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**Unformatted text preview: **r in class.
Sol.
I’m not going to do every one of the commutation relations, but a few will suﬃce to demonstrate the
relations for the creation and annihilation operators. For convenience, I’m going to set y = 0.
Consider:
ˆ
ˆ
φ(x), φ(0) d3 pd3 q
(2π )6 = [π (x), π (0)] =
ˆ
ˆ − d3 pd3 q
(2π )6 1
[ap , aq ]eip·x + [ˆp , aq ]e−ip·x + [ˆp , aq ]eip·x + [ˆp , aq ]e−ip·x
ˆˆ
a† ˆ †
a ˆ†
a† ˆ
4Ep Eq
Ep Eq
[ap , aq ]eip·x + [ˆp , aq ]e−ip·x − [ˆp , aq ]eip·x − [ˆp , aq ]e−ip·x
ˆˆ
a† ˆ †
a ˆ†
a† ˆ
4 Since both of these expressions are equal to zero, we can add or subtract them on a term-by term basis,
and the two will cancel. either the commutators with exactly 1 creation operator or the ones with 2
creation or two annihilation operators are identically zero.
Finally, consider:
ˆ
φ(x), π (0)
ˆ =
= i
2
iδ (x) d3 pd3 q
[ap , aq ]eip·x − [ˆp , aq ]e−ip·x − [ˆp , aq ]eip·x + [ˆp , aq ]e−ip·x
ˆˆ
a† ˆ †
a ˆ†
a† ˆ
(2π )6 and so:
ˆ
d3 x φ(x), π (0)
ˆ =i
=
= i
2
i
2 d3 q
[a0 , aq ] − [ˆ† , aq ] − [ˆ0 , aq ] + [ˆ† , aq ]
ˆˆ
a0 ˆ †
a ˆ†
a0 ˆ
(2π )3
d3 q
a ˆ†
[a† , aq ] − [ˆ0 , aq ]
ˆ0 ˆ
(2π )3 which immediately gives the expected commutator for [ˆp , aq ]. Note that we used the relation:
a† ˆ
d3 xeip·x = (2π )3 δ (p)
Likewise, we’ve already established that if these terms are non-zero, then the [ˆ, a] and [ˆ† , a† ] comaˆ
aˆ
mutators are zero.
3. Consider the following mode (in the real quantized scalar ﬁeld):
|p = ap |...

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