By Berne B.J., Harp G.D.

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18) To ﬁnd the equation satisﬁed by K, consider the time t = tf + , when the position is xn+2 . 20) where xn+2 = xn+1 − η. 28) This has justiﬁed our choice of normalization N( ) in Eq. 17). 29) as required. 30) In a scattering problem, the initial wavefunction is a plane wave. In a bound-state problem, the initial wavefunction is unknown, so one has an eigenvalue problem (solving ψf = ψi ), which is very difﬁcult, in general. In a problem where the potential term is small, it is most convenient to calculate K in perturbation theory.

191) Now the little group of this vector is clearly O(n − 1). 194) which do not leave φ invariant. We then reparametrize the n-component ﬁeld φk as 0 .. n−1 ki . 195) v 0 i=1 v+η in terms of the (n − 1) ﬁelds ξi , and the ﬁeld η. The action of ki on the vector vi = vδin is given by (ki v)j = v(Lin )j l δin = −ivδij Thus to lowest order, one has ξ1 ξ2 . φ = .. 200) and the (n − 1) ﬁelds ξi are massless Goldstone bosons. Note that the number of Nambu–Goldstone bosons is equal to the difference in the number of generators of the original symmetry O(n) and the ﬁnal symmetry O(n − 1): 1 1 n(n − 1) − (n − 1)(n − 2) = (n − 1) 2 2 This is an example of a general theorem that we now prove.

We then reparametrize the n-component ﬁeld φk as 0 .. n−1 ki . 195) v 0 i=1 v+η in terms of the (n − 1) ﬁelds ξi , and the ﬁeld η. The action of ki on the vector vi = vδin is given by (ki v)j = v(Lin )j l δin = −ivδij Thus to lowest order, one has ξ1 ξ2 . φ = .. 200) and the (n − 1) ﬁelds ξi are massless Goldstone bosons. Note that the number of Nambu–Goldstone bosons is equal to the difference in the number of generators of the original symmetry O(n) and the ﬁnal symmetry O(n − 1): 1 1 n(n − 1) − (n − 1)(n − 2) = (n − 1) 2 2 This is an example of a general theorem that we now prove.