# a glaring deficiency in the QFT lagrangian formalism

Summary: In a quantum field theory there is no way to fully constrain the motion of a test-particle using either the equations of motion, or the Noether current, in the presence of gravity. This is not a problem either at the classical level, or in flat space: At the classical level, the initial position and initial velocity (6 conditions) fully constrain the dynamics. At the quantum level, the feynman, advanced, or retarded propagators, plus the freedom to add a constant (4 conditions) fully constrain the free-space dynamics. There is however, an infinite number of curved space propagators and no guidance from the Lagrangian EOM or Noether symmetries on which to choose. The problem is underconstrained and undetermined! External information is needed to choose the dynamics.

## The Classical Lagrangian beautifully constrains the motion by the EOM

Let's begin with the classical Lagrangian, $L_m=mc^2\sqrt{1-v^2/c^2}$.

The equations of motion describe the path that a point particle traverses through the space according to, $\dot p_i=0$, where $p_i=m\gamma \dot x_i$. These equations of motion entail both linear momentum conservation as well as angular momentum conservation. The problem is fully constrained and easily solved.

Now let's add gravity to the picture: $L=L_m+L_g$ where $L_g=\phi(r)=GMm/r$. All spherical symmetries and time translation symmetries still manifest in the equations of motion (for example $\frac d{dt}(r^2p_\theta)=0$) except that now there is no symmetry in the r-direction. Instead, it is replaced by the equation of motion $\dot p_r=\nabla \phi$. The problem is still fully constrained and has a unique solution once a finite number of boundary conditions are specified.

So classically all symmetries give conservation laws, except when gravity breaks a symmetry we instead have an equation of motion detailing a new conservation that exchanges potential energy for kinetic energy. So the symmetry in one direction is replaced by an energy balance in that direction. This concludes our discussion of the classical case.

## The Quantum Lagrangian cannot constrain the motion by the EOM - We must also include Noether symmetries by hand

In the quantum field theory, the matter lagrangian is replaced by fields $\mathscr L_m=\sqrt{-g}g^{\mu\nu}\partial_\mu\phi\partial_\nu\phi$. There is only one (!) equation of motion and that is the KG equation

$\partial_\mu\partial^\mu\phi(x)=0$

which can be thought of as a different equation at every spacetime point.

Now the KG equation has an infinite number of solutions $\phi(x)$ corresponding to an infinite number of boundary conditions. However, just as in the classical case there is symmetry, and the complete set of symmetries (rotational, translatation, etc.) are the same as in the classical case and constrain the form of $\phi(x)$ to a finite set.

Therefore, one can solve for example, $\langle\phi(x)\phi(y)\rangle$ but if all we know is that it solves KG in x and y then there will be an infinite number of solutions. Imposing Poincare symmetry (by hand, since they are not part of the EOM) reduces it to 4 solutions: Feynman, Advanced, Retarded propagators plus a constant.

## Why it matters that the two are different

Here we encounter the glaring deficiency. In the classical case, the symmetries were part of the equations of motion, and things like angular momentum were conserved explicitly and led to a finite number of initial conditions needed to solve. In the quantum picture the equations of motion are NOT manifestly symmetric, and one has to impose symmetries by way of Noether's currents to get a finite number of initial conditions.

And here is the rub: at least I could add constraints on $\langle\phi(x)\phi(y)\rangle$ using symmetries of the Minkowski vacuum by hand. When I add gravity, there is no analogue for $\dot p_r=\nabla \phi$ to replace the radial symmetry that existed in Minkowski space. In other words, the expression $\langle\phi(x)\phi(y)\rangle$ is not constrained by either the equations of motion, or the Noether currents! It is under constrained! There are an infinite number of solutions, and no way to even discuss the change of momentum in the radial direction. What am I missing? and is there a way to talk about the motion of a (quantum) test-particle in the presence of a gravitational field if I cannot constrain the Green's functions with either eom or using Noether?

• I'm not exactly sure what you're asking, but it seems that it worries you that there are many solutions to the KG equation. Well, there are many solutions to the classical particle Lagrangian too, until you add initial conditions. And you want to use the KG equation to find the movement of a test particle, but it is a field equation. – Javier Apr 11 '15 at 21:53
• Indeed, I've edited it to be more clear. There is a big difference in that classically there are a finite number of boundary conditions: 6, initial position vector, and initial velocity vector. At the quantum level there are a finite number of boundary conditions on the Green's function once you impose symmetry: 4, the feynman, advanced and retarded propagators, and the freedom to add a constant. – alphanzo Apr 11 '15 at 22:02
• That's true, but I'm not sure what the question is. Also don't forget that there are classical fields too: You need boundary conditions for the Maxwell and Einstein equations (and any other classical field equation). – Javier Apr 11 '15 at 22:04
• yes, that is quite true, however, the problem persists. In the case of Maxwell's equation there is a stress-energy tensor and it has all the symmetries of the Poincare group, so it can be used to impose Poincare symmetry at the quantum level, or to tell you how they break down by replacing a symmetry with a (electromagnetic) force. There is no gravitational force in quantum gravity, and there is no way to choose a Green's function out of infinitely many on symmetry arguments as there was in flat spacetime. – alphanzo Apr 11 '15 at 22:18
• the question is whether there is a way to talk about energy conservation in gravity, at least effectively, so that there might be a (pseudo) tensor which defines a quantum analog of $\dot p_r=\nabla\phi$ in terms of fields. This way there will be a finite number of Green's functions instead of infinite, which is what one would expect. – alphanzo Apr 11 '15 at 22:33