From Classical Mechanics to Quantum Field Theory

A Concise Introduction to Quantum Field Theory 195

The Hamiltonian is given by

Hˆ=^1

2 m

(

ˆp^2 +m^2 ω^2 xˆ^2

)

,

which in terms of creation and annihilation operators (see Example 1.2.4 of the
first part)

a=

√

mω

2

ˆx+

i

√

2 mω

ˆp; a†=

√

mω

2

xˆ−

i

√

2 mω

pˆ

reads as

Hˆ=ω

(

a†a+

1

2

)

.

Now, because of the commutation relations

[H,aˆ †]=ωa† [H,aˆ ]=−ωa [a, a†]=1,

we have that we can associate to any eigenstate|E〉of the HamiltonianHˆ,Hˆ|E〉=
E|E〉, two more eigenstatesa†|E〉anda|E〉with eigenvaluesE+ωandE−ω,
respectively. Indeed,

Haˆ †|E〉=a†Hˆ|E〉+ωa†|E〉=(E+ω)a†|E〉,

Haˆ |E〉=aHˆ|E〉−ωa|E〉=(E−ω)a|E〉.

The quantum HamiltonianHˆis a positive operator since for any physical state
|ψ〉∈H,〈ψ|Hˆ|ψ〉=ω〈ψ|a†a+^12 I|ψ〉=ω(||a|ψ〉||^2 +^12 |||ψ〉||^2 )≥0. However, the
ladder of quantum statesan|E〉generated by any eigenvalue|E〉ofHˆwill contain
negative energy states unless before reaching negative eigenvalues the state of the
ladder vanishes, i.e. an^0 |E〉= 0. Thus, the only possibility which is compatible
with the positivity of the HamiltonianHˆis the existence of a final ground state
such thata|E 0 〉= 0. But then, the energy of this ground stateH|E 0 〉=^12 ω|E 0 〉
is non-vanishing, unlike the energy of the classical vacuum configuration which
vanishes. The non-trivial value of the quantum vacuum energy has remarkable
consequences for the physics of the vacuum in the quantum field theory.
We shall denote from now on the ground state|E 0 〉by| 0 〉. Higher energy states
are obtained by applying the creation operatora†to the ground state| 0 〉:

|n〉=

√^1

(n+1)!

(a†)n| 0 〉; H|n〉=

(

n+

1

2

)

ω|n〉 (3.8)

for any positive integern∈N.Thestate|n〉has unit norm, i.e.‖|n〉‖^2 =1and
satisfies that

a|n〉=

√

n|n− 1 〉.