Hermitian operators in spherical coordinates

[noospace]

Hi,

I have a general question. How do I show that an operator expressed in spherical coordinates is Hermitian? e.g. suppose i have the operator $i \partial /\partial \phi$. If the operator was a function of x I know exactly what to do, just check

$\int_\mathbb{R} \psi_l^\ast \hat{A} \psi_m dx = \int_\mathbb{R} (\hat{A} \psi_l)^\ast \psi_m dx$.

Do i need to check that

$\int\int\int Y_{lm}^\ast \hat{A} Y_{lm} r^2\sin\theta d\theta\d\phi dr= \int\int\int (\hat{A}Y_{lm})^\ast Y_{lm} r^2\sin\theta d\theta\d\phi dr$

or is there a simpler way?

[EDIT: I figured it out. The operator I have has no r or theta dependence so the integral is trivial.]

 

[Jhero]

Hi there,

To show that an operator expressed in spherical coordinates is Hermitian, you can follow a similar approach as you would for an operator expressed in Cartesian coordinates. The key is to remember that the Hermitian property of an operator is equivalent to its adjoint being equal to its complex conjugate.

In this case, to show that the operator i \partial /\partial \phi is Hermitian, you can start by writing out its adjoint, which is -i \partial /\partial \phi. Then, you can use the definition of the inner product in spherical coordinates to show that the integral

\int\int\int Y_{lm}^\ast \hat{A} Y_{lm} r^2\sin\theta d\theta\d\phi dr

is equal to its complex conjugate. This will prove that the operator is Hermitian.

As you mentioned, if the operator has no r or theta dependence, the integral will be trivial and the operator will be automatically Hermitian. However, in more complex cases where the operator does have r and/or theta dependence, you will need to perform the integral and show that it is equal to its complex conjugate.

I hope this helps. Please let me know if you have any further questions.