To achieve the desired normalization we pretend space is divided into cubes of volume V = L and impose on the field the periodic boundary condition: A ( x + L , y + L , z + L , t ) = A ( x , y , z , t ) or equivalently ( k x , k y , k z ) = 2 π π L ( n x , n y , n z ) where n can assume any integer value. This allows us to consider the field in any one of the imaginary cubes and to define the mode function: A k ( r ) = 1 V e k e i k ⋅ ⋅ r which satisfies the Helmholtz equation, transversality, and the "box normalization": ∫ ∫ V d 3 r | A k ( r ) | 2 = 1 where e k is chosen to be a unit vector which specifies the polarization of the field mode. The condition k · e k = 0 means that there are two independent choices of e k, which we call e k 1 and e k 2 where e k 1 · e k 2 = 0 and e k 1 = e k 2 = 1. Thus we define the mode functions: A k λ λ ( r ) = 1 V e k λ λ e i k ⋅ ⋅ r , λ λ = { 1 2 in terms of which the vector potential becomes : A k λ λ ( r , t ) = 2 π π ℏ ℏ c 2 ω ω k V [ a k λ λ ( 0 ) e i k ⋅ ⋅ r + a k λ λ † † ( 0 ) e − − i k ⋅ ⋅ r ] e k λ λ or: A k λ λ ( r , t ) = 2 π π ℏ ℏ c 2 ω ω k V [ a k λ λ ( 0 ) e − − i ( ω ω k t − − k ⋅ ⋅ r ) + a k λ λ † † ( 0 ) e i ( ω ω k t − − k ⋅ ⋅ r ) ] where ω k = kc and a k λ, a k λ are photon annihilation and creation operators for the mode with wave vector k and polarization λ. This gives the vector potential for a plane wave mode of the field. The condition for (k x, k y, k z) shows that there are infinitely many such modes. The linearity of Maxwell's equations allows us to write: A ( r t ) = ∑ ∑ k λ λ 2 π π ℏ ℏ c 2 ω ω k V [ a k λ λ ( 0 ) e i k ⋅ ⋅ r + a k λ λ † † ( 0 ) e − − i k ⋅ ⋅ r ] e k λ λ for the total vector potential in free space. Using the fact that: ∫ ∫ V d 3 r A k λ λ ( r ) ⋅ ⋅ A k ′ λ λ ′ ∗ ∗ ( r ) = δ δ k , k ′ 3 δ δ λ λ , λ λ ′ we find the field Hamiltonian is: H F = ∑ ∑ k λ λ ℏ ℏ ω ω k ( a k λ λ † † a k λ λ + 1 2 )