Let $$L=-\frac{d^2}{dx^2}+x^2$$ be the Hermite operator. If a function $f$ is in the Schwarz space, $\mathcal{S}(\mathbb{R})$, then $$L(f)=-\frac{d^2f}{dx^2}+x^2f.$$
Moreover, define the annihilation and creation operators, respectively, as $$A(f)=\frac{df}{dx}+xf\quad\text{and}\quad A^\ast(f)=-\frac{df}{dx}+xf.$$
I am asked to show that for all $f,g\in\mathcal{S}(\mathbb{R})$
- $(Af,Af)=(A^\ast A,f)\geq 0$
- $A^\ast A = L-I$
Where the inner product of $f,g\in\mathcal{S}(\mathbb{R})$ is defined to be $$\int_\mathbb{R} f(x)\overline{g(x)}\,dx$$
I'm having difficulty computing $(A^\ast A f,f)$ and recovering $(Af,Af)$ from it. I see that if we let the operator $A^\ast$ act on $Af$ we have \begin{align} (A^\ast Af,f)&=\Big(A^\ast\Big(\frac{df}{dx}+xf\Big),f\Big)\\ &=\Big(\Big(-\frac{d}{dx}+x\Big)\Big(\frac{df}{dx}+xf\Big),f\Big)\\ (?)&=\Big(\frac{d^2f}{dx^2}+x^2f,f\Big)\\ &=(Lf,f)\geq 1 \quad\text{(by an earlier exercise)}. \end{align} Which demonstrates the inequality, but I'm missing the identity.
Furthermore, computing $A^{\ast} A$ yields $$A^\ast A=\frac{d^2}{dx}+x^2\ne L-1.$$
Since neither of my calculations are correct, I feel I must be missing something fundamental.
Does anyone have a suggestion on how to approach these inner products and operators, or care to provide a solution.
For reference, this is exercise 23 in chapter 5 of Stein and Shakarchi's Fourier Analysis.
It seems that you've mistaken the computation of $A^*A$. In particular, $d/dx$ does not commute with $x$(which is the reason for Heisenberg's uncertainty principle in physics), i.e., $x\cdot\dfrac d{dx}-\dfrac d{dx}\cdot x\not=0$, so $A^*A\not=-(d/dx)^2+x^2$. In fact we have $$\dfrac{d}{dx}(xf)=f+x\dfrac{d}{dx}f\Rightarrow \frac d{dx}\cdot x=1+x\cdot\dfrac d{dx}$$ So \begin{align} A^*A&=\left(-\dfrac d{dx}+x\right)\left(\dfrac d{dx}+x\right)\\ &=-\dfrac{d^2}{dx^2}-\dfrac d{dx}\cdot x+x\cdot\dfrac d{dx}+x^2\\ &=-\dfrac{d^2}{dx^2}-1-x\cdot\dfrac d{dx}+x\cdot\dfrac d{dx}+x^2\\ &=L-1 \end{align} To see $\left<Af,Af\right>=\left<A^*Af,f\right>$, it's better to look at it in a more general way: we may prove \begin{align} \left<Af,g\right>&=\int_\mathbb{R}\left(\dfrac d{dx}+x\right)f\cdot \bar{g}dx\\ &=\int_\mathbb{R}xf(x)\overline{g(x)}dx+f\bar{g}\Big\vert^{\infty}_{-\infty}-\int_\mathbb{R}f\dfrac d{dx}\bar{g}dx\\ &=\int_\mathbb{R}f\left(-\dfrac d{dx}+x\right)\bar{g}dx\\ &=\left<f,A^*g\right> \end{align} Where we used integration by part.