# Euclidean Geometry and Quantum Mechanics: Angular Momentum

In classical quantum mechanics the rotation operator $$\mathcal{D}(\vec{\phi})$$ is generated by the hermitian angular momentum operators $$\vec{J}$$ obeying the commutation relations $$[J_i,J_j]=i\hbar\epsilon_{ijk}J_k \tag{1}$$ The way I understand it these commutation relations arise from the euclidean geometry of space, let me make this more precise: The commutator relation (1) is valid (up to some constant $$i\hbar$$) for the generators of the 3D rotation matrices.$$^1$$ For rotations in euclidean space this relation arises naturally. In our abstract hilbert space we construct rotations by the same rule,"borrowing" it from euclidean space.

This is not trivial or obvious but physically makes sense, because we suppose in classical quantum mechanics that the actual physical space is euclidean.

When my line of thought is correct I would suspect changes to (1) in relativistic quantum mechanics as there the physical space is not euclidean any more. Is this indeed correct?

$$^1$$ This fact can be established by analyzing how a geometrical vector $$a$$ in euclidean space changes under a rotation around an axis $$n$$ $$a\overset{R}{\rightarrow}a'$$ Choosing a set of coordinates one finds that there is a linear mapping between the two vectors $$a'$$ and $$a$$, that can be expressed by this matrix. Approximating this matrix for small rotation angels one arrives at the generators I'm talking about.

The homogeneous symmetry group of special relativity is $$O(1,3)$$ (the part connected to unity of that group is $$SO_{+}(1,3)$$). This is not a completely different symmetry group than the homogeneous symmetry group of Galilean mechanics which is $$O(3)$$ (or, again, the part connected to unity is $$SO(3)$$). In other words, $$SO(3)$$ is contained in $$SO_{+}(1,3)$$. Thus, I wouldn't expect so much for the angular momenta relations to change in relativity as I'd expect them to expand.

An element of $$SO_{+}(1,3)$$ can be expressed as $$\Lambda = \exp\big(-\frac{i}{2}\theta^{\alpha\beta}S_{\alpha\beta}\big)$$

where $$\theta_{\alpha\beta}$$ entails the information of boosts and rotations whereas $$S_{\alpha\beta}$$ are the generators of infinitesimal transformations who can be shown to satisfy the commutation relations of $$SO_{+}(1,3)\simeq SL(2,\mathbb{C})$$. One can now hope to see the usual angular momenta among these $$S_{\alpha\beta}$$ and they'd be correct to be hopeful.

If one defines

$$S^{i}\equiv\frac{1}{2}\epsilon^{ijk}S_{jk}, K^i\equiv S^{0i}$$

then one can see that

$$[S^i,S^j]=i\epsilon^{ijk}S^{k}$$ $$[S^i,K^j]=i\epsilon^{ijk}K^{k}$$ $$[K^i,K^j]=-i\epsilon^{ijk}S^{k}$$

It can be shown that $$S^i$$ are the generators of rotations whereas $$K^{i}$$ are the generators of boosts and the full Lie algebra of relativity comes down to giving the commutation relations between boosts and between a boost and a rotation in addition to the non-relativistic commutation relations between rotations.

So, as I said, we get a bigger symmetry group with more generators (and thus, naturally, more commutation relations). But we also obtain the usual generators of rotation with the same commutation relations in that algebra in the form of $$[S^i, S^j]=i\epsilon^{ijk}S^{k}$$.

For the sake of simplicity, I have limited the discussion to the homogeneous parts of the symmetry groups. If you include the non-homogeneous parts then you'd see that the generators of orbital angular momenta also play a role in generating rotations but it doesn't really change the discussion much qualitatively.