- #31
samalkhaiat
Science Advisor
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In QED, interactions are written in terms of A_{\mu}, i.e., \mathcal{H}_{int} = A_{\mu}J^{\mu}. So, your statement is empty unless you show us one example in which the interactions are written in terms of the tensor F_{\mu\nu}.A. Neumaier said:
It is good to have good memory! I recall that you once claimed that Weinberg does not mention virtual particles in his book, and I showed you that your claim was false. So, let us see if there is really “No vector potential” on the couple of pages before p.252!
I can understand why Weinberg writes a_{\mu} instead of A_{\mu}. However, EVERY equation satisfied by his a_{\mu} is also satisfied by the vector potential A_{\mu}. Is this not enough to say that a_{\mu} is the vector potential?
1)a_{\mu}(x) = \int \frac{d^{3}p}{(2 \pi)^{3/2}\sqrt{2p^{0}}} \sum_{\sigma = \pm 1}\left( e_{\mu}(\vec{p},\sigma) e^{ipx}a(\vec{p},\sigma) + \mbox{H.C.}\right) . \ \ \ \ (5.9.23)
This is the vector potential expanded in terms of two independent polarization vectors, i.e., the vector potential with two of its components have been gauged away (see below).
2)\partial^{2}a_{\mu}(x) = 0 . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (5.9.24)
In certain gauge (see below), the vector potential can be made to satisfy this equation.
3)a^{0}(x) = 0 . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (5.9.28)
This is the vector potential in the Temporal gauge.
4)\vec{\nabla} \cdot \vec{a} = 0. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (5.9.29)
This is the vector potential in the Coulomb gauge.
5)U(\Lambda) a_{\mu}(x) U^{-1}(\Lambda ) = \Lambda^{\nu}{}_{\mu} a_{\nu}(\Lambda x) + \partial_{\mu} \Omega ( x , \Lambda ) . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (5.9.31)
This is exactly how the vector potential transforms under the Lorentz group : since F = dA and F is a rank-2 antisymmetric covariant tensor, then A must be a covariant vector up to a gauge transformation. Eq(5.9.31) shows one more ugly feature of the Coulomb gauge quantization. This is the fact (which Weinberg brushes over in his book) that in the Coulomb gauge the gauge function \Omega is not a c-number but an operator. Nothing is more ugly than treating a^{0}(x) and \nabla \cdot \vec{a}(x) as c-numbers and the gauge function as an operator.
6) And finally f_{\mu\nu} = \partial_{\mu}a_{\nu} - \partial_{\nu}a_{\mu} , \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (5.9.34) is nothing but the definition of the field tensor in terms of the vector potential.
So, Weinberg’s a_{\mu} is the vector potential A_{\mu}. To be fair, Weinberg does not claim that a_{\mu} is not the vector potential. Eq(5.9.19) shows that Weinberg arrived at a_{\mu} from the 4-vector representation (1/2 ,1/2) by ignoring the fact that there is no massless 4-vector field of helicity \pm 1. In fact he shows that there exists no such representation [see eq(5.9.22)], but carry on with the construction by saying on page 250, line 7 “Let’s temporarily close our eyes to this difficulty, and go ahead anyway, …”.
Now, are you surprised “that Eqs. (5.9.34), (5.9.24), (5.9.28), and (5.9.29) show that f^{\mu\nu} satisfies the vacuum Maxwell equations”? Moreover, didn’t I say (at the end of #25) that, in the Coulomb gauge, the Maxwell equations do hold as operator equations?
The theorem say that “one cannot hope to formulate QED in term of F_{\mu\nu}, j^{\mu} and local charged fields without essentially going to the Gupta-Bleuler formulation. Fields describing charged particles can be defined as local fields only in a Hilbert space equipped with an indefinite metric and only if the Maxwell's equations are abandoned as operator equations.” The result is “obtained without ever introducing the electromagnetic potential A_{\mu},” and therefore “The conclusion of the theorem applies also to formulations in which A_{\mu} is never introduced.”
The framework of the “covariant operator formalism of gauge theories” is that of Strocchi-Wightman: \left( \mathcal{V}, \mathcal{V}_{ph}, U, (\Omega , \varphi_{m}^{(j_{1},j_{2})} , D)\right) consisting of an indefinite metric Hilbert space (Krein space) \mathcal{V}, a physical subspace \mathcal{V}_{ph} = \big\{ \Psi \in \mathcal{V}: \langle \Psi , \Psi \rangle \geq 0 \big\}, a unitary representation (a , \Lambda) \mapsto U(a , \Lambda ) such that U \mathcal{V}_{ph} \subset \mathcal{V}_{ph}, a U-stable vacuum \Omega such that \langle \Omega , \Omega \rangle = 1, a U-stable domain D \subset \mathcal{V}_{ph} and a family of operator-valued distributions satisfying the rest of Wightman axioms.
The physical Hilbert space (i.e., Hilbert space with positive-definite metric) is defined by \mathcal{H}_{ph} = \overline{\mathcal{V}_{ph}/ \mathcal{V}_{0}}, where \mathcal{V}_{0} = \big\{ \psi \in \mathcal{V}_{ph}| \langle \psi , \psi \rangle = 0 \big\}. The elements of \mathcal{H}_{ph} are equivalence classes [ \Psi ] = \big\{ \Psi + \psi | \psi \in \mathcal{V}_{0} , \Psi \in \mathcal{V}_{ph}\big\}.
The space \mathcal{H}_{ph} carries one good news for you. Indeed, the Maxwell equations do hold as operator equations in \mathcal{H}_{ph}. However, this comes with heavy price, because all elements of \mathcal{H}_{ph} have zero electric charge, and [J^{0}(f)] generates trivial automorphism, i.e., \mathcal{H}_{ph} has no room for electrons.
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