Proof that $\exp(\hat{L}) \hat{a} \exp(-\hat{L}) = \hat{a} + [\hat{L}, \hat{a}] + [\hat{L}, [\hat{L}, \hat{a}]]/2\dots$ .

Question

I've tried to just multiply series from left side, it works. Firstly I'm not sure that its rigorous enough, secondly there's another way.

Hint from the textbook says I should consider $\hat{a}(\zeta)=e^{\zeta \hat{L} }\hat{a}e^{-\zeta \hat{L} }$ and find DE for it, but I'm not sure how can I do that.

Note: $\exp(\hat{a})$ is just $1+\hat{a}+\hat{a}\hat{a}/2+\dots$ as Taylor expansion.

Answer 1

Following the hint, we can take the derivative of $\hat a(\zeta)$ with respect to $\zeta$ and obtain $$\frac{d}{d\zeta} \hat a(\zeta) = [\hat L,\hat a(\zeta)].$$ This is a first-order ODE, and $\hat a(\zeta)$ is the unique solution of this ODE with the initial condition $\hat a(0) = \hat a$.

Now show that the series $\hat a + [\zeta \hat L,\hat a] + [\zeta \hat L,[\zeta \hat L,\hat a]]/2 + \cdots$ also satisfies this ODE and initial condition; this shows that it must equal $\hat a(\zeta)$ for all $\zeta \in \mathbb{R}$.

Answer 2

There is another proof that I find more enlightening and simpler.

If you have two elements $a,b$ in a Banach algebra $A$ it holds in general that: $$\exp({a}) b \exp({-a})=\exp(\mathrm{ad}_a)(b) = 1+ [a,b]+\frac12[a,[a,b]]+\frac16 [a,[a,[a,b]]]+...$$

The proof works like this:

1. Denote for $a\in A$ $L_a: A\to A$, $b\mapsto a\cdot b$ and $R_a: A\to A$, $b\mapsto b\cdot a$ the left and right multiplication operators on $A$.
2. Note that $R_a$ and $L_a$ are bounded linear maps on $A$ (with norm $≤\|a\|$). This means that the operators $\exp(L_a)$ and $\exp(R_a)$ exist.
3. $L_x$ and $R_y$ commute for all $x,y$, since $$L_x(R_y(b))=x(ay)=(xa)y=R_y(L_x(b))$$ This means $\exp(L_x)\cdot\exp(R_y)=\exp(L_x+R_y)$.
4. You have $$\exp({a})b=\sum_{n}\frac{a^n}{n!}b=\sum_n \frac{(L_a)^n(b)}{n!}=\exp(L_a) (b)\\ b\exp(-a)=\sum_n b\frac{(-a)^n}{n!}=\sum_n \frac{(-R_a)^n(b)}{n!}=\exp(-R_a)(b)$$
5. Putting it all together: $$\exp(a)b\exp(-a)=\exp(L_a)\left(\exp(-R_a)(b)\right)=\exp(L_a-R_a)(b)$$ since $R_a$ and $L_a$ commute! But $$(L_a-R_a)(b)=ab-ba=[a,b]=\mathrm{ad}_a(b)$$ So $$\exp({a}) b \exp({-a})=\exp(\mathrm{ad}_a)(b) = 1+ [a,b]+\frac12[a,[a,b]]+\frac16 [a,[a,[a,b]]]+...$$