H. A. TANAKA LECTURE 6: MORE PROPERTIES OF THE DIRAC EQUATION NEXT TIME • Please turn in PS 1 by 1700 on Thursday (28 Sep) • Box K around the corner • Please be sure to staple your pages • If you want to submit electronically, please arrange with Randy. • [email protected] • Midterm: • Material so far: • drawing Feynman diagrams for EM, weak, strong interactions. Allowed/ forbidden interactions • special relativity, relativistic kinematics • Phase space • There will be no detailed calculations . . . .. • necessary detailed equations will be provided. TRANSFORMING THE DIRAC EQUATION µ µ i @ m =0 i @0 0 m 0 =0 µ ) µ µ i @0 (S ) m(S )=0 µ
⌫ µ @x i @⌫ (S ) m(S )=0 @xµ0
S is constant in space time, so we can move it to the left of the derivatives
⌫ µ @x i S @⌫ mS =0 @xµ0
Now slap S-1 from the left ⌫ µ 1 µ @x (i @µ m) =0 S i S @⌫ mS =0 ! @xµ0 Since these equations must be the same, S must satisfy ⇥ ⇥ 1 µ ⇥x = S S µ ⇥x ⇥ PARITY OPERATOR • For the parity operator, 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 = S S = S S ⇥xµ 0 = 0 0 0 = 0 1 1 1 ⇥ = S S 0 10 i 0 i 0 0 0 i i 2 1 2 = = = = = S S 0 1 3 1 3 ⇥ = S S 0 2 ( ) =1 0 µ, ⇥ =2gµ⇥ 0 i = i 0 SP = { } ⇥ LORENTZ COVARIANT QUANTITIES • Recall that for four vectors aµ, bµ • aµ, bµ transform as Lorentz vectors
• aµbµ is a scalar (invariant 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: ⇥ = ⇥† • Consider a Lorentz transformation with S acting on the spinor 0 0 • We can also show generally that S† S =
0 0 • ⇥⇥¯ ⇥†S† S⇥ = ⇥† ⇥ = ⇥⇥¯ This gives us • so this is a Lorentz invariant ¯ • Using the parity operator . ¯ 0 0 0 0 0 ¯ γ0 ⇥⇥ (⇥†S† )(SP ⇥)=⇥† ⇥ = ⇥† ⇥ = ⇥⇥ • Recall SP = P • We see that ¯ is invariant under parity THE γ5 OPERATOR γ5 01 • Define the operator as: 10 ⇥ 5 = i 0 1 2 3
• It anticommutes with all the other γ matrices: µ, 5 =0
• use anti-commutation ⇥ relations to move γµ to the other side • γµ will anti-commute with for µ≠ν • γµ will commute when µ=ν • Consider the quantity ⇥ ¯ 5⇥ • Can show that this is invariant under Lorentz transformation. • What about under parity? ¯ 5 0 5 0 5 0 5 ¯ 5 ⇥ ⇥ (⇥†SP† ) (SP ⇥)= (⇥†SP† ) SP ⇥ = ⇥† ⇥ = ⇥ ⇥ ⇥ 5 ⇥ ¯ ⇥ is a “pseudoscalar” OTHER COMBINATIONS • We can use γµ to make vectors and tensor quantities: ⇤⇤¯ scalar 1 component ⇤ ¯ 5⇤ pseudoscalar 1 component ⇤ ¯ µ⇤ vector 4 components ⇤ ¯ µ 5⇤ pseudovector 4 components i ⇤⇥¯ µ⇥ ⇤ antisymmetric tensor 6 components ⇥µ⇥ = ( µ ⇥ ⇥ µ) 2 • Lorentz indices and γ5 tell you how it transforms • γ5 introduces a sign (adds a “pseudo”)
* • Every combination of ψ iψj is a linear combination of the above. • Interactions can be classified as “vector”,“pseudovector”, etc. ANGULAR MOMENTUM AND DIRAC • 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
• Starting with the Dirac equation, • determine the Hamiltonian by solving for the energy • Hints: µ 0 2 ( pµ mc)⇥ =0 ( ) =0
• Answer: H = c 0 ( p + mc) · ORBITAL ANGULAR MOMENTUM • We want to evaluate • Recall: [H, L ] j k L = x p Li = ijkx p 0 0 a b H = c ( p + mc) H = c ⇥ab p + mc · • which parts do not commute 0 a b j k [H, Li]=[c ⇥ab p + mc , ⇤ijkx p ]
0 a b [c ⇥ab p , ⇤ijkxjpk] [mc, ijkxjpk]
0 b j k c ⇥ab⇤ijk[p ,x p ] [A, BC]=[A, B]C + B[A, C]
0 a bj k 0 j k 0 c⇥ab⇤ijk ( i ⇥ p ) i c ⇥ijk p i c (⇥ p⇥)
Orbital angular momentum is not conserved “SPIN”: ⇥ 0 S⇥ = ⇥ = • Consider the operator: 2 2 0 ⇥ acting on Dirac spinors • Satisfies all properties of an angular momentum operator. • Let’s consider the commutator of this with the hamiltonian ~c [ 0~ p~ + 0mc, ⌃~ ] H = c 0 ( p + mc) 2 · · • once again, consider in component/index notation ~c [H, Si]= [ 0 apb + 0mc, ⌃i] 2 ab • which part doesn’t commute?
c [AB, C]=[A, C]B + A[B,C] ⇥ pb[ 0 a, i] 2 ab 0 a i 0 c b 0 a i [ , ] ⇥abp [ , ] a 0 0 i 2
j 0[ , ] 0 j a i aij [ a, ]0 aij j 0 i T HE “TOTAL” SPIN OPERATOR • Define the operator:
2 0 S = S S S = = · 2 0 ⇥ • 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 1700 on Thursday (28 September) • Box K around the corner • Please read Chapter 5