Computational Physics

470 Computational methods for lattice field theories

with
[ˆπn,φˆl]=−iδnl. (15.15)

The hats are put aboveφandπto emphasise that they are now operators. The
Hamiltonian can be diagonalised by first Fourier transforming and then applying
operator methods familiar from ordinary quantum mechanics to it. The result is [ 2 , 6 ]

H=

1

2

∫π

−π

dkωkaˆ†kaˆk (15.16)

whereaˆ†kis a creation operator: it creates a Fourier mode

φn=eikn (15.17)

andaˆkis the corresponding destruction or annihilation operator. In the ground state
(the ‘vacuum’) there are no modes present and the annihilation operator acting on
the ground state gives zero:
ak| 0 〉=0. (15.18)

The Fourier modes represent energy quanta of energy
ω; thenumber operator
nk=a†kakacting on a state|ψ〉counts the number of modes (quanta) with wave
vectork, present in that state. The Hamiltonian(15.16)operator then adds all the
energy quanta which are present in the state.
In fact,aˆkis given in terms of the Fourier transforms of theφˆandπˆoperators:

aˆk=

1

√

4 πωk

[ωkφˆk+iπˆk], (15.19)

analogous to the definition of creation and annihilation operators for the harmonic
oscillator. The frequencyωis related tokby

ωk=ω−k=

√

m^2 + 2 ( 1 −cosk). (15.20)

For smallkwe find the continuum limit:

ωk=

√

m^2 +k^2 (15.21)

which is the correct dispersion relation for a relativistic particle with massm(in units
wherec=
=1).
We see that the Hamiltonian(15.16)of a particular configuration is simply given
as the sum of the energies of a set of one-particle Hamiltonians (remember these
particles are nothing but Fourier-mode excitations of the field): the particles do not
interact. Therefore, the field theory considered so far is calledfree field theory. The
eigenstates of the free field theory are

|k 1 ,...,kM〉=ˆa†k 1 ...aˆ†kM| 0 〉 (15.22)