H. A. TANAKA LECTURE 6: MORE PROPERTIES OF THE DIRAC EQUATION TRANSFORMING THE DIRAC EQUATION: µ µ iγ ⇤µ⇥ mc ⇥ =0 iγ ⇤ ⇥ mc ⇥ =0 − ⇥ µ − µ iγ ⇤ (S⇥) mc (S⇥)=0 µ − ⇥ µ ⇤x iγ µ ⇤⇥ (S⇥) mc (S⇥)=0 ⇤x − S is constant in space time, so we can move it to the left of the derivatives ⇥ µ ⇤x iγ S µ ⇤⇥ ⇥ mc (S⇥)=0 ⇤x − Now slap S-1 from both sides ν @x⌫ iγ ⇤ν ⇥ mc ⇥ =0 1 µ − S− i~γ S @⌫ mc S =0 ! @xµ0 − Since these equations must be the same, S must satisfy ⇥ ⇥ 1 µ ⇥x γ = S− γ S µ ⇥x ⇥ EXAMPLE: THE PARITY OPERATOR • For the parity operator, we want to invert the spatial coordinates while keeping the time coordinate unchanged: ∂x ∂x1 10 0 0 0 =1 = 1 0 10 0 ∂x0 ∂x1 − P = − ⇥ 00 10 ∂x − ∂x2 3 ⇧ 00 0 1 ⌃ = 1 = 1 ⇧ − ⌃ ∂x2 − ∂x3 − ⇤ ⌅ • We then have Recalling We find that γ0 satisfies our needs ⇥ 0 1 0 ⇥ 1 µ ⇥x γ = S γ S γ = S− γ S − ⇥xµ γ0 = γ0γ0γ0 = γ0 1 1 1 ⇥ γ = S− γ S − 0 10 i 0 i 0 0 0 i i 2 1 2 γ = γ = γ γ γ = γ γ γ = γ γ = S− γ S 0 1 − 3 − 1 3 − ⇥ γ = S− γ S 0 2 − (γ ) =1 0 γµ, γ⇥ =2gµ⇥ γ0γi = γiγ0 SP = γ { } ⇥ − LORENTZ COVARIANT QUANTITIES • Recall that for four vectors aµ, bµ • aµ, bµ transform as Lorentz vectors (obviously) µ • a bµ is a scalar (does not change under Lorentz transformations • aµbν is a tensor (each has a Lorentz transformation) • From the previous discussion, we know: • Dirac spinors have four components, but don’t transform as Lorentz vectors • How do combinations of Dirac spinors change under Lorentz Transformations? HOW DO WE CONSTRUCT A SCALAR? 0 ¯ 0 • We can use γ : define: ⇥ = ⇥†γ • Consider a Lorentz transformation with S acting on the spinor 0 0 • We can also show generally that S†γ S = γ 0 0 • This gives us ⇥⇥¯ ⇥†S†γ S⇥ = ⇥†γ ⇥ = ⇥⇥¯ ⇒ • so this is a Lorentz invariant • We can construct the parity operator to check how ψψ ¯ transforms under the parity operation. 0 • Recall SP = γ • We can investigate how ψψ ¯ transforms under parity 0 0 0 0 0 ⇥⇥¯ (⇥†S† γ )(S ⇥)=⇥†γ γ γ ⇥ = ⇥†γ ⇥ = ⇥⇥¯ ⇒ P P ψψ¯ doesn’t change sign under parity it is a Lorentz scalar THE γ 5 OPERATOR 5 • Define the operator γ as: 01 5 0 1 2 3 10 γ = iγ γ γ γ ⇥ • It anticommutes with all the other γ matrices: γµ, γ5 =0 ⇥ µ • use the canonical anti-commutation relations to move γ to the other side µ • γ will anti-commute with for µ≠ν µ • γ will commute when µ=ν ¯ 5 • We can then consider the quantity ⇥ ⇥ • Can show that this is invariant under Loretnz transformation. • What about under parity? 5 0 5 0 5 0 5 5 ⇥¯ ⇥ (⇥†S† )γ γ (S ⇥)= (⇥†S† )γ S γ ⇥ = ⇥†γ γ ⇥ = ⇥¯ ⇥ ⇥ P P − P P − − ⇥¯ 5 ⇥ is a “pseudoscalar” OTHER COMBINATIONS µ • We can use γ to make vectors and tensor quantities: ⇤⇤¯ scalar 1 component ⇤¯ 5⇤ pseudoscalar 1 component ⇤¯ µ⇤ vector 4 components ⇤¯ µγ5⇤ pseudovector 4 components i ⇤⇥¯ µ⇥ ⇤ antisymmetric tensor 6 components ⇥µ⇥ = (γµγ⇥ γ⇥ γµ) 2 − • You can tell the transformation properties by looking at the Lorentz indices 5 • γ introduces a sign (adds a “pseudo”) * • Every combination of ψ iψj is a linear combination of the above. • We will see that the above are the basis for creating interactions with Dirac particles. Interactions will be classified as “vector” or “pseudovector”, etc. Angular Momentum and the Dirac Equation: • Conservation: • In quantum mechanics, what is the condition for a quantity to be conserved? [H, Q]=0 • Free particle Hamiltonian in non-relativistic quantum mechanics: p2 p2 1 H = [H, p]=[ ,p]= [p2,p]=0 2m 2m 2m • thus we conclude that the momentum p is conserved • If we introduce a potential: p2 1 [H, p]=[ + V (x),p]= [p2 + V (x),p] =0 2m 2m • thus, momentum is not conserved Hamiltonian for the Dirac Particle: • Starting with the Dirac equation, (γµp mc)⇥ =0 µ − • determine the Hamiltonian by solving for the energy • Hints: (γ0)2 =0 • Answer: H = cγ0 (γ p + mc) · Is orbital angular momentum conserved? • We want to evaluate [H, L ] • Recall: j k L = x p Li = ijkx p × 0 0 a b H = cγ (γ p + mc) H = cγ ⇥abγ p + mc · 0 a b j k [H, Li]=[cγ ⇥abγ p + mc , ⇤ijkx p ] • which parts do not commute? 0 a b [cγ ⇥abγ p , ⇤ijkxjpk] [mc, ijkxjpk] • 0 b j k cγ ⇥ab⇤ijk[p ,x p ] [A, BC]=[A, B]C + B[A, C] 0 a bj k 0 j k 0 c⇥ab⇤ijkγ γ ( i⇥ p ) icγ ⇥ijkγ p icγ (⇥ p⇥) − − − × Orbital angular momentum is not conserved Spin ⇥ 0 • Consider the operator:S⇥ = Σ⇥ = acting on Dirac spinors 2 2 0 ⇥ • Note that it satisfies all the properties of an angular momentum operator. • Let’s consider the commutator of this with the hamiltonian H = cγ0 (γ p + mc) ~c · [γ0~ p~ + γ0mc, ⌃~ ] 2 · • once again, consider in component/index notation ~c [H, Si]= [γ0δ γapb + γ0mc, ⌃i] 2 ab • which part doesn’t commute? c ⇥ pb[γ0γa, Σi] [AB, C]=[A, C]B + A[B,C] 2 ab a i c b 0 a i 0 σ σ 0 ⇥abp γ [γ , Σ ] [ , ] 2 σa 0 0 σi − j 0[σ , σ ] 0 σj a i aijγ [σa, σ ]0 aij σj 0 − i − The “total” spin operator: • Define the operator: σ 0 2 S = Σ = S = S S 2 0 σ · ⇥ • Calculate its eigenvalue for an arbitrary Dirac spinor: • If the eigenvalue of the operator gives s(s+1), where s is the spin, what is the spin of a Dirac particle? NEXT TIME • Please turn in problem set 1 by 1600 on Thursday (6 Oct) • Please read Chapter 5.