# Embedding particles (with spin) into fields (Schwartz qft)

I am reading Schwartz's QFT and am a little lost in ch 8.2 in which we are trying to "embed particles into fields".

I think I am missing some very (obvious?) high level intuition for what is going on here.

What I understand so far is that we want to find a set of basis states for a given particle that form a representation of the Poincare group. In particular this basis should be irreducible (this means that there is no subset of the states that are also representations) and unitary (this means that the matrix elements $$\mathcal{M}$$ are invariant under transformations.)

There is a brief mention of a proof by Wigner that the only unitary irreducible representations of the Poincaré group are infinite. They can be classified by $$m$$, which is continuous ($$m^2=p^2$$) and $$J$$ (the spin we know from quantum mechanics). Further, given an allowed value for $$J$$ there are $$2J+1$$ states for a given mass.

Everything feels good up to this point until Schwartz states that, "For spin 0, the embedding is easy, we just put the one degree of freedom into a $$J=0$$ scalar field." How is this "easy", and where did the one degree of freedom go? Is this just a statement that the field is a scalar and not a tensor? Could we equally well embed $$J=0$$ in a degenerate tensor $$a_\mu =(a,a,a,a)$$, for example?

• Schwartz is going for a sort of trial and error approach, where you just try to match mathematical structures as best you can and see what happens. But IMO the intuition is very much NOT obvious, as evidenced by the fact that Weinberg's incomprehensible bible is the only book that really explains it. Sep 29, 2020 at 15:28
• You’re not alone, I spent days trying to understand that section of Schwartz when I first saw it. Coming back to it later didn’t help either. The only thing that cleared it up was reading Weinberg. Oct 14, 2020 at 1:13

I think it's to do with how these states transform. So consider transforming a scalar you will always stay as a scalar. But when you transform your 'a' vector the components will transform such that we have multiple degrees of freedom.

Yes, you can embed a scalar field in different ways. For example, you could start with a "metric"

$$$$g_{\mu\nu} = e^{\phi/M} \eta_{\mu\nu}$$$$ and write down an action for $$g_{\mu\nu}$$. This two-component tensor will embed the scalar degree of freedom, since $$\phi$$ transforms as a scalar under Lorentz transformations. The apparent extra degrees of freedom due to the fact that $$g_{\mu\nu}$$ has more than one component, will be cancelled by gauge symmetries (redundancies) since the action will only include dynamics for $$\phi$$.

However you would be making your life extraordinarily and unnecessarily complicated in order to achieve a simple result. A simple scalar field, not embedded in some kind of higher dimensional tensor, is sufficient to describe spin-0 particles.

Ultimately the "embedding" follows looking at creation and annhilation operators. In order to describe a spin-0 particle, you want a creation operator to look like this $$$$a_p^\dagger | 0 \rangle = N | p \rangle$$$$ where $$N$$ is a normalization constant. Notice there are no spin indices or spin degrees of freedom on $$a$$. A scalar field can "embed" these creation and annihilation operators via the Fourier transform $$$$\phi(x,t) = \int \frac{d^3 p}{(2\pi)^3 \sqrt{2 \omega_p}} \left( a_p e^{i (\omega t - p \cdot x)} + a_p^\dagger e^{-i (\omega t - p \cdot x)} \right)$$$$ and since the creation and annihilation operators don't have any internal spin indices, there is no need for $$\phi$$ to have any indices (aka it is fine for it to be a scalar field).

Schwartz answers your question in the section on Massive Spin 1 embeddings (Section 8.2.2) with the line "if $$A_\mu$$ transformed as four scalars, $$\partial_\mu A_\mu$$ would not be Lorentz invariant". To make a Lorentz invariant theory with a degenerate tensor you would need to form Lorentz invariants in your Lagrangian. Namely things like $$a^\mu a_\mu = \begin{array}{c} (a & a & a & a) \end{array} \left( \begin{array}{c} a \\ a \\ a \\ a \end{array} \right) = -3 a^2$$

So you end up with a scalar field in your Lagrangian anyway. For the degree of freedom part of the question, I refer you to Counting Degrees of Freedom in Field Theories as the answer there is much more detailed than I could care to write down :)