# Question about $e^{\frac{-itA}{\hbar}}(\hat{Q}+ \hat{P})e^{\frac{itA}{\hbar}}$

This arises in the context of trying to rigorously understand quantum dynamics but it's a functional analysis issue. For simplicity suppose we are in dimension $$1$$.

Let $$\hat{Q}$$ be the operator that's multiplication by $$x$$ and $$\hat{P} = \frac{h}{i}\frac{d}{dx}$$. Suppose $$A$$ is a denseley defined self-adjoint operator on $$L^2(\mathbb{R})$$ and that $$e^{\frac{-itA}{\hbar}}(\hat{Q}+ \hat{P})e^{\frac{itA}{\hbar}} = a\hat{Q} + b\hat{P}$$ where $$a,b \in\mathbb{R}$$

The author then says it follows from "exponentiation" that $$e^{\frac{-itA}{\hbar}}e^{\frac{i}{\hbar}(\hat{Q}+ \hat{P})}e^{\frac{itA}{\hbar}} = e^{\frac{i}{\hbar}(a\hat{Q} + b\hat{P})}$$ I can't think of a rigorous explanation for what exponentiation means and why the first identity implies the second. In particular I'm uncomfortable with the fact that conjugating by a unitary operator goes through the exponentiation. I'm assuming this is a functional calculus thing. I was wondering if anyone knew of a rigorous explanation for this.

• In most case this should be enough : let $D(A)\subset L^2$ be a dense subspace and $A$ be self-adjoint $D(A) \to L^2$ (that is for $f,g \in D(A)$, $\langle Af,g \rangle = \langle f, Ag \rangle$) and $(A-cI)^{-1}$ be compact for some $c$, then the SVD gives $(A-cI)^{-1} f= \sum_n \lambda_n \langle f,\phi_n \rangle \phi_n$, $A f= \sum_n (\lambda_n^{-1}+c) \langle f,\phi_n \rangle \phi_n$ where $\lambda_n^{-1}+c$ are real and $e^{i t A} f = \sum_n e^{it(\lambda_n^{-1}+c)} \langle f,\phi_n \rangle \phi_n$ is well-defined and unitary $L^2 \to L^2$ – reuns Apr 16 at 5:15
• What do you mean by SVD? – Questions Apr 16 at 5:27

I'm not sure this is the level of rigor you need. But in physics this identity is normally motivated as follows (let me set $$h=1$$ for readability): $$U = e^{it A}\;\Longrightarrow\; U U^{\dagger} = \mathbb{1}\,.$$ Then \begin{aligned} U^\dagger\, e^{i(\hat{Q}+\hat{P})} U &= \sum_{n=0}^\infty U^\dagger\, \frac{i^n}{n!}(\hat{Q}+\hat{P})^n \,U \\&= \sum_{n=0}^\infty\frac{i^n}{n!}\, U^\dagger\, (\hat{Q}+\hat{P})\,UU^\dagger (\hat{Q}+\hat{P}) \cdots U U^\dagger (\hat{Q}+\hat{P}) \,U \\& =\sum_{n=0}^\infty\frac{i^n}{n!}\, \big(U^\dagger\, (\hat{Q}+\hat{P})\,U\big)^n = \\&= e^{i \,U^\dagger\, (\hat{Q}+\hat{P})\,U}\,. \end{aligned} From which the desired identity follows. In the second line I simply inserted $$\mathbb{1}$$ in between each factor in the form of $$UU^\dagger$$.
I guess one should also motivate that the exponential of the operator $$\hat{Q} + \hat{P}$$ is well defined as a power series. For the operator $$\hat{Q}$$ alone this is just multiplication by $$e^x$$ and it clearly matches the sum $$\sum_n x^n/n!$$. For $$\hat{P}$$ the convergence of the exponential is the same as the convergence for the Taylor expansion $$e^{ia\hat{P}} f(x) = \sum_{n=0}^\infty \frac{a^n}{n!}f^{(n)}(x) = f(x+a)\,.$$ Assuming $$f \in C^\infty(\mathbb{R}) \cap L^2(\mathbb{R})$$. In order to exponentiate $$\hat{Q} + \hat{P}$$ you may use Baker–Campbell–Hausdorff formula. $$e^{\hat{Q}}e^{\hat{P}} = e^{\hat{Q}+\hat{P} + \frac12 i}\,,$$ where I used $$[\hat{Q},\hat{P}] = i \mathbb{1}$$ and the fact that all other terms are commutators with $$\mathbb{1}$$ which vanish.