# Generators in Field Theory and Derivatives

Let's consider a representation of the multiplicative group $$(0,\infty)$$ on Minkowski space $$\mathbb{R}^4$$ by dilations.

\begin{align} \rho:(0,\infty)&\rightarrow\text{GL}(\mathbb{R}^4)&\\ a &\mapsto \rho(a):\mathbb{R}^4\rightarrow\mathbb{R}^4\\ x&\mapsto \rho(a)x:=ax=e^{\ln(a)}x. \end{align}

I would then say that a generator of this representation is $$D=-i$$. Indeed $$\rho(e^y)x=e^{iyD}x$$ However, field theory books tend to say the generator is $$D=-ix^\mu\partial_\mu$$. This is because for an infinitesimal transformation $$x\mapsto x+\epsilon x\approx e^{i\epsilon D}x$$ (whatever that may mean since D is unbounded and thus one should be careful not to Taylor expand such exponentials and instead use Stone's theorem and the measurable functional calculus). What is going on here?

A similar problem may be seen when reading Tong's Quantum Field Theory notes vs. Ramond's Field Theory: A Modern Primer regarding the generators of the Lorentz group. The first says $$(M_{\mu\nu})_{\rho\sigma}=\eta_{\rho\mu}\eta_{\sigma\nu}-\eta_{\sigma\mu}\eta_{\rho\nu},$$ while the second $$L_{\mu\nu}=i(x_\mu\partial_\nu-x_\nu\partial_\mu).$$ I kept the original notation although maybe my problem relies on the fact that I don't understand the difference between the $$M$$s and the $$L$$s.

• I have suggested the required edit according to how I interpreted your meaning, please make sure if that is how you intended it to look, I believe you had too many alignment characters – Triatticus Jan 3 at 13:07
• A quick hint, I don’t have time to expand my answer right now. The $L_{\mu\nu}$ representation acts on scalar fields $\phi(x)$. You may check it satisfies the Lorentz algebra by calculating $[L_{\mu\nu}, L_{\sigma\rho}]$. The representation depends on the “objects” you are provided. If you have a spinor field, you don’t use $L_{\mu\nu}$. – Luthien Jan 3 at 14:50
• When field theory books say the generator is $D=-i x^\mu \partial_\mu$ they are probably referring to a the transformation being performed on a scalar field, not on a 4-vector – Luthien Jan 3 at 15:04
• @Triatticus Thanks for your help. However, I wanted to have the $x$ on the last line aligned with the $\mathbb{R}^4$ in the second one. Do you know how to do this? – Iván Mauricio Burbano Jan 4 at 0:37

As I said in my comments, the $$M_{\mu\nu}$$ and the $$L_{\mu\nu}$$ of your question are two different representations of the Lorentz group. Roughly speaking, it means that the transformations that they generate are meant to act on different objects.

The $$M_{\mu\nu}$$ is the 4-vector representation. It's a $$4\times4$$ matrix, whose components are indeed given by $$(M_{\mu\nu})^\sigma_{\rho}$$. Usually, this is the starting point of lectures and books on the subject: you start with your 4-vectors, with your Minkowski space, you work with $$4\times4$$ matrices and so on. Then you find the generators and you see they satisfy an algebra, the Lorentz algebra. You can find an expression for $$(M_{\mu\nu})^\sigma_{\space\rho}$$ (that I will call $$(J_{\mu\nu})^\sigma_{\space\rho}$$) in the following way.

For an infinitesimal Lorentz transformation we have that $$\Lambda^\mu_{\space\nu}\approx\delta^\mu_{\space\nu}+\omega^\mu_{\space\nu}$$, where $$\omega^\mu_{\space\nu}$$ contains the parameters of the transformation and $$\omega_{\mu\nu}$$ is antisymmetric. But you also have $$\Lambda=1-i\omega_{\rho\sigma}J^{\rho\sigma}$$, where $$J^{\rho\sigma}$$ can be set antisymmetric, as it is contracted with an antisymmetric tensor. By equating we get that $$-i\omega_{\rho\sigma}(J^{\rho\sigma})^\mu_{\space\nu}/2=\omega^\mu_{\space\nu}$$ and consequently $$$$(J^{\rho\sigma})^\mu_{\space\nu}=i(\eta^{\sigma\mu}\delta^\rho_{\space\nu}-\eta^{\rho\mu}\delta^{\sigma}_{\space\nu})$$$$
This expression is valid for the 4-vector representation.

The representation $$L_{\mu\nu}$$ of your question is the scalar field representation. I am provided with a field, a function of space time: I am acting on an Hilbert space, not on the Minkowski space anymore. Your $$L_{\mu\nu}$$ thus is not a $$4\times4$$ matrix but an operator, as you may indeed see.

If you are interested on the derivation of $$L_{\mu\nu}$$ or further details let me know, I will expand my answer.

To conclude, when you say that "However, field theory books tend to say the generator is $$D=−ix^\mu\partial_\mu$$", it's because you are probably dealing with a scalar field.

Edit:

Keep in mind that for infinitesimal parameters ($$\omega_{\rho\sigma}\to 0$$) we have for a Lorentz transformation $$$$\Lambda=1-i\frac{\omega_{\rho\sigma}}{2}J^{\rho\sigma}$$$$

Now, let's consider a scalar field, i.e. a field such that $$\phi'(x')=\phi(x)$$, where the transformation on the space-time point reads $$x'^\mu=\Lambda^\mu_{\space\nu}x^\nu$$. For an infinitesimal transformation we have that $$x'^\mu=x^\mu+\omega^{\mu}_{\space\nu} x^\nu$$ at first order. Substituting in the field we get $$$$\phi'(x)=\phi(x)-\omega^\rho_{\space\nu}x^\nu\partial_\rho\phi(x)=(1-\omega^{\rho\nu}x_\nu\partial_\rho)\phi(x)$$$$ Now we can use the fact that $$\omega_{\rho\sigma}$$ is antisymmetric (so only the antisymmetric part of $$x_\nu\partial_\rho$$ is kept) and we get $$$$\phi'(x)=[1-\omega^{\rho\nu}(x_\nu\partial_\rho-x_\rho\partial_\nu)/2]\phi(x)$$$$ You may now recognize the generator, $$J_{\mu\nu}$$ of the Lorentz transformation, that we indicate often with $$L_{\mu\nu}$$ for a scalar field, as $$L_{\mu\nu}=x_\nu\partial_\rho-x_\rho\partial_\nu$$.

• In short: $$\left. {\frac{d}{d\theta}}\phi\left(M_{\mu\nu}(\theta)x\right) \right|_{\theta=0} = -i \left( L_{\mu\nu} \phi \right)(x)$$ – md2perpe Jan 3 at 19:30
• @md2perpe yes, I will expand a bit the proof on my answer if the OP feels like needing it :) – Luthien Jan 3 at 19:40
• Thanks for your answer! I would really like to know more about the derivation of the $L_{\mu\nu}$. In particular, I would like to view it from the perspective of classical field theory. On the other hand, how does this relate to the fact that in QFT we want a unitary representation $U$ of the double cover $\Lambda:SL(2,\mathbb{C})\rightarrow L_+^\uparrow$ of the Lorentz group and a finite dimensional representation $D$ (I guess of the Lorentz group) such that $U(\alpha)^\dagger\phi^a(x)U(\alpha)=D^a_b(\Lambda(\alpha))(\phi^b(\Lambda(\alpha)^{-1}x))$? Any further reading suggested? – Iván Mauricio Burbano Jan 4 at 0:45
• @md2perpe Your answer I am sure has to do with my previous comment. What is $M_{\mu\nu}(\theta)$ in the equation? – Iván Mauricio Burbano Jan 4 at 0:49
• @IvánMauricioBurbano I've edited my question to include $L_{\mu\nu}$. Regarding your question in the comment, I think you're reading some part of the book/lectures related to fields that are already quantized (moreover, they are not scalar, and you are using passive transformations of the spacetime points). – Luthien Jan 4 at 1:27