Quantum Field Theory I

60 CHAPTER 2. FREE SCALAR QUANTUM FIELD
The ”miracle”is not overyet. The free field haveturned out to be equivalent
to the system of independent oscillators, and this system will now turn out to
be equivalent to still another system, namely to the system of free non-teracting
relativistic particles. Indeed, the free Hamiltonian written in terms of a ⃗p (t) and
a + ⃗p (t) becomes14 ∫
H =
∫
=
( 1
d 3 x
2 π2 + 1 2 |∇ϕ|2 + 1 )
2 m2 ϕ 2
d 3 (
p
(2π) 3 ω ⃗p a + ⃗p (t)a ⃗p(t)+ 1 ] [a )
⃗p (t),a + ⃗p
2
(t)
wherethelasttermisaninfiniteconstant(since[a ⃗p (t),a + ⃗p (t)] = (2π)3 δ 3 (⃗p−⃗p)).
This is ourfirst example of the famous (infinite) QFT skeletonsin the cupboard.
This one is relatively easy to get rid of (to hide it away) simply by subtracting
the appropriate constant from the overall energy, which sounds as a legal step.
Another way leading to the same result is to realize that the canonical quantization
does not fix the ordering in products of operators. One can obtain
different orderings at the quantum level (where the ordering does matter) starting
from the different orderings at clasical level (where it does not). One may
therefore choose any of the equivalent orderings at the classical level to get the
desired ordering at the quantum level. Then one can postulate that the correct
ordering is the one leading to the decent Hamiltonian. Anyway, the standard
form of the free scalar field Hamiltonian in terms of creation and annihilation
operators is
∫
d 3 p
H =
(2π) 3 ω ⃗p a + ⃗p (t)a ⃗p(t)
This looks pretty familiar. Was it not for the explicit time dependence of the
creation and annihilation operators, this would be the Hamiltonian of the ideal
gas of free relativistic particles (relativistic because of the relativistic energy
ω ⃗p = √ ⃗p 2 +m 2 ). The explicit time dependence of the operators, however, is
not an issue — the hamiltonian is in fact time-independent, as we shall see
shortly (the point is that the time dependence of the creation and annihilation
operators turns out to be a + ⃗p (t) = a+ ⃗p eiω ⃗pt and a ⃗p (t) = a ⃗p e −iω ⃗pt respectively).
Still, it is not the proper Hamiltonian yet, since it has nothing to act on.
But once we hand over an appropriate Hilbert space, it will indeed become the
old friend.
14 The result is based on the following algebraic manipulations
∫ d 3 x π 2 = − ∫ √
d 3 xd 3 pd 3 p ′ ω⃗p ω )( )
⃗p ′
(a
(2π) 6 2 ⃗p (t)−a + −⃗p (t) a ⃗p ′ (t)−a + −⃗p ′ (t) e i(⃗p+⃗p′ ).⃗x
= − ∫ √
d 3 pd 3 p ′ ω⃗p ω )( )
⃗p ′
(a
(2π) 3 2 ⃗p (t)−a + −⃗p (t) a ⃗p ′ (t)−a + −⃗p ′ (t) δ 3 (⃗p+⃗p ′ )
= − ∫ )
d 3 p ω ⃗p
(a
(2π) 3 2 ⃗p (t)−a
)(a + −⃗p (t) −⃗p (t)−a + ⃗p (t)
∫
d 3 x |∇ϕ| 2 +m 2 ϕ 2 = ∫ d 3 xd 3 pd 3 p ′ −⃗p.⃗p ′ +m 2
(2π) 6 2 √ ω ⃗p ω ⃗p ′
= ∫ d 3 p ω ⃗p
(2π) 3 2
)( )
(a ⃗p (t)+a + −⃗p (t) a ⃗p ′ (t)+a + −⃗p ′ (t) e i(⃗p+⃗p′ ).⃗x
(a ⃗p (t)+a + −⃗p (t) )(a −⃗p (t)+a + ⃗p (t) )
where in the last line we have used (⃗p 2 +m 2 )/ω ⃗p = ω ⃗p .
Putting everything together, one obtains the result.