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

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 $$\phi(\Lambda^{-1}x)$$.

• I'm not sure I understand the problem. The definition of a representation surely doesn't depend on which pen strokes we use to write our equations, but rather on what those pen strokes mean. You've already said that $\exp\left[-\frac{i}{2}\omega_{\mu\nu} \mathcal L^{\mu\nu}\right]\phi(x) = \phi\big(\Lambda^{-1} x\big)$, so clearly $\phi$ does transform in the trivial representation where $D(\Lambda) = 1$. Commented Oct 2, 2021 at 15:21
• Your first and your last equations are compatible because in the first you compare $\phi'$ and $\phi$ at distinct points $x$ and $\Lambda^{-1}x$ while in the last you compare $\phi$ and $\phi'$ at the same point $x$.
– Gold
Commented Oct 2, 2021 at 16:55

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 $$\theta$$ $$\phi(\Lambda^{-1}x) = \exp\left(\theta\left(y\partial_x - x\partial_y\right)\right)\phi(x)$$
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 $$U = \exp(\omega_{\mu\nu}\mathcal{L}^{\mu\nu})$$
• Why do you call $\alpha,\beta$ indices of the $D$ matrix as "spin indices"? For example, if $\phi$ be a 4-vector field, these will be Lorentz indices. Is there something I missed? Commented Oct 10, 2021 at 1:38