Let $(\Lambda,a)\in\text{ ISO}_o(3,1)$ be a finite (proper) Poincare transformation and Let $U(\Lambda,b)$ be the corresponding unitary operator implementing this transformation on the Hilbert space of (free) one particle states.

Consider now an infinitesimal Poincare transformation, $(\Lambda,a)=(1+\omega,\varepsilon)$ where 1 is the identity operator and $\varepsilon^a$, $\omega_{ab}=-\omega_{ba}$ are small parameters.

In most textbooks I have found, the authors simply claim that the corresponding unitary operator for such an infinitesimal transformation is (see Weinberg V1 pg 59 or Srednicki pg 17 for example) $$U(1+\omega,\varepsilon)=1+\frac{1}{2}i\omega_{ab}\mathcal{J}^{ab}-i\varepsilon_a\mathcal{P}^a+\cdots$$

where $\mathcal{J}^{ab}$ and $\mathcal{P}^a$ are the generators of Lorentz transformations and translations respectively. I assume that this is some sort of taylor expansion in the infinitesimal parameters $\omega$ and $\varepsilon$, where we only keep linear terms in the infinitesimal parameters.

There are a couple of facets of this expansion which I am unsure about.

  1. Where does the factor of $\frac{1}{2}$ and $i$ come from?
  2. It is my impression that the signs in front of the second and third terms are based on your metric convention. If this is true, why is it so?

Regarding (1), I get the feeling that we have some freedom of choice. We choose the $1/2$ because it makes the exponential expression for a finite transformation look nicer. The factor of $i$ ensures that $U$ is hermitian. If this is true, $\textit{why}$ do we have this freedom of choice?


1 Answer 1


For (1), the idea is to expand in all independent quantities, which are $\omega_{01}, \omega_{02},\omega_{03}, \omega_{12},\omega_{13}, \omega_{23}$, and $\epsilon_{0}$, $\epsilon_{1}$,$\epsilon_{2}$,$\epsilon_{3}$. These are just numbers however, and hence need matrix "coefficients". So when you do the expansion you would get something like this $$ U=1 + \omega_{01}\tilde{J}^{01} + \omega_{02}\tilde{J}^{02}+\omega_{03}\tilde{J}^{03}+\omega_{12}\tilde{J}^{12}+\omega_{13}\tilde{J}^{13}+\omega_{23}\tilde{J}^{23} + \epsilon_{0}\tilde{P}^0 +\epsilon_{1}\tilde{P}^1+\epsilon_{2}\tilde{P}^2+\epsilon_{3}\tilde{P}^3+\mathcal{O}(\omega^2,\epsilon^2)$$

Now $U$ is supposed to be unitary, $U^\dagger=U^{-1}$. Note that we have $U^{-1}(1+\omega,\epsilon)=U(1-\omega,-\epsilon)$. When you expand both sides, this means you must have, $$ \tilde{J}^{12\dagger}=-\tilde{J}^{12}$$ and the same for all the other generators, which means that they are anti-hermitian. It is then a common convention in physics to pull out a factor of $i$, to make them Hermitian, i.e you would define generators $J^{12}$ such that $$\tilde{J}^{12}=iJ^{12}$$. Then $$\tilde{J}^{12\dagger}=i^\ast J^{12\dagger}=-iJ^{12\dagger}$$ but we want $\tilde{J}^{12\dagger}=-\tilde{J}^{12}=-i J^{12}$ and hence the new generators $J$ must be hermitian, i.e. $$J^{12\dagger}=J^{12}$$ Hermitian operators in quantum mechanics are associated with observables, so this is just a convenient convention to make them manifest. This answers your second question.

As for the first part, note that what you wrote is really what I wrote. $$\omega_{\mu\nu}J^{\mu\nu}= \omega_{00}J^{00}+\omega_{01}J^{01}+\omega_{01}J^{10}+\dots$$ The terms with $\omega_{\mu\mu}$ (no sum on $\mu$) vanish by the antisymmetry of $\omega$, and similarly by the antisymmetry, you get $\omega_{10}J^{10}=(-1)^2\omega_{01}J^{01}=\omega_{01}J^{01}$ and hence you actually get $2\omega_{01}J^{01}$, instead of $\omega_{01}J^{01}$ as I had in the expansion at the top. The factor of $1/2$ in your expansion is there to compensate for this.

The minus sign before the momentum generators is basically just convention. I vaguely recall a reason why its convenient so I'll add it here when I get more time later.

Update: I didn't state clearly but because the $\omega$ are antisymmetric it follows that the $J$ must be too because $$\omega_{\mu\nu}J^{\mu\nu}=\omega_{\nu\mu}J^{\nu\mu}=-\omega_{\mu\nu}J^{\nu\mu}$$ where for the first equality I rename $\mu\rightarrow \nu$ and vice-versa and then in the second equality use the antisymmetry of $\omega$. Equating the first and last expression then tells you that $J^{\mu\nu}=-J^{\nu\mu}$.


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