# Weinberg's argument for antiparticles: is this really a necessary condition?

In Chapter 5 to "The Quantum Theory of Fields" by Weinberg he gives a nice point of view on antiparticles. He says the following (I've called the last equation (*) because it is not numbered in Weinberg's book):

It may be that the particles that are destroyed and created by these fields carry non-zero values of one or more conserved quantum numbers like the electric charge. For instance, if particles of species $$n$$ carry a value $$q(n)$$ for the electric charge $$Q$$, then $$[Q,a(\mathbf{p},\sigma,n)]=-q(n)a(\mathbf{p},\sigma,n), \\ [Q,a^\dagger(\mathbf{p},\sigma,n)]=q(n)a^\dagger(\mathbf{p},\sigma,n).$$ In order that $$\mathscr{H}(x)$$ should commute with the charge operator $$Q$$ (or some other symmetry generator) it is necessary that it be formed out of fields that have simple commutation relations with $$Q$$: $$[Q,\psi_\ell(x)]=-q_\ell\psi_\ell(x)\tag{5.1.33}$$ for then we can make $$\mathscr{H}(x)$$ commute with $$Q$$ by constructing it as a sum of products of fields $$\psi_{\ell_1}\psi_{\ell_2}\cdots$$ and adjoints $$\psi_{m_1}^\dagger \psi_{m_2}^\dagger\cdots$$ such that $$q_{\ell_1}+q_{\ell_2}+\cdots - q_{m_1} - q_{m_2}-\cdots =0\tag{*}.$$

That this is sufficient I understand. Construct $$\mathscr{H}(x)$$ as $$\mathscr{H}(x)=\sum_{NM}\sum_{\ell_1\cdots \ell_N}\sum_{\bar{\ell}_1\cdots \bar{\ell}_M}g_{\ell_1\cdots \ell_N \bar{\ell}_1\cdots \bar{\ell_M}}\psi_{\ell_1}(x)\cdots \psi_{\ell_N}(x)\psi^\dagger_{\bar{\ell}_1}(x)\cdots \psi^\dagger_{\bar{\ell}_M}(x).$$

Then we can show that (omitting the argument $$x$$ to simply the notation): $$[Q,\psi_{\ell_1}\cdots \psi_{\ell_N}\psi^\dagger_{\bar{\ell}_1}\cdots \psi^\dagger_{\bar{\ell}_M}]=\left(\sum_{i=1}^N\psi_{\ell_1}\cdots \psi_{\ell_{i-1}}[Q,\psi_{\ell_i}]\psi_{\ell_{i+1}}\cdots \psi_{\ell_N}\right)\psi^\dagger_{\bar{\ell}_1}\cdots \psi^\dagger_{\bar{\ell_M}}\\ \quad +\psi_{\ell_1}\cdots \psi_{\ell_N}\left(\sum_{i=1}^M\psi^\dagger_{\bar{\ell}_1}\cdots\psi^\dagger_{\bar{\ell}_{i-1}}[Q,\psi^\dagger_{\bar{\ell}_i}]\psi^\dagger_{\bar{\ell}_{i+1}}\cdots \psi^\dagger_{\bar{\ell_M}}\right)$$

Therefore it is clear that if (5.1.33) holds and (*) holds, $$[Q,\mathscr{H}(x)]=0$$. This shows that (5.1.33) together with (*) is sufficient to ensure charge conservation.

But Weinberg speaks as if it were necessary. He says it himself that for $$Q$$ to commute with $$\mathscr{H}(x)$$ it is necessary that $$\mathscr{H}(x)$$ be formed out of fields for which (5.1.33) holds.

Why is this true? I can't see how $$[Q,\mathscr{H}(x)]=0$$ implies that the fields appearing in the construction of $$\mathscr{H}(x)$$ should satisfy (5.1.33) and (*).

Since Weinberg is, after all, a physics book rather than a book of rigorous math, I'm not convinced that one should attempt to understand "necessary" here in its rigorous logical meaning rather than a colloquial meaning. In any case, whether the Hamiltonian is "necessarily" formed from such fields is an ill-defined question to begin with:

Suppose we start with a Hamiltonian of the form suggested by Weinberg, with $$N$$ fields $$\psi_i$$ and $$M$$ $$\psi^\dagger_j$$ with simple commutation relations. If we now "rotate in field space", replacing the fields $$\psi_i$$ by fields $$\psi'_i$$ , where the latter are the result of rotating the $$N$$-vector of $$\psi_i$$ by some angle, you can plug these into the Hamiltonian (probably making it very ugly) to obtain the Hamiltonian of a theory that is completely equivalent to our "theory of simple fields" from the beginning but has no such simple commutation relations.

Conversely, if we start from fields with non-simple relations but $$[H,Q] = 0$$, then elementary representation theory for $$\mathrm{U}(1)$$ (the symmetry group of the charge numbers we are considering here) suggests that the vector space of fields must decompose into one-dimensional representations. Simply switch basis so that we use the basis vectors of these irreps as your fields and we have arrived at a formulation of the theory with simple commutation relations.

So if one reads Weinberg as "it is necessary that for a Hamiltonian that is the sums of products of fields that have conserved charges, a following choice of fields exists", then "necessary" is correct. If one reads him as "the only way to write down such a theory is with fields with simple commutation relations", then it's wrong.

This is completely analogous to changing coordinates on configuration space in non-field-theories. Consider a theory where $$x$$-momentum is conserved but $$y$$- and $$z$$-momenta are not. Then certainly no one would suggest that using $$(x,y,z)$$-coordinates rather than e.g. spherical coordinates is necessary to have a Hamiltonian where $$x$$-momentum is conserved - it's just a nicer choice of coordinates.

• Thanks for the answer ! Indeed I was thinking that Weinberg's claim was indeed about the existence of such a choice of fields not about that being valid for any choice. So let me see if I get your point. The idea is that $Q$ furnishes a hermitian representation of the abelian Lie algebra $\mathfrak{u}(1)$. Then we build a representation on the space of fields with $Q$ acting as $[Q,\psi_\ell]$. As this is a representation of an abelian Lie algebra it decomposes as a direct sum of one-dimensional representations for which (5.1.33) holds. Is that your point?
– Gold
Jan 12, 2020 at 22:28
• @user1620696 Yes, exactly. Jan 12, 2020 at 22:39