By J. F. Davis
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Additional resources for A survey of the spherical space form problem
It turns out to be convenient to deﬁne the generating functional Z [ J ] to obtain the matrix elements of the T -products efﬁciently. We couple an external ﬁeld J (t) (also called the source) with the coordinate x(t) as x(t)J (t) in the Lagrangian, where J (t) is deﬁned on the interval [ti , t f ]. Deﬁne the action with the source as tf S[x(t), J (t)] = ti dt [ 12 m x˙ 2 − V (x) + x J ]. 119) The transition amplitude in the presence of J (t) is then given by x f , t f |x i , ti J tf = x exp i ti dt ( 12 m x˙ 2 − V (x) + x J ) .
Let us consider the unitary operator Uˆ (a) = e−ia pˆ . 1. The operator Uˆ (a) deﬁned as before satisﬁes Uˆ (a)|x = |x + a . 2 CANONICAL QUANTIZATION 15 Proof. It follows from [x, ˆ p] ˆ = i that [x, ˆ pˆ n ] = in pˆ n−1 for n = 1, 2, . .. Accordingly, we have [x, ˆ Uˆ (a)] = x, ˆ n (−ia)n n pˆ = a Uˆ (a) n! which can also be written as ˆ xˆ Uˆ (a)|x = Uˆ (a)(xˆ + a)|x = (x + a)U(a)|x . This shows that Uˆ (a)|x ∝ |x + a . Since Uˆ (a) is unitary, it preseves the norm of a vector. Thus, Uˆ (a)|x = |x + a .
It is analytic over the whole s-plane except at the simple pole at s = 1. 161) 38 QUANTUM PHYSICS we obtain β π ζ−d2 /dτ 2 (0) = 2 log ζ(0) + 2ζ (0) = − log(2β). 162) 2 . 163) The inﬁnite product in this equation is well known but let us pretend that we are ignorant about this product. The partition function is now expressed as Tr e−β H = 2β ∞ 1+ p=1 2 βπ pπ −1/2 π ω tanh(βω/2) 1/2 . 151), we have proved the formula ∞ 1+ n=1 βω nπ 2 = π sinh(βω) βω = sinh(π x) . 165) What about the inﬁnite product expansion of the cosh function?