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**Unformatted text preview: **erent reason. Let’s consider the aa terms:
d3 pd3 q
(2π )6 =− d3 x =− RHS d3 pd3 q
(2π )3 =− µ
Ep qi
ap aq e−i(pµ +qµ )x
4Ep Eq Ep qi
ap aq e−i(Ep +Eq )t δ (p + q )
4Ep Eq d3 p Ep (−pi )
ap a−p e−2iEp t
(2π )3
2
4E
p This integral is odd in direction i. The ap a−p term is clearly even, as is everything that contains
Ep . The only component that isn’t is pi , which is odd, and thus the entire contents of the integral
is odd. Since we integrate d3 p over all p-space, this integral vanishes. We get the same result for
a∗ a∗ since they are complex conjugates of one another.
pq As with the energy term, we are left with:
d3 pd3 q
(2π )6 µ
Ep qi
ap a∗ e−i(pµ −qµ )x + C.C.
q
4Ep Eq = d3 x = d3 pd3 q
δ (p − q )
(2π )3 = Pi d3 p Ep pi
ap a∗ + C.C.
p
(2π )3 2Ep Ep qi
ap a∗ e−i(Ep −Eq )t + C.C.
q
4Ep Eq d3 p
p i ap a∗
p
(2π )3 =
exactly as expected. 2. This is from Tong, but I really like it:
The Fourier decomposition of a real scalar ﬁeld and its conjugate momentum in the Schroedinger
picture is given by:
d3 p
(2π )3 ˆ
φ(x) = π (x)
ˆ =i 1
ap eip·x + ap e−ip·x
ˆ
ˆ†
2Ep d3 p
(2π )3 Ep
ap eip·x − ap e−ip·x
ˆ
ˆ†
2 Show that the the commutation relations:
ˆ
ˆ
φ(x), φ(y ) = [ˆ (x), π (y )] = 0
π
ˆ ˆ
φ(x), π (y )
ˆ = iδ (x − y) imply that:
[ˆp , aq ] =
aˆ
ap , aq
ˆ ˆ† ap , aq = 0
ˆ† ˆ† = (2π )3 δ (p − q ) a point I glossed ove...

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