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**Unformatted text preview: **p0 +q0 )t
4Ep Eq
2
−Ep + |p|2 + m2 ap a−p e−i(p0 +q0 )t =0
This vanishes since the term in the parentheses is necessarily zero by our dispersion relation:
2
Ep = |p|2 + m2 The a∗ a∗ term vanishes identically – and necessarily – since the two terms are complex conjugates
pq
to one another. This leaves only the aa∗ terms. I’m only going to write out the terms with the
negative exponents, and leave C.C. to correspond to the complex conjugate.
d3 pd3 q
(2π )6 µ
1
Ep Eq + p · q + m2 ap a∗ e−i(pµ −qµ )x + C.C.
q
4Ep Eq = 1
2 d3 x = 1
2 d3 pd3 q
(2π )3 = 1
2 d3 p
(2π )3 = E 1
2 d3 p 1
2
2Ep ap a∗ + C.C
q
(2π )3 2Ep 1
δ (p − q ) Ep Eq + p · q + m2 ap a∗ e−i(Ep −Eq )t + C.C.
q
4Ep Eq 1
2
4Ep 2
Ep + |p|2 + m2 ap a∗ + C.C
q The Complex conjugate simply doubles the result. Multiplying it up we get:
d3 p
∗
Ep ap ap
(2π )3 E= (1) (b) What is the momentum for this ﬁeld?
Sol.
This result is much simpler, but involves an additional trick. First, we note the general relation:
˙
d3 xφ(x)φ,i (x) Pi = − that we derived from the T i0 component of the stress-energy tensor from our Noether relation.
Plugging in, we get 4 terms:
Pi = − d3 x d3 pd3 q
(2π )6 µ
µ
µ
µ
Ep qi
ap aq e−i(pµ +qµ )x − ap a∗ e−i(pµ −qµ )x − a∗ a∗ ei(pµ −qµ )x + a∗ a∗ ei(pµ −qµ )x
q
pq
pq
4Ep Eq Here, the aa and a∗ a∗ terms vanish for a di...

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