# Coulomb potential as an operator

1. Feb 9, 2016

### ShayanJ

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

2. Feb 9, 2016

### andresB

My first thought is that the momentum operator act as a derivative on a function of the operator r

3. Feb 9, 2016

### A. Neumaier

Since the $x_j$ commute and $r^2=\sum x_j^2$, $x$ and $r$ commute. You can differentiate the relation defining $r$ to find the partial derivatives, and then use $[p_k,f(x)]=-i\hbar \partial f(x)/\partial x_k$. (Don't use $i$ as index if it appears as $\sqrt{-1}$, too.)

Last edited: Feb 10, 2016
4. Feb 9, 2016

### ShayanJ

I've been able to prove $[p_k,f(x)]=-i\hbar \frac{df(x)}{dx_k}$ only for the case of $f(x)$ being a polynomial in $x_k$. But here $\frac 1 r$ doesn't satisfy this condition. Is there a general proof?

Last edited: Feb 9, 2016
5. Feb 9, 2016

### blue_leaf77

Actually the differential operator should be of partial one.
The commutator suggested by Neumaier is applicable to any smooth function $f(x)$, not only polynomials.

6. Feb 9, 2016

### ShayanJ

$f(x)$ is a function of positon operator, not a smooth function of a real variable!

7. Feb 9, 2016

### blue_leaf77

Who said that $f(x)$ is a function of real variable?

8. Feb 9, 2016

### ShayanJ

I got the impression that you meant it was trivial since you didn't provide any proof, so I thought that's what you mean. Do you know where can I find a proof?

9. Feb 9, 2016

### blue_leaf77

Consider
$$[p,x^n] = -i\hbar nx^{n-1}$$
and the Taylor expansion of $f(x)$.

10. Feb 9, 2016

### ShayanJ

Actually that's why I didn't think about using the formula in the first place.

11. Feb 9, 2016

### blue_leaf77

It seems that Taylor expansion of a function operator cannot be taken lightly, I have actually also addressed exactly the same problem on this regard almost a year ago There I asked if a function $f(x)=(1+x)^{-1}$ can be expanded in power terms, as the same function with $x$ real numbers can only have Taylor expansion for a bounded x $-1<x<1$.

12. Feb 9, 2016

### ShayanJ

Is there any proof that doesn't use Taylor expansion of operator functions?

13. Feb 9, 2016

### strangerep

Well, it starts from Taylor expansion of ordinary functions. But yes, one can extend to very general (complex analytic) functions.

To get started, restrict to the case where an (ordinary, polynomial) operator $A$ has an inverse $A^{-1}$. I.e., $A A^{-1} = 1$. Then take the commutator with $p$ on both sides, and use the Leibniz rule on the LHS. What do you find?

[And thank you to @A. Neumaier for teaching me this generalized material ages ago.]

14. Feb 9, 2016

### ShayanJ

For any operator B, I find $[B,A]A^{-1}=A[A^{-1},B]$! So?

15. Feb 9, 2016

### strangerep

Take it further...

When $A = A(x)$ (ordinary polynomial), and $B = p$, and you've already found that $[p, A(x)] = A'(x)$ (modulo factors of $i$ and $\hbar$, depending on your conventions),... then what is $$[p, A^{-1}(x)] ~=~ ?$$

16. Feb 9, 2016

### ShayanJ

I don't see how $[p,A]A^{-1}=A[A^{-1},p]$ implies $[p,A(x)]\propto A'(x)$!

17. Feb 9, 2016

### strangerep

You already had that back in post #4. Now use that to find $[p,A^{-1}] = ?$

18. Feb 9, 2016

### ShayanJ

$[p,A(x)^{-1}]=i\hbar A(x)^{-1} A'(x) A^{-1}(x)$

19. Feb 9, 2016

### strangerep

You keep stopping too soon.

Let $F(x) := A^{-1}(x)$. From the above, what is $[p, F(x)] = \; ?$

20. Feb 9, 2016

### andresB

Can you be sure that A^-1 commutes with A' ?