Pictures Paths Particles Processes

November 7, 2013 279
The propagator is given by Eq.(10.99), where now
f(u) = µ −
p∑
j=1
γ j
(
u j + 1 u j )
. (10.110)
We shall now arrange for the only the highest possible power of 1 − u to survive
in this expression. We first put u = exp(ik∆), so that the function f(u) becomes
f(u) = µ −
B r
≡
p∑
j=1
p∑
We now seek to find the γ’s such that
j=1
2γ j cos(jk∆) = µ − ∑ r≥0
(k∆) 2r B r ,
2(−) r
(2r)! j2r γ j . (10.111)
B 1 = B 2 = · · · = B p−1 = 0 , B p = − 1 . (10.112)
∆2p−1 In that case, we can take arbitrary constants c r , with c p = 1, and always have
with
p∑
c r B r =
r=1
Q(j) =
p∑
γ j Q(j) = B p , (10.113)
j=1
p∑
r=1
2(−) r
(2r)! c rj 2r . (10.114)
The polynomial Q(j) is even and of degree 2p in j, and Q(0) = 0. We can now,
for any preassigned q with 1 ≤ q ≤ p, choose the numbers c r such that
Q(0) = · · · = Q(q − 1) = Q(q + 1) = · · · = Q(p) = 0 , Q(q) ≠ 0 , (10.115)
upon which
Obviously, the polynomial Q(j) is given by
from which we derive
Q(j) = 2(−)p
(2p)!
γ q = B p /Q(q) . (10.116)
∏
0 ≤ n ≤ p
n ≠ q
(j 2 − n 2 )
, (10.117)
γ q =
(−) q−1 (2p)!
∆ 2p−1 (p − q)!(p + q)!
, 1 ≤ q ≤ p . (10.118)