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I have the following in my lecture notes

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In a past evaluation I was asked to combine the rotation operators about the $x$, $y$ and $z$ axes to get the rotation operator about a generic axis $u$ with $u$ the unit vector in the direction of the axis Therefore I must write (4.30) in terms of $R_z(\theta)= exp(-i\theta L_z)$, $R_x(\theta)= exp(-i\theta L_x)$, $R_y(\theta)= exp(-i\theta L_y)$. I have already tried it myself but I don't come up with anything, actually I arrive at eq (4.31) with an "=" which is wrong. I also googled it but I was unable to find it. Can someone shed some light on how it's done?

some_math_guy
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    WP. – Cosmas Zachos Sep 29 '22 at 11:17
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    Related: https://physics.stackexchange.com/q/29100/2451 , https://physics.stackexchange.com/q/597050/2451 and links therein. – Qmechanic Sep 29 '22 at 11:24
  • Please don't use screenshots of text. Type them up. – Sean E. Lake Sep 29 '22 at 11:32
  • Are you asking how to define the L s as logarithms of the R s? Your group theory review of the rotation matrices failed to cover this? You expect to supplant it with google searches? – Cosmas Zachos Sep 29 '22 at 14:24
  • @CosmasZachos I am asking how to solve it, since I don't have experience with rotation manipulations although I have seen at least one of these matrices in linear algebra so I am not sure what is the best approach here. I wasn't sure if you need to pass through the rotation matrices necessarily, or if it is enough to taylor expand and do some manipulations with the expansions – some_math_guy Sep 29 '22 at 14:45
  • How to solve what? Work out (4.30) out of the three R's? You should NEVER do that... Whoever suggested it? You have mastered the CBH expansion, right? – Cosmas Zachos Sep 29 '22 at 14:50
  • @CosmasZachos The exercise just said combine the 3 R's to get the R_u. And it's the first time I hear about that formula you are linking . Moreover I haven't seen Lie groups and I don't seen most undergraduate courses on QM introduce them or expect that you know about them but maybe it is just fancy names for commutator manipulations that have been presented informally as it is usual in physics judging by the application section of the wikipedia page that looks simple, although I am unfamiliar with that expansion formula – some_math_guy Sep 29 '22 at 15:05
  • I wrote an explicit answer. – Cosmas Zachos Sep 29 '22 at 15:06

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From (4.29), you work out $L_z$ near the identity, $R_z(0)={\mathbb I}$, $$ L_z=\lim_{\theta \to 0}~ i~\partial_\theta R_z(\theta), $$ and likewise for the x and y axes rotations. You can check they are all Hermitean.

You then immediately have $$ R_{\mathbf u}(\theta)= \exp \bigl( -i\theta (u_xL_x + u_yL_y+u_zL_z)\bigr )\\ = \exp \Biggl ( \theta \left (u_x \lim_{ a \to 0} \partial_a R_ x( a)+ u_y \lim_{ b \to 0} \partial_b R_ y( b)+ u_z \lim_{ c \to 0} \partial_c R_ z( c) \right) \Bigg) . \tag{4.30} $$

Cosmas Zachos
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  • The question is to express 4.30 in terms on $R_x(\theta), R_y(\theta), R_z(\theta)$ I am not sure why can't I get from 4.30 that $R_u(\theta)=R_x(\theta) R_y(\theta) R_z(\theta)$, although 4.31 warns against it, then there must be some other way to do it – some_math_guy Sep 29 '22 at 15:12
  • I just did; what is it you don't understand? Since $L_x,L_y,L_z$ do not commute among themselves, (4.30) *cannot* equal to that product of Rs, as you see above. Try the CBH expansion I linked for you for small θ to see that. – Cosmas Zachos Sep 29 '22 at 15:17
  • Let $R_u = f(R_x,R,y,R,x) $ If (4.31) is not true, so f is not a multiplication of the 3 arguments, I guess there must be other expresion for f; otherwise why would the exercise ask to find an expression for f? – some_math_guy Sep 29 '22 at 15:27
  • I am not sure if I can take the log of each of the R's as you suggested, since we are dealing with operators – some_math_guy Sep 29 '22 at 15:28
  • I supplemented my answer. Don't take the log. Take the limit near the identity, which I presume angular momentum generators were defined for you in classical mechanics. – Cosmas Zachos Sep 29 '22 at 15:34
  • Unfortunately I haven't seen those in classical mechanics. What does $~\partial_\theta R_z(\theta)$ mean ? That notation msut be from classical mechanics I guess – some_math_guy Sep 29 '22 at 15:58
  • Derivation of the operator with respect to the c-number variable θ. – Cosmas Zachos Sep 29 '22 at 15:59