# Complete derivation of generator of rotations

I have been look all across the internet and every book I could find trying to get a full derivation of the generator of rotations and more specifically angular momentum as a generator of rotations. I tried finding the generator of rotations but there comes a point at which almost all sources I have read define the generator. For example, Sakurai states

$$U_\epsilon \approx 1 - iG\epsilon \\ G \to \frac{p_x}{\hbar}; \quad \epsilon \to \mathrm{d}x \\ \implies \mathcal{D}(\hat{\mathbf{n}},\mathrm{d}\phi) \approx 1 - i \frac{\mathbf{J}\cdot \hat{\mathbf{n}}}{\hbar} \mathrm{d}\phi \\ \implies \mathcal{D}_z = \exp{\frac{-iJ_z\phi}{\hbar}}$$

I would really like to see a rigorous derivation of this. Can anyone help?

• Are you missing an equals sign or something (maybe $\approx$) in the first line? – David Z Feb 18 '15 at 2:53
• For internal rotation, this is the definition of intrinsic angular momentum and thus no proof for it. For spatial rotation, if you accept $i\partial_{\phi}$ as a coordinate representation of $J_z$ (or orbital angular momentum), indeed you can check that the operator $D_z$ rotates states in z-axis. However, sakurai states reasons (not proofs) why those are generators by considering the group properties of 3-D rotation – mastrok Feb 18 '15 at 3:42

Transformations of the quantum state are expressible as unitary transformations. The first order expansion of a unitary transformation around identity necessarily gives us $1- i \epsilon \hat K$ where $\hat K$ is a hermitean operator. Just by considering the Galilean transforms of usual 3D coordinates (i.e. non-quantum transformations), we are able to derive the commutation relations of these infinitesimal generators which should apply also in the case of transformations of quantum states.
We then have ten infinitesimal generators of the Galilean group and their commutation relations. Say we call $\hat H$ the infinitesimal generator of time translation and we postulate the position operator $\hat Q$ and velocity operator $\hat V$ and just by requiring $$\frac{\rm d}{{\rm d}t}\langle\hat{Q}\rangle = \langle\hat V\rangle$$
we obtain $$\hat{V} = i [\hat H, \hat Q]$$ and by similar requirements we can recover a large set of commutation relations between the position and velocity operators and the infinitesimal generators. Say we call $G$ the operator which gives the particle an infinitesimal "boost" in the velocity space. We can derive that $$[\hat G, \hat Q]=0$$ In return, by physical considerations of degrees of freedom we can identify $\hat Q=\hat G$. Similarly, under the assumptions of no internal degrees of freedom the operator $\hat Q \times \hat P$ van be identified with the generator of rotations $\hat J$ because of the same commutation relations with the rests. Assuming internal degrees of freedom gives us the commutation relations a spin operator (i.e. $\hat S = \hat J - \hat Q \times \hat P$) should fulfill.
But beware, there are caveats. When the physical situation does not posses translation symmetry (i.e. there is a potential field in which a particle is moving), the general relation between $\hat{V}$ and $\hat{P}$ is not what you would expect $$\hat P = M \hat V + A(\hat Q)$$ where $A(\hat Q)$ corresponds up to factor to the vector potential of the magnetic field. So $\hat Q \times \hat P$ is not the angular momentum in the classical sense but rather in the sense of the phase space of Hamiltonian mechanics.