1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Derive infinitesimal rotation operator

  1. Apr 16, 2015 #1
    1. The problem statement, all variables and given/known data
    Derive the infinitesimal rotation operator around the z-axis.

    2. Relevant 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:
    $$ \approx \psi(r,\theta, \phi) - \epsilon \frac{\partial \psi(r,\theta, \phi)}{\partial \phi} + \frac{\epsilon^2}{2} \frac{\partial^2 \psi(r,\theta, \phi)}{\partial \phi^2} ...$$
    3. 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?
     
  2. jcsd
  3. Apr 16, 2015 #2

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    It's a Taylor series.
     
  4. Apr 16, 2015 #3
    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?
     
  5. Apr 17, 2015 #4

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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 ?
     
  6. Apr 17, 2015 #5
    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 ...
     
  7. Apr 17, 2015 #6

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Think more along the lines of ##f(a+\varepsilon)##. What would its Taylor expansion be about ##x=a##?
     
  8. Apr 17, 2015 #7
    Thanks, that would give the coefficients, but then why wouldn't we have ϕ+ϵ in f(a), like
    $$ \approx ψ(r,θ,ϕ+\epsilon)−ϵ\frac{∂ψ(r,θ,ϕ+\epsilon)}{∂ϕ}+\frac{ϵ^2}{2}\frac{∂^2ψ(r,θ,ϕ+\epsilon)}{∂ϕ^2} $$
     
  9. Apr 19, 2015 #8

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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.​

    --
     
  10. Apr 19, 2015 #9
    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?
     
  11. Apr 19, 2015 #10

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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?
     
  12. Apr 19, 2015 #11
    well, when you put it that way .... :-) I do muddle myself up sometimes, thanks for your patience.
     
  13. Apr 20, 2015 #12

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Almost, but you can see from dimensional considerations that it should be something like
    $$
    ψ(r,θ,ϕ+\epsilon) = ψ(r,θ,ϕ)+ϵ\frac{∂ψ(r,θ,ϕ)}{∂ϕ}+\frac{ϵ^2}{2}\frac{∂^2ψ(r,θ,ϕ)}{∂ϕ^2} \ + \mathcal O (\epsilon^3)
    $$
     
  14. Apr 20, 2015 #13
    I wouldn't have thought of that, I like it - thanks also.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Derive infinitesimal rotation operator
  1. Derivative Operators (Replies: 3)

  2. Rotation operator (Replies: 1)

  3. Rotation operators (Replies: 5)

Loading...