# Coordinate Transformation of Scalar Fields in QFT

By definition scalar fields are independent of coordinate system, thus I would expect a scalar field $\psi [x]$ would not change under the transformation $x^\mu \to x^\mu + \epsilon^\mu$. Correct?

Now when I look in the book "Introduction to QFT" by Peskin and Schroeder they state (in an example) that the scalar field $\psi[x]$ under an infinitesimal coordinate transformation $x^\mu \to x^\mu -a^\mu$ transforms as $\psi[x] \to \psi[x+a] = \psi[x] +a^\mu\partial_\mu\psi[x]$.

One possible solution I thought was that a scalar field $f:M\to\mathcal{R}$ from a variety (in this case spacetime) is invariant under coordinate transformations, but that the coordinate representation $\psi[x]$ does change under coordinate transformations (obviously).

Is this anywhere near correct?

• Not sure I agree. Scalars should be invariant under rotations, but why translations? e.g. temperature can be thought of as a scalar field. Why would you expect temperature to be invariant under translations? – innisfree Sep 15 '13 at 13:25
• Scalar fields translations under translation $x\rightarrow x^\prime = x-a$ as: $\phi(x)\rightarrow \phi^\prime(x^\prime) = \phi(x) = \phi(x^\prime+a)=\phi(x^\prime) + a^\mu\partial_\mu\phi(x^\prime)$. – Will Sep 15 '13 at 14:59
• I was under the impression that these transformations leaves the action invariant...or, more lazily, the Lagrangian. – Alex Nelson Sep 15 '13 at 15:03
• Scalar fields transform* – Will Sep 15 '13 at 15:12

Here's what's really going on. In classical field theory, a basic set of objects that we often consider are scalar fields $\phi:M\to \mathbb R$ where $M$ is a manifold. Now we can ask ourselves the following question:

Is there some natural notion of how a scalar field defined on a given manifold "transforms" under a coordinate transformation?

I claim that the answer is yes, and I'll attempt to justify my claim both mathematically, and physically. The bottom line is that we ultimately have to define the way in which fields transform under certain types of transformations, but any old definition will not necessarily be useful in math or physics, so we must make well-motivated definitions and then show that they are useful for modeling physical systems.

Mathematical perspective. (manifolds and coordinate charts)

Recall that a coordinate system (aka coordinate chart) on an $n$-dimensional manfiold $M$ is a (sufficiently smooth) mapping $\psi:U\to \mathbb R^n$ where $U$ is some open subset of $M$. We can use such a coordinate system to define a coordinate representation $\phi_\psi$ of the scalar field $\phi$ as \begin{align} \phi_\psi = \phi\circ\psi^{-1}:V\to\mathbb R \end{align} where $V$ is the image of $U$ under $\psi$. Now let two coordinate systems $\psi:U_1\to \mathbb R^n$ and $\psi_2:U_2\to\mathbb R^n$ be given such that $U_1\cap U_2\neq \emptyset$. The coordinate representation of $\phi$ in these two coordinate systems is $\phi_1 = \phi\circ \psi_1^{-1}$ and $\phi_2 = \phi\circ \psi_2^{-1}$.

Now consider a point $x\in U_1\cap U_2$, then $x$ is mapped to some point $x_1\in \mathbb R^n$ under $\psi_1$ and to some point $x_2\in \mathbb R^n$ under $\psi_2$. We can therefore write \begin{align} \phi(x) &= \phi \circ \psi_1^{-1} \circ \psi_1(x) = \phi_1(x_1) \\ \phi(x) &= \phi \circ \psi_2^{-1} \circ \psi_2(x) = \phi_2(x_2) \end{align} so that \begin{align} \phi_1(x_1) = \phi_2(x_2) \end{align} In other words, the value of the coordinate representation $\phi_1$ evaluated at the coordinate representation $x_1 = \psi_1(x)$ of the point $x$ agrees with the value of the coordinate representation $\phi_2$ evaluated at the coordinate representation $x_2 = \psi_2(x)$ of the same point $x$. This is one way of understanding what it means for a scalar field to be "invariant" under a change of coordinates.

If, in particular, the manifold $M$ we are considering is $\mathbb R^{3,1} = (\mathbb R^4, \eta)$, namely four-dimensional Minkowski space, then we could consider the following two coordinate systems: \begin{align} \psi_1(x) &= x \\ \psi_2(x) &= \Lambda x+a \end{align} where $\Lambda$ is a Lorentz transformation and $a\in \mathbb R^4$, then the coordinate representations $\phi_1$ and $\phi_2$ of $\phi$ are, as noted above, related by \begin{align} \phi_1(x) = \phi_2(\Lambda x + a) \end{align} If we switch notation a bit and write $\phi_1 = \phi$ and $\phi_2 = \phi'$, then this reads \begin{align} \phi'(\Lambda x +a) = \phi(x) \end{align} which is the standard expression you'll see in field theory texts.

Physical perspective.

Here's a lower-dimensional analogy. Imagine a temperature field $T:\mathbb R^2 \to \mathbb R$ on the plane that assigns a real number that we interpret as the temperature at each point on some two-dimensional surface. Suppose that this temperature field is generated by some apparatus under the surface, and suppose that we translate the apparatus by a vector $\vec a$. We could now ask ourselves:

What will the temperature field produced by the translated apparatus look like? Well, each point in the temperature distribution will be translated by the amount $\vec a$. So, for example, if the point $\vec x_0$ has temperature $T(\vec x_0) = 113^\circ\,\mathrm K$, then after the apparatus is translated, the point $\vec x_0 + \vec a$ will have the same temperature $113^\circ\,\mathrm K$ as the point $\vec x_0$ before the apparatus was translated. The mathematical way of writing this is that if $T'$ denotes the translated temperature field, then $T'$ is related to $T$ by \begin{align} T'(\vec x+\vec a) = T(\vec x) \end{align} The a similar argument could be made for a scalar field on Minkowski space, but instead of simply translating some temperature apparatus, we could imagine boosting or translating something producing some Lorentz scalar field, and we would be motivated to define the transformation law of a scalar field under Poincare transformation as \begin{align} \phi'(\Lambda x+a) = \phi(x) \end{align}

• Very clear, yet rigorous explanation. Hits the OP's uncertainty dead-on, I think. – Casimir Oct 17 '15 at 10:15
• @Casimir Glad to hear it. – joshphysics Oct 17 '15 at 17:15
• Just a suggestion: You could expand your answer to include a note on the Taylor expansion mentioned by the OP ($\Psi(x) \to \Psi(x + a) = \Psi(x) + a^\mu \partial_\mu \Psi(x) + \mathcal{O}(a^2)$). – Casimir Oct 17 '15 at 17:35
• Is the notion of "being a scalar" w.r.t. a group transformation related to "being invariant" under a group transformation? Usually when I say that a function is invariant under a transformation of coordinates I mean that: $F(x')=F(x)$ with obvious notation. Does this imply that $F'(x)=F(x)$ if additionally the function is a scalar? – DR10 Dec 9 '15 at 16:13
• @DR10 Yes, the notion of being a scalar w.r.t a group action is precisely that it's invariant under that group action. A function being a scalar means, in your notation, that $F'= F$ provided $F'$ is the transformed function. Now, if one takes the action of the group on functions to be (by definition) $F'(x) = F(T_g^{-1}(x))$ ($T$ is the action on coordinates), then the property that the function is a scalar together with this transformation law gives $F(T_g^{-1}(x)) = F(x)$, which in your notation is $F(x') = F(x)$. Related: physics.stackexchange.com/a/155887/19976 – joshphysics Dec 9 '15 at 19:24

You might be confusing what we mean when we say that scalar fields are invariant. Under a Lorentz transformation ($x\rightarrow x^\prime = \Lambda x$) the scalar field ($\phi(x)$) is defined to transform as $$\phi(x)\rightarrow\phi^\prime(x^\prime) = \phi(x) = \phi(\Lambda^{-1} x^\prime)$$ So we see that in the new coordinates $x^\prime$, our scalar field has transformed to $\phi^\prime(x^\prime) = \phi(\Lambda^{-1}x^\prime)$. Similarly, under a translation $x\rightarrow x^\prime = x - a$ we have $$\phi(x)\rightarrow \phi^\prime(x^\prime) = \phi(x) = \phi(x^\prime+a)$$ or if $a$ is taken to be infinitesimal, we find to first order in $a$ $$\phi(x)\rightarrow \phi^\prime(x^\prime) = \phi(x^\prime+a) = \phi(x^\prime) + a^\mu\partial_\mu\phi(x^\prime)$$ In both of these examples, the field hasn't transformed (the transformation was trivial), only the way we represent the point. We want to write the point $x$ in terms of our new coordinate system ($x^\prime$) in each case, which is where this expansion comes from.

• I don't think this addresses the conceptual misunderstanding of the OP. – innisfree Sep 15 '13 at 17:56