# How can we show that spin is the generator of rotation? [duplicate]

Starting with the behaviour of wavefunctions, $$\psi(\vec{r})$$ under rotation, $$\psi'\left(\vec{r}\right)=\psi\left(R^{-1}{\vec r}\right)$$ it is possible to show that the orbital angular momentum $$\hat{L}_i=\varepsilon_{ijk}\hat{x}_j\hat{p}_k,$$ are the generators of rotation. But how can we show that spin $$\vec{S}$$, or for a particle with spin, $$\vec{J}=\vec{L}+\vec{S}$$ is the generator of rotation?

If you have a composite system with Hilbert space $$\mathscr H_A\otimes \mathscr H_B$$, then the action of a rotation $$R$$ on elements of the form $$\psi\otimes \phi$$ is $$R\big(\psi\otimes \phi\big) = \big(R_A\psi\big) \otimes \big(R_B \phi\big)$$ where $$R_A$$ and $$R_B$$ are the corresponding rotation operators on $$\mathscr H_A$$ and $$\mathscr H_B$$. Writing $$R = \exp\left[i\epsilon J\right]$$, $$R_A = \exp\left[i\epsilon J_A\right],$$ and $$R_B=\exp\left[i\epsilon J_B\right]$$ and expanding to first order, we find that

$$J = J_A \otimes \mathbb 1 + \mathbb 1 \otimes J_B$$

which gives the form of the infinitesimal generators of rotations on the composite space.

For a spin-$$s$$ particle in $$d$$-dimensional space, the Hilbert space takes the form $$L^2(\mathbb R^d) \otimes \mathbb C^{2s+1}$$. The infinitesimal generators of rotations on $$L^2(\mathbb R^d)$$ are conventionally called $$L_n$$, while the infinitesimal generators of rotations on $$\mathbb C^{2s+1}$$ are conventionally called $$S_n$$. Therefore, in this context we have that

$$J_n = L_n \otimes \mathbb 1 + \mathbb 1 \otimes S_n$$

In a slight abuse of notation, the tensor product notation is conventionally dropped to leave $$J_n = L_n + S_n$$, where the $$L$$ operators are understood to act on the spatial wavefunction and the $$S$$ operators are understood to act on the $$2s+1$$-dimensional spinor.

It is worth mentioning that the above explanation is a bit backward from an intuitive and historical standpoint. In writing down the theory of the spin-1/2 particle, we don't really start by magically writing down the actions of rotations (that is, (projective) representations of $$\mathrm{SO}(3)$$) on $$L^2(\mathbb R^d)\otimes \mathbb C^{2s+1}$$ and then derive the corresponding generators. Rather, we figure out what the different possible representations of $$\mathrm{SO}(3)$$ look like, define the infinitesimal generators, and then exponentiate them to find the action of rotations on the corresponding spaces.

• Thanks for your detailed answer. I have found a more down-to-earth answer here physics.stackexchange.com/questions/727655 Jul 23 at 16:37
• @Solidification If there is an aspect of my answer which you find too esoteric, feel free to mention it and I can try to clarify. Jul 23 at 16:46
• This answer is also useful for this question: physics.stackexchange.com/q/773156/226902 Jul 23 at 17:52