# Self adjoint operators in spherical polar coordinates

• JALAJ CHATURVEDI
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

## Answers and Replies

Gold Member
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)##