Exponential operator expansion

Question

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!

Answer 1

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 $$ \sum_{j=2}^\infty \sum_{i=0}^{j-2} \sum_{k=0}^{j-2-i} \frac{A^i B A^k B A^{j-2-i-k}}{j!} $$

Answer 2

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.