The Dirac Equation May 18, 2015 0.1 Introduction In non-relativistic quantum mechanics, wave functions are descibed by the time-dependent Schrodinger equation : 1 @ − r2 + V = i (1) 2m @t p2 This is really just energy conservation ( kinetic energy ( 2m ) plus potential energy (V) equals total energy (E)) with the momentum and energy terms replaced by their operator equivalents @ p ! −ir; E ! i (2) @t In relativistic quantum theory, the energy-momentum conservation equation is E2 −p2 = m2 (note that we are working in the standard particle physics units whereh ¯ = c = 1). Proceeding with the same replacements, we can derive the Klein-Gordon equation : @2 E2 − p2 − m2 ! − + r2 − m2 = 0 (3) @t2 In covariant notation this is µ 2 −@ @µ − m = 0 (4) Suppose is solution to the Klein-Gordon equation. Multiplying it by −i ∗ we get @2 i ∗ − i ∗r2 + i ∗m2 = 0 (5) @t2 Taking the complex conjugate of the Klein-Gordon equation and multiplying by −i we get @2 ∗ i − i r2 ∗ + i m2 = 0 (6) @t2 If we subtract the second from the first we obtain @ @ @ ∗ i ∗ − + r · [−i ( ∗r − r ∗)] = 0 (7) @t @t @t This has the form of an equation of continuity @ρ + r · j = 0 (8) @t with a probability density defined by @ @ ∗ ρ = i ∗ − (9) @t @t and a probability density current defined by j = i ( ∗r − r ∗) (10) Now, suppose a solution to the Klein-Gordon equation is a free particle with energy E and mo- mentum p µ = Ne−ipµx (11) 1 Substitution of this solution into the equation for the probability density yields ρ = 2EjNj2 (12) The probability density is proportional to the energy of the particle. Now, why is this a problem? If you substitute the free particle solution into the Klein-Gordon equation you get, unsurpisingly, the relation E2 − p2 = m2 (13) so the energy of the particle could be p E = ± p2 + m2 (14) The fact that you have negative energy solutions is not that much of a problem. What is a problem is that the probability density is proportional to E. This implies a possibility for negative probability densities...which don't exist. This (and some others) problem drove Dirac to think about another equation of motion. His starting point was to try to factorise the energy momentum relation. In covariant formalism 2 2 2 µ 2 E − p − m ! p pµ − m (15) µ where p is the 4-momentum : (E; px; py; pz). Dirac tried to write µ 2 κ λ p pµ − m = (β pκ + m)(γ pλ − m) (16) where κ and λ range from 0 to 3. This notation looks a bit confusing. It may help if we write both sides out long-hand. On the left-hand side we have µ 2 2 2 2 2 2 2 p pµ − m = p0 − p · p = p0 − p1 − p2 − p3 − m (17) On the right-hand side we have κ λ 0 1 2 3 0 1 2 3 (β pκ + m)(γ pλ − m) = (β p0 − β p1 − β p2 − β p3 + m)(γ p0 − γ p1 − γ p2 − γ p3 − m) (18) Expanding the right-hand side κ λ κ λ 2 λ κ (β pκ + m)(γ pλ − m) = β γ pκpλ − m + mγ pλ − mβ pκ (19) µ 2 This should be equal to p pµ − m , so we need to get rid of the terms that are linear in p. We can do this just be choosing βκ = γκ. Then µ 2 κ λ 2 p pµ − m = γ γ pκpλ − m (20) Now, since κ, λ runs from 0 to 3, the right hand side can be fully expressed by κ λ 2 0 2 2 1 2 2 2 2 2 3 2 2 γ γ pκpλ − m = (γ ) p0 + (γ ) p1 + (γ ) p2 + (γ ) p3 (21) 0 1 1 0 0 2 2 0 + (γ γ + γ γ )p0p1 + (γ γ + γ γ )p0p2 (22) 0 3 3 0 1 2 2 1 + (γ γ + γ γ )p0p3 + (γ γ + γ γ )p1p2 (23) 1 3 3 1 2 3 3 2 2 + (γ γ + γ γ )p1p3 + (γ γ + γ γ )p2p3 − m (24) 2 This must be equal to µ 2 2 2 2 2 2 pµp − m = p0 − p1 − p2 − p3 − m (25) One could make the squared terms equal by choosing (γ0)2 = 1 and (γ1)2 = (γ2)2 = (γ3)2=-1, but this would not eliminate the cross-terms. Dirac realised that what you needed was something which anticommuted : i.e. γµγν + γνγµ = 0 if µ 6= ν. He realised that the γ factors must be 4x4 matrices, not just numbers, which satisfied the anticommutation relation fγµ; γνg = 2gµν (26) where 01 0 0 0 1 µν B0 −1 0 0 C g = B C (27) @0 0 −1 0 A 0 0 0 −1 We also define, for later use, another γ matrix called γ5 = iγ0γ1γ2γ3. This has the properties (you should check for yourself) that (γ5)2 = 1; fγ5; γµg = 0 (28) We will want to take the Hermitian conjugate of these matrices at various times. The Hermitian conjugate of each matrix is γ0y = γ0 γ5y = γ5 γµy = γ0γµγ0 = −γ for µ 6= 0 (29) If the matrices satisfy this algebra, then you can factor the energy momentum conservation equa- tion µ 2 κ λ pµp − m = (γ pκ + m)(γ pλ − m) = 0 (30) The Dirac equation is one of the two factors, and is conventionally taken to be λ γ pλ − m = 0 (31) Making the standard substitution, pµ ! i@µ we then have the usual covariant form of the Dirac equation µ (iγ @µ − m) = 0 (32) @ @ @ @ where @µ = ( @t ; @x ; @y ; @z ), m is the particle mass and the γ matrices are a set of 4-dimensional matrices. Since these are matrices, is a 4-element column matrix called a \bi-spinor". This bi-spinor is not a 4-vector and doesn't transform like one. 0.2 The Bjorken-Drell convention Any set of four 4x4 matrices that obey the algebra above will work. The one we normally use includes the Pauli spin matrices. Recall that each component of the spin operator S for spin 1/2 particles is defined by its own Pauli spin matrix : 1 1 0 1 1 1 0 −i 1 1 1 0 S = σ = S = σ = S = σ = (33) x 2 1 2 1 0 y 2 2 2 i 0 z 2 3 2 0 −1 3 In terms of the Pauli spin matrices, the γ matrices have the form 1 0 0 σµ 0 1 γ0 = γµ = γ5 = (34) 0 −1 −σµ 0 1 0 Each element of the matrices in Equations 34 are 2x2 matrices. 1 denotes the 2 x 2 unit matrix, and 0 denotes the 2 x 2 null matrix. 0.3 The Probability Density and Current In order to understand the probability density and probability flow we will want to derive an equation of continuity for the probability. The first step is to write the Dirac equation out longhand : @ @ @ @ iγ0 + iγ1 + iγ2 + iγ3 − m = 0 (35) @t @x @y @z We want to take the Hermitian conjugate of this : @ @ @ @ [iγ0 + iγ1 + iγ2 + iγ3 − m ]y (36) @t @x @y @z Now, we must remember that the γ are matrices and that is a 4-component column vector. Thus 0 1 0 @ 1 1 1 0 0 0 @t @ 0 1 0 0 @ 2 [γ0 ]y = [B C B @t C]y (37) B0 0 −1 0 C B @ 3 C @t @ A @ @t A 0 0 0 −1 @ 4 @t y 0 @ 1 1 0 1y @t 1 0 0 0 @ 2 0 1 0 0 = B @t C B C (38) B @ 3 C B0 0 −1 0 C @ @t A @ A @ 4 0 0 0 −1 @t 01 0 0 0 1 @ y @ y @ y @ y B0 1 0 0 C = 1 2 3 4 B C (39) @t @t @t @t @0 0 −1 0 A 0 0 0 −1 @ y = γ0 (40) @t Using the Hermitian conjugation properties of the γ matrices defined in the previous section we can write Equation 36 as @ y @ y @ y @ y −i γ0y − i γ1y − i γ2y − i γ3y − m y (41) @t @x @y @z which, as γµy = −γµ for µ 6= 0 means we can write this as @ y @ y @ y @ y −i γ0 − i (−γ1) − i (−γ2) − i (−γ3) − m y (42) @t @x @y @z 4 We have a problem now - this is no longer covariant. That is, I would like to write this in the form y µ (−i@µγ − m) where the derivative @µ now operates to the left. I can't because the minus signs on the spatial components coming from the −γµ spoils the scalar product. To deal with this we have to multiply the equation on the right by γ0, since we can flip the sign of the γ matrices using the relationship −γµγ0 = γ0γµ. Thus @ y @ y @ y @ y −i γ0γ0 − i (−γ1γ0) − i (−γ2γ0) − i (−γ3γ0) − m yγ0 (43) @t @x @y @z or @ y @ y @ y @ y −i γ0γ0 − i (γ0γ1) − i (γ0γ2) − i (γ0γ3) − m yγ0 (44) @t @x @y @z Defining the adjoint spinor = yγ0, we can rewrite this equation as @ @ @ @ −i γ0 − i (γ1) − i (γ2) − i (γ3) − m (45) @t @x @y @z and finally we get the adjoint Dirac equation µ (i@µγ + m) = 0 (46) Now, we multiply the Dirac equation on the left by : µ (iγ @µ − m) = 0 (47) and the adjoint Dirac equation on the right by : µ (i@µγ + m) = 0 (48) and we add them together, which eliminates the two mass terms µ µ (γ @µ ) + ( @µγ ) = 0 (49) or, more succinctly, µ @µ( γ ) = 0 (50) .