Coulomb potential as an operator

[ShayanJ]

I've found a document that does some calculations related to LRL vector operator.
Somewhere in it, the author uses the equation below:
${\Large \mathbf{L} \cdot \frac{\mathbf{x}}{r}=\frac{1}{r}(\mathbf{L} \cdot \mathbf{x})+\mathbf{x} \cdot (\mathbf{L} \frac{1}{r})+\frac{1}{r} (\mathbf{x} \cdot \mathbf{L}) }$
But I don't know how he got this. I can't prove it. This is really annoying me. Can anyone help?
Thanks

 

[blue_leaf77]

Well you can imagine $\mathbf{L}$ as a differential operator (which it is in the position basis), and then place a state to right most. Writing in each component,
$$
L_i\frac{x_i}{r}\psi
$$
Now this one has the form of $\hat{D}(fgh)$ where $D$ is a differential operator and $f$ ,$g$, and $h$ are all functions of the variable with which $D$ differentiates. So, it's just the product rule of derivative.

 

[vanhees71]

Well, a pragmatic way to solve these riddles is to work in the position representation (aka wave mechanics). Then it's clear that the definition of a function of the position operator is simply defined via
$$\langle \vec{x}|f(\hat{\vec{x}}) \psi \rangle=\langle f^{\dagger}(\hat{\vec{x}}) x|\psi \rangle = \langle f^*(\vec{x}) \vec{x} \psi \rangle = f(\vec{x}) \langle \vec{x}|\psi \rangle.$$
Further you can prove that
$$\langle \vec{x}|\hat{p} \psi \rangle=-\mathrm{i} \vec{\nabla} \langle \vec{x}|\psi \rangle.$$
To prove this you only need to know that the momentum operator generates translations, i.e., you have
$$\exp(-\mathrm{i} \hat{p} \vec{\xi}) |\vec{x} \rangle=|\vec{x} + \vec{\xi} \rangle.$$
So from this you find
$$\langle \vec{x}+\vec{\xi}|\psi \rangle = \langle \vec{x}|\exp(+\mathrm{i} \vec{\xi} \cdot \vec{x})|\psi \rangle.$$
Taking the gradient wrt. $\vec{\xi}$ and then setting $\vec{\xi}=0$ gives
$$\vec{\nabla}_x \langle \vec{x}|\psi \rangle=\mathrm{i} \langle \vec{x}|\hat{p} \psi \rangle.$$
Now it's pretty easy to calculate all kinds of commutators etc. just using the position representation. It's also more convenient now to use this representation in terms of differential operators and products of position functions as just derived, i.e., you write
$$\psi(\vec{x})=\langle \vec{x}|\psi \rangle, \quad \hat{O} \psi(\vec{x})=\langle \vec{x}|\hat{O} \psi \rangle$$
where $\hat{O}$ is some operator-valued function of $\hat{\vec{x}}$ and $\hat{\vec{p}}$. You must only be careful with operator ordering in products of $\hat{\vec{x}}$ and $\hat{\vec{p}}$. In this notation we have without operator-ordering trouble
$$\hat{p} \psi(\vec{x})=-\mathrm{i} \vec{\nabla} \psi(\vec{x}), \quad f(\hat{\vec{x}}) \psi(\vec{x})=f(\vec{x}) \psi(\vec{x}).$$
There is also no commutator problem in the definition of orbital angular momentum since components of position and momentum wrt. a Cartesian coordinate system in different directions commute, i.e., you simply have
$$\hat{\vec{L}} \psi(\vec{x})=\hat{\vec{x}} \times \hat{\vec{p}} \psi(\vec{x})=-\mathrm{i} \vec{x} \times \vec{\nabla} \psi(\vec{x}).$$
The Hamiltonian for the Kepler problem thus reads
$$\hat{H}=-\frac{1}{2m} \Delta - \frac{Z e^2}{4 \pi r}.$$
Now you can prove all the formulae mentioned in this thread by using well-known rules of partial derivatives.

In the treatment of the Kepler problem using the Runge-Lenz vector as in the posted paper above, it's more convenient to stay in Cartesian coordinates and just use brute force calculus to derive all the necessary commutators and other formulae needed. That's just dull work, which you can as well do using a computer algebra system like Mathematica ;-)).