H&M Chapter 5 Review of Dirac Equation

Transcription

1 HM Chapter 5 Review of Dirac Equation Dirac s Quandary Notation Reminder Dirac Equation for free particle Mostly an exercise in notation Define currents Make a complete list of all possible currents Aside on Helicity Operator Solutions to free particle Dirac equation are eigenstates of Helicity Operator Aside on handedness

2 Dirac s Quandary Can there be a formalism that allows wave functions that satisfy the linear and quadratic equations simultaneously: H =(~ ~p + m H 2 =(~p ~p + m 2 If such a thing existed then the linear equation would provide us with energy eigenvalues that automatically satisfy the relativistic energy momentum relationship

3 Dirac s Quandary (2 Such a thing does indeed exist: Wave function is a 4 component object α i = # $ σ i σ i # β = I $ I ( ( α is thus a 3-vector of 4x4 matrices with special commutator relationships like the Pauli matrices. While β is a diagonal 4x4 matrix as shown. # σ = # (; σ $ 2 = i # (; σ $ i 3 = # (; I = ( $ $

4 The rest is history In the following I provide a very limited reminder of notation and a few facts. If these things don t sound familiar, then I encourage you to work through ch. 5 carefully.

5 Rewrite a little: H =(~ ~p + m H =( ~ ~p + 2 m =( ~ ~p + m ( H ~ ~p = m Now it looks covariant: ( p ~ ~p = m µ p µ = m p = m Where we have introduced: p µ =(H, ~p =, 5 i = i and for later use: 5 = i 2 3

6 They satisfy: { µ, } =2 µ In particular: ( 2 =, ( 2 =( 2 2 =( 3 2 = So they cannot be hermitian. In fact ( =, ( i = i Also: { 5, µ } =, ( 5 2 =, ( 5 = 5 6

7 Notation Reminder ( Sigma Matrices: # σ = # (; σ $ 2 = i # (; σ $ i 3 = # (; I = ( $ $ Gamma Matrices: $ γ = I $ σ ; γ j =,2,3 j =,2,3 $ = ; γ 5 = I I( σ j =,2,3 ( I ( State Vectors: #. (. ψ = (; ψ ψ T γ =.....( ( $. (

8 Notation Reminder (2 Obvious statements about gamma matrices γ ( j =,2,5T = γ ( j =,2,5 ; γ ( j =,3T = γ ( j =,3 γ ( j =,5T = γ ( j =,5 ; γ ( j =,2,3T = γ ( j =,2,3 Probability density ψ ψ T γ Scalar product of gamma matrix and 4-vector ψγ ψ =ψ T ψ =# /A µ A µ = A A 2 A 2 3 A 3 Is again a scalar formed from a 4-vector

9 Dirac Equation of free particle m = i@ µ µ + m = Ansatz: = e ip x u(p or (i /@ + m = Explore this in restframe of particle: ψ + 2 = 2me imt $ $ ; ψ 2 = 2me imt ; ( ( Normalization chosen to describe 2E particles, as usual.

10 Particle vs antiparticle in restframe ψ + 2 = 2me imt $ $ ; ψ 2 = 2me imt ; ( ( Recall, particle antiparticle means E,p E, p Let s take a look at Energy Eigenvalues: particle " # $ mi mi u = Eu for p = we get this equation to satisfy by the energy eigenvectors It is thus obvious that 2 of the solutions have E <, and are the lower two components of the 4-component object u.

11 Particle Anti-particle ψ + 2 = 2me imt $ ( ; ψ 2 = 2me imt $ ( ; ψ + 2 = 2me imt $ ( ; ψ 2 = 2me imt $ ( ; Negative energy solution Positive energy solution

12 Not in restframe this becomes: The lower 2 components are thus coupled to the upper 2 via this matrix equation, leading to free particle and antiparticle solutions as follows.

13 (Anti-Particle not in restframe ψ + 2 = Ne ipx σ p E + m ( ( ; ψ 2 = Ne ipx σ p E + m ( ( ψ + 2 = Ne +ipx σ p E +m ( ( ; ψ 2 = Ne +ipx σ p E +m ( ( I suggest you read up on this in HM chapter 5 if you re not completely comfortable with it.

14 What we learned so far: Dirac Equation has 4 solutions for the same p: Two with E > Two with E < The E < solutions describe anti-particles. The additional 2-fold ambiguity describes spin ±/2. You will show this explicitly in Exercise HM 5.4, which is part of HW next week. We thus have a formalism to describe all the fundamental spin-/2 particles in nature.

15 Helicity Operator The helicity operator commutes with both H and P. Helicity is thus conserved for the free spin-/2 particle. The unit vector here is the axis with regard to which we define the helicity. For (,,, i.e., the z-axis, we get the desired ±/2 eigenvalues. # σ 3 = $ (

16 Dirac Equation for particle and anti-particle spinors It s sometimes notationally convenient to write the antiparticle spinor solution (i.e. the p, E as e ip x u (,2 (p an explicit antiparticle spinor that satisfies a modified Dirac equation: e i( p x u (3,4 ( p =e ip x v (,2 (p ( γ µ p µ mu = ( γ µ p µ + mv = particle antiparticle The v-spinor then has positive energy. We won t be using v-spinors in this course. HM Equation (5.33 and (5.34

17 Antiparticles We will stick to the antiparticle description we introduced in chapter 4: e - e - p,e -p,-e Initial state e + e - is an initial state e - e - with the positron being an electron going in the wrong direction, i.e. backwards in time.

18 Some more reflections on γ µ There are exactly 5 distinct γ matrices: $ γ = I $ σ ; γ j =,2,3 j =,2,3 $ = ; γ 5 = I I( σ j =,2,3 ( I ( γ 5 iγ γ γ 2 γ 3 Where Every one of them multiplied with itself gives the unit matrix. As a result, any product of 5 of them can be expressed as a product of 3 of them.

19 Currents Any bi-linear quantity can be a current as long as it has the most general form: ψ( 4x4ψ By finding all possible forms of this type, using the gamma-matrices as a guide, we can form all possible currents that can be within this formalism.

20 The possible currents ψψ ψγ 5 ψ ψγ µ ψ ψγ 5 γ µ ψ scalar pseudo-scalar vector pseudo-vector or axial-vector ψ i ( 2 γ µ γ ν γ ν γ µ ψ tensor Note: we define µ = i 2 ( µ µ

21 Electromagnetic Interactions Minimal substitution µ! µ + ea µ id µ Dirac: (i/@ m =! (i /D m = (i/@ + e /A m = (electron Had we started with positron we would have (i/@ e /A m C = 2

22 There must be a correspondence: complex conjugate the first (i/@ + e /A m = ( i µ + e µ A µ m = multiply by a matrix D D( i µ + e µ A µ md D = such that D µ D = µ and define C = D So that (i/@ e /A m C = It is convenient to define D = C, C = D so that C = C T and C = T C 22

23 EM-Current We have seen the vector: µ It is µ ( µ = (/@ + /@ = i (m mµ = and = = > has the interpretation of a density. For it to be a charge density of electrons, multiply by e j µ = e µ =(,~j j µ C = e C µ C = e T C µ C T = e T ( µ T T = e µ = j µ A Cµ j µ C =( A µ( j µ =A µ J µ EM is invariant under charge conjugation 23

24 Massless spin-/2: 2-component spinors H =(~ ~p + m need only α matrices { i, j } =2 ij i = i may take i = ± i (Pauli Massless Dirac: two decoupled equations for 2-component spinors E = ~ ~p E =+~ ~p Helicity: divide through by E = ~p 2 ~ ˆp = 2 2 ~ ˆp =+ 2 helicity = 2 =+ 2 invariant under boosts (since massless particles travel at speed of light 24

25 But we can also have solutions with negative energy E = ~p pz E = ~ ~p with, e.g., p along z-axis: E = p z Take 2 2 Negative energy: antiparticle. Use E, p in Feynman rules for anti-particle of energy/momentum +E,+p. So for negative energy solution 2 ~ ˆp =+ 2 helicity =+ 2 Particle is left-handed, L its anti-particle is right handed, R Parity(P: L R, so theory with only one type of 2-component spinor breaks P OK for weak interactions! 25

26 Weyl representation 2-component in 4-component notation: u =, i = i i corresponds to i = i i, I =, I 5 I =. I Projection operator 2 ( 5 u = I = projection operator: ( 2 ( 5 2 = 2 ( 5 26

27 Peak ahead into weak interactions There is a charged current, coupling electron to neutrino J µ = e µ 2 ( 5 and the hermitian conjugate J µ = 2 ( 5 µ e This is called a V A current, a vector minus a pseudo-vector (or axial vector 27