PHY489/1489 THE DIRAC EQUATION SO FAR: • Quick introduction to: • special relativity, relativistic kinematics • quantum mechanics, golden rule, etc. • Phase space: • how to set up and integrate over phase space to determine integrated and differential rates/cross sections • Now we move to the particles themselves • start with the Dirac equation • describes “spin 1/2” particles • quarks, leptons, neutrinos RELATIVISTIC WAVE FUNCTION

• In non-relativistic quantum mechanics, we have the Schrödinger Equation:

⇥ p2 H = i H = p i ⇥t 2m ⇥ ⇤ 2 2 ⇥ = i 2m⇥ ⇥t

• Inspired by this, Klein and Gordon (and actually Schrödinger) tried: @2 E2 = p2 + m2 ( 2 + m2) = ) r @t2 @2 + 2 = m2 @t2 r “Manifestly Lorentz Invariant” ✓ ◆ @ @ @ @ µ 2 @ (@ , @ , @ , @ )= , , , ( @ @µ + m ) =0 µ ) 0 1 2 3 @t @x @y @z ✓ ◆ ISSUES WITH KLEIN-GORDON • Within the context of quantum mechanics, this had some issues: 2 • As it turns out, this allows negative probability densities: =0 | | • Dirac traced this to the fact that we had second-order time derivative • “factor” the E/p relation to get linear relations and obtained: p pµ m2c2 =0 (↵p + m)(p m) µ )  • and found that: ↵ =  µ, ⌫ = µ⌫ + ⌫ µ =2gµ⌫ { }

• Dirac found that these relationships could be held by matrices, and that the corresponding wave function must be a “vector”. µp m =0 (iµ@ m) =0 µ ) µ THE DIRAC EQUATION IN ITS MANY FORMS

(iµ@ m) =0 a a µ µ ⇥ µ

(i@/ m) =0 a a µ = a 0 a 1 a 2 a 3 ⇤ ⇥ µ 0 1 2 3 @ @ @ @ @ (@ , @ , @ , @ )= , , , µ ) 0 1 2 3 @t @x @y @z ✓ ◆

i(0@ 1@ 2@ 3@ ) m =0 0 1 2 3 ⇥ ⇤ @ @ @ @ i 0 1 2 3 m =0 @t @x @y @z  ✓ ◆ • “Dirac’s spoken vocabulary consists DIRAC QUOTES: of “yes”, “no”, and “I don’t know.”

• Q: What do you like best about America? • Q: Professor Dirac, I did not understand • A: Potatoes. how you derived the formula on the top left • A: That is not a question. It is a statement. Next question, please. • Q: What is you favourite sport?

• A: Chinese chess. • I was taught at school never to start a sentence without knowing the end of it. • Q: Do you go to the movies? • A: Yes. • Q: When? • A: 1920, perhaps also 1930.

• Q: How did you find the Dirac Equation? • A: I found it beautiful. “GAMMA” MATRICES: 01 0 i 10 µ =(0, 1, 2, 3) ⌅ =(1, 2, 3)= , , 10 i 0 0 1 ⇤ ⇥ ⇥ ⇥⌅ 10 0 0 01 0 0 10 0 = = 00 10⇥ 0 1 • Note that this is a particular ⇥ ⇧ 00 0 1 ⌃ representation of the matrices ⇧ ⌃ ⇤ ⌅ • 0001 Any set of matrices satisfying the 0010 0 1 anti-commutation relations works 1 = ⇥ = 0 100 1 0 • There are an infinite number of ⇧ 1000⌃ ⇥ ⇧ ⌃ possibilities: this particular one ⇤ ⌅ (Björken-Drell) is just one example 000 i 2 2 00i 0 0 = = µ ⇥ µ ⇥ ⇥ µ µ⇥ 0 i 00⇥ 2 0 , = + =2g ⇥ { } ⇧ i 00 0⌃ ⇧ ⌃ ⇤ ⌅ 0010 000 1 0 3 3 = = 100 0⇥ 3 0 ⇥ ⇧ 0100⌃ ⇧ ⌃ ⇤ ⌅ IN FULL GLORY

A @ 0 @ @ + i @ @t @z @x @y m 000 1 0 0 @ @ i @ @ 2 0 @t @x @y @z 1 0 m 00 3 2 0 i @ @ @ @ 0 1 0 1 = 0 1 @z @x i @y @t 0 00m 0 3 0 6 B @ @ @ @ C 7 6 B + i 0 C B 000m C7 B 4 C B 0 C 6 B @x @y @z @t C B C7 B C B C 4 @ A @ A5 @ A @ A p @ B µ , µ p m p 0 0 A = p p · m 0 ✓ · 0 ◆✓ B ◆ ✓ ◆

p + m p p m p p2 m2 (p )2 0 0 0 = 0 p p +· m p p · m 0 · p2 m2 (p )2 ✓ · 0 ◆✓ · 0 ◆ ✓ 0 · ◆

From problem 4.20c 2 ( a)( b)=a b + i (a b) ( p) = p p · · · · ⇥ · ·

2 2 2 • This is just the KG equation four times p0 m p 0 A 0 2 2 2 = • Wavefunctions that satisfy the Dirac 0 p0 m p B 0 ✓ ◆✓ ◆ ✓ ◆ equation also satisfy KG SOLUTIONS TO THE DIRAC EQUATION • Particle at rest • General plane wave solution ik x (x) e · ⇠ • A particle at rest has only time dependence. @ @ @ 10 @t A @t A 0 @ = im @ (i mc) =0 0 1 B B @t ✓ ◆✓ @t ◆ ✓ @t ◆

• The equation breaks up into two independent parts: @ @ = im = im @t A A @t B B

imt +imt A = e A(0) B = e B(0) DIRAC'S DILEMMA • ψB appears to have negative energy imt +imt A = e A(0) B = e B(0)

• Why don’t all particles fall down into these states (and down to -∞)?

• Dirac’s excuse: all states in the universe up to a certain level (say E=0) are filled. • Pauli exclusion prevents collapse of states down to E = -∞ • We can “excite” particles out e e of the sea into free states e kF =0 This leaves a “hole” that looks like a particle with opposite properties hole (positive charge, opposite spin, etc.)

Dirac originally proposed that this might be the proton EXCUSE TO TRIUMF • 1932: Anderson finds “positrons” in cosmic rays • Exactly like electrons but positively charged: Fits what Dirac was looking for

Dirac predicts the existence of anti-matter and it is found

SOLUTIONS TO THE DIRAC EQUATION • Note that all particles have the same mass

1 0 imc2t/ 0 imt 1 (t)=e (t)=e 1 0 ⇥ 2 0 0 1 ⇧ 0 ⌃ B 0 C “spin up” ⇧ ⌃ “spin down” B C ⇤ ⌅ @ A positive energy solutions (particle) 0 0 0 0 (t)=e+imt (t)=e+imt 3 0 1 1 4 0 1 1 B 0 C B 0 C “spin down”B C “spin up” B C @ A @ A “negative” energy solutions (anti-particle) PEDAGOCIAL SORE POINT • Discussion we had thus far about difficulties with relativistic equations, (negative probabilities, negative energies) is historic • The framework for dealing with quantum mechanics and special relativity (i.e. quantum field theory) had not been developed • The old tools of NR quantum mechanics had reached their limit and new ones were necessary. • In particular, the idea of a “wavefunction” had to be revisited • Until this was done, there were many difficulties! • Once QFT was developed, all of these problems go away. • Both KG and Dirac Equations are valid in QFT • No negative probabilities, no negative energies PLANE WAVE SOLUTIONS TO THE DIRAC EQUATION • Consider a solution of the form: ik x (x)=e · u(k) column vector

Column vector of 4 space-time elements with space- dependence time dependence

• and place it in the Dirac equation: (iµ@ m) =0 µ

µ ik x µ ( k m)e · u(k) =0 ( kµ m)u(k) =0 µ MORE EXPLICITLY: • By definition: µ 0 0 1 1 2 2 3 3 k0 k kµ = k k k k k k· · 0 ⇥ • So that the Dirac equation reads: Note 2x2 notation µ u ( kµ m)u(k)=0 u(k) A uB ⇥

k0 mc k uA k k · mc u ✓ · 0 ◆✓ B ◆

k u = · u (k0 m)uA (k )uB =0 A k m B · 0 k (k )uA (k0 + m)uB =0 uB = · uA · k0 + m DETERMING u • this means we can identify k ↔ ±p k k k k uA = · uB uB = · uA · · =1 k0 m k0 + m k0 + m k0 m µ µ p p ip p = k p = z x y · px + ipy pz ± ⇥ • We can now construct the column vector u: electrons Used positive k solutions 0 Use positive1 solutions 0 1 u = N u2 = N 0 1 1 0 p /(E + m) 1 (p ip )/(E + m) z x y B (p + ip )/(E + m) C B pz/(E + m) C B x y C B C @ A @ A positrons Used negative k solutions pz/(E + m) (px ipy)/(E + m) (p + ip )/(E + m) p /(E + m) u = N x y u = N z 3 0 1 1 4 0 0 1 B 0 C B 1 C B C B C @ A @ A NORMALIZATION:

• Choose “normalization” of the wavefunctions • Note that multiples of the solutions are still solution • normalization convention simply fixes this arbitrary choice:

u1 T u2 u†u =2E u† (u ) u = ⇥ u† =(u1,u2,u3 u4) u3 ⇧ u ⌃ ⇧ 4 ⌃ ⇤ ⌅ • for u1

2 2 p u†u1 = N 1+ =2E N = pE + m 1 (E + m)2  LORENTZ COVARIANCE • The Dirac equation “works” in all reference frames. • What exactly does this mean? (iµ@ m) =0 “Lorentz Covariant” µ • i, m, γ are constants that don’t change with reference frames.

• ∂µ and ψ will change with reference frames, however.

• ∂µ is a derivative that will be taken with respect to the space-time coordinates in the new reference frame. We’ll call this ∂’µ • how does ψ change? • ψ’ = Sψ where ψ’ is the spinor in the new reference frame • Putting this together, we have the following transformation of the equation when evaluating it in a new reference frame µ µ (i @µ m) =0 (i @µ0 m) 0 =0 )µ i @0 (S ) m(S )=0 What properties does S need µ to make this work? PROPERTIES OF S • In general, we know how to relate space time coordinates in one reference with another (i.e. Lorentz transformation), we can do the same for the derivatives x⇥ • Using the chain rule, we get: µ µ = µ ⇥ x x x

• where we view x as a function x’ (i.e. the original coordinates as a function of the transformed or primed coordinates). • Note the summation over ν • if the primed coordinates moving along the x axis with velocity β:

0 0 0 1 ⇧x x = ⇥(x + x ) (⇤ =0,µ= 0) 0 = 1 1 0 ⇧x x = ⇥(x + x ) 2 2 0 x = x ⌃x 3 3 (⌅ =0,µ= 1) 1 = ⇥ x = x ⌃x

etc. 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 @ m) =0 S i S µ @⌫ mS =0 µ ! @x 0 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 = 000 = 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⇥ 0i = i0 SP = { } ⇥ NEXT TIME • Read 4.6-4.9 and Chapter 5 • Lots of notation, lots of stuff going on . . . . • please stop by if you have questions!