Take the operator depending on two parameters:
<br />
f(s, t) \equiv \exp[s A + t B]<br />
What are its partial derivatives with respect to s and t?
EDIT:
Scratch that for now. One way of solving it is through the use of Feynamn label ordering rule (R. P. Feynman, Phys. Rev.
84, 108 (1951)). If you go through that paper, the following steps ought to make sense:
<br />
\exp(A + B) = \exp\left(\int_{0}^{1}{(A_{s} + B_{s}) \, ds}\right) = \exp\left(\int_{0}^{1}{A_{s } \, ds}\right) \, \exp\left(\int_{0}^{1}{B_{s} \, ds}\right)<br />
Then, for the second exponential, we can perform Taylor expansion:
<br />
\exp\left(\int_{0}^{1}{B_{s} \, ds}\right) = 1 + \sum_{n = 1}^{\infty}{\frac{1}{n!} \, \left(\int_{0}^{1}{B_{s} \, ds}\right)^{n}} = 1 + \sum_{n = 1}^{\infty}{\frac{1}{n!} \, \int_{0}^{1}{\int_{0}^{1}{\ldots \int_{0}^{1}{dt_{n} \, dt_{n - 1} \, \ldots \, dt_{1} \, B_{t_{n}} \, B_{t_{n - 1}} \, \ldots \, B_{t_{1}}}}}<br />
We can always make a permutation of the dummy indices \{t_{1}, \ldots, t_{n - 1}, t_{n}\} so that we always have the condition t_{n} \ge t_{n - 1} \ge \ldots \ge t_{1} and the order in which the operators are written is automatically properly ordered. But, then, the intervals of integration for the dummy variable t_{k - 1}, \; 2 \le k \le n is [0; t_{k}] and [0, 1] for t_{n}. There are n! such permutations which cancels the factor 1/n!. Thus we can write:
<br />
\exp\left(\int_{0}^{1}{B_{s} \, ds}\right) = 1 + \sum_{n = 1}^{\infty}{\int_{0}^{1}{\int_{0}^{t_{n}}{\ldots \int_{0}^{t_{2}}{dt_{n} \, dt_{n - 1} \, \ldots \, dt_{1} \, B_{t_{n}} \, B_{t_{n - 1}} \, \ldots \, B_{t_{1}}}}}}<br />
As for the factor \exp\left(\int_{0}^{1}{A_{s} \, ds}\right), we divide it into n + 1 segments:
<br />
\exp\left(\int_{0}^{1}{A_{s} \, ds}\right) = \exp\left(\int_{t_{n}}^{1}{A_{s} \, ds}\right) \, \exp\left(\int_{t_{n - 1}}^{t_{n}}{A_{s} \, ds}\right) \, \ldots \, \exp\left(\int_{0}^{t_{1}}{A_{s} \, ds}\right)<br />
and insert each factor on the proper place to ensure normal ordering. We have:
<br />
\exp\left(\int_{0}^{1}{(A_{s} + B_{s}) \, ds}\right) = \exp\left(\int_{0}^{1}{A_{s} \, ds}\right) + \sum_{n = 1}^{\infty}{\int_{0}^{1}{\int_{0}^{t_{n}}{\ldots \int_{0}^{t_{2}}{dt_{n} \, dt_{n - 1} \, \ldots \, dt_{1} \, \exp\left(\int_{t_{n}}^{1}{A_{s} \, ds}\right) \, B_{t_{n}} \, \exp\left(\int_{t_{n - 1}}^{t_{n}}{A_{s} \, ds}\right) \, B_{t_{n - 1}} \, \ldots}}}}<br />
<br />
\times \, \exp\left(\int_{t_{1}}^{t_{2}}{A_{s} \, ds}\right) \, B_{t_{1}} \, \exp\left(\int_{0}^{t_{1}}{A_{s} \, ds}\right)<br />
Everything is normal-ordered in the expression on the rhs and we can take away the labels on the operators. It is convenient to introduce:
<br />
f(t) \equiv e^{-t \, A} \, B \, e^{t \, A} <br />
The expression can be rewritten as:
<br />
\exp(A + B) = e^{A} \, \left( 1 + \sum_{n = 1}^{\infty}{\int_{0}^{1}{\int_{0}^{t_{n}}{\ldots \int_{0}^{t_{2}}{dt_{n} \, dt_{n - 1} \, \ldots \, dt_{1} \, f(t_{n}) \, f(t_{n - 1}) \, \ldots \, f(t_{1})}}}}\right)<br />
Next, use a corollary of the Baker-Cambell-Hausdorff Theorem as well as the fact that the commutator of
A and
B is a constant
k, to prove:
<br />
f(t) = B - t \, k<br />
Notice that:
<br />
\int_{0}^{t_{2}}{dt_{1} \, f(t_{1})} = B \, t_{2} - \frac{k}{2} \, t^{2}_{2} = t_{2} \, \left( B - \frac{k}{2} \, t_{2} \right)<br />
<br />
\int_{0}^{t_{3}}{dt_{2} \, (B - t_{2} k) \, (B \, t_{2} - \frac{k}{2} \, t^{2}_{2})} = B^{2} \, \frac{t^{2}_{3}}{2} - \frac{k}{2} \, B \, t^{3}_{3} + \frac{k^{2}}{2 \cdot 4} \, t^{4}_{3} = \frac{t^{2}_{3}}{2} \, \left( B - \frac{k}{2} \, t_{3} \right)^{2}<br />
Continue in this way until you see some pattern. Prove it by mathematical induction. Then, do some algebraic simplification to get the final results..
OPs, please don't ban me.
