# Creation and annihilation operators in coordinate space

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?

• $d/dx$ is not Hermitean, but anti-Hermitean. The second piece of the last line takes a minus sign when you do the conjugate. Jun 12, 2021 at 11:19
• A particular form of operators $a$ and $a^{\dagger}$ depends also on the scalar product definition. Jun 12, 2021 at 14:05

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 $$a\not=a^\dagger$$.

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

• Thanks @Young Kindaichi. Could you explain to me though why $\left(\frac{d}{d x}\right)^{\dagger}=-\frac{d}{d x}$?
– Y2H
Jun 12, 2021 at 11:25
• @Y2H Please find the edit. Jun 12, 2021 at 11:30

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).