# Meaning of Lorentz Generators

I'm trying to understand infinitesimal Lorentz transformations in quantum field theory. I've studied some Lie theory from mathematicians, but I'm having trouble adjusting conceptually to how Lie algebras are actually used in theoretical physics.

1. The book I'm reading introduces the Hermitian generators: $$L_{\mu \nu} = i(x_{\mu} \partial_{\nu} - x_{\nu} \partial_{\mu})$$ and then uses these to express an infinitesimal Lorentz transformation $\Lambda^{\mu}_{\;\;\nu}$. The problem is, I tend to think of symbols with Greek indices (like $\Lambda^{\mu}_{\;\;\nu}$ and $L_{\mu \nu}$) as being Lorentz tensors, i.e. like matrices, whereas $L_{\mu \nu}$ seems to be some kind of differential operator acting on fields? How can we then write the tensor $\Lambda^{\mu}_{\;\;\nu}$ in terms of these $L_{\mu \nu}$?

2. The book goes on to say that the $L_{\mu \nu}$ form the Lie algebra of $SO(3,1)$ and that the most general representation of this Lie algebra is of the form: $$M_{\mu \nu} = L_{\mu \nu} + S_{\mu \nu}$$ where $S_{\mu \nu}$ are Hermitian operators and satisfy the same commutation relations as the $L_{\mu \nu}$ and commute with them. However, I've been taught to think of a general representation of a Lie algebra as acting on some arbitrary Hilbert space. So how am I supposed to think of $S_{\mu \nu}$ as a tensor, and how does it make sense to add $S_{\mu \nu}$ to $L_{\mu \nu}$?

• $\Lambda^{\mu}_{\;\nu} = \exp \left( \frac{i}{2} \Omega^{\alpha \beta} \left[ J_{\alpha \beta} \right] ^{\mu} _{\;\nu} \right)$ where $\exp$ is matrix exponential and $\Omega^{\alpha \beta}$ is a collection of 6 parameters (antisymmetric tensor) describing the concrete Lorentz transformation. – Prof. Legolasov Apr 13 '17 at 7:40
• @SolenodonParadoxus You should also define $[J_{\alpha\beta}]^\mu{}_\nu$ for a more complete explanation. – Danijel Apr 13 '17 at 7:42
• @SolenodonParadoxus This isn't really an explanation: so $(J_{\alpha \beta})_{\mu \nu}$ are still some kind of differential operators, but $\Lambda^{\mu}_{\;\;\nu}$ is some kind of matrix? How does $\exp$ convert between these? – Maxim Apr 13 '17 at 7:46
• – Christoph Apr 13 '17 at 9:18

1. The $L_{\mu\nu}$ are infinitesimal generators of the Lorentz transformation on the space of fields/functions. If you want to view them as operators on a Hilbert space, just consider the Hilbert space of square-integrable functions. The commutators of the $L_{\mu\nu}$ are the commutation relations of the Lie algebra $\mathfrak{so}(1,3)$, so they form a representation of said algebra.
The Greek indices actually do mean that $L_{\mu\nu}$ is a Lorentz tensor - just check what happens to it under the transformation $x\mapsto \Lambda x$. They actually generate Lorentz transformations in the sense that if you view the $L_{\mu\nu}$ as vector fields on Minkowksi space $\mathbb{R}^{1,3}$, then the integral curves of these vector fields are of the form $x(t) = \Lambda(t)x_0$, where $\Lambda(t)$ is a Lorentz transformation acting on the arbitrary starting point $x_0$ of the integral curve. This is also explained in more detail in the Wikipedia article on the Lie algebra of the Lorentz group.
2. You're supposed to read that sum as a sum on a tensor product. You've got the "natural" representation of the $L_{\mu\nu}$ on the space of functions, and now, in quantum mechanics, you can additional have "internal" spin degrees of freedom, for instance a spinor-valued field like a Dirac spinor field. Then you take your space of real-valued functions $F$ and the spinor space $\mathbb{C}^4$, where the $L$ act on $F$ and the action of the Lorentz algebra on $\mathbb{C}^4$ is given by the $S$, and form the combined space $F\otimes\mathbb{C}^4$ of spinor-valued fields/functions. Then the action of the Lorentz algebra on this space is given by $L_{\mu\nu}\otimes 1 + 1\otimes S_{\mu\nu}$, which is often sloppily written as $L+S$.