Lecture 6/7 (February 10/12, 2014) DIRAC EQUATION. The non-relativistic Schrödinger equation was obtained by noting that the Hamiltonian 2

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1 P470 Lecture 6/7 (February 10/1, 014) DIRAC EQUATION The non-relatvstc Schrödnger equaton was obtaned by notng that the Hamltonan H = P (1) m can be transformed nto an operator form wth the substtutons H ( = 1, c= 1) () t P m resultng ψ ( x, t ) = ψ ( x, t ) t (3) Ths equaton s lnear n the tme-dervatve and quadratc n the space dervatve, and s manfestly non-covarant under Lorentz transformaton. Clearly, the lefthand sde and the rght-hand sde of Equaton 3 transforms dfferently under Lorentz transformaton. We brefly revew Lorentz transformaton. Space-tme coordnates (t, x, y, z) are denoted by the contravarant 4-vector The covarant 4-vector, x μ, s o 1 3 (,,, ) (,,, ) x x x x x t x y z g μυ s the metrc tensor ( ) ( ) x x x x x t x y z g x ν,,,, -, -, - o 1 3 = ν g ν = (4)

2 P470 The Lorentz transformaton relates x μ to x μ : ( x ) = Λ x (5) ν ν For an nertal system S movng along z wth ν c = β, we have where ( ) 1 γ = 1 β γ 0 0 βγ Λ= βγ 0 0 γ (6) Any set of four quanttes whch transform lke x μ under Lorentz transformaton s called a four-vector. The total energy E and the momentum P form a four-vector P = P P P P = P o 1 3 (,,, ) ( E, ) Scalar product of two 4-vectors are nvarant under Lorentz transformaton (7) ν AB = g AB (8) ν Some examples: xx = xx = t x PP = PP = E P xp = xp = te xp The H, P transcrpton s Lorentz covarant t P = x One can readly obtan the contnuty equaton from the Schrödnger equaton:

3 P470 3 m where ρ = ψ * ψ 0, J = ( ψ * ψ ψ ψ *) ρ s the probablty densty, and J s the current densty. ρ + J = 0 (9) t Note that the contnuty equaton can be wrtten n a covarant form j = 0, j =, J ρ ( ) (10) The early attempt for formulatng a Lorentz covarant equaton started from the relatvstc energy-momentum relaton E = P + m wth the E, P transcrpton, one obtans t t ψ = + ( ) m ψ (11) whch s the relatvstc Schrödnger equaton, or the Klen-Gordon equaton. Equaton 11 can be wrtten as where ( ) + m ψ = 0 (1) = = = (13) t x x s the d Alembertan operator, a Lorentz nvarant operator. From the Klen-Gordon equaton, one can readly derve the followng contnuty equaton:

4 P470 4 ρ + J = 0 t φ φ* ρ = φ* φ t t j = J = ( φ* φ φ φ* ) ( φ* φ φ φ* ) ( 14) ( 15) For a plane-wave soluton to Klen-Gordon equaton We have φ = Ne ( Et P x) ρ = EN, J= PN (16) (17) Snce ( ) 1 E P m =± +, ρ can be negatve for the E < 0 solutons. The orgn of ths problem s the second dervatve n tme for the term t. Drac tred to fnd a relatvstc covarant equaton wth a postve-defnte probablty densty. He assumed an equaton lnear n t and n : Hψ = ( α P+ βm) ψ (18) α P= α P + α P + α P It s evdent that α s and β cannot be just numbers, because the equaton would not be nvarant even under spatal rotaton. Drac proposed that the equaton be H ψ = P + m ψ. consdered as a matrx equaton. It should satsfy ( ) Applyng H to Equaton 18 leads to H ψ = α P + ( αα j + α jα) PP j + ( αβ + βα) mp + β m ψ j

5 P470 5 Therefore { α α j} δj { α β}, = ;, = 0 α = β = 1 What are the other propertes of α, β? Frst, α, β are Hermtan, snce H s Hermtan (see Equaton 18). In addton, Tr(α ) = 0, Tr(β) = 0. Ths can be shown by the antcommutator relatons n Equaton 19. Recall Tr(AB) = Tr(BA) ( α ) = ( α β ) = ( βα β ) = ( α ββ ) Tr Tr Tr Tr ( α ) Tr ( α ) 0 = Tr = (0) Smlarly, Tr(β) = 0. Therefore, α, β are traceless, Hermtan matrces. The egenvalues of α, β can only be +1 or -1 snce α = β = 1. α x= λx α x= λ x= x λ =± 1 (1) The trace of a matrx s equal to the sum of all ts egenvalues. Therefore, for a traceless matrx havng +1, -1 as possble egenvalues, t must have an even number of dmenson n. n cannot be, snce only three traceless n = ndependent matrces exst ( σx, σ y, σ z, for example), and we need four traceless matrces to accommodate α1, α, α3, β. n = 4 s the smallest possble dmenson for α, β. The Paul-Drac representaton s gven as 0 σ I 0 α = σ 0 β = 0 I () The Drac equaton α P mβ + ψ = Hψ can be expressed n a more covarant fashon by multplyng β from the left: (3) ( β H ( βα ) P m) ψ = 0

6 P470 6 Introducng γ = ( β, βα ) P =, therefore, Equaton 3 becomes ( γ P m) ψ 0 = (4) Explctly, = = x ( γ m) ψ 0 recall 0 0 o I σ γ = ; γ 0 I = σ 0 (5) (6) γ μ s satsfy the antcommutaton relaton γ μ γ ν + γ ν γ μ = g μν I (7) o ( γ ) I, ( γ ) = = I Although γ o s Hermtan, (γ o ) + = γ o, γ s ant-hermtan: (γ ) + = -γ (8) However, (γ μ ) + = γ o γ μ γ o and γ μ = γ o (γ μ ) + γ o (9) To obtan the adjont Drac equaton, one takes Hermtan conjugate of Equaton 5 + ( ) m ψ γ ψ = (30) Snce γ μ s not Hermtan for μ = 1,, 3, t s useful to multply Equaton 30 from the rght by γ o to yeld (usng Equaton 9) + o + o ( ) m( ) 0 ψ γ γ ψ γ = (31) Defne o ψ = ψ γ (3) the adjont Drac equaton s ψγ + mψ = 0 (33)

7 P470 7 or where P = γ P and P acts to ts left n Equaton 34. ψ ( P + m) = 0 (34) One s now ready to check the contnuty equaton and the probablty densty and current densty for the Drac equaton. Multplyng Equaton 5 by ψ to ts left, and Equaton 33 by ψ to ts rght, we obtan ψγ ψ ψ mψ = 0 (35) ψγ ψ + mψψ = 0 Addng these two equatons yelds ( ) 0 ψγ ψ = (36) Defne j = ψγ ψ (37) the contnuty equaton j = 0 follows. Note that o o + ρ = j = ψγ ψ = ψ ψ = ψ 0 4 = 1 Hence the probablty densty s ndeed postve defnte for the Drac equaton. As wll be shown later j μ transforms lke a 4-vector, and ρ = j o s not a Lorentz nvarant. Rather, t s the tme component of a 4-vector. It s not too surprsng that ρ s not a Lorentz nvarant, snce a probablty densty satsfes the followng: 3 ρ dx= const (38) Snce d 3 x undergoes Lorentz contracton, t s natural that ρ must also change under the Lorentz transformaton n order to preserve the ntegral.