4 Dirac Equation. and α k, β are N N matrices. Using the matrix notation, we can write the equations as imc

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1 4 Dirac Equation To solve the negative probability density problem of the Klein-Gordon equation, people were looking for an equation which is first order in / t. Such an equation is found by Dirac. It is difficult to take the square root of 2 c m 2 c 4 for a single wave function. One can take the inspiration from E&M: Maxwell s equations are first-order but combining them gives the second order wave equations. Imagining that ψ consists of N components ψ l, 1 ψ l c t + 3 N k=1 n=1 αln k ψ n x + imc k N β ln ψ n =, 53 where l = 1, 2,..., N, and x k = x, y, z, k = 1, 2, 3. ψ 1 ψ 2 ψ =., 54 and α k, β are N N matrices. Using the matrix notation, we can write the equations as 1 ψ imc + α ψ + βψ =, 55 c t where α = α 1 ˆx+α 2 ŷ +α 3 ẑ. N components of ψ describe a new degree of freedom just as the components of the Maxwell field describe the polarization of the light quantum. In this case, the new degree of freedom is the spin of the particle and ψ is called a spinor. We would like to have positive-definite and conserved probability, ρ = ψ ψ, where ψ is the hermitian conjugate of ψ so is a row matrix. Taking the hermitian conjugate of Eq. 55, ψ N n=1 1 ψ c t + ψ α imc ψ β =. 56 Multiplying the above equation by ψ and then adding it to ψ 55, we obtain 1 ψ ψ c t + ψ t ψ + ψ α ψ + ψ α ψ + imc ψ βψ ψ β ψ =. 57 The continuity equation t ψ ψ + j = 58 8

2 can be obtained if α = α, β = β, then with From Eq. 55 we can obtain the Hamiltonian, 1 c t ψ ψ + ψ αψ = 59 Hψ = i ψ t = j = cψ αψ. 6 c i + βmc2 One can see that H is hermitian if α, β are hermitian. ψ. 61 To derive properties of α, β, we multiply Eq. 55 by the conjugate operator, 1 c t α imc 1 β c t + α + imc β ψ = [ 1 2 c 2 t 2 αi α j j + m2 c 2 β 2 imc ] 2 βαi + α i β i ψ = 62 We can rewrite α i α j j as 1 2 αi α j + α j α j. Since it s a relativistic system, the second order equation should coincide with the Klein-Gordon equation. Therefore, we must have Because if we take the determinant of the above equation, α i α j + α j α i = 2δ ij I 63 βα i + α i β = 64 β 2 = I 65 βα i = α i β = Iα i β, 66 det β det α i = 1 N det α i det β, 67 we find that N must be even. Next, we can rewrite the relation as Taking the trace, α i 1 βα i = β no summation. 68 Tr [ α i 1 βα i] = Tr [ α i α i 1 β ] = Tr[β] = Tr[ β], 69 we obtain Tr[β] =. Similarly, one can derive Tr[α i ] =. 9

3 Covariant form of the Dirac equation Define γ = β, γ j = βα j, j = 1, 2, 3 Multiply Eq. 55 by iβ, 1 iβ c t + α + imc iγ x + iγj x mc j γ µ = γ, γ 1, γ 2, γ 3, γ µ = g µν γ ν 7 β ψ = ψ = iγ µ µ mc ψ = 71 Using the short-hand notation: γ µ µ, γ µ A µ A, i mc ψ = 72 From the properties of the α j and β matrices, we can derive γ = γ, hermitian 73 γ j = βα j = α j β = α j β = βα j = γ j, anti-hermitian74 γ µ = γ γ µ γ, 75 γ µ γ ν + γ ν γ µ = 2g µν I. Clifford algebra. 76 Conjugate of the Dirac equation is given by i µ ψ γ µ mc ψ = i µ ψ γ γ µ γ mc ψ = 77 We will define the Dirac adjoint spinor ψ by ψ ψ γ. Then The four-current is i µ ψγ µ + mc ψ =. 78 j µ c = ψγµ ψ = ρ, j, µ j µ =. 79 c 1

4 Properties of the γ µ matrices We may form new matrices by multiplying γ matrices together. Because different γ matrices anticommute, we only need to consider products of different γ s and the order is not important. We can combine them in ways. Plus the identity we have 16 different matrices, I γ, iγ 1, iγ 2, iγ 3 γ γ 1, γ γ 2, γ γ 3, iγ 2 γ 3, iγ 3 γ 1, iγ 1 γ 2 iγ γ 2 γ 3, iγ γ 3 γ 1, iγ γ 1 γ 2, γ 1 γ 2 γ 3 iγ γ 1 γ 2 γ 3 γ 5 = γ 5. 8 Denoting them by Γ l, l = 1, 2,, 16, we can derive the following relations. a Γ l Γ m = a lm Γ n, a lm = ±1 or ± i. b Γ l Γ m = I if and only if l = m. c Γ l Γ m = ±Γ m Γ l. d If Γ l I, there always exists a Γ k, such that Γ k Γ l Γ k = Γ l. e TrΓ l = for Γ l I. Proof: Tr Γ l = TrΓ k Γ l Γ k = TrΓ l Γ k Γ k = TrΓ l. f Γ l are linearly independent: 16 k=1 x kγ k = only if x k =, k = 1, 2,, 16. Proof: 16 x k Γ k Γ m = x m I + x k Γ k Γ m = x m I + x k a km Γ n = Γ n I. k=1 k m k m Taking the trace, x m TrI = k m x ka km TrΓ n = x m =. for any m. This implies that Γ k s cannot be represented by matrices smaller than 4 4. In fact, the smallest representations of Γ k s are 4 4 matrices. Note that this 4 is not the dimension of the space-time. the equality is accidental. g Corollary: any 4 4 matrix X can be written uniquely as a linear combination of the Γ k s. X = 16 k=1 x k Γ k TrXΓ m = x m TrΓ m Γ m + k m x k TrΓ k Γ m = x m TrI = 4x m x m = 1 4 TrxΓ m 11

5 h Stronger corollary: Γ l Γ m = a lm Γ n where Γ n is a different Γ n for each m, given a fixed l. Proof: If it were not true and one can find two different Γ m, Γ m such that Γ l Γ m = a lm Γ n, Γ l Γ m = a lm Γ n, then we have Γ m = a lm Γ l Γ n, Γ m = a lm Γ l Γ n which contradicts that γ k s are linearly independent. Γ m = a lm Γ m, a lm i Any matrix X that commutes with γ µ for all µ is a multiple of the identity. Proof: Assume X is not a multiple of the identity. If X commutes with all γ µ then it commutes with all Γ l s, i.e., X = Γ l XΓ l. We can express X in terms of the Γ matrices, X = x m Γ m + k m x k Γ k, Γ m I. There exists a Γ i such that Γ i Γ m Γ i = Γ m. By the hypothesis that X commutes with this Γ i, we have X = x m Γ m + k m x k Γ k = Γ i XΓ i = x m Γ i Γ m Γ i + k m x k Γ i Γ k Γ i = x m Γ m + k m ±x k Γ k. Since the expansion is unique, we must have x m = x m. Γ m was arbitrary except that Γ m I. This implies that all x m = for Γ m I and hence X = ai. j Pauli s fundamental theorem: Given two sets of 4 4 matrices γ µ and γ µ which both satisfy {γ µ, γ ν } = 2g µν I, there exists a nonsingular matrix S such that γ µ = Sγ µ S 1. Proof: F is an arbitrary 4 4 matrix, set Γ i is constructed from γ µ and Γ i is constructed from γ µ. Let S = 16 i=1 Γ i F Γ i. Γ i Γ j = a ij Γ k Γ i Γ j Γ i Γ j = a 2 ij Γ2 k = a2 ij Γ i Γ i Γ j Γ i Γ j Γ j = Γ j Γ i = a 2 ijγ i Γ j = a 3 ijγ k Γ i Γ j = a ijγ k 12

6 For any i, Γ i SΓ i = j Γ i γ j F Γ jγ i = j a 4 ij Γ k F Γ k = j Γ k F Γ k = S, a 4 ij = 1. It remains only to prove that S is nonsingular. S = 16 i=1 Γ i GΓ i, By the same argument, we have S = Γ i S Γ i. for G arbitrary. S S = Γ i S Γ iγ isγ i = Γ i S SΓ i, S S commutes with Γ i for any i so S S = ai. We can choose a because F, G are arbitrary, then S is nonsingular. Also, S is unique up to a constant. Otherwise if we had S 1 γ µ S1 1 = S 2 γ µ S2 1, then S2 1 S 1 γ µ = γ µ S2 1 S 1 S2 1 S 1 = ai. Specific representations of the γ µ matrices Recall H = cαi +βmc 2. In the non-relativistic limit, mc 2 term dominates the total energy, so it s convenient to represent β = γ by a diagonal matrix. Recall Trβ = and β 2 = I, so we choose β = I I where I = α k s anticommute with β and are hermitian, A α k k = A k, 82 A k : 2 2 matrices, anticommute with each other. These properties are satisfied by the Pauli matrices, so we have σ α k k 1 i 1 = σ k, σ 1 =, σ 2 =, σ 3 = 83 1 i 1 From these we obtain I γ = β =, γ i = βα i = I σ i σ i, γ 5 = iγ γ 1 γ 2 γ 3 = I. I 84 This is the Pauli-Dirac representation of the γ µ matrices. It s most useful for system with small kinetic energy, e.g., atomic physics. Let s consider the simplest possible problem: free particle at rest. ψ is a 4- component wave-function with each component satisfying the Klein-Gordon equation, ψ = χ e i p x Et, 85 13

7 where χ is a 4-component spinor and E 2 = p 2 c 2 + m 2 c 4. Free particle at rest: p =, ψ is independent of x, Hψ = i cα + mc 2 γ ψ = mc 2 γ ψ = Eψ. 86 In Pauli-Dirac representation,γ = diag1, 1, 1, 1, the 4 fundamental solutions are 1 χ 1 =, E = mc2, χ 2 = 1, E = mc2, χ 3 = 1, E = mc2, χ 4 =, E = mc2. 1 As we shall see, Dirac wavefunction describes a particle pf spin-1/2. χ 1, χ 2 represent spin-up and spin-down respectively with E = mc 2. χ 3, χ 4 represent spin-up and spin-down respectively with E = mc 2. As in Klein-Gordon equation, we have negative energy solutions and they can not be discarded. For ultra-relativistic problems most of this course, the Weyl representation is more convenient. ψ 1 ψ PD = ψ 2 ψ 3 = ψa ψ1 ψ3, ψ ψ A =, ψ B ψ b = ψ 4 ψ 4 In terms of ψ A and ψ B, the Dirac equation is Let s define x ψ A + iσ ψ B = mc ψ A, x ψ B iσ ψ A = mc ψ B. 88 ψ A = 1 2 φ 1 + φ 2, ψ B = 1 2 φ 2 φ

8 and rewrite the Dirac equation in terms of φ 1 and φ 2, x φ 1 iσ φ 1 = mc φ 2, x φ 2 + iσ φ 2 = mc φ 1. 9 On can see that φ 1 and φ 2 are coupled only via the mass term. In ultra-relativistic limit or for nearly massless particle such as neutrinos, rest mass is negligible, then φ 1 and φ 2 decouple, x φ 1 iσ φ 1 =, x φ 2 + iσ φ 2 =, 91 The 4-component wavefunction in the Weyl representation is written as φ1 ψ Weyl =. 92 Let s imagine that a massless spin-1/2 neutrino is described by φ 1, a plane wave state of a definite momentum p with energy E = p c, φ 2 x φ 1 = i 1 c t φ 1 = E c φ 1, iσ φ 1 = 1 σ pφ 1 φ 1 e i p x Et. 93 Eφ 1 = p cφ 1 = cσ pφ 1 or σ p p φ 1 = φ The operator h = σ p/ p is called the helicity. Physically it refers to the component of spin in the direction of motion. φ 1 describes a neutrino with helicity 1 left-handed. Similarly, σ p p φ 2 = φ 2, h = +1, right-handed. 95 The γ µ s in the Weyl representation are I σ γ =, γ i i = I σ, γ 5 = I. 96 I Exercise: Find the S matrix which transform between the Pauli-Dirac representation and the Weyl representation and verify that the γ µ matrices in the Weyl representation are correct. 15