Nonperturbative calculation of symmetry breaking in quantum field theory

A new version of the delta expansion is presented, which, unlike the conventional delta expansion, can be used to do nonperturbative calculations in a self-interacting scalar quantum field theory having broken symmetry. We calculate the expectation value of the scalar field to first order in δ, where δ is a measure of the degree of nonlinearity in the interaction term. Typeset using REVTEX 1 In a private discussion with D. Bessis we learned about the remarkable non-Hermitian Hamiltonian H = p/2 + ix. (1) The eigenvalues of this Hamiltonian are all real and positive! (We have no proof of this property, but there is ample numerical evidence.) This Hamiltonian is intriguing because it suggests a novel way to apply the delta expansion to quantum field theories having a broken symmetry. In this paper we introduce our new version of the delta expansion and use it to calculate the (nonzero) value of 〈φ〉 in a scalar quantum field theory [see Eq. (31)]. The delta expansion is a Taylor series in powers of δ, where δ measures the degree of nonlinearity of an interaction term [1]. For a scalar quantum field theory, the conventional approach [2] has been to introduce the parameter δ into the Euclidean Lagrangian density by L = (∇φ)/2 +mφ/2 + g(φ). (2) The advantage of the delta expansion is that it is nonperturbative in the coupling constant g and mass m and has a nonzero radius of convergence. The conventional delta expansion has been used to study renormalization [3], supersymmetry [4], local gauge invariance [5], stochastic quantization [6], and finite-temperature field theory [7]. A drawback of the Lagrangian density in Eq. (2) is that it becomes a g|φ| theory rather than a gφ theory when δ = 1 2 . The g|φ| theory is symmetric under φ → −φ and cannot exhibit symmetry breaking. In general, the conventional delta expansion is unsuitable for studying theories with symmetry breaking. In the past it was assumed that to satisfy the physical requirement that the Hamiltonian be bounded below for all δ, the parameter δ should appear in the Lagrangian density (2) as the exponent of φ and not of φ. This assumption appears to be even more reasonable if one expands the interaction term in Eq. (2) in powers of δ: g(φ) = gφ ∞ ∑ k=0 δ k! (logφ). (3)