Malcolm John The Dirac Equation C4: particle physics These notes consider the properties of the Dirac equation introducing gamma-matrices, the Dirac current, helicity and chirality. 1 The Klein-Gordon problem Thus for a free particle with a plane wavefunction, = e i(p·x−Et) ∗ = e−i(p·x−Et) We remember the nonrelativistic energy-momentum relation ˙ − i(p·x−Et) for a free particle of mass, m, is = iEe ˙∗ = iEe−i(p·x−Et) E = p2=2m : (1) the probability density and probability current are deduced, With quantum mechanical operators, ρ = 2E ; j = 2p (9) @ E ! i~ ; p ! −i~r ; (2) @t and together in the covariant from: Eq. 1 becomes an operator equation acting on a complex jµ = 2pµ ; where @ jµ = 0 : (10) wavefunction, (x; t), µ @ ~ The problem is revealed when the plane wavefunction is sub- i + r2 = 0 ; (3) @t 2m stituted into Eq. 5 to give the energy eigenvalues, which is the time-dependent Schrödinger equation in the ab- q E = ± jpj2 − m2 : (11) sence of an external potential. We understand ρ = j j2 as the probability density and j j2d3 x as the probability of finding the In addition to the expected positive energy solution are an particle in elemental volume d3 x. equal number of negative energy solutions; and by Eq. 9 these negative energy solutions have negative probability. For the relativistic particle case, one might consider repeat- ing the quantum mechanical operator substitution for the rela- tivistic energy-momentum relation, 2 The Dirac revelation (1928) E2 = jpj2c2 + m2c4 : (4) This gives the Klein Gordon equation, 2 Paul Dirac sought to overcome the problem of negative proba- 2 @ 2 2 2 4 −~ = −~ c r + mc bility in relativistic quantum mechanics by defining an equation @t2 (5) linear in both time and spatial derivatives, @2 h i or = r2 − m ; @t2 @ E = i = (α · p + βm) p = −ir : @t in natural units, ~ = c = 1. It is of interest to form the con- | {z } (12) tinuity equation by multiplying Eq. 5 by −i ∗ and its complex Hamiltonian conjugate by −i , To show this Dirac equation satisfies Klein Gordon (which is 2 2 2 ∗ ∗ ∗ j j −i ¨ = −i r2 + im just E = p + m ), the Hamiltonian is applied twice, (6) ¨∗ 2 ∗ ∗ 2 −i = −i r + im ; @ 2 − = −iαx@x − iαy@y − iαz@z + βm (13) @t2 and subtracting one from another, @2 h α2@2 α2@2 α2@2 β2m2 (14) 2 = x x + y y + z z + i ∗ ¨ − i ¨∗ = i ∗r2 − i r2 ∗ : (7) @t + (αxαy + αyαx)@x@y − m(αxβ + βαx)@x @ h i Using the product rule a∗a˙ − aa˙∗ = a∗a¨ − aa¨∗ (in reverse), + (α α + α α )@ @ − m(α β + βα )@ @x y z z y y z y y y on both time and space parts independently gives the continu- i +(αzαx + αxαz)@z@x − m(αzβ + βαz)@z ity equation for the wavefunction probability, 2 2 requiring of the terms: αi = β = 1; (αiα j + α jαi) = 0; and @ h ∗ ∗i ∗ ∗ i ˙ − i ˙ + r · −i r + i r = 0 : (α β + βα ) = 0. Or more succinctly with β ≡ α , @t i i 0 | {z } | {z } (8) ρ j αµαν + αναµ = 2δµν : (15) Malcolm John - 2020 This condition cannot be satisfied by numbers, αµ must be ma- current and probability density defined as, trices. The matrices are shown to be traceless by, j = yα ρ = y : (19) y Tr(αµ) = Tr(αναµαν) The probability density, y = j j2 + j j2 + j j2 + j j2 is = − Tr(αyα α ) using Eq. 15 1 2 3 4 ν ν µ always positive, as required. Each four-vector component is = − Tr(αµ) = 0 ; generated from the multiplication of a spinor with its transpose conjugate, sometimes through a matrix. In this sense, a spinor where the first equality is proved by, is sometimes described as a “square-root" of a four-vector. y y Tr(Ai j) = Tr(B jk Bki Ai j) = Tr(BkiAi jB jk) |{z} 3 Intrinsic spin = 1 if B is hermitian. This is appropriate as we require energy to be observable (real eigenvalues) so the Hamiltonian in Eq. 12, and thus αi and β, A satisfying aspect of the Dirac equation is that it requires the are hermitian. From the eigenvalue formula, particles it describes to have an intrinsic property that behaves like angular momentum. The proof start by recalling standard αµv = λv quantum mechanics commutations, vy αyα v = λ∗λ vyv µ µ |{z} |{z} [pi; p j] = 0; [xi; x j] = 0; [pi; x j] = −i~δi j: (20) = 1 jλj2 = 1 ) λ = ±1 : Also the commutation relation between Pauli spin matrices, PN But the traceless condition, Tr(αµ) = i λi = 0, requires the [σ ; σ ] = 2i σ : (21) dimension of αµ to be even. For N=2 Pauli spin matrices, i j i jk k ! ! ! 0 1 0 −i 1 0 We require angular momentum to commute with the Hamil- σ = σ = σ = (16) x 1 0 y i 0 z 0 −1 tonian because angular momentum quantum states are well established to have define states of definite energies. So, satisfy Eq. 15 but we need four such matrices (αx; αy; αz; β) so N = 4 is required. Specifically the set of four hermitian, trace- [H; L] = [α · p + βm; r × p]: (22) less 4 × 4 matrices define the Dirac representation, The βm term commutes trivially, and for the rest, we exemplify ! ! I 0 0 σi the x component, β = αi = (17) 0 −1 σi 0 h i [α · p; Lx] = αx px + αy py + αz pz; ypz − zpy : (23) where σi are the Pauli spin matrices, satisfies Eq. 15. This is not unique, the Weyl representation is also valid, see Sec. 8. The only non-zero parts are terms of the form [xi; pi] = −i~, [H; Lx] = αy pz[py; y] − αz py[pz; z] = −i~(αy pz − αz py) The 4 × 4 structure of the Dirac equation give four solu- = −i~(α ^ p) tions, . To check if this first-order equation indeed resolves x the problems of Klein Gordon, the probability density and cur- ) [H; L] = −i~ α ^ p , 0 ! (24) rent is calculated in a similar manner as in Eq. 6. However, The Dirac equation appears not to commute with angular mo- as this equation has matrices, the Hermitian conjugate is used mentum, which cannot be a correct description of fermion rather than just the complex conjugate. states with definite angular momentum having definite energy. @ @ y i y − (−i) = @t @t y This is solved by connecting to SU(2), which described fun- −iαx@x − iαy@y − iαz@z + mβψ 1 damental spin- 2 objects (spinors), the rotations of which are y y y y − i@x αx + i@y αy + i@z αz + m β : generated by the Pauli spin matrices. Here we have 4 × 4 so try, ! then using the product rule on each term, σ 0 S = 1 ~ ; (25) 2 0 σ @ @ y X y + + yα @ + @ yα = 0 @t @t i i i i (σ are Pauli spin matrices). Setting ~ = 1, S x; S y; S z are: | {z } i | {z } 00 1 1 00 −i 1 01 0 1 B C B C B C @ y y 1 B1 0 C 1 Bi 0 C 1 B0 −1 C + r · α = 0 (18) = B C ; B C ; B C : @t 2 B 0 1C 2 B 0 −iC 2 B 1 0 C B C B C B C @ 1 0A @ i 0 A @ 0 −1A ∂ρ which is of the form = −r · j with the Dirac probability @t The commutation with the Hamiltonian proceeds as above: Malcolm John - 2020 1 identify the non-trivial part and exemplify the x component. mechanical angular momentum with S = 2 ~, [H; S] = [α · p + βm; S] S 2 = S (S + 1) ; |{z} (29) 1 ! ! ! ! S = ~ ; I 0 σ 0 σ I z 2 1 − 1 = 0 : 2 −I σ 0 2 σ 0 −I and it is intrinsic to the Dirac Hamiltonian. = [α · p; S] e.g. ) [H; S x] = [αx px + αy py + αz pz; S x] 4 Gamma matrices = px[αx; S x] + py[αy; S x] + pz[αz; S x] : Dealing with each of the three terms as well as employing the We continue by rearranging the Dirac equation of Eq. 12 and commutation relation for Pauli spin matrices, multiplying by the matrix β, ! ! ! ! 1 0 σx σx 0 1 σx 0 0 σx [αx; S x] = 2 − 2 " # σx 0 0 σx 0 σx σx 0 @ 0 1 0 1 β i + iα · r − βm = 0 0 σ2 0 σ2 1 B xC − 1 B xC @t = 2 @B 2 AC 2 @B 2 AC = 0 : σx 0 σx 0 " # 2 iβ@t + i βαx @x + i βαy @y + i βαz @z − β m = ! ! ! ! # |{z} |{z} |{z} # 1 0 σy σx 0 1 σx 0 0 σy [αy; S x] = 2 − 2 0 1 2 3 σy 0 0 σx 0 σx σy 0 γ γ γ γ I ! ! 1 0 σyσx 1 0 σxσy = 2 − 2 σyσx 0 σxσy 0 to arrive at a manifestly covariant form of the Dirac equation, ! 1 0 [σyσx] = 2 h i [σyσx] 0 µ iγ @µ − mI = 0 (30) ! 1 0 −2iσz = 2 = −iαz : −2iσz 0 where the identity matrix I is shown to make it explicit this is a ! matrix equation, acting on the four-component Dirac spinor. 1 0 [σzσx] [αz; S x] = 2 (by the same reasoning) [σzσx] 0 ! 1 0 +2iσy = 2 = iαy : +2iσy 0 It is important to understand that, though written with a µ super- script, γµ are not fourvectors.