Saturday 13 June 2020

quantum mechanics - 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 (*).




Answer



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.


No comments:

Post a Comment

Understanding Stagnation point in pitot fluid

What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...