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?


1 Answer 1


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.

  • $\begingroup$ Thanks for your detailed answer. I have found a more down-to-earth answer here physics.stackexchange.com/questions/727655 $\endgroup$ Commented Jul 23, 2023 at 16:37
  • $\begingroup$ @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. $\endgroup$
    – J. Murray
    Commented Jul 23, 2023 at 16:46
  • $\begingroup$ This answer is also useful for this question: physics.stackexchange.com/q/773156/226902 $\endgroup$
    – Quillo
    Commented Jul 23, 2023 at 17:52

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