0.1 Klein-Gordon Equation 0.2 Dirac Equation

energy state of momentum ⃗ k has momentum − ⃗ k, we change to − ⃗ k in the
integral over ˜b. Then we redefine ṽ s (− ⃗ k) to be v s ( ⃗ k) and ˜b s (− ⃗ k) to b † s (⃗ k). The
latter takes account of the fact that after quantization a and a † annihilate
and create particles respectively. Annihilation of a particle in a negative
energy state creates an antiparticle there. so the roles of the operator and
its conjugate are interchanged. The expansion for ψ now looks like
∫ (
ψ(x) = a s ( ⃗ k)u s ( ⃗ k)e i⃗k·⃗x + b † s( ⃗ k)v s ( ⃗ )
k)e −i⃗ k·⃗x
(18)
˜dk ∑ s
It is easy to substitute this expansion into the Hamiltonian to obtain
∫ (
H = ω a † s (⃗ k)a s ( ⃗ k) − b s ( ⃗ )
k)b † s (⃗ k)
˜dk ∑ s
(19)
using the fact that u s and v s are eigenstates of H D that satisfy the above
orthonormality relations. It should be possible to prove that the Lagrangian
has a form something like
∫ ( i
)
L = ˜dk (a † ȧ + bḃ† − ȧ † a − ḃb† ) − 1
2
2ω(a † a + bb † ) .
Perhaps you can prove it with the aid of the following identities involving the
solutions u s ( ⃗ k) and v s ( ⃗ k), which we also list for future reference. Involving
bars:
Involving ± ⃗ k:
u s ( ⃗ k)u s ′( ⃗ k) = 2mδ ss ′
v s ( ⃗ k)v s ′( ⃗ k) = −2mδ ss ′
u s ( ⃗ k)v s ′( ⃗ k) = 0
Involving γ i k i :
u † s (⃗ k)v s ′(− ⃗ k) = 0
v † s (⃗ k)u s ′(− ⃗ k) = 0 (20)
u s ( ⃗ k)γ i k i u s ′( ⃗ k) = 2 ⃗ k 2 δ ss ′
v s ( ⃗ k)γ i k i v s ′( ⃗ k) = 2 ⃗ k 2 δ ss ′
u s ( ⃗ k)γ i k i v s ′( ⃗ k) = 2ω√
⃗k 2
δ ss ′
(21)
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