Coulomb potential as an operator

[A. Neumaier]

Can you provide any reference on this?

Not directly. Taking a limit in an operator equation (e.g., to go from polynomials to power series) needs an appropriate topology on operators; these are discussed in books on functional analysis; similarly for the use of Cauchy's theorem (which is complex analysis applied to operators). I recommend that you read the first volume (and the beginning of the second) of the Math Physics book series by Reed and Simon.

What algebraic equation do you mean?

$f(x)^2(x_1^2+x_2^2+x_3^2)=1$. Apply the commutator with $p_i$ on both sides using the product rule, and collect terms.

By the way, due to limitations of the current copying software here on PO, when you copy part of a post containing equations you often need (as in your previous post) to edit the formulas to make them come out correctly.

 

[rubi]

By the functional calculus, $f(x)$ is the operator that multiplies by $f(x)$. You can then do the calculation directly:
$[p,f]\Psi = -i (f\Psi)^\prime + i f\Psi^\prime = -i f^\prime\Psi - i f \Psi^\prime + i f \Psi^\prime = -i f^\prime\Psi$

 

[bob012345]

Ju

I want to calculate the commutator ${\Large [p_i,\frac{x_j}{r}]}$ but I have no idea how I should work with the operator ${\Large\frac{x_j}{r} }$.
Is it $x_j \frac 1 r$ or $\frac 1 r x_j$? Or these two are equal?
How can I calculate ${\Large [p_i,\frac 1 r]}$?
Thanks

Just curious, what are you going to do with this?

 

[A. Neumaier]

By the functional calculus, $f(x)$ is the operator that multiplies by $f(x)$. You can then do the calculation directly:
$[p,f]\Psi = -i (f\Psi)^\prime + i f\Psi^\prime = -i f^\prime\Psi - i f \Psi^\prime + i f \Psi^\prime = -i f^\prime\Psi$

nice and simple, thanks!

 

[ShayanJ]

Ju

Just curious, what are you going to do with this?

I'm trying to calculate the commutator of the Hydrogen Hamiltonian with the LRL vector.

 

[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 ;-)).