Squaring an operator

Question

There is an excercise of squaring an operator in my book of quantun mechanics. The operator is $$\hat{A}=\frac{\mathrm{d}}{\mathrm{d}x}+x$$ And I should compute $\hat{A}^2$. He gives me a result $$\hat{A}^2=\frac{\mathrm{d^2}}{\mathrm{d}x^2}+2x\frac{\mathrm{d}}{\mathrm{d}x}+x^2+1$$Isn't there a mistake? If I understand it correctly, there should not be that 2 before $x\frac{\mathrm{d}}{\mathrm{d}x}$?!

Answer 1

Apply the operator to $f(x)$. Then: $$\hat Af=\frac {df}{dx}+xf$$

Apply the operator to $\hat Af$. Then: \begin{align} \hat A^2f &= \hat A\left(\frac {df}{dx}+xf\right) \\ &= \left(\frac {d }{dx}+x \right)\left(\frac {df }{dx}+xf \right) \\ &= \frac {d }{dx}\left(\frac {df }{dx}+xf \right)+x\left(\frac {df }{dx}+xf \right) \\ &= \left(\frac {d }{dx}\frac {df }{dx}+\frac {d xf}{dx} \right)+\left(x\frac {df }{dx}+x^2f \right) \\ &= \frac {d^2f }{dx^2}+x\frac {d f}{dx} +f +x\frac {df }{dx}+x^2f \\ &= \frac {d^2f }{dx^2}+2x\frac {d f}{dx} +x^2f +f \\ &= \left(\frac {d^2 }{dx^2}+2x\frac {d }{dx} +x^2 + 1 \right)f \end{align} So, $$ \boxed { \hat A^2 = \frac {d^2 }{dx^2}+2x\frac {d }{dx} +x^2 + 1. }$$