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Looking at the QED-Lagrangian

$$\mathcal L = -\bar\psi(\not\!p + e\not\!\!A + m)\psi -\frac14 F_{\mu\nu} F^{\mu\nu} $$

I was wondering: While $\bar\psi\not\!\!A\psi$ describes the interaction between the electron field (or rather generally, a charged fermion) and the photon field (a neutral boson), the $\bar\psi\not\!p\psi$ describes the kinematic propagation with a fixed operator $\not\!p=-i\hbar\not\!\partial$. Would it make any sense to instead consider $\not\!p$ another bosonic particle field that also has to be determined, a "momenton" so to speak? Clearly that would also require another term similar to the $-\frac14 F_{\mu\nu} F^{\mu\nu}$ one, probably relating to general relativity in the ideal case (although one yielding the already established $\not\!p=-i\hbar\not\!\partial$ in vacuum would be a start and probably straightforward to construct). And I'm also tempted to rescale everything with $\frac1m$ so the term $\bar\psi\frac1m\not\!p\psi$ would include a coupling constant... Or is that "momenton" basically the graviton?

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It doesn't, really.

You are definitely free to introduce a new vector field $p^\mu$ and add an interaction $\bar \psi \!\!\not\! p\,\psi$ to your Lagrangian, together with a kinetic term for $p^\mu$. Whether you do this or not, you must include the standard kinetic term $\bar \psi \!\!\not\! \partial\,\psi$ anyway. In this case the new term $\bar \psi \!\!\not\! p\,\psi$ does not replace the old term $\bar \psi \!\!\not\! \partial\,\psi$ but rather they appear together.

The reason the term $\bar \psi \!\!\not\! \partial\,\psi$ must always be present is the following. This term is a kinetic term, without it the equations of motion of $\psi$ do not include derivatives. In absence of kinetic terms, the field $\psi$ is essentially frozen in spacetime, with no dynamics of its own. Its Euler-Lagrange equations are algebraic equations, so that $\psi$ rigidly follows the dynamics of the rest of fields.

In the standard terminology, a field $\psi$ with no kinetic term is auxiliary, it basically acts as a Lagrange multiplier. The path integral for auxiliary fields can always be done explicitly: you just solve the classical equations of motion, and plug the result back into the Lagrangian. One says that $\psi$ is integrated out.

So, all in all, the replacement $\partial\to p$ renders the field $\psi$ non-dynamical, so it becomes an entirely different theory, one much more trivial than the original one (instead of being more general).

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Not sure where this idea is coming from, but perhaps the following will help illustrate why $\partial \to p$ is not a natural or useful operation. In quantum mechanics, operators and differential operators are emphatically not on the same footing. A regular operator is a map $\mathcal H\to\mathcal H$, a differential operator is a map $\mathrm{End}(\mathcal H)\to\mathrm{End}(\mathcal H)$. These are entirely different objects, so the proposed replacement is mostly meaningless from a mathematical point of view.

The apparent formal similarity between $\bar \psi \!\!\not\! p\,\psi$ and $\bar \psi \!\!\not\! A\,\psi$ is due to bad notation: the derivative should be denoted by $\partial$, not $p$. The differential operator maps operators into operators, a regular operator maps states to states. Again, these are completely different operations, and the notation really should reflect that. Put it differently, in $\not p\psi$, $p$ is acting on $\psi$; in $\not A\psi$, $A$ is multiplying $\psi$ (i.e., it denotes a composition). A more explicit notation would be $p(\psi)$ vs $A\circ\psi$, in which case the formal similarity breaks down.

In regular QM, the position $x$ is the eigenvalue of $\hat X$, and $\partial$ is just the matrix element of $p$ in the basis of eigenvectors of $\hat X$. In QFT, there is no $\hat X$, and $\partial$ does not denote the matrix element of some abstract operator. Instead, $x$ denotes a label on operators, with no deeper origin. The $x$-dependence of $\psi$ is fundamental, not a consequence of something else, unlike in regular QM, where $\psi(x)=\langle x|\psi\rangle$.

The analogue of $\partial$ in regular QM is $\frac{\mathrm d}{\mathrm dt}$, which is a differential operator that acts on operators. There is no abstract operator for which $\frac{\mathrm d}{\mathrm dt}$ is a matrix element. In field theory, $\frac{\mathrm d}{\mathrm dt}$ is replaced by $\partial$ and, again, there is no operator for which it is a matrix element. See also this PSE post.

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  • $\begingroup$ I think I understand your point - but then again $\partial$ means we're already in a spatial representation with implicit coordinates everywhere, i.e. $\psi(x)$ etc. But if instead you consider those projections into spacetime, then before that $p$ cannot yet just be $\partial$ - does this make sense? I'm not sure I properly explain what I mean... $\endgroup$ – Tobias Kienzler Nov 22 at 15:06
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    $\begingroup$ @TobiasKienzler Recall that in QFT there is no position operator $\hat X$, so the coordinates $x$ do not represent a choice of basis. We are not in the $|x\rangle$ representation, so $\partial$ is not the matrix element of an abstract operator $p$. In QFT, the coordinates $x$ are intrinsic to the formalism, not just a convenient choice. This is the origin of the letter "F" in QFT: the degrees of freedom are fields, i.e., position-dependent. $\psi(x)$ is not $\langle x|\psi\rangle$. $\psi(x)$ itself is the fundamental object, not derivable from something more fundamental. $\endgroup$ – AccidentalFourierTransform Nov 22 at 15:21
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In my opinion, the idea is worth further investigation, if we think of it as a search for a formalism that treats momentum as the same (or similar) class of object as a quantum field, rather than trying to just implement it within the standard formulations of QFT.

Google search produced the following work that seems “spiritually close” to OP's suggestion (but not an exact match), the duality between coordinates and fields suggested by Faraggi & Matone:

Abstract:

We introduce a “prepotential” $\mathcal{F}$ in quantum mechanics and show that the coordinate $x$ is proportional to the Legendre transform of $\mathcal{F}$ with respect to the probability density. Inversion of the Schrödinger equation leads us toconsider a $x-ψ$ duality related to a modular symmetry. The scaling of $x$ is determined by the “beta–function”, suggesting that in quantum mechanics the space coordinate is a macroscopic variable of a statistical system with $\hbar$ playing the role of scale. The formalism is extended to higher dimensions and to the Klein–Gordon equation.

In this work a method borrowed from Seiberg–Witten theory is applied to invert the the (single particle) Schrödinger wavefunction $ψ=ψ(x)$ to $x=x(ψ)$. In this formalism the quantum dynamics is described by prepotential $\mathcal{F}$, which satisfies a non–linear third–order differential equation which replaces the Schrödinger equation. Coordinate $x$ is the Legendre transform of $\mathcal{F}$ with respect to the probability density.

The hope is that this dual description of quantum mechanics could produce a new structures once second quantization is performed. This approach offers a way to describe spacetime coordinates in terms of quantum quantities providing possible insight into structure of spacetime at high energies.

Similarly, we could hope that there exists description of coordinates on momentum space in terms of solutions of Schrödinger (or Klein–Gordon, or Dirac) equations and after the second quantization even in terms of quantum fields (the “momenton” of OP's title). This description may indeed be relevant for quantum gravity (but this would not be the “graviton” in the sense of perturbatively quantized QG).

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  • $\begingroup$ Sounds very interesting, thanks! I'll have to read it later :) On first skimming your answer I first thought about Wigner/Weyl, which on second thought is unrelated but I'll just mention it here anyway for future reference 😅 $\endgroup$ – Tobias Kienzler Nov 27 at 7:56
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Here you go (kinetic part only): $$ \mathcal L = -\bar\psi\not\!p\psi + g(p_\mu+i\hbar\partial_\mu)\bar\psi(p^\mu+i\hbar\partial^\mu)\psi. $$ It recovers the original Lagrangian when the coupling constant $g$ approaches $\infty$.

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