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Derive infinitesimal rotation operator

  • #1
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Homework Statement


Derive the infinitesimal rotation operator around the z-axis.

Homework Equations


My book gives this equation (which I follow) with epsilon the infinitesimal rotation angle:
$$ \hat{R}(\epsilon) \psi(r,\theta, \phi) = \psi(r,\theta, \phi - \epsilon) $$
but I just don't get the next line in the book:
$$ \approx \psi(r,\theta, \phi) - \epsilon \frac{\partial \psi(r,\theta, \phi)}{\partial \phi} + \frac{\epsilon^2}{2} \frac{\partial^2 \psi(r,\theta, \phi)}{\partial \phi^2} ...$$

The Attempt at a Solution


I was expecting something like
$$ =(1-\epsilon) \psi(r,\theta, \phi) $$
I can see the books Eq is a series, but just can't see how they get to it, with partial diffs as well?
 

Answers and Replies

  • #2
vela
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It's a Taylor series.
 
  • #3
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Thanks, I can almost see that - I assume this uses partials because we only vary ϕ ?
I can't quite see how -ϵ is included?
 
  • #4
BvU
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Well, would you be happier with $$
\hat{R'}(\epsilon) \psi(r,\theta, \phi) = \psi(r,\theta, \phi + \epsilon)
$$when you write that as a Taylor series ?
 
  • #5
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Sorry, no. I suspect there is something about the ϕ - ϵ term in a function that I just don't know - and couldn't find by searching. (the text has ϵ clockwise, so I'm happy with the -ϵ). To revert to Wiki basics: f(x) at a point a =:
upload_2015-4-18_10-36-20.png

So f is ψ(r,θ,ϕ) in this case, and x is ϕ ? a should be ϵ ? Already I can see this isn't working ...
 
  • #6
vela
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Think more along the lines of ##f(a+\varepsilon)##. What would its Taylor expansion be about ##x=a##?
 
  • #7
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Thanks, that would give the coefficients, but then why wouldn't we have ϕ+ϵ in f(a), like
$$ \approx ψ(r,θ,ϕ+\epsilon)−ϵ\frac{∂ψ(r,θ,ϕ+\epsilon)}{∂ϕ}+\frac{ϵ^2}{2}\frac{∂^2ψ(r,θ,ϕ+\epsilon)}{∂ϕ^2} $$
 
  • #8
BvU
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With $$
\frac{∂ψ(r,θ,ϕ)}{∂ϕ}=\lim_{\epsilon\downarrow 0} \ \frac {ψ(r,θ,ϕ+\epsilon) - ψ(r,θ,ϕ)}{\epsilon}
$$
you get a plus sign in (*)
$$
ψ(r,θ,ϕ+\epsilon) = ψ(r,θ,ϕ)+ϵ\frac{∂ψ(r,θ,ϕ)}{∂ϕ}+\frac{ϵ^2}{2}\frac{∂^2ψ(r,θ,ϕ)}{∂ϕ^2} \ + \mathcal O (\epsilon^3)
$$
and a minus sign in
$$
ψ(r,θ,ϕ-\epsilon) = ψ(r,θ,ϕ)-ϵ\frac{∂ψ(r,θ,ϕ)}{∂ϕ}+\frac{(-ϵ)^2}{2}\frac{∂^2ψ(r,θ,ϕ)}{∂ϕ^2} \ + \mathcal O (\epsilon^3)
$$

(*)
The Taylor expansion of ##f(a+\varepsilon)## about ##x=a## is not with ##\frac{∂f(a+\varepsilon)}{∂x}## and ##\frac{∂^2f(a+\varepsilon)}{∂x^2}## etc.
but with ##\frac{∂f(a)}{∂x}## and ##\frac{∂^2f(a)}{∂x^2}## etc.​

--
 
  • #9
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Sorry, I'm still stuck here. If I took the the Taylor series of ψ(r,θ,ϕ) w.r.t. ϕ, I would get
$$ ψ(r,θ,ϕ)+\frac{∂ψ(r,θ,ϕ)}{∂ϕ}+\frac{∂^2 ψ(r,θ,ϕ)}{∂ϕ^2}+... $$
Then if I look at Taylor series of f(a+ε), I can see clearly how ε works in the coefficients, but we seem to be using f(a+ε) for the coefficients, and f(a) for the functions/differentials?
 
  • #10
vela
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Take ##f(x) = f(a) + f'(a)(x-a) + \frac{1}{2!}f''(a)(x-a)^2 + \frac{1}{3!}f'''(a)(x-a)^3 + \cdots## and use the substitution ##x = a + \varepsilon##. What do you get?
 
  • #11
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well, when you put it that way .... :-) I do muddle myself up sometimes, thanks for your patience.
 
  • #12
BvU
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...the Taylor series of ψ(r,θ,ϕ) w.r.t. ϕ, I would get
$$ ψ(r,θ,ϕ)+\frac{∂ψ(r,θ,ϕ)}{∂ϕ}+\frac{∂^2 ψ(r,θ,ϕ)}{∂ϕ^2}+... $$
Almost, but you can see from dimensional considerations that it should be something like
$$
ψ(r,θ,ϕ+\epsilon) = ψ(r,θ,ϕ)+ϵ\frac{∂ψ(r,θ,ϕ)}{∂ϕ}+\frac{ϵ^2}{2}\frac{∂^2ψ(r,θ,ϕ)}{∂ϕ^2} \ + \mathcal O (\epsilon^3)
$$
 
  • #13
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I wouldn't have thought of that, I like it - thanks also.
 

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