It will be assumed throughout that this state is unique and normalized as follows,
‖|∅〉λ=〈∅|∅〉= 1 (19.55)As an immediate result of the expressions for the hamiltonian and momentum operators (19.51),
it follows that the ground state has zero energy and zero momentum,
H|∅〉=P|∅〉= 0 (19.56)A different convention for the ordering of the oscillators inHmight have resulted in a non-zero
energy of the ground state.
19.6.2 One-photon states
Applying any raising operatora†α(k) to the vacuum will raise the energy above zero. We define
these states as follows,
|k,α〉≡a†α(k)|∅〉 (19.57)The total energy and momentum of such states may be computed by applying the operatorsHand
Pthereto, and we find,
H|k,α〉=∫ d (^3) k′
(2π)^3
|k′|
∑
β
a†β(k′)[aβ(k′),a†α(k)]|∅〉 (19.58)
where we have used the relationaβ(k′)|∅〉 = 0 in exhibiting the commutator. Using now the
canonical commutator, we find,
H|k,α〉 = |k||k,α〉
P|k,α〉 = k|k,α〉 (19.59)
The normalization of the one-particle states is that of the continuum, as is suitable for states
indexed by the continuous momentumk,
〈k,α|k′,β〉= (2π)^3 δ(3)(k′−k)δαβ (19.60)
Given the fact that the state|k,α〉has total energyE= ̄h|k|and momentump= ̄hk, its relativistic
invariant mass vanishes. We interpret these states as containing asingle photon, of energy ̄h|k|and
momentum ̄hk. We shall see below what the nature is of states with higher numbers of photons.
They cannot, in general, have vanishing mass.
19.6.3 Multi-photon states
Applying now several raising operators to| 0 〉, we obtain the following types of states,
|ψ〉=|k 1 ,α 1 ;···;kn,αn〉≡a†α 1 (k 1 )···a†αn(kn)|∅〉 (19.61)