What is the role of the $F_{\mu\nu}$ tensor in QED?

Question

In Classical Electrodynamics the physically important quantity is the electromagnetic tensor $F\in\Omega^2(M)$ where $M$ is spacetime.

It turns out that since $dF = 0$ by Poincare's lemma (assuming that $M$ is contractible), there should be $A\in \Omega^1(M)$ such that $F = dA$.

This $A$ is then used to simplify things because it is simpler to compute $A$ and $A$ directly yields $F$. Furthermore, $F$ is more fundamental, since $F$ may exist even if $M$ is not contractible while in this case $A$ is not guaranteed to exist since Poincare's lemma doesn't apply. More than that as I've said, the physical thing is $F$, while $A$ is not. One way to see this is that we can add any $d\phi$ to $A$, because $d(A+d\phi)=dA$ and $F$ is not altered and hence the physics is not changed.

Now let's get to QED. In QED it turns out that it seems (at least by the treatment the books I'm reading use) that the important object is $A$. The field associated to the photon is $A$, the field one quantizes is $A$ and $A$ yields the Feynman rules.

The field $F$ appears in the lagrangian, but it is directly written in terms of $A$ and everything is done with $A$. Hardly ever the field $F$ seems to be used in QED.

Why is that? If even from classical EM, we know that the physical thing is $F$ and $A$ is just something to make life easier in computations that doesn't carry physical meaning itself, why in QED it seems the important object is $A$? In that case, what ends up being the role of $F$ in QED?

Answer 1

Even in classical electrodynamics, when formulating the theory as an action principle, you should write the action in terms of $A$. Otherwise, you can't recover Maxwell's equations. With an action $\sim F^2$, if $F$ is the field with respect to which we should take the variation of the action, we can only get zeroth-order in time equations for $F$, which is obviously absurd because of Maxwell's correction to Ampere's law, $\nabla \times \mathbf B = \mathbf j + \partial_t \mathbf E$ -- first order in time.

We also can't get $\nabla\cdot\mathbf B = 0$ or $\partial_t \mathbf B = - \nabla \times \mathbf E$ without adding them as constraints to the Lagrangian, but because these constraints are precisely $dF = 0$, that's the same as declaring $F = dA$ on contractible subsets.

Now, the transformation $A \mapsto A + d\phi$ is called a gauge transformation. $F$ and hence the action $\sim F^2$ is invariant under gauge transformations, so we don't actually need that a global gauge potential exists. One exists on every contractible subset, if the domains of $A_1$ and $A_2$ overlap, then on the overlap $A_1$ and $A_2$ are gauge transformations of each other, and it doesn't matter which we use.

We can "lift" this to a better formulation of electrodynamics by using fiber bundles. In this view, there is only one gauge potential, but it lives on a "princial bundle" over spacetime. The Lagrangian is then a function from this bundle. By choosing a "section" of the bundle, we get the potential in a specific gauge as an object on spacetime, but such sections are not guaranteed to exist globally, for topological reasons.

The precise way to do this requires a book-length treatment, more or less. I can recommend

• Baez and Muniain, Gauge Fields, Knots, and Gravity.
• Naber, Topology, Geometry and Gauge fields, 2 vols.

with a background in differential geometry from, e.g.,

• Lee, Manifolds and Differential Geometry

You may also find some useful material in

• Penrose, The Road to Reality.
• Nakahara, Geometry, topology, and physics (I didn't like this book very much, but lots of people mention it...)

Answer 2

So as to the first question, why is $A_\bullet$ so gosh-darn fundamental in Quantum Mechanics?, this is because momentum is so gosh-darn fundamental in QM, and the QED Lagrangian specifies that the canonical momentum necessarily contains some vector potential in it.

Maybe the most obvious place where this causes a real quantum effect is in so-called AB-rings ("Aharonov-Bohm" rings). An AB-ring is simply a path for an electron to travel through, which is shaped as a ring-shaped path, but also with two terminals added on the ring: an entrance and an exit, for which we want to know the transmission amplitude. We can describe each of these entrance and exit nodes with a scattering matrix much like a beam splitter's; but for the "arms" that go around the circle, the vector potential becomes extremely important because it adds a quantum mechanical phase to each arm. We can subtract out some of this phase, but what we absolutely can't remove is the quantum phase added from traversing around the whole circle: this is proportional to the flux of the field through the loop.

Here is the result: by tuning this vector potential, even if the magnetic field is held constant at zero on the actual loop, we see interference effects due to the vector potential's modulation of the phases of these two arms. And that's why in some sense this vector potential field is now seen to also be very "fundamental": this description is much easier if you acknowledge $A_{\bullet}$ than if you try to work it out from a $F_{\bullet\bullet}$-centric perspective.

Answer 3

Another point not raised in the other two good answers is that the gauge transformation you say makes $F$ more physical than $A$ is actually an essential ingredient in the making of Quantum Field Theories (QFT). The simplest example is Quantum Electrodynamics (QED). Let's start with the Lagrangian for a free fermion field of mass $m$ (the $\gamma_\mu$'s are four matrices acting on the spinor $\psi$ and its conjugate $\bar{\psi}$ and Einstein summation will be used throughout).

$$L_F = \bar{\psi}(i\gamma_\mu\partial^\mu-m)\psi$$

A transformation $\psi(x)\to e^{-iq\Lambda}\psi(x)$, with a constant $\Lambda$, leaves the Lagrangian invariant, as $\bar{\psi}(x)\to e^{iq\Lambda}\psi(x)$. You immediately recognise these transformations are a group representation of $U(1)$. But the recipe that has proven useful to write QFT Lagrangian is to require a local gauge symmetry, i.e. invariance under $\psi(x)\to e^{-iq\Lambda(x)}\psi(x)$.

$L_F$ is not invariant under this local gauge symmetry though:

$$L_F \to L_F -iq\bar{\psi}(\gamma_\mu\partial^\mu\Lambda)\psi$$

But then when we add the interaction between the fermionic field and the electromagnetic field,

$$L_I = -q\bar{\psi}\gamma_\mu\psi\,A^\mu,$$

it should be obvious that the term $-iq\bar{\psi}(\gamma_\mu\partial^\mu\Lambda)\psi$ can be cancelled by a gauge transformation of $A^\mu$:

$$A^\mu\to A^\mu+\partial^\mu\Lambda$$

Morale: the gauge transformation of $A$ plays a crucial role in QED. More complex gauge transformations of $\psi$, representations of groups more complex than $U(1)$ are the foundation of the Standard Model. For example, $SU(3)$ for QCD. And some arcane representation of $SU(2)\times U(1)$ for the electroweak theory. In passing, in that last case, an explicit mass term for the fermion would break local gauge invariance, so I would need to leave that out, and bring in the Higgs mechanism but this is another story I bring in only so that I don't push falsehoods!

Answer 4

"...why in QED it seems the important object is $A$?.."

This is true, and the reason for this is that static EM interactions have $1/r^2$ behavior as $r$ goes to infinity. Really, the statement means that the propagator of EM interaction must have $1/p^{2}$ form, which (taken into account two polarizations and repelling of same sign charges) can be described only by massless vector-like field. The $F_{\mu\nu}$ tensor would give different behavior of the static potential.

"...In that case, what ends up being the role of $F$ in QED?.."

1) $A_{\mu}$ as a Lorentz 4-vector is completely unphysical (excepting some situations), and not only because of the gauge variance. More important reason is that it can't represent particles defined mathematically to be irreducible representations of the Poincare group with zero mass. This is crucial as long as the Poincare symmetry is taken to be the symmetry of our world. The true Poincare-covariant field representing photons is $F_{\mu\nu}$.

2) As the classical ED, QED is based on the gauge symmetry (which is rather the gauge redundancy in the quantum theory, but this is not important at the moment). Therefore, all observed quantities depending on the EM field are always expressed in terms of $F_{\mu\nu}$. In particular, if we start from the complicated gauge theory including $F_{\mu\nu}$, matter fields and $A_{\mu}$ itself, and then integrate out matter fields, the resulting effective action will contain only $F_{\mu\nu}$.

Answer 5

the role of is to write Maxwell's Equation in four dimensions, you now the four-vector of energy-momentum $ P^{\mu} $, like this we consider the four-vector of potential $ A^{\mu} = (\phi, \textbf{A}) $, The quantity $ A $ is called the vector potential or gauge field.where the field strength is defined by:\ $ F^{\mu\nu} = \partial^{\mu}A^{\nu} - \partial^{\nu}A^{\mu} $

Maxwell’s equations may be written in terms of the antisymmetric tensor $ F^{\mu\nu} $ defined by:\ $ F^{\mu\nu} = \partial^{\mu}A^{\nu} - \partial^{\nu}A^{\mu} = $ $ \begin{pmatrix} 0 & - E_{x} & - E_{y} & - E_{z}\\ E_{x} & 0 & - B_{z} & B_{y}\\ E_{y} & B_{z} & 0 & -B_{x}\\ E_{z} & -B_{y} & B_{x} & 0\\ \end{pmatrix} $ \ It is apparent that the electromagnetic field is a tensor field. \ Therefore we find: \begin{equation} F_{0i} = \partial_{O}A_{i} - \partial_{i}A_{0} = \frac{\partial A_{i}}{\partial t} +\partial_{i}\phi = E_{i} \end{equation} \begin{equation} F_{ij} = \partial_{i}A_{j} - \partial_{j}A_{i} = \varepsilon_{ijk}(\nabla \times A) = \varepsilon_{ijk}B_{k} \end{equation} \ Ex:\ \begin{equation} F^{01} = \frac{\partial A^{1}}{\partial x_{0}} - \frac{\partial A^{0}}{\partial x_{1}} = - E_{x} \end{equation} As this exempel you can rewrite all Maxwelle's equations. We can also write equation of motion by: The equations of motion which follow from Lagrangian \begin{equation} L = - \frac{1}{4}F_{\mu\nu}F^{\mu\nu} \end{equation} that we mentioned are:\ \begin{equation} \partial_{\mu}(\frac{\partial L}{\partial(\partial_{\mu} A_{\nu})}) = - \partial_{\mu}F^{\mu\nu} = 0. \end{equation} $ A $ results from a correction to Maxwell's equations in the form: \begin{equation} B = \nabla \times A \end{equation} and \begin{equation} E = - \nabla \phi - \frac{\partial A}{\partial t} \end{equation} \begin{equation} L_{QED} = \bar{\psi}[\gamma^{\mu}(i\partial_{\mu} - qA_{\mu}) - m]\psi - \frac{1}{4}F_{\mu\nu}F^{\mu\nu} - J^{\mu}A_{\mu} \end{equation}

Answer 6

You're confusing being "physical", in the sense of being gauge-invariant, with being canonical (I.e. being the $q$ in the Euler-Lagrange equation). While $F$ is invariant, $A$ is canonical. We need an $F^2$ term in the action because without derivatives of $A$ its momentum vanishes and we can't formulate an interesting Hamiltonian formalism.

It's not surprising we can change the canonical field with an action-preserving transformation, if you know anything about canonical transformations in classical mechanics. It's also not surprising we need a dynamical $A$-field, as without an on-shell control of how it varies the gauge transformation won't work. What I mean by that is we can't very well say $A$ is a constant put in by hand, as arbitrary gauge transformations won't preserve that.