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

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    $\begingroup$ 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 $D(\Lambda) = 1$. $\endgroup$
    – J. Murray
    Commented Oct 2, 2021 at 15:21
  • $\begingroup$ 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 $x$. $\endgroup$
    – Gold
    Commented Oct 2, 2021 at 16:55

1 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 $$ U = \exp(\omega_{\mu\nu}\mathcal{L}^{\mu\nu})$$

  • $\begingroup$ So what is the conclusion? Are scalar fields 1-dimensional representation of Lorentz group or infinite-dimensional? $\endgroup$ Commented Oct 3, 2021 at 9:02
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    $\begingroup$ They are infinite dimensional. In general, tensor fields combine an infinite-dimensional representation with a finite dimensional representation. They all transform under the same infinite-dimensional function representation, then add on the spin representations. A scalar field is the special case that doesn't have any extra spin representation. $\endgroup$ Commented Oct 3, 2021 at 12:50
  • $\begingroup$ 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? $\endgroup$ Commented Oct 10, 2021 at 1:38
  • $\begingroup$ @mithusengupta123 No, you didn't miss anything. I just needed a general term for all the possible indices that a general vector field could have, and "spin indices" seemed the best. The reason we use Lorentz indices for vector fields is that it is a convenient way to encode a finite-dimensional spin-1 representation. To get other spins, you use other representations with other indices. $\endgroup$ Commented Oct 10, 2021 at 12:45

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