k-1)((A+\lambda B)/N)}{d\over d\lambda}[e^{(A+\lambda B)/N}]e^{(N-k)((A+\lambda B)/N)}$$ We are allowed to make $$N$$ as large as we wish, and in the limit we may expand the derivative to first order in $$1\over N$$ to get $${d\over d\lambda}[e^{(A+\lambda B)/N}]\rightarrow {d\over d\lambda}[1+{(A+\lambda B)/N}]= B/N$$ returning everything and going to the continuum limit ( sum to integral): $${d\over d\lambda}e^{A+\lambda B} = \int_0^1 e^{\tau (A+\lambda B)}Be^{(1-\tau) (A+\lambda B)} d\tau$$ So to first order in $$\lambda$$ :

$$e^{A+\lambda B}\sim e^A +\lambda \int_0^1 e^{\tau A}Be^{(1-\tau) A}d\tau+\cdots$$

Higher orders are straightforward, but cumbersome.

The actual expansion is \eqalign{\exp(A + \lambda B) &= \sum_{j=0}^\infty \frac{(A+\lambda B)^j}{j!}\cr &= \sum_{j=0}^\infty \sum_{k=0}^j \frac{P_{j,k}(A,B)}{j!} \lambda^k} where $$P_{j,k}$$ is the sum of the products of $$j$$ factors (in all orders) of which $$k$$ are $$B$$ and the other $$j-k$$ are $$A$$. Thus the coefficient of $$\lambda^0$$ in the expansion of $$\exp(A+\lambda B)$$ is $$\exp(A)$$, but the coefficient of $$\lambda^1$$ is $$\sum_{j=1}^\infty P_{j,1}(A,B)/j! = \sum_{j=1}^\infty \sum_{i=0}^{j-1} A^i B A^{j-i-1}/j!$$ It certainly isn't anything as simple as $$\exp(A) B$$.

EDIT: You can write $$\sum_{j=1}^\infty \sum_{i=0}^{j-1} \frac{A^i B A^{j-i-1}}{j!} = \sum_{i=0}^\infty A^i B f_i(A)$$ where $$f_i(x) = \sum_{j=i+1}^\infty \frac{x^{j-i-1}}{j!} = \frac{\exp(x) (i! - \Gamma(i+1,x))}{x^{i+1} i!}$$ and $$\Gamma(\cdot, \cdot)$$ is the incomplete Gamma function.

EDIT: For the coefficient of $$\lambda^2$$, you want to consider all products with two $$B$$'s. So $$\sum_{j=2}^\infty \sum_{i=0}^{j-2} \sum_{k=0}^{j-2-i} \frac{A^i B A^k B A^{j-2-i-k}}{j!}$$

• Thanks for the answer! Is there a way to decompose the "$A$" part? Oct 16, 2020 at 4:11
• Thank you again for the helpful edit! May I kindly ask you to also add, how you obtained the first order coefficient as well as the result in your EDIT, and how to obtain the second order coefficient :-)? Oct 16, 2020 at 17:11