Quantum Field Theory I

2.2. CANONICAL QUANTIZATION 61
Forthe relativisticquantumtheory(and thisiswhatweareafter)theHamiltonian
is not the whole story, one rather needs all 10 generators of the Poincarè
group. For the space-translations P ⃗ = ∫ d 3 x π ∇ϕ one obtains 15
∫
⃗P =
d 3 p
(2π) 3 ⃗p a+ ⃗p (t)a ⃗p(t)
while for the rotations and the boosts, i.e. for Q ij = ∫ d 3 x π ( x j ∂ i −x i ∂ j) ϕ
and Q 0i = ∫ (
d 3 x ∂L
∂ ˙ϕ x i ∂ 0 −x 0 ∂ i) ϕ−x i L the result is 16
∫
Q ij d 3 p
= i
(2π) 3 a+ ⃗p (t)( p i ∂ j −p j ∂ i) a ⃗p (t)
∫
Q 0i = i
d 3 p
(2π) 3 ω ⃗pa + ⃗p (t)∂i a ⃗p (t)
where ∂ i = ∂/∂p i in these formulae.
The reason why these operators are regarded as good candidates for the
Poincarè group generators is that at the classical level their Poisson brackets
obey the corresponding Lie algebra. After the canonical quantization they are
supposed to obey the algebra as well. This, however, needs a check.
The point is that the canonical quantization does not fix the ordering of
terms in products. One is free to choose any ordering, each leading to some
version of the quantized theory. But in general there is no guarantee that a
particular choice of ordering in the 10 generators will preserve the Lie algebra.
Therefore one has to check if his or her choice of ordering did not spoil the
algebra. The alert reader may like to verify the Lie algebra of the Poincarè
group for the generators as given above. The rest of us may just trust the
printed text.
15⃗ P =
∫ d
3 p
(2π) 3 ⃗p
2 (a ⃗p(t)−a + −⃗p (t))(a −⃗p (t)+a + ⃗p (t))
= ∫ d 3 p ⃗p
(2π) 3 2 (a ⃗p (t)a −⃗p (t)−a + −⃗p (t)a −⃗p (t)+a ⃗p (t)a + ⃗p (t)−a+ −⃗p (t)a+ ⃗p (t))
Now both the first and the last term vanish since they are integrals of odd functions, so
⃗P = ∫ d 3 p ⃗p
(2π) 3 2 (a+ ⃗p (t)a ⃗p (t)+a ⃗p (t)a + ⃗p (t)) = ∫ d 3 p
(2π) 3 ⃗p(a + ⃗p (t)a ⃗p (t)+ 1 2 [a ⃗p(t),a + ⃗p (t)])
and the last term again vanishes as a symmetric integral of an odd function.
16 Q ij = ∫ d 3 xd 3 pd 3 p ′
2(2π) 6
√
ω ⃗p /ω ⃗p ′(a ⃗p (t)−a + −⃗p (t))(a ⃗p ′ (t)+a+ −⃗p ′ (t))x j p ′i e i(⃗p+⃗p′ ).⃗x −i ↔ j
We start with a ⃗p a + −⃗p ′ = [a ⃗p ,a + −⃗p ′ ]+a + −⃗p ′ a ⃗p = (2π) 3 δ(⃗p+ ⃗p ′ )+a + −⃗p ′ a ⃗p , where a ⃗p stands for
a ⃗p (t) etc. The term with the δ-function vanishes after trivial integration. Then we write
x j p ′i e i(⃗p+⃗p′ ).⃗x as −ip ′i ∂ ′j e i(⃗p+⃗p′ ).⃗x , after which the d 3 x integration leads to ∂ ′j δ 3 (⃗p+⃗p ′ ) and
then using ∫ dk f (k)∂ k δ(k) = −∂ k f (k)| k=0 one gets
Q ij = −i ∫ √
d 3 p
2(2π) 3 ∂ ′j ω ⃗p /ω ⃗p ′(a + −⃗p ′ a ⃗p −a + −⃗p a ⃗p ′ +a ⃗pa ⃗p ′ −a + −⃗p a+ −⃗p ′ )p ′i | ⃗p ′ =−⃗p +i ↔ j
Now using the Leibniz rule and symetric×antisymmetric cancelations one obtains
Q ij = −i ∫ d 3 p
2(2π) 3 p i ((∂ j a + ⃗p )a ⃗p −a + −⃗p ∂j a −⃗p +a ⃗p ∂ j a −⃗p −a + −⃗p ∂j a + ⃗p )+i ↔ j
At this point one uses per partes integration for the first term, the substitution ⃗p → −⃗p for
the second term, and the commutation relations (and ⃗p → −⃗p ) for the last two terms, to get
Q ij = i ∫ d 3 p
(2π) 3 a + ⃗p (pi ∂ j −p j ∂ i )a ⃗p −i ∫ d 3 p
4(2π) 3 (p i ∂ j −p j ∂ i )(2a + ⃗p a ⃗p +a ⃗p a −⃗p −a + ⃗p a+ −⃗p )
In the second term the dp i or dp j integral is trivial, leading to momenta with some components
infinite. Thisterm wouldbe therefore importantonly ifstates containing particleswith infinite
momenta are allowed in the game. Such states, however, are considered unphysical (having
e.g. infinite energy in the free field case). Therefore the last term can be (has to be) ignored.
(One can even show, that this term is equal to the surface term in the x-space, which was set
to zero in the proof of the Noether’s theorem, so one should set this term to zero as well.)
Boost generators are left as an exercise for the reader.