Is it possible to find the ground state of interacting $\phi^{4}$ theory with quartic interaction analytically?

Question

Is it possible to find the ground state of interacting $\phi^4$ theory with quartic interaction analytically?

Answer 1

In 2 and 3 spacetime dimensions, this is possible. In fact, the whole $\phi^4$ theory exists as a mathematically well-defined Wightman QFT in 2 and 3 spacetime dimensions.

The complete definition of these models is given this excellent textbook.

In 4 spacetime dimensions, this has never been done, and in fact there are indications that the interacting $\phi^4$ theory doesn't even exist in 4 dimensions – these come from taking continuum limits of lattice models numerically and observing that the interaction doesn't survive this limit.

Answer 2

A rigorous analytic construction of interactive $\phi^4$ theory in 4 spacetime dimensions is not known; neither is it known to be impossible.

Almost certainly, a construction via lattice approximations (mimicking those used in lower dimensions) is doomed to fail. This is called the triviality problem. The reason is that to approach the continuum limit one needs to pass through a nonphysical singularity called the Landau pole. A comprehensive review article on triviality is

• D.J. Callaway, Triviality pursuit: can elementary scalar particles exist? Physics Reports, 167 (1988), 241-320.

He says: ''Other attempts to construct a nontrivial $\phi^4$ theory are often a bit more abstract in nature. Rigorous discussions of triviality (see section 2) often require that a $\phi^4$ field theory is defined as an infinite-cutoff limit of a ferromagnetic lattice theory. It has been argued that this is an assumption whose removal changes the nature of the problem dramatically. Indeed, no argument appears to prevent the existence of an interesting nontrivial ultraviolet limit of an antiferromagnetic lattice $\phi^4$ theory, even in $d>4$. This remains an interesting open problem.''

About nonlattice approximations essentially nothing is known; the discussion in Section 8 of

• G. Gallavotti and V. Rivasseau, Field theory in dimension 4: a modern introduction to its unsolved problems, Annales de l'I.H.P. Physique théorique 40 (1984), 185-220.

is still valid today.

In addition, the known triviality results are irrelevant for approaches without an explicit cutoff (which excludes lattice approximations) that would have to be moved to zero or infinity to establish covariance. In particular, the Landau pole is not an obstacle for fully covariant approximation methods such as causal perturbation theory: Since no cutoff is present, nothing needs to cross the pole; thus the construction involves no singularity.

There is no known closed form solution, even in 1+1 dimensions, and it is unlikely that there will ever be one. A classical or quantum system admitting a closed form solution is called an integrable model. It is characterized by having enough additional symmetries (aka conserved quantities by Noether's theorem) or hidden symmetries (involving quantum groups rather than symmetry groups), so that one can obtain a solution by exploiting this additional structure. This is not the case for $\phi^4$ theory.

Many integrable models exist but compared to the general case they are very rare, just as homogeneous spaces (manifolds with a transitive symmetry group) are rare compared to arbitrary manifolds. See also this thread.

Answer 3

An interesting system is the homogeneous spin $1/2$ interacting gas in the low density and temperature limit. It can be described by the Hamiltonian $$ H = \sum_{k,\sigma} \varepsilon_k c^\dagger_{k\sigma} c_{k\sigma} + \frac{1}{2V} \sum_{k_1 \sigma_1} \sum_{\sigma_2} \sum_{Q \sigma_3} \sum_{k_4 \sigma_4} \tilde{U}(|\vec{k}_1 - \vec{k}_4|) c^\dagger_{k_1 \sigma_1} c^\dagger_{Q-k_1 \sigma_2}c_{Q-k_4 \sigma_3} c_{k_4 \sigma4}, $$ where V is a volume which must be assumed to be infinity, $c^\dagger_{k \sigma}$ creates a particle with wave vector $\vec{k}$ and spin $\sigma$, $\tilde{U}(|\vec{k}|)$ is the interaction potential between two fermions, which is assumed to be radial and $\varepsilon_k = \hbar^2k^2/(2m)$. Further simplifications, made by BCS allow us to describe the system with the reduced BCS Hamiltonian, given by $$ H = \sum_{k,\sigma} \varepsilon_k c^\dagger_{k\sigma} c_{k\sigma} + \frac{g}{V} \sum_{k_1} \sum_{k_4} c^\dagger_{k_1 \uparrow} c^\dagger_{-k_1 \downarrow}c_{-k_4 \downarrow} c_{k_4 \uparrow}. $$ For the reduced BCS Hamiltonian, an exact solution has been found by Richardson and used in research. I read that here and here. I want to emphasize that this Hamiltonian doesn't have the complete interaction, as the first Hamiltonian. Also in the thermodynamic limit the exact solution agrees with the mean-field approximation. The first paper I shared gives more details about the solution.