0.1 Klein-Gordon Equation 0.2 Dirac Equation

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i.e., α and β i are a set of 4 anticommuting matrices whose square is unity. In terms of derivatives the Dirac equation is: i∂ 0 ψ = −iβ i ∂ i ψ + αmψ To give it a covariant appearence (deceptive because the γ’s are numerical matrices that do not change under LT, but see below), introduce γ i = αβ i and γ 0 = α, multiply the equation by α and obtain The γ µ satisfy −iγ µ ∂ µ ψ + mψ = 0 (2) {γ µ , γ ν } = −2g µν (3) where {} means anticommutator. A possible explicit form is ( ) ( ) 1 0 0 σ γ 0 = γ i i = 0 −1 −σ i 0 Each entry is a 2 by 2 matrix, making γ 4 by 4. An equivalent explicit form is ( ) ( ) 0 1 0 σ γ 0 = γ i i = 1 0 −σ i (5) 0 The first form is the standard form, good for taking the NR limit. The second form is the chiral form, good for neutrinos and things. Changing involves reshuffling components of ψ. Both forms can be represented as direct products of two sets of Pauli matrices τ i for the skeleton and σ i for the entries: the first set is (τ 3 , iτ 2 σ i ) and the second is (τ 1 , iτ 2 σ i ). Both forms (and others you can easily invent) satisfy Eq.(3). In both forms, the hermitian conjugate of γ 0 is γ 0 and of γ i is −γ i . A very useful relation valid for both forms is γ µ† = γ 0 γ µ γ 0 . 0.3 Free Particle Solutions Look for solution in the form of an eigenstate of the energy-momentum vector: ψ = ψ 0 e ipx , (4) 2
where ψ 0 is a constant 4-component column vector (Dirac spinor). ψ 0 must satisfy (γ µ p µ + m)ψ 0 = 0. If ψ 0 has upper components χ and lower components χ ′ (both ordinary 2-component spinors), then the equation takes the form ( ) ( ) −E + m ⃗σ · ⃗p χ −⃗σ · ⃗p E + m χ ′ = 0 (6) using the standard form for the γ-matrices and replacing p 0 by E. There is a non-trivial solution if the determinant is zero. To evaluate things with σ use the identity σ i σ j = δ ij + iɛ ijk σ k . The determinant is (−(E) 2 + ⃗p 2 + m 2 ) 2 which has double roots at E = ± ω with ω = + √ ⃗p 2 + m 2 , implying two positive energy states and two negative energy states for each value of ⃗ k. Negative energy states are still present, but Dirac eliminated them by saying that they were full, so the exclusion principle forbade transitions to them. A hole in the negative energy ’sea’ behaves like positive particle of the same mass: positron. DISCUSS: motion of holes. A solution of Eq.(6) is obtained by taking χ to be an arbitrary twocomponent spinor. The equation then requires that χ ′ = ⃗σ · ⃗pχ/(E + m). Note that the equation implies that χ ′ ≈ vχ in the NR limit. (χ called large components and χ ′ small components.) We will return to these states when we treat the Dirac equation as a field equation. The physical meaning of states and operators clearer if we multiply the Dirac equation by γ 0 and identify p 0 with the Hamiltonian: H = p 0 = γ 0 γ i p i + γ 0 m, where p i can be interpreted either as the momentum operator or its eigenvalue. Clearly the spatial momentum commutes with H, but what about the angular momentum? Calculation shows that the orbital angular momentum ⃗r × ⃗p does not commute with H. By inspecting the commutator, we can find that adding a term gives an operator that does commute with H: J i = ɛ ijk r j p k + i 4 ɛ ijkγ j γ k The explicit form of the second term is ( ) S i = 1 σi 0 2 0 σ i 3
Page 1: [revised Oct 15, 2010] 0.1 Klein-Go
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 and 14: expressed in terms of the fields. A
Page 15: This Lagrangian leads to solutions

0.1 Klein-Gordon Equation 0.2 Dirac Equation

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