Coulomb potential as an operator

  • Thread starter ShayanJ
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Main Question or Discussion Point

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
 

Answers and Replies

  • #2
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My first thought is that the momentum operator act as a derivative on a function of the operator r
 
  • #3
A. Neumaier
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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.)
 
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  • #4
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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?
 
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  • #5
blue_leaf77
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##[p_k,f(x)]=-i\hbar \frac{df(x)}{dx_k}##
Actually the differential operator should be of partial one.
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?

I have no idea what you're talking about!
The commutator suggested by Neumaier is applicable to any smooth function ##f(x)##, not only polynomials.
 
  • #6
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The commutator suggested by Neumaier is applicable to any smooth function
## f(x) ## is a function of positon operator, not a smooth function of a real variable!
 
  • #7
blue_leaf77
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Who said that ##f(x)## is a function of real variable?
 
  • #8
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Who said that ##f(x)## is a function of real variable?
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
blue_leaf77
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Consider
$$
[p,x^n] = -i\hbar nx^{n-1}
$$
and the Taylor expansion of ##f(x)##.
 
  • #10
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Consider
$$
[p,x^n] = -i\hbar nx^{n-1}
$$
and the Taylor expansion of ##f(x)##.
A while ago I asked about this and micromass didn't seem to agree with you.(See here!)
Actually that's why I didn't think about using the formula in the first place.
 
  • #11
blue_leaf77
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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
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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##.
Is there any proof that doesn't use Taylor expansion of operator functions?
 
  • #13
strangerep
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Is there any proof that doesn't use Taylor expansion of operator functions?
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? :oldwink:


[And thank you to @A. Neumaier for teaching me this generalized material ages ago.]
 
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  • #14
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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? :oldwink:
For any operator B, I find ## [B,A]A^{-1}=A[A^{-1},B] ##! So?
 
  • #15
strangerep
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For any operator B, I find ## [B,A]A^{-1}=A[A^{-1},B] ##! So?
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
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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 $$[p, A^{-1}(x)] ~=~ ?$$
I don't see how ## [p,A]A^{-1}=A[A^{-1},p] ## implies ##[p,A(x)]\propto A'(x)##!
 
  • #17
strangerep
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I don't see how ## [p,A]A^{-1}=A[A^{-1},p] ## implies ##[p,A(x)]\propto A'(x)##!
You already had that back in post #4. Now use that to find ##[p,A^{-1}] = ?##
 
  • #18
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## [p,A(x)^{-1}]=i\hbar A(x)^{-1} A'(x) A^{-1}(x)##
 
  • #19
strangerep
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You keep stopping too soon.

Let ##F(x) := A^{-1}(x)##. From the above, what is ##[p, F(x)] = \; ?##
 
  • #20
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Can you be sure that A^-1 commutes with A' ?
 
  • #21
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Can you be sure that A^-1 commutes with A' ?
They're both functions of x so I think yes, which gives ## [p,A^{-1}(x)]=i\hbar [A(x)^{-1}]^2 A'(x) ##.

You keep stopping too soon.

Let ##F(x) := A^{-1}(x)##. From the above, what is ##[p, F(x)] = \; ?##
So I have ## [p,F(x)]=i\hbar [F(x)]^2 A'(x) ##. But what is ##A'(x)## in terms of ##F(x)##?
 
  • #22
strangerep
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Can you be sure that A^-1 commutes with A' ?
Sorry -- I should have specified that ##A^{-1}## is 2-sided inverse operator. But in the cases we're considering, where ##A## is a function of a single operator ##x##, it's a reasonably safe assumption, at least for purposes of physics (with appropriate caveats on the domain of definition).
 
  • #23
strangerep
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So I have ## [p,F(x)]=i\hbar [F(x)]^2 A'(x) ##. But what is ##A'(x)## in terms of ##F(x)##?
Oh well, I guess we can all be a bit blind sometimes. Certainly, it happens to me.

You have essentially proven that $$[p, A^{-1}(x)] ~=~ \frac{d A^{-1}(x)}{dx} ~.$$ This generalizes the earlier derivative formula to a larger class of functions than mere polynomials.

From this, you can go further, to functions of the form ##f(x) = g(x)/h(x)##, and generalize the derivative formula even further.
 
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  • #24
A. Neumaier
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## [p,A(x)^{-1}]=i\hbar A(x)^{-1} A'(x) A^{-1}(x)##
You have essentially proven that ##[p, A^{-1}(x)] ~=~ \frac{d A^{-1}(x)}{dx}##. This generalizes the earlier derivative formula to a larger class of functions than mere polynomials.

From this, you can go further, to functions of the form ##f(x) = g(x)/h(x)##, and generalize the derivative formula even further.
In more explicit terms: Once you know that ##[p,f(x)]=-i\hbar f'(x)## for some function ##f## of an operator vector ##x## with commuting components you get it for ##g(x):=f(x)^{-1}## in place of ##f(x)##, too. By linear combination if you start with ##f(x)=x^n## with a multiexponent ##n## (where it follows from the definition by induction) you get it first for all polynomials, then for their inverses, then for all rational functions. Then for limits of rational functions, for integrals, and Cauchy's integral theorem gives it for all functions analytic on the joint spectrum of ##x## (with exception of any singularities). This is what you need to handle the Coulomb potential.
 
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  • #25
strangerep
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In more explicit terms: [...]
Yes -- that's where I was heading. But I didn't want to "deprive" Shyan of the enjoyment I experienced (some time ago) of discovering these things by working them out. :biggrin:
 

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