# Field transformations

I'm reading Maggiore's book "A modern introduction to quantum field theory" and I'm very confused by what he did in chapter 2.6 page 31 eq. (2.80). He basically wants to find the generators of the infinite representation of the Lorentz group (comparing how the Weyl spinor field change at equal coordinate). He writes:

$$\psi_L \rightarrow \psi'_L(x') = \Lambda_L \psi_L(x)$$

which is the finite dimensional representation. The coordinates transform as $x^{\mu} \rightarrow x'^{\mu} = x^{\mu}+\delta x^{\mu}$. Then in (2.80) he compares the field at equal coordinate:

\begin{align} \delta_0 \psi_L &= \psi'_L(x) - \psi_L(x) \\ &= \psi'_L(x'^{\mu}-\delta x^{\mu}) - \psi_L(x) \\ &= \psi'_L(x') - \delta x^{\mu}\partial_{\mu}\psi_L(x) - \psi_L(x) \\ &= (\Lambda_L - 1)\psi_L(x) - \delta x^{\mu}\partial_{\mu}\psi_L(x) \end{align}

I don't understand why after the third equality the "primes" magically disappear from $\delta x^{\mu}\partial_{\mu}\psi_L(x)$. I would have written $\delta x^{\mu}\partial_{\mu}\psi'_L(x')$! Any idea?

• They're the same thing to linear order in $\delta x$, so it doesn't matter which you pick. – Holographer Jan 25 '15 at 19:42
• I was thinking the same thing, but how can you show it? Since $\Lambda_L$ is independent of the coordinates if I plug the first equation in, I get a $\Lambda_L$ in front of the derivative. – Worldsheep Jan 25 '15 at 20:24

He has just used $$\psi_L \rightarrow \psi'_L(x') = \Lambda_L \psi_L(x)$$ and factored $\psi_L(x)$ from the first and third term in the third line of the expression.
• I mean the term with the derivative, not the first one! If I use the relation you are suggesting I get $\Lambda_L$ in front of the derivative. – Worldsheep Jan 25 '15 at 20:20
If you look at equation (2.79) in the book, you could expand $\Lambda_{L}$ in orders of the parameters $\omega_{\mu \nu}$. If you consider infinitesimal transformations, so we have infinitesimal parameters $\omega_{\mu\nu}$, you get $$\psi'_{L}(x') = (\mathbf{1} - \omega_{\mu\nu}S^{\mu\nu})\psi_{L}(x).$$ Also $\delta x$ is proportional to $\omega_{\mu \nu}$ and by taking use of the previous result one gets
$$\delta x^{\rho} \partial_{\rho} \psi'_{L}(x') =\delta x^{\rho} \partial_{\rho} \psi_{L}(x).$$