Is force described by Grassmann number possible?

Question

Bosonic operators can often couple to external classical fields, just like charge density $\rho$ can couple to a classical electric field $E$. The fermionic counterpart is that a Grassmann number field/force $\eta$ couples to a fermionic operator $A$. A minimal example would be $$H=\epsilon c^\dagger c+\eta(c^\dagger+c).$$ Certainly, one can consider more complex cases and the responses to the Grassmannian force.

However, does such a Grassmannian force and coupling, as the fermionic counterpart of classical fields, exist in any concrete phenomenon or experiment? Is it realizable at all?

Answer 1

For what it's worth, here is a rough outline of a possible realization of OP's scenario:

• We can introduce a Grassmann-odd source/force/current $\bar{\eta}$ for any Grassmann-odd field $\psi$.

• These Grassmann-odd sources $\bar{\eta}(\phi)$ could in principle be made up of fields $\phi$ from a larger encompassing theory. (At least some of these $\phi$s must apparently be Grassmann-odd as well.)

• We could then study correlators $\langle \bar{\eta}(\phi)\ldots \bar{\eta}(\phi)\rangle$ of such Grassmann-odd sources.

• Obviously, observables must be Grassmann-even, i.e. contain an even number of Grassmann-odd fields, cf. e.g. this related Phys.SE post.

Answer 2

The Grassmann field (or force) would physically have to be a representation of another fermion field (probably in some condense-matter scenario). According to the Hamiltonian of the OP, this field is modeled as a classical field. However, the fermion field associated with $c$ is not conserved. So it is absorbed into, or excited from, this classical fermion field.

In other words, the theory is a just a semi-classical model of the physical scenario and does not necessarily represent the most accurate way to model it. Such classical models are often used to simplify calculations, depending on what one wants to learn from it.

Answer 3

To a large extent this depends on what one means by "force", which is really only a clear notion in classical physics. In quantum field theories, one has lots of interactions, and what people often mean by "forces" are only a subset of these which come from exchanges of vector bosons. But there are "forces" as well from all sorts of particles, and in fact we already know such a "fermionic force" which exists.

There's a Higgs force, though it's only at very small distances, $V \propto \exp(-r/m_H)$. Most pertinently to this question, there is a force mediated by neutrinos, fermions which we know exist. Here's an old paper and here's a recent one. This neutrino force falls off as $r^{-5}$ compared to a classical massless vector boson $r^{-2}$. But still one should reasonably call this a "fermionic force". All the fermions give rise to something similar, but neutrinos have the largest effect since they are by far the lightest.

Answer 4

A Hamiltonian should be Hermitian, while OP's Hamiltonian $$H=\epsilon c^\dagger c+\eta(c^\dagger+c)$$ is not Hermitian. One can easily check that this part $$(\eta(c^\dagger+c))^\dagger = (c^\dagger+c)\eta^\dagger$$ is not Hermitian.

If you enforce reality condition on $\eta$ such as $\eta^\dagger = \eta$, then $$(\eta(c^\dagger+c))^\dagger = (c^\dagger+c)\eta = - \eta(c^\dagger+c)$$ therefore this term is anti-Hermitian. Note that the minus sign comes from the Grassmann-odd nature of the field.

The only way for this to work is Hamiltonian: $$ H=\epsilon c^\dagger c+\eta(c^\dagger+c)+(c^\dagger+c)\eta^\dagger $$ conditional on that $\eta^\dagger \neq \eta$.