In order to prove that $\hat{x}$ is hermitian, it is enough to show that
\begin{equation} \langle\psi|x|\phi\rangle = \langle\psi|x|\phi\rangle^* \end{equation} Now, \begin{eqnarray} \langle\psi|\hat{x}|\phi\rangle &=& \int_{-\infty} ^{\infty} \psi(x)^*\hat{x}\phi(x) dx\\ \langle\psi|\hat{x}|\phi\rangle^* &=& \int_{-\infty} ^{\infty} (\phi(x)^*\hat{x}^*\psi(x))^* dx\\ &=& \int_{-\infty} ^{\infty} (\psi(x)^*\hat{x}^*\phi(x))^* dx \end{eqnarray} Now both must be equal for $\hat{x}$ to be hermitian. But I don't understand why. $\hat{x}^* = \hat{x}$.
I read this post. One of the answer wrote that $\hat{x}^* = \hat{x}$ because eigenvalue of $\hat{x}$ is real and that is why $\hat{x}^* = \hat{x}$. But isn't that logic circular? because we know that hermitian operators have real eigenvalues. So using the fact that $\hat{x}^* = \hat{x}$ means we are already assuming $\hat{x}$ to be hermitian


1 Answer 1


Why do you still have a hat on your $x$ in the integral? Inside the intergral it's just the integration variable $x$, as in "$dx$", which is real number.

I suspect that you confused by
$$ \langle x |\hat x|x'\rangle = x' \langle x |x'\rangle= x'\delta (x-x') $$ and te definition of the wavefunctions $$ [\phi(x)]^* = \langle \phi|x\rangle, \quad \psi(x)= \langle x|\psi\rangle. $$ so that inserting two complete set of position eigenstates $$ I = \int dx |x\rangle \langle x|, \quad I = \int dx' |x'\rangle \langle x'|, $$ we have $$ \langle\psi|\hat x|\phi\rangle = \int dx dx' \langle\psi|x\rangle\langle x| \hat x'|x'\rangle\langle x'| \phi\rangle\\ = \int dx dx' \psi(x)^* x' \phi(x') \delta (x-x')\\ = \int dx \psi(x)^* x\phi(x). $$


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