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ˆ†
Compute the following:
ˆ ˆ ˆ† ˆ
(a) O = ap ap ap (in as simple a form as you can make it)
Sol.
I’ll freely admit that “simple” in this case is ambiguous. It would have been better, for, instance,
to say, “in terms of a single operator plus whatever you need to add to it,” but even that leaves
some ambiguity. Instead, I’ll show you what I really wanted. Recall that:
ap , a† = (2π )3 δ (p − q )
ˆ ˆq
and thus:
ˆ = (2π )3 δ (0)ˆp + a† ap ap
′
a
ˆp ˆ ˆ
This is especially useful because the second term operatoring on an arbitrary state will almost
certainly be zero.
(b) p|ap ap |p
ˆ† ˆ
Sol.
We can use the same relation above to expand this out:
(2π )6 δ (0)2 0|0 + 2(2π )3 δ (0) 0|a† ap |0 + 0|a† ap a† ap |0
ˆp ˆ
ˆp ˆ ˆp ˆ
The latter two terms are identically zero, since:
ap |0 = 0
ˆ
so
p|ap ap |p = (2π )6 δ (0)2
ˆ† ˆ
which is simply the volume of the space, squared
ˆ
(c) p|O|p
Sol.
As there is an odd number of operators:
ˆ
p|O|p = 0 item Finally, consider the quantization of the complex scalar ﬁeld. Previously, we showed that there
is a conserved (classical) current:
j µ = i(ψ ∗ ∂ µ ψ − ψ∂ µ ψ ∗ )
In QFT, of course, this current becomes an operator.
(a) Write out (simplifying wherever possible) the “charge” operator, ˆ0 . You may ﬁnd the operators
j
from problem 2 to be a useful guide.
Sol.
The quantized version of the operator is:
˙ˆ
˙ˆ
ˆ0 = i(ψψ † − ψ † ψ )
j
You can switch the order of both terms, but they need to be consistent. Recall:
ˆ
ψ (x) = d3 p
(2π )3 ˆ
ψ † (x) = d3 p
(2π )3 and µ
µ
1
c† e−ipµ x + ˆp eipµ x
ˆp
b
2Ep 1
ˆ† e−ipµ xµ + cp eipµ xµ
bp
ˆ
2Ep so
ˆ0
j d3 pd3 q
(2π )6 =
= Ep
4Ep Eq µ c† e−ipµ x − ˆp eipµ x
ˆp
b µ
µ
− ˆ† e−ipµ x − cp eipµ x
bp
ˆ µ µ ˆ† e−iqµ xµ + cq eiqµ xµ
bq
ˆ c e−iqµ x + ˆq eiqµ x
ˆq
b µ Four of these terms identically cancel by the commutation relations (and an arbitrary switching
of p and q ):
c† ˆ† − ˆ† c† = 0
ˆp bq bp ˆq
and similarly with the cˆ terms. This leaves:
ˆb
ˆ0
j = d3 pd3 q
(2π )6 Ep
4Ep Eq µ c† cq e−i(pµ +qµ )x + cp c† ei(pµ +qµ )x
...

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