Self adjoint operators in spherical polar coordinates

[JALAJ CHATURVEDI]

Hi, I have a general question. How do I show that an operator expressed in spherical coordinates is self adjoint ? e.g. suppose i have the operator i ∂/∂ϕ. If the operator was a function of x I know exactly what to do, just check
<ψ|Qψ>=<Qψ|ψ>
But what about dr, dphi and d theta

 

[Cryo]

You need to know what your inner product is. The adjointness is defined relative to inner product, i.e. an operator may or may not be self-adjoint depending on which inner product you use.

Normally, for full 3d spherical polars ##\langle \psi_1 | \psi_2 \rangle = \int d^3 r \psi_2^{*} \left(\boldsymbol{r}\right)\psi_1 \left(\boldsymbol{r}\right)=\int_0^\infty r^2 dr \int_0^\pi \sin\theta d\theta \int_0^{2\pi} \psi_2^{*} \left(r,\,\theta,\,\phi\right)\psi_1 \left(r,\,\theta,\,\phi\right)##

It therefore makes sense to define the inner product for functions of ##\phi## as ##\langle \Phi_1 | \Phi_2\rangle=\int_0^{2\pi} d\phi \Phi_1^{*}\left(\phi\right)\Phi_2\left(\phi\right)##. Now an operator ##\hat{O}## on this Hilbert space (function of ##\phi## in the domain ##0\dots2\pi##) is self-adjoint if:

##\int_0^{2\pi} d\phi \Phi_1^{*}\left(\phi\right)\hat{O}\Phi_2\left(\phi\right)=\int_0^{2\pi} d\phi \Phi_2\left(\phi\right) \hat{O} \Phi_1^{*}\left(\phi\right)##