0.1 Klein-Gordon Equation 0.2 Dirac Equation

More documents

Recommendations

Info

$' \^' I$

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) 14
This Lagrangian leads to solutions a( ⃗ k) = a 1 ( ⃗ k)e −iωt and b( ⃗ k) = b 1 ( ⃗ k)e −iωt where a 1 ( ⃗ k) and b 1 ( ⃗ k) are constants. The free <strong>Dirac</strong> field can also be written as a superposition of the time dependent solutions: ∫ ψ(x) = ˜dk (a s ( ⃗ k)u s ( ⃗ k)e ikx + b † s (⃗ k)v s ( ⃗ ) k)e −ikx with 4-dimensional k and x in analogy to the similar forms for the scalar and electromagnetic fields. The Hamiltonian has the same expression in terms of the a’s and b’s as in the previous time independent expansion, With the aid of the orthogonality relations between u’s and v’s, The expansion of ψ can be inverted to express a s ( ⃗ k) and b † s (⃗ k) in terms of ψ(x): ∫ a s ( ⃗ k) = d 3 xu † s( ⃗ k)e −ikx ψ(x) ∫ b † s (⃗ k) = d 3 xv s † (⃗ k)e ikx ψ(x) 15
Page 1 and 2: [revised Oct 15, 2010] 0.1 Klein-Go
Page 3 and 4: where ψ 0 is a constant 4-componen
Page 5 and 6: and for boosts Σ 0i = − 1 2 ( 0
Page 7 and 8: In higher order, the equivalence of
Page 9 and 10: is conserved. The operator P define
Page 11 and 12: and a similar series for G with coe
Page 13: expressed in terms of the fields. A

0.1 Klein-Gordon Equation 0.2 Dirac Equation

Create successful ePaper yourself

Delete template?

Save as template?