210
17. Spontaneously Broken Global Symmetry
ˆ
situation is the existence of some field (here φ
′ (y)) with non-vanishing vacuum
expectation value, just as in (17.53).
From (17.58), we can write (17.59) as
0 = ‹0| ˆ
φ
′ (y) 0
(17.60)
∫
| ›
= ‹
ˆ
0|[ d
3
x ˆ j 0 (x), φ(y)]|0›.
(17.61)
Since, by assumption, ∂ ˆ
μ j
μ = 0, we have as usual
∫
∫
∂
ˆ
d
3
x ˆ j
3
0 (x) + d x divj(x) = 0,
(17.62)
∂x 0
whence
∫
∫
∂
3
ˆ
ˆ
3
ˆ
ˆ
d x 0 [j 0 (x), φ(y)] 0 =
d x 0 [divj(x), φ(y)] 0 (17.63)
∂x 0
‹ |
| ›
−
‹ |
| ›
∫
= − dS · ‹0| ˆ
ˆ
[j(x), φ(y)]|0›.
(17.64)
If the surface integral vanishes in (17.64), (17.61) will be independent of x 0 .
The commutator in (17.64) involves local operators separated by a very large
space-like interval, and therefore the vanishing of (17.64) would seem to be
unproblematic. Indeed so it is – with the exception of the case in which the
symmetry is local and gauge fields are present. A detailed analysis of exactly
how this changes the argument being presented here will take us too far afield
at this point, and the reader is referred to Guralnik et al. (1968) and Bernstein
(1974). We shall treat the ‘spontaneously broken’ gauge theory case in chapter
19, but in less formal terms.
Let us now see how the independence of (17.61) on x 0 leads to the necessity
for a massless particle in the spectrum. Inserting a complete set of states in
(17.61), we obtain
∫
∑
ˆ
0 =
d
3
x
n
{‹0| ˆ j 0 (x)|n›‹n|φ(y)|0› − ‹0| ˆ
φ(y)|n›‹n| ˆ j 0 (x)|0›}
(17.65)
∫
∑
=
d
3
ˆ
x
{‹
ˆ
0| ˆ j 0 (0)|n›‹n|φ(y)|0›e
−ipn·x
− ‹0|φ(y)
n
|n›‹n| ˆ j 0 (0)|0›e
ipn·x
}
(17.66)
using translation invariance, with p n the 4-momentum eigenvalue of the state
|n›. Performing the spatial integral on the right-hand side we find (omitting
the irrelevant (2π)
3 )
∑
0 =
δ
3 (p n )[
n
‹0| ˆ
| ›‹ | ˆ | ›e
ipn0x0
− ‹ ˆ
j 0 (0) n n φ(y) 0
0|φ(y)|n›‹n| ˆ j 0 (0)|0›e
−ipn0x0 ].
(17.67)
/
/
/