Creation and annihilation operators in coordinate space

Question

I am trying to express the creation and annihilation operators of a single quantum harmonic oscillator in coordinate space. The problem is that, when I use $P \to -i\hbar d/dx$, I get $a=a^\dagger$:

$$a=\left(\frac{m \omega}{2 \hbar}\right)^{1 / 2} X+i\left(\frac{1}{2 m \omega \hbar}\right)^{1 / 2} P \\ = \left(\frac{m \omega}{2 \hbar}\right)^{1 / 2} X + \hbar \left(\frac{1}{2 m \omega \hbar}\right)^{1 / 2} \frac{d}{dx}.$$

Since all coefficients are real and $X$ is Hermitian, it follows that $a^\dagger = a$. What am I doing wrong here?

Answer 1

Note that : $$\left(\frac{d}{dx}\right)^\dagger=-\frac{d}{dx}$$

$$a=(\cdots )X+(\cdots )\frac{d}{dx}$$ $$a^\dagger=(\cdots )X-(\cdots )\frac{d}{dx}$$ As expected $a\not=a^\dagger$.

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Edit: It's not a rigorous proof $$P=P^\dagger $$ $$P\rightarrow -i\hbar \frac{d}{dx}$$ $$-i\hbar \frac{d}{dx}=\left(-i\hbar \frac{d}{dx}\right)^\dagger=+i\hbar\left(\frac{d}{dx}\right)^\dagger$$ $$\Rightarrow \left(\frac{d}{dx}\right)^\dagger=-\frac{d}{dx} $$

Answer 2

Let $D:=d/dx$. I'll give another proof $D^\dagger=-D$. using indefinite integrals (over an unstated space, such as $\Bbb R$ or $\Bbb R^3$):$$\langle u|D^\dagger|v\rangle=\overline{\langle v|D|u\rangle}=\overline{\int v^\ast Dudx}=\int v Du^\ast dx=-\int u^\ast Dvdx=-\langle u|D^\dagger|v\rangle,$$where the penultimate $=$ uses integration by parts (its surface term is $0$ due to the at-infinity behaviours of $u,\,v,\,Du,\,Dv$ for $u,\,v$ both $L^2$-normalizable).