Is the product of two hermitian matrices always hermitian?

[A. Neumaier]

So, with the (correct) version of the post above we may write more general for $P=\hat{p}_r$:

$\hat{p}_r~~=~~ i\hbar\left(\dfrac{\partial}{\partial r} +\dfrac{1}{r}\right)$

For an arbitrary power $\hat{p}_r^n$ we can write

$\hat{p}^n_r~~=~~ (i\hbar)^n\left(\dfrac{\partial}{\partial r} +\dfrac{n}{r}\right)\left(\dfrac{\partial}{\partial r}\right)^{n-1}$

or alternatively:

$\hat{p}^n_r~~=~~ (i\hbar)^n\left(\dfrac{1}{r^n}\dfrac{\partial}{\partial r}r^n\right)\left(\dfrac{\partial}{\partial r}\right)^{n-1}$

But these powers have different domains, which causes the problems discussed in the present thread.

 

[Neitan Lei]

Intro to QM, David Griffiths, p269
View attachment 248344

 

[A. Neumaier]

View attachment 248827

This shows explicitly that the domain changed; it is no longer an operator on the physical Hilbert space.

 

[vanhees71]

It is the Lebesgue integral over $R^3$, but only if you say that $x$ denote Cartesian coordinates.

The integral is over Euclidean $\mathbb{R}^3$ and as such the volume element is independent of the choice of coordinates,
$$\mathrm{d}^3 x = \epsilon_{ijk} \partial_i \vec{x} \partial_j \vec{x} \partial_k \vec{x} \mathrm{d}^3 q.$$
Of course, it's this specific integral measure to be used in the Hilbert space, because we are dealing with a representation/realization of the Galilei group, where space is Euclidean.