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It is often claimed in quantum field theory texts that to have a sensible Lorentz invariant theory, the fields introduced must be in representations of the Lorentz group. This fact has always seemed obvious to me until recently when I pondered on it and realised I couldn't convince myself of it.

For example, I could define a field $\psi$, which is a scalar under rotations and gains a factor of $e^{\vec{v} \cdot \vec{s}}$ under boosts, where $\vec{s}$ is the vector that points from my house to my local grocery store (potentially time dependent), as well a field $\phi$ which transforms the same way but with a minus sign in the exponent. Then

$$\int d^4x \psi \phi$$

is a Lorentz invariant action, though it is clearly absurd. My question is, why am I allowed to have scalars, and Dirac spinors, but not grocerystoreions?

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    $\begingroup$ If some group transformation leaves the (linear) EOM invariant, it sends solutions to other solutions, so your solutions must form a representation. For your example, you should also be careful because the composition of two boosts in different directions can be expressed as a boost and rotation, so what you wrote may lead to inconsistencies and not actually be a representation. I think any actual representation you can construct via subgroup and quotient group shenanigans would already be accounted for in the rep theory. $\endgroup$ Commented Dec 29, 2023 at 15:37

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The physical symmetries of a theory are defined by the action. (Here I'm thinking of classical field theory and ignoring the possibility of quantum anomalies). The logic is that the dynamics of a system are determined by the action; a symmetry is a transformation of the degrees of freedom that leaves the system invariant; therefore, a symmetry is a transformation of the fields that leaves the action invariant [as pointed out in the comments, actually the action need not actually be strictly invariant to give the same dynamics; for example if a transformation simply rescales the action by an overall constant, then you will get the same equations of motion and therefore the same dynamics, at least classically. I will use the phrase "leave the action invariant" somewhat loosely to mean "gives you an identical physical system"]. Any transformation of the fields you come up with that leaves the action invariant, is a symmetry of the theory.

In your case, the action has a Lorentz symmetry under which $\phi$ and $\psi$ transform as scalars in the usual way. In general, if you're able to construct a Lorentz invariant action, then you'd expect the individual pieces of the action must transform in some representation, in order to combine to form a scalar. Having said that, the representation can be "non-obvious" (more precisely, nonlinearly realized), if the symmetry is spontaneously broken. In the 1860s wasn't obvious that it would be helpful to formulate Maxwell's equations in terms of fields that transform nicely under Lorentz transformations, given that the world we live in day-to-day apparently has a preferred frame where the Sun is stationary.

Now, you're right that often the above story is usually presented "backwards," where we start with field transformations, and then build a Lagrangian from there. That is because we are usually not in the position of knowing the fundamental Lagrangian and deriving the symmetries (the logical order of operations), but instead we have some idea what the degrees of freedom and symmetries are and we are guessing at what the underlying Lagrangian is (illogical but that's how science works!).

Now, how can we interpret your "symmetry"? I've said the action determines what the symmetries are, and your transformation is apparently a symmetry of the action. In fact, your action does have a kind of scaling symmetry, where the fields transform as $\phi \rightarrow \lambda \phi$ and $\psi \rightarrow \lambda^{-1} \psi$ while the coordinates stay fixed, and the transformation you've defined is essentially a subgroup of the full symmetry group where you simultaneously do a Lorentz boost and a rescaling, with a related parameter. However, this scaling symmetry will not be present if you add normal kinetic terms for $\phi$ and $\psi$, like $(\partial \phi)^2$ and $(\partial \psi)^2$. (Unless you generalize the symmetry to allow the coordinates to rescale, like in the conformal group).

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    $\begingroup$ Thanks, so to be clear, what I defined is a symmetry of the action, but we don't call it Lorentz symmetry since that name is reserved for the symmetry in which they transform as Lorentz scalars? After some thinking I also realised that the transformation I stated doesn't have the correct composition rules for the Lorentz group either, so I suppose that's an additional reason why you can't call it a Lorentz transformation. $\endgroup$ Commented Dec 29, 2023 at 4:19
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    $\begingroup$ @LeucaPatmore Basically, yes. We would call a symmetry of the action a Lorentz symmetry if the fields transformed in a representation of the Lorentz group. The "symmetry group" exists in some abstract Platonic sense, and then the question is, when you write down an action, are there symmetries of the action that form representations of that abstract group. $\endgroup$
    – Andrew
    Commented Dec 29, 2023 at 15:24
  • $\begingroup$ Would be good to clarify what exactly you mean by "leaving action invariant" to account for things like an overall factor or shift by a constant. $\endgroup$
    – mavzolej
    Commented Dec 30, 2023 at 14:43
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    $\begingroup$ @mavzolej Thanks, I added a comment. $\endgroup$
    – Andrew
    Commented Dec 30, 2023 at 14:55
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  1. Briefly speaking, symmetries in physics are typically induced by a group action, which is equivalent to a (possibly non-linearly realized) representation of the group, cf. e.g. this Phys.SE post.

  2. The group representation is often irreducible, cf. e.g. this Phys.SE post.

  3. Sometimes the group action depends on fiducial/reference/gauge objects. It might even be impossible to not use fiducial/reference/gauge objects. It is often assumed that the theory should behave covariant/invariant under change of such fiducial/reference/gauge choices.

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In physics fields live on manifolds (spacetime). Technically they are types of fiber bundles on manifolds. roughly the assigning of one of a certain type of object at each point on the manifold. Symmetries are groups that act on the manifold and its associated bundles. How the group acts on the bundles is mathematically related to how the group acts on the manifolds. In the case of vector fields (tangent bundles), this is not too hard to see. Tangent vectors are defined as directional derivatives like

$$ \partial/\partial x_i $$

So we can see that if a transformation on the manifold changes the $x_i$ coordinate in a certain way then it will also change the $\partial/\partial x_i$ vector in a related way.

This idea that the transformations of the fields must be mathematically related to the transformations of the manifolds generalizes to more complex fields/bundles like spinor bundles. Here the answer is a little less satisfactory however because the spinor bundle is essentially defined so that when you transform the manifold the spinor transforms as a representation of the spin group, so the result is essentially baked into the definition.

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