In Peskin & Schroeder they state that:

...if $\Phi_a$ is an $n$ component multiplet, the Lorentz transformation law is given by an $n \times n$ matrix $M(\Lambda)$: $$\Phi_a(x) \rightarrow M_{ab}(\Lambda)\Phi_b(\Lambda^{-1}x) \tag{3.8}$$

I am trying to understand what is physically going on in this equation.

First, to my understanding, $\Phi_a$ here represents a system with $n$ internal degrees of freedom. For example, if we are considering a particle with non-zero spin and no other properties then we will have $n = 3$, is this correct?

Second, is the overall idea that when we take the Lorentz transform of a point $x$ then we also have to keep track of how the vector components of $\Phi_b$ are also transformed? This is the part that is confusing me.


1 Answer 1


The object $\Phi(x)$ here is field that takes values in some representation space of the Lorentz group. In particular, you should think of $\Phi(x)$ as a column vector in some $\mathbb{C}^N$ or $\mathbb{R}^N$ such that a Lorentz transformation $\Lambda$ acts on that space by means of an $N\times N$ matrix $M(\Lambda)$. The field $\Phi(x)$ then encodes the creation and annihilation operators of relativistic particle states classified by Wigner's analysis of the irreducible unitary representations of the Lorentz group.

What is going on is that you want to construct interactions that satisfy cluster decomposition and that are Lorentz invariant. The easiest way to accomplish this is to have them defined by an interaction density ${\cal H}(x)$ that transforms as a Lorentz scalar and is built from the creation and annihilation operators. The issue is that the creation and annihilation operators have a complicated transformation law with respect to Lorentz transformations. As a result, we introduce these objects $\Phi(x)$ that effectively repackage the creation and annihilation operators and demand that by definition they should transform simply under Lorentz transformation.

By transforming simply I mean exactly $\Phi(x)\to M(\Lambda) \Phi(\Lambda^{-1}x)$ as you pointed out. The reason is that it is then very simple to combine these objects into some ${\cal H}(x)$ that transforms as a scalar.

Since the transformation of $a(p,\sigma)$ and $a^\dagger(p,\sigma)$ is fixed by the representation theory of the Poincaré group, demanding $\Phi(x)$ transforms in a certain way effectively introduces a constraint connecting unitary representations of the Poincaré group on particle states and representations of the Lorentz group on fields. At this state one introduces fields that are linear in $a(p,\sigma)$ and $a^\dagger(p,\sigma)$ and these constraints effectively allow for the determination of the coefficients in the expansion.

In particular, imagine you want to describe a massive particle of spin $s$. Now recall that rotations are a subset of Lorentz transformations. This means that if the field transforms in some representation of the Lorentz group, restricting the representation to the rotation subgroup you get one inherited representation of rotations. This representation of rotations as usual will decompose into irreducibles, which are the various spin $j$ representations encountered when studying angular momentum in QM. What one finds out is that in order for the field to be able to encode a massive spin $s$ particle it is necessary that the spin $s$ representation appear among the irreducible factors in that decomposition.

For example, a vector field lives in the so-called $(\frac{1}{2},\frac{1}{2})$ representation of the Lorentz group. The rotation representation associated to that is reducible and decomposes as $0\oplus 1$, i.e., it encodes either spin zero or spin one. So a single representation of the Lorentz group (in this case the vector) may be capable of encoding particles of more than one spin (in this case spin zero and one).

Going back to the interactions, of course one does not then build ${\cal H}(x)$ by hand. Rather, one introduces the framework of canonical quantization, where $\Phi(x)$ is promoted to the basic variable which has a dynamic dictated by a Lagrangian form $\mathbf{L}$. Then the interaction density ${\cal H}(x)$ follows from the Lagrangian. The interacting field is supposed then to have the same transformation behavior as the free field, but it no longer has a linear expansion in creation and annihilation operators everywhere. Instead, the interacting field $\Phi(x)$ is supposed to match the in/out fields encoding the creation and annihilation operators by means of the scattering assumption $$\Phi(x)\to \sqrt{Z}\Phi_{\rm in/out}(x)\quad \text{as $t\to \pm \infty$}.$$

Therefore the Physics described by the dynamics encoded in some Lagrangian form of these objects $\Phi(x)$ can be thought of as a convenient encoding of the scattering of relativistic particles. Of course, there might be more to it than just scattering.

This viewpoint is the one taken by Weinberg in The Quantum Theory of Fields. Chapter 5 is the one that introduces fields transforming in Lorentz representations and how they encode relativistic particles.

  • $\begingroup$ It’s been a long time since I studied QFT. The Lagrangian density is a Lorentz scalar. How can the Hamiltonian density also be one? For a massless scalar field, doesn’t it look like $\frac12[(\partial_t\phi)^2+(\nabla\phi)^2]$? That’s not Lorentz-invariant. $\endgroup$
    – Ghoster
    Commented Feb 17, 2023 at 19:58
  • $\begingroup$ Maybe the issue is the terminology that I used, I'm referring to the Hamiltonian density for the interaction part, not the full Hamiltonian. I'll rather call it interaction density to avoid this issue. $\endgroup$
    – Gold
    Commented Feb 17, 2023 at 20:05
  • $\begingroup$ Thank you for your answer. I will have to revisit it once I have learned some more physics to fully understand it, but it seems the basic idea is we require the field to take values in a representation of the Lorentz group. This requirement automatically enforces Lorentz invariance. $\endgroup$
    – CBBAM
    Commented Feb 17, 2023 at 20:23
  • $\begingroup$ You're welcome. Yes, we demand that fields take values in a representation of the Lorentz group. One may of course present this in different ways. Most QFT textbooks like P&S will start from fields, whereas Weinberg starts from particles. But in either case, that field are valued in representation spaces of the Lorentz group is part of the definition. $\endgroup$
    – Gold
    Commented Feb 17, 2023 at 20:25

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