# Is total angular momentum or orbital angular momentum the generator of the rotation operator in quantum mechanics?

Suppose that we have a wavefunction , $$\psi(\textbf{r})$$. An infinitesimal rotation will induce a change in the wavefunction \begin{align*} \psi\rightarrow\psi'=\hat{U}_{\hat{\textbf{n}}}(\delta\theta)\psi(\textbf{r}') \end{align*} Where $$\hat{U}_{\hat{\textbf{n}}}(\delta\theta)$$ is an arbitrary rotation around axis $$\hat{\textbf{n}}$$ by infinitesimally small angle $$\delta\theta$$. The transformation must maintain the wavefunction's value if both the wavefunction and the position are transformed \begin{align*} \psi'(\textbf{r}')=\psi(\textbf{r}) \end{align*} Where $$r'$$ represents the transformed coordinates, $$\textbf{r}'=R_{\hat{\textbf{n}}}(\theta)\textbf{r}\approx\textbf{r}-\delta\theta\hat{\textbf{n}}\times\textbf{r}$$ for rotation matrix $$R_{\hat{\textbf{n}}}(\theta)$$. Therefore, \begin{align*} \psi'(\textbf{r})&=\hat{U}_{\hat{\textbf{n}}}(\delta\theta)=\psi(R^{-1}_{\hat{\textbf{n}}}(\theta)\textbf{r})\\ &\approx\psi(\textbf{r}-\delta\theta\hat{\textbf{n}}\times\textbf{r})\\ &=\psi(\textbf{r})-\delta\theta(\hat{\textbf{n}}\times\hat{\textbf{R}})\cdot\nabla\psi(\textbf{r})+\mathcal{O}\left((\delta\theta)^{2}\right)\\ &=\psi(\textbf{r})-\frac{i}{\hbar}\delta\theta(\hat{\textbf{n}}\times\hat{\textbf{R}})\cdot\hat{\textbf{P}}\psi(\textbf{r})+\mathcal{O}\left((\delta\theta)^{2}\right)\\ &=\psi(\textbf{r})-\frac{i}{\hbar}\delta\theta\hat{\textbf{n}}\cdot(\hat{\textbf{R}}\times\hat{\textbf{P}})\psi(\textbf{r})+\mathcal{O}\left((\delta\theta)^{2}\right)\\ &=(1-\frac{i}{\hbar}\delta\theta\hat{\textbf{n}}\cdot\hat{\textbf{L}})\psi(\textbf{r})+\mathcal{O}\left((\delta\theta)^{2}\right) \end{align*} Where $$\hat{\textbf{R}}=(\hat{X},\hat{Y},\hat{Z})$$ is the vector containing the position operators and \begin{align*} \hat{\textbf{L}}=\hat{\textbf{R}}\times\hat{\textbf{P}} \end{align*} Is the (oribtal) angular momentum operator. Therefore, in the limit as $$\delta\theta\rightarrow0$$ \begin{align*} \hat{U}_{\hat{\textbf{n}}}(\delta\theta)=(\mathbb{1}-\frac{i}{\hbar}\delta\theta\hat{\textbf{n}}\cdot\hat{\textbf{L}}) \end{align*} which can be exponentiated after using the composition property to obtain \begin{align*} \hat{U}_{\hat{\textbf{n}}}(\theta)=\exp\left(-\frac{i}{\hbar}\theta\hat{\textbf{n}}\cdot\hat{\textbf{L}}\right) \end{align*} This suggests that the orbital angular momentum operator is the generator of the rotation operator, however in Sakurai they seem to insert the total angular momentum, \begin{align*} \hat{J}=\hat{L}+\hat{S} \end{align*} for spin angular momentum $$\hat{S}$$ into the rotation operator. I can't see what I've missed out in my derivation. So my question is, total angular momentum or orbital angular momentum the generator of the rotation operator in quantum mechanics? And total angular momentum is in fact the generator (as I expect it will be) what have I missed in my derivation?

• Looks like your $\psi$ is a scalar, so it describes a particle without spin. You need to use a spinor-valued wave function to describe a particle with spin $1/2$, vector-valued to describe spin $1$, etc. Sep 14, 2022 at 14:48

You miss the spin part, so it is correct that you do not derive $$\hat{S}$$ in the full angular momentum operator.
As we know spin is a basic property of particles, and a particle with spin $$l$$ ($$l$$ is a half-integer or an integer) needs $$2l+1$$ components to describe its wavefunction fully. Consider an electron, whose spin is $${1 \over 2}$$, and it needs two functions, $$\psi_{{1 \over 2}}(r)$$ and $$\psi_{-{1 \over 2}}(r)$$, to correspondingly describe its spin-up and spin-down part. The whole wavefunction is the superposition of two, $$\psi_{{1 \over 2}}(r)+\psi_{-{1 \over 2}}(r)$$.
Now consider some rotation operator $$\hat{U}_{\hat{n}}(\phi)$$. Consider a particle with spin $$l$$ and one of its component $$\psi_{m}(r)$$ where $$m \in \{l,l-1,l-2,\cdots,-l\}$$, under the action of $$\hat{U}(\phi)$$, it transforms by the equation $$\hat{U}_{\hat{n}}(\phi)\psi_{m}(r) = \sum_{m'}{D^{(l)}_{mm'}(\hat{n},\phi)\psi_{m'}\big((R(\hat{n},\phi))^{-1}r\big)}$$ where $$D^{(l)}$$ is the $$2l+1$$ irreducible representation of $$\text{SU}(2)$$ and $$R \in \text{SO}(3)$$ is the $$3$$-dimensional rotation. If you expand the equation with $$\phi$$ around $$0$$, you will get the full angular momentum operator $$\hat{J}=\hat{L}+\hat{S}$$ since the components of $$\hat{S}$$ after the multiplication of $$i$$ are the generators of the representation $$D^{(l)}$$ (the Lie algebra of $$D^{(l)}$$ is anti-Hermitian and $$\hat{S}$$ is Hermitian, and they differ by a factor of $$i$$). As a special case, when $$l=0$$, we have $$\hat{J}=\hat{L}$$, which is your result.
• "The whole wavefunction is the superposition of two, $\psi_{{1 \over 2}}(r)+\psi_{-{1 \over 2}}(r)$." - no, that is incorrect. That particular superposition is much too restrictive. Sep 14, 2022 at 16:27
• Why is it too restrictive? I do not specify any condition on $\psi_{{1 \over 2}}(r)$ and $\psi_{-{1 \over 2}}(r)$. They only need to satisfy $\int{d^3r\psi^{\dagger}_{{1 \over 2}}(r)\psi_{{1 \over 2}}(r)}+\int{d^3r\psi^{\dagger}_{-{1 \over 2}}(r)\psi_{-{1 \over 2}}(r)}=1$. Sep 14, 2022 at 16:40
• Sorry for the late response, and I cannot find a source where the transformation is rigorously derived. However, it can be reasonably argued as follow: Consider the non-normalized function $\Psi_m(r;r')=\psi_m(r')\delta(r-r')$, so the wavefunction $\psi_m(r)=\int{d^3r'\Psi_m(r;r')}$. If $\Psi_m(r;r')$ transforms in the same way as the equation when being rotated, then so does $\psi_m(r)$. Sep 19, 2022 at 8:30