By Borovik A. V.

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12) is shown in [46] in a clear way. 13b) n=NR where the charges are regularized in terms of the ζ function regularization. Here, it is important to note that one should regularize the charge with the gauge invariant way since the Hamiltonian has still the invariance of a large gauge trans1 formation n → n + LgA 2π . 14a) Q5 = QR − QL . 15a) LgA1 . 15b) π Since the charge of the vacuum must be zero, we set Q = 0 where we should neglect the λ2 term. 12) by making the time derivative of Q5 . Namely, the chiral current is not conserved any more due to the anomaly.

We note that all of the momenta and any of the physical observables do not depend on the regulator ε when we solve the PBC equations as we discuss below. In this case, the eigenvalue equation becomes N H | k1 , · · · , kN = ∑ Ei | k1 , · · · , kN . 6) ki = L L j=i 2 where ni ’s are integer, and runs as ni = 0, ±1, ±2, · · · , N0 where 1 N0 = (N − 1). 2. 37 Vacuum State First, we want to make a vacuum. We write the PBC equations for the vacuum which is filled with negative energy particles [9, 15] ki = 2πni 2 − L L N ∑ tan−1 i= j ki =k j g ki |k j | − k j |ki | .

This point can be seen quite nicely in the Bethe ansatz solution in the massless Thirring model, and we will discuss it later. 1. Intuitive Discussion Here, we present an intuitive discussion of the chiral symmetry breaking in the NJL models and show that there should not appear any massless boson at all [4]. The treatment here is far from rigorous, but we believe that the essential physics of the spontaneous symmetry breaking phenomena and bosons associated with the symmetry breaking in fermion field theory models should be clarified since there is still a misunderstanding in this problem.