Lecture # November 5 Scatterng of two dentcal partcle Relatvtc Quantum Mechanc: The Klen-Gordon equaton Interpretaton of the Klen-Gordon equaton The Drac equaton Drac repreentaton for the matrce α and β Covarant form of the Drac equaton Chapter 3, page 595-608 Chapter 5, page 679-694 694 Branden & Joachan,, Quantum Mechanc Scatterng of two dentcal partcle n the center-of of-ma frame (a) Detector (b) Detector θ π θ θ Partcle cattered n the drecton (θ,φ) Partcle cattered n the drecton (π θ,π+φ) Snce the partcle are dentcal collonal procee (a) and (b) can not be dtnguhed. In center-of-ma ytem µ ( r ) ψ ( r ) Eψ ( r ) + V = where µ=m/ the reduced ma, r=r -r.
Scatterng of two dentcal partcle pnle boon (a) Detector (b) Detector θ Partcle cattered n the drecton (θ,φ) π θ θ Partcle cattered n the drecton (π θ,π+φ) In clacal mechanc the dfferental cro ecton for catterng n the drecton (θ,φ) would be mply the um of dfferental cro ecton for obervaton of partcle and n that drecton. If the ame were true n quantum mechanc we would obtan the clacal reult cl k r f ( θ, φ) kr = f ( θ, π ) + f ( π θ, φ + π ) ; ψ ( r) r e + e r Scatterng of two dentcal partcle pnle boon (a) Detector (b) Detector θ Partcle cattered n the drecton (θ,φ) π θ θ Partcle cattered n the drecton (π θ,π+φ) In clacal mechanc the dfferental cro ecton for catterng n the drecton (θ,φ) would be mply the um of dfferental cro ecton for obervaton of partcle and n that drecton. If the ame were true n quantum mechanc we would obtan the clacal reult cl k r f ( θ, φ) kr = f ( θ, π ) + f ( π θ, φ + π ) ; ψ ( r) r e + e r
Scatterng of two dentcal partcle: pnle boon cl = f ( θ, π ) + f ( π θ, φ + π ) ncorrect reult Spnle boon: wave functon mut be ymmetrc under r r r r. Clearly, the functon ψ ( r) whch atfe boundary condton k r f ( θ, φ) kr ψ k ( r) r e + e r : not ymmetrc under r - r nterchange. However, we can make a + + + ymmetrc combnaton ψ ( r) = ψ k ( r) + ψ k ( r); ψ ( r) = ψ ( r). The correpondng aymptotc form ( ) ( ) [ (, ) (, )] kr + k r k r e ψ r r e + e + f θ φ + f π θ φ + π r nce r r : ( r, θ, φ) ( r, π θ, φ + π ). Scatterng of two dentcal partcle: pnle boon Therefore, the dfferental cro ecton = f ( θ, π ) + f ( π θ, φ + π ) * = f ( θ, π ) + f ( π θ, φ + π ) + Re f ( θ, π ) f ( π θ, φ + π ) and the total cro ecton σ tot = f ( θ, π ) + f ( π θ, φ + π ) 3
Scatterng of two dentcal pn ½ fermon Note: we aume that partcle nteract through central force. Total wave functon mut be antymmetrc wth repect of nterchangng two partcle. Cae : Snglet tate, S=0 the patal wave functon mut be ymmetrc. The correpondng catterng ampltude = f ( θ ) + f ( π θ ) Cae : Trplet tate, S= the patal wave functon mut be antymmetrc. The correpondng catterng ampltude t = ft ( θ ) ft ( π θ ) Scatterng of two dentcal pn ½ fermon If the pn of both partcle are randomly orented the dfferental cro ecton gven by 3 t = + 4 4 3 = f + f + 4 4 f f ( θ ) ( π θ ) t ( θ ) t ( π θ ) If the nteracton pn-ndependent,.e. f ( θ ) = f ( θ ) = f ( θ ) * = f ( θ ) + f ( π θ ) Re f ( θ ) f ( π θ ) t Agan, the reult dfferent from the clacal reult. 4
Relatvtc Quantum Mechanc The Klen-Gordon equaton Interpretaton of the Klen-Gordon equaton The Drac equaton Drac repreentaton for the matrce α and β Covarant form of the Drac equaton Why Relatvtc Quantum Mechanc? The Schrödnger equaton: correctly decrbe the phenomena only f partcle velocte are v c. It not nvarant under a Lorentz change of the reference frame (requred by the prncple of relatvty). Need: a relatvtc generalzaton! 5
The Klen-Gordon equaton So, how to come up wth uch an equaton? Free partcle of pn zero For the relatvtc partcle wth ret ma m and momentum p, 4 E = m c + p c. Ung the correpondence rule E Eop = ; p pop = 4 one can wrte: Ψ = ( m c c ) / Ψ. Problem:. It not clear how to nterpret the operator on rght-hand de. If expanded n power ere t lead to dfferental operator of nfnte order.. The tme and pace coordnate do not appear n a ymmetrc way (no relatvtc nvarance?). Free partcle of pn zero The Klen-Gordon equaton So we remove the quare root and try agan (there wll be conequence of removng th quare root!) 4 E = m c + p c E Eop = ; p pop = Ψ = Ψ 4 ( m c c ). The Klen-Gordon equaton Note: t econd-order dfferental equaton wth repect to the tme unlke the Schrödnger equaton. 6
Probabltc nterpretaton of nonrelatvtc quantum mechanc * P( r, t) = Ψ ( r, t) = Ψ ( r, t) Ψ( r, t) Poton probablty denty Probablty conerved: P(, t) d = 0 r r Ψ( r, t) Ung the Schrödnger equaton = + V ( r, t) Ψ( r, t) m P( r, t) we obtan + j( r, t) = 0, where j can be nterpreted a probablty current denty j r m ( ) ( ) * * (, t) = Ψ Ψ Ψ Ψ. Interpretaton of the Klen-Gordon equaton: Problem We try to contruct a poton probablty denty P(r,t) and probablty current denty j(r,t) whch atfy the contnuty equaton: Ψ P( r, t) + j( r, t) = 0. * Ψ 4 multply by Ψ = ( m c c ) Ψ multply by Ψ * Ψ 4 * Ψ = ( m c c ) Ψ * * * * Ψ Ψ = c Ψ Ψ Ψ Ψ ( ) 7
Interpretaton of the Klen-Gordon equaton: Problem If we requre that the expreon from j(r,t) and P(r,t) had correct non-relatvtc lmt we defne j r m ( ) ( ) * * (, t) = Ψ Ψ Ψ Ψ. P( r, t) Then, we obtan the equaton + j( r, t) = 0. wth P Ψ Ψ mc * * ( r, t) = Ψ Ψ. P(r,t) not potve-defnte and can not be nterpreted a poton probablty denty. Interpretaton of the Klen-Gordon equaton: Problem Free partcle Klen-Gordon equaton: Plane wave oluton: E = ω p = k Ψ = Ψ ( kr ωt ) Ψ ( r, t) = Ae. 4 ( m c c ). ω = m c + c k 4 We get addtonal negatve-energy 4 E = ± m c + c p oluton and energy pectrum not bound from below. Then, arbtrary large energy can be extracted from the ytem f external perturbaton lead to tranton between potve and negatve energy tate. mc 0 E = -mc mc 8
Interpretaton of the Klen-Gordon equaton In 934 W. Paul and V. Wekopf renterpreted Klen-Gordon equaton a a feld equaton and quantzed t ung the formalm of quantum feld theory. Klen-Gordon equaton Relatvtc wave equaton for pnle partcle n the framework of many-partcle theory; negatve energy tate are nterpreted n term of antpartcle. Stll, t poble to defne potve-defnte poton probablty denty wthn the framework of the relatvtc theory? Drac equaton Note: we wll tll get negatve-energy tate Drac equaton P.A. Drac (98) We tart from the wave equaton n the form Ψ = H Ψ. Ψ Ψ Ψ = Spatal coordnate (x =x, x =y, x 3 =z) of a pace-tme pont (event) Ψ N and the tme coordnate (x 4 =ct) have to enter on the ame footng. Therefore, the hamltonan H mut be lnear n pace dervatve a well. Free partcle α, α, α3 and β Smplet Hamltonan: ) Mut be ndependent of r and t ) Mut be lnear n p and m H = c + β mc = pop op ) are ndependent of r, t, p, and E ) do not have to commute wth each other p 9
How to determne α and β? Ψ = c Ψ + β mc Ψ or Eop c op β mc Ψ = 0 p The oluton of the Drac equaton alo mut be a oluton of the Klen-Gordon equaton E 4 c p m c Ψ = 0. op op We ue t to determne the retrcton on the value of α and β by matchng the coeffcent n Eop c pop β mc Eop c pop βmc Ψ = 0 and 4 E c p m c Ψ = 0. op op Note: we drop the ndex op n the dervaton below. How to determne α and β? Some tranformaton E c p β mc E c p β mc Ψ = 0 ( p) β ( p) ( p )( p) ( p) E Ec E mc c E + c + c β mc ( p) β β β 3 4 mc E + mc + m c Ψ = ( p ) ( p )( p) ( p) ( p) ( p )( p) ( p) ( p) 0 3 4 E E c β mc c mc β β β m c + + + + + Ψ = 0 3 4 E c mc β + β β m c Ψ = 0 E 3 3 ( ) ( ) c α p c α α + α α p p mc α β + βα p β m c Ψ = 0 3 4 j j j = < j = 0
How to determne α and β? Now we can match the coeffcent of 3 3 3 4 E c α p c ( αα j + α jα ) p p j mc ( αβ + βα ) p β m c Ψ = 0 = < j = and α = β = 4 E c p m c Ψ = 0. α β + βα = 0, =,, 3 α α + α α = 0 j j j Properte of α and β α = α = α = 3 β = { α β} { α β} { α β}, = 0 3, = 0, = 0 { α α} { α α3} { α α }, = 0, = 0, = 0 3 Therefore, α, α, α 3,and β antcommute n par and ther quare are equal to unty. Clearly, they mut be matrce. The egenvalue of α and β are ±. βα = α β ( ) α = βα β = α ββ = α β Tr α Tr βα β Tr α β Tr βα β = ( ) = ( ) = ( ) = 0 Tr α ( α ) = Tr ( β ) = 0 = α β = β
Tr What f the lowet rank of α and β? Drac repreentaton of α and β. What the lowet rank of the repreentaton for α and β? ( α ) Tr ( β ) 0 = = Therefore, rank N mut be even. For x matrce we can not fnd a repreentaton of more than 3 antcommutng matrce. Therefore, the lowet repreentaton ha N=4. Drac repreentaton: 0 σ I 0 α = β =, σ 0 0 I where σ (=,,3) are Paul matrce 0 0 0 σ x = σ y = σ z = 0 0 0 Drac equaton Ψ Ψ For N=4, the wave functon Ψ = Ψ N and decrbe pn ½ partcle. a four-component pnor We note that th reult may be foreeen a n non-relatvtc quantum mechanc pn ½ partcle are decrbed by -component pnor and each pn ½ partcle ha an antpartcle wth the ame ma and pn, whch lead to 4-component wave functon. Hgher rank matrx repreentaton correpond to partcle wth pn greater than ½.
Covarant form of the Drac equaton Ψ = c Ψ + β mc Ψ xµ ( x, ct) 3 Ψ β + α Ψ β mcψ = 0 x4 = x 3 mc βα + β + Ψ = 0 = x x4 mc γ µ + Ψ = 0 γ = βα γ 4 = β xµ γ, γ = δ, µ, ν =,,3, 4 { } µ ν µν From the Drac repreentaton 0 σ I 0 γ = γ 4, for the α and β = σ 0 0 I 3