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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?

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    $\begingroup$ $d/dx$ is not Hermitean, but anti-Hermitean. The second piece of the last line takes a minus sign when you do the conjugate. $\endgroup$ – FrodCube Jun 12 at 11:19
  • $\begingroup$ A particular form of operators $a$ and $a^{\dagger}$ depends also on the scalar product definition. $\endgroup$ – Vladimir Kalitvianski Jun 12 at 14:05
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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 $$P=P^\dagger $$ $$P\rightarrow -i\hbar \frac{d}{dx}$$ $$-i\hbar \frac{d}{dx}=\left(-i\hbar \frac{d}{dx}\right)^\dagger=+i\hbar\left(\frac{d}{dx}\right)^\dagger$$ $$\Rightarrow \left(\frac{d}{dx}\right)^\dagger=-\frac{d}{dx} $$

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

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