Difference between $\phi(x)\to\phi'(x)=\phi(\Lambda^{-1}x)$ and $\phi(x)\to\phi'(x) =e^{-\frac{i}{2}\omega_{\mu\nu}\mathcal{L}^{\mu\nu}}\phi(x)$

Question

A set of objects $\phi^\alpha$, with $\alpha=1,2,...n$, transforms as a representation $D(\Lambda)$ of dimension $n$ of the Lorentz group if, under a Lorentz transformation: $$\phi^\alpha(x)\to\phi^{\prime\alpha}(x) = \left[D(\Lambda)\right]^{\alpha}_{~\beta}~ \phi^\beta(\Lambda^{-1}x) $$ where $$ \left[D(\Lambda)\right]^{\alpha}_{~\beta}= \left[\exp\left(-\frac{i}{2}\omega_{\mu\nu}J_D^{\mu\nu}\right)\right]^{\alpha}_{~\beta} $$ where $J_D^{\mu\nu}$ are the generators in the representation $D$.

Now for a scalar field $\phi$, $$\phi'(x)=\phi(\Lambda^{-1}x)$$ so that comparing with the definition of a representation (first equation), we have, $$\left[D(\Lambda)\right]=1.$$ Therefore, scalar fields are one-dimensinal repreentation of the Lorentz group. So far so good!

Now, since $$\phi(\Lambda^{-1}x)=\exp\left[-\frac{i}{2}\omega_{\mu\nu}\mathcal{L}^{\mu\nu}\right]\phi(x)~~ {\rm with}~~ \mathcal{L}^{\mu\nu}=i(x^\mu\partial^\nu-x^\nu\partial^\mu),$$ we can also show, $$\phi(x)\to\phi^{\prime}(x) = \exp\left[-\frac{i}{2}\omega_{\mu\nu}\mathcal{L}^{\mu\nu}\right] \phi(x).$$ How does one interpret this? This does not conform to our definition of representation given in the first equation: contrary to the first and the third equations above, we have $\phi(x)$ on the RHS instead of $\phi(\Lambda^{-1}x)$.

Answer 1

First consider the simpler example of $$\phi'(x) = \phi(x+a)$$ If $\phi$ is analytic we can write out the Taylor series $$\phi'(x) = \phi(x)+a\frac{d\phi}{dx}(x) + \frac{a^2}{2}\frac{d^2\phi}{dx^2}(x)+\dots$$ Or, slightly more suggestively $$\phi'(x) = \left(1+a\frac{d}{dx}+\frac{1}{2}a^2 \frac{d^2}{dx^2}+\dots\right)\phi(x)$$ $$= \exp\left(a \frac{d}{dx}\right)\phi(x)$$ where we have recognized that the Taylor series expansion of the derivatives matches the Taylor series expansion of the exponential function. Hence, we can write $$\phi(x+a) = \exp\left(a\frac{d}{dx}\right) \phi(x)$$

This connection between translation and the exponential of a derivative operator generalizes to other space transformations, like rotations and boosts. Consider a rotation around $\hat{z}$ by an infinitesimal angle $\epsilon$, which sends $x^\mu$ to $\Lambda^{-1}x^\mu = x^\mu - \epsilon (0,-y,x,0)^\mu$. The first-order term of the "Taylor expansion" of this is $$\phi(x') = \phi(x)+ \epsilon \frac{d\phi(x')}{d\epsilon}$$ $$ = \phi(x) + \epsilon \frac{d x'^\nu}{d\epsilon} \frac{\partial \phi}{\partial x^\nu}(x)$$ $$ = \phi(x) + \epsilon \left(y \frac{\partial \phi}{\partial x} - x \frac{\partial \phi}{\partial y} \right)$$. This suggests (and you can prove it rigourously) that when $\Lambda$ is a rotation around the $z$ axis by angle $\theta$ $$\phi(\Lambda^{-1}x) = \exp\left(\theta\left(y\partial_x - x\partial_y\right)\right)\phi(x)$$

Taking the exponential of differential operators in this way may feel new, but it's really not any different than taking the exponential of anti-hermitian matrices to get unitary matrices like you do when dealing with spin. The difference is that instead of acting on finite dimensional vectors like spinors, these operators act on the space of functions.

For a general spin field we combine these differential operators with the spin-index operators, so we would write $$\phi'(x) = D_\beta^\alpha U \phi^\beta(x)$$ where $$ D = \exp(\omega_{\mu\nu}J^{\mu\nu})$$ and $$ U = \exp(\omega_{\mu\nu}\mathcal{L}^{\mu\nu})$$