A NEW LOOK AT MASSLESS BOSON FIELDS,SPONTANEOUSLY BROKEN SYMMETRY AND GOLDSTONE THEOREM

Abstract: The sponstaneous breakdown of a continuous symmetry implies the infinite de-generacy of continuous vacuum states, while Goldstone theorem implies the existenceof zero mass Goldstone bosons. When a particular vacuum state satisfying a_k|vac>=0 , b_k|vac>=0 is chosen, other degenerate vacuum states|vac>' of the broken symmetry are usuallyviewed as the superposition of states containing different number of zero energy andzero momentum Goldstone bosons, i.e., with a_k and a_k⁺ representing annihilation and creation operators of Goldstone bosonscarrying momentumk, while b_k representing the annihilation operator of other par-ticles. But actually such a consideration is not correct, because the zero frequencypart of a free hermitian massless scalar field φ（T） can be written as V-(1/2)∫_Vd³xφ(xt)=Q+Pt where V is the size of the normalization box and Q, P are space-time independentoperators. Due to cannonical quantization, they satisfy [Q, P]=i. Thus no zerofrequency annihilation and creation operators a_k=0, a_k⁺=0 can be defined. The freeHamiltonian becomes with N_k=a_k⁺a_k，k=|K|.Thus P and N_k form a complete commuting set of opera-tors describing the eigenstates. Since the eigenvalues of P are continuous all theseeigenstates are not normalizable and all calculation has to be done with the aid ofdistribution (i.e., the delta function.) It can be shown that there is no transition between states with different eigen-values of P, if φ(x) is to remain massless. Thus the vacuum state is not degenerate,but can be chosen as any one state in an mfinitely wide energy band. As a resultthe eigenvalues of P are not measurable. The free Lagrangian of φ(x) has the continuous symmetry φ(x)→φ(x)+η where η is any number, the generator being Z=∫_Vd³xφ(x) We discover that e^iηz |vac> is not orthogonal to |vac>; thus there is no sponstaneously breaking of this sym-metry. Using distribution, the difficulty of (﹤vac|e^iηz)φ(x)(e^-iηz |vac>)-﹤vac|φ(x)|vac>=η﹤vac|vac> can easily be resolved. When φ(x) is complex, each chosen vacuum state is infinitelybut discretely degenerate. Thus there is still no spontaneously broken symmetry. Using these vacuum states, it is easy to verify that in proving Goldstone theorem,the inserted zero momentum and zero energy states do not represent the coherentsuperposition of states containing different number of Goldstone bosons, and do notnecessarily orthogonal to the chosen vacuum state.