# Derive infinitesimal rotation operator

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1. Apr 16, 2015

### ognik

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. Apr 16, 2015

### vela

Staff Emeritus
It's a Taylor series.

3. Apr 16, 2015

### ognik

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. Apr 17, 2015

### BvU

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. Apr 17, 2015

### ognik

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 =:

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. Apr 17, 2015

### vela

Staff Emeritus
Think more along the lines of $f(a+\varepsilon)$. What would its Taylor expansion be about $x=a$?

7. Apr 17, 2015

### ognik

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. Apr 19, 2015

### BvU

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 (*)
$$ψ(r,θ,ϕ+\epsilon) = ψ(r,θ,ϕ)+ϵ\frac{∂ψ(r,θ,ϕ)}{∂ϕ}+\frac{ϵ^2}{2}\frac{∂^2ψ(r,θ,ϕ)}{∂ϕ^2} \ + \mathcal O (\epsilon^3)$$
$$ψ(r,θ,ϕ-\epsilon) = ψ(r,θ,ϕ)-ϵ\frac{∂ψ(r,θ,ϕ)}{∂ϕ}+\frac{(-ϵ)^2}{2}\frac{∂^2ψ(r,θ,ϕ)}{∂ϕ^2} \ + \mathcal O (\epsilon^3)$$

(*)
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.​

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9. Apr 19, 2015

### ognik

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. Apr 19, 2015

### vela

Staff Emeritus
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. Apr 19, 2015

### ognik

well, when you put it that way .... :-) I do muddle myself up sometimes, thanks for your patience.

12. Apr 20, 2015

### BvU

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. Apr 20, 2015

### ognik

I wouldn't have thought of that, I like it - thanks also.