# Transformation of fields

In Introduction to QFT by Peskin & Schroeder, p. 18, the authors have described infinitesimal change in field as a result of infinitesimal translation as

$$x^\mu\rightarrow x'^{\mu} =x^\mu-a^\mu \tag{2.16a}$$

Which then leads to (1st order variation in the parameter)

$$\phi(x)\rightarrow \phi(x+a)=\phi(x)+ a^\mu\partial_\mu\phi(x)\tag{2.16b}$$

My question is: Why is there a plus sign in second equation? Shouldn't the equation be

$$\phi(x)\rightarrow \phi(x-a)=\phi(x)- a^\mu\partial_\mu\phi(x)~ ?$$

You should be very very careful with field transformations. In a quantum field theory, only the field transforms -- coordinates DO NOT TRANSFORM, even for spacetime transformations. However, describing spacetime transformations as acting on the coordinates is often a useful tool to describe how the fields themselves transform. But you must always remember that at the end of the day, the action on the coordinates is a tool.

Every single symmetry transformation is described by a field transformation $$\phi(x) \to \phi'(x)$$ To describe the symmetry, we must then explain what is $\phi'(x)$ in terms of $\phi(x)$. For instance, for a $U(1)$ gauge transformation, we write $$\phi'(x) = e^{i Q \alpha(x)} \phi(x)~.$$

When the field transformation corresponds to a spacetime transformation, it is convenient to describe $\phi'(x')$ as opposed to $\phi'(x)$. For instance, for scalar operators, we write $$\phi'(x') = \phi(x)~.$$ Note that this is just a tool for us to determine what is $\phi'(x)$. For instance, for translations, we have $x'=x-a$ so that $$\phi'(x') = \phi'(x-a) = \phi(x) \quad \implies \quad \phi'(x) = \phi(x+a)~.$$ For Lorentz transformations, $x' = \Lambda x$ so that $$\phi'(x') = \phi'(\Lambda x) = \phi(x) \quad \implies \quad \phi'(x) = \phi(\Lambda^{-1} x)~.$$

The author is using here an active transformation. Imagine a laboratory in which you transform the field by displacing, you'd get then a new field of the form $\phi_{\rm new}(x)$. There are two things here to notice

1. The field is still being evaluated at the old coordinates

2. If you transform the coordinates by a translation, then $\phi_{\rm new}(x_{\rm new}) = \phi(x)$, where

$$x_{\rm new} = x-a$$

Or in the notation of the author

$$\phi'(x - a) = \phi(x)$$

which is equivalent to

$$\phi'(x) = \phi(x + a) \stackrel{\mathcal{O(a)}}{=} \phi(x) + a^\mu\partial_\mu\phi(x)$$

• Does $x_{new}$ in $\phi_{new}(x_{new})$ represent new co-ordinate system or just new co-ordinate in same frame ? – solphy101 Sep 25 '17 at 12:13
• @solphy101 Yes, it is the new coordinate system – caverac Sep 25 '17 at 12:14
• But Aren't we doing translation without changing co-ordinate system as you said in your first point? – solphy101 Sep 25 '17 at 12:17
• @solphy101 The equation $\phi_{\rm new}(x_{\rm new}) = \phi(x)$ is just saying that if you were to translate both the field and the coordinates, you'd get the same old field evaluated as the old coordinates – caverac Sep 25 '17 at 12:27