Note 2 Lng fong L Contents Ken Gordon Equaton. Probabty nterpretaton......................................2 Soutons to Ken-Gordon Equaton............................... 2 2 Drac Equaton 3 2. Probabty nterpretaton..................................... 4 2.2 Souton to Drac equaton................................... 4 2.3 Drac equaton under Lorentz transformaton.......................... 5 Ken Gordon Equaton Casscay, the energy momentum reaton s of the form E = p 2 2m + V (~r) For the quantzaton of ths system, we do the repacement E @ @t, p r and act on a wavefuncton @ @t = [ 2m r2 + V (~r)] Schrodnger equaton Ths equaton does not work for reatvstc system because spata coordnate x and tme t are not on equa footng. In other words, ths s not nvarant under Lorentz transformatons. For reatvstc free partce, we have nstead E 2 = ~p 2 + m 2 The correspondng wave equaton s then Ths s known as Ken-Gordon equaton.. Probabty nterpretaton From Ken-Gordon equaton and ts compex conjugate, we can derve the contnuty equaton, where (r 2 + m 2 ) = @ 2 0 ( + m 2 ) = 0; where = @ 2 0 r 2 = @ @ = @ 2 (@ 2 0 r 2 + m 2 ) = 0 (@ 2 0 r 2 + m 2 ) = 0 @ @t + r j = 0 = ( @ 0 @ 0 ); ~j = ( r r )
Then P = R d 3 x s conserved,.e. dp dt = V @ @t d3 x = V r j d 3 x = I S j ds = 0 f j = 0; on S Snce P s conserved, we woud ke to nterpret t as probabty. However t s easy to see that P s not postve as requred by the de nton of probabty. For exampe, f ~e Et (x) ; then = 2E j (x)j 2 0. e. we get negatve probabty and t s not vabe. On the other hand f we take the probabty densty to be = whch s postve as n the case of Schrodnger equaton, then t s easy to see that t s not conserved, d d 3 x 6= 0 dt Thus t s not possbe to have probabty nterpretaton for Ken-Gordon equaton..2 Soutons to Ken-Gordon Equaton The Ken Gordon equaton ( + m 2 ) (x) = ( r 2 + @ 2 0 + m 2 ) (x) = 0; s a d erenta equaton wth constant coe cents and has pan wave souton, (x) = e px f p 2 0 P 2 m 2 = 0 or p 0 = p ~p 2 + m 2 (a) Postve energy souton: P 0 = p = p ~p 2 + m 2 ; ~p arbtrary (+) (x) = exp p t + p x (b) Negatve energy souton: P 0 = p = p ~p 2 + m 2 ( ) (x) = exp p t p x Note that postve energy soutons (+) (x) togather wth the negatve energy souton ( ) (x) form a compete set of soutons. The most genera souton s a near superposton of postve energy and negatve energy souutons, (x) = = d 3 k p (2) 3 2 k [a(k)e ~ k~x k t + a(k) + e ~ k~x+ k t ] d 3 k p (2) 3 2 k [a(k)e kx + a(k) + e kx ] () where kx = k t ~ k ~x
2 Drac Equaton Drac(928) want to construct a reatvstc wave equaton rst order n tme dervatve just ke Schrodnger equaton whch has conserved probabty and postve. By speca reatvty the wave equaton s aso rst order n spata coordnates. He assume an Ansatz where ; are assumed to be matrces. Then E = p + 2 p 2 + 3 p 3 + m = ~ ~p + m (2) E 2 = ( p + 2 p 2 + 3 p 3 + m) 2 = 2 ( j + j )p p j + 2 p 2 + ( + )m To get reatvstc energy momentum reaton, we requre From Eq( 3) we get j + j = 2 j (3) + = 0 (4) 2 = (5) 2 = (6) Togather wth Eq(5) we see that ; a have egenvaues. Eq( 3) aso mpes Takng the trace Thus Smary, 2 = 2 =) 2 = 2 T r 2 = T r ( 2 ) = T r 2 2 = T r (2 ) T r ( ) = 0 (7) T r () = 0 From Eqs(6,7) we get the mportant resut that ; a have even dmenson. Reca that Pau matrces ; 2 ; 3 are a traceess and ant-commutng. But here we need 4 such matrces. Thus ; a have to be 4 4 matrces. One conveent chocs s that used by Bjoken and Dre where matrces take the form 0 = 0 ; = 0 0 Drac equaton s obtaned from Eq(2) by the repacenments, E @ @t ; ~p r For conveent, de ne a new set of matrces ( ~ r + m) = @ @t ( ~ r @ t + m) = 0 0 = ; = and n Bjorken and Dre notaton, 0 = 0 0 = 0 0 (8)
Drac equaton s then ( @ 0 @ 0 + m) = 0; or ( @ + m) = 0 Ths s usuay referred to as Drac equaton n covarant form. Note that the ant-commutatons are now n a smper form, f ; g = 2g 2. Probabty nterpretaton We can now show that Drac equaton gve a correct form for the peobabty. From the Drac equaton n the hermtan form we get @ y = (f ~ r + m) g) y @t and ( @ y @t Integrate over space, we get d dt + y @ @t ) = y ( ~ r + m) f( ~ r + m) g y d 3 x( y ) = f y (~ r) f(~ r) g y gd 3 x = r y (~ r) d 3 x = 0 The probabty R d 3 x( y ) s conserved and postve. 2.2 Souton to Drac equaton We ook for souton n the pane wave form, (x) = e px u where u and are 2 components coumn vector. Then Drac equaton becomes u ( /p m) = 0 where /p = p Usng the representaton gven n Eq(8), we get m ~ ~p ~ ~p m u u = p 0 (p0 m)u (~ ~p) = 0 (~ ~p)u + (p 0 + m) = 0 (9) These are homogeneous near equatons of u and : Non-trva souton exsts f p 0 m ~ ~p ~ ~p p 0 + m = 0 It s easy to see that ths determnanta condton gves p 2 0 = ~p 2 + m 2 or p 0 = p ~p 2 + m 2
(a) Postve energy souton p 0 = E = p ~p 2 + m 2, Substtute ths nto Eqs(9) we get, We can wrte the souton n the form, = e px u = ~ ~p E + m u = e px N Here s an arbtrary 2 components vector and N s normazaton constant to be determned ater. (b) Negatve energy souton p 0 = E = p ~p 2 + m 2, Smary, the souton can be wrtten as, = e px N The standard notaton for these 4-component coumn vector, spnors are, u(p:s) = N s v(p; s) = e px N s N = p E + m Drac conjugate One of the unua features of Drac equaton n momentum space ( /p m) (p) = 0 s not hermtan. Ths s because n the Hermtan conjugate y (p)( /p y m) = 0 0 s are not hermtan, But we can wrte y 0 = 0 y = y = 0 0 Then y (p)( 0 0 p m) = 0 or y (p) 0 ( p m) = 0 ( /p m) = 0 where = y 0 Drac conjugate 2.3 Drac equaton under Lorentz transformaton Unke the Ken-Gordon equaton whch s nvarant under Lorentz transformaton, Drac equaton s not. We now study how Drac equaton ( @ m) (x) = 0 behaves under Lorentz transformaton x x 0 = x
In the new coordnate system, the Drac equaton s of the form ( @ 0 m) 0 (x 0 ) = 0 (0) Note that we have used the same matrces (In genera, d erent sets of -matrces are reated by smarty transformaton - Pau s theorem). Assume that 0 (x 0 ) and (x) are reated by a near transformaton, 0 (x 0 ) = S (x) We need to nd the operator S: Invert the Lorentz transformaton x = x 0 =) @ @x 0 = @ @x @x @x 0 = @ @x Then Eq(0) becomes ( @ m)s (x) = 0 or ((S S) @ m) (x) = 0 In order for ths equaton to be equvaent to the orgna Drac equaton, we requre (S S) = or (S S) = () To construct S, we consder n ntesma transformaton = g + + O( 2 ) wth j j << Pseudo-othogonaty mpes Wrte S as S = Identfyng coe cent of " ; we get g (g + )(g + ) = g + = 0; =) antsymmetrc 4 + O( 2 ) then S = + 4 : 4 4 matrces. Then Eq() yeds, ( + 4 ) ( It s straghtforward to check that gven by 4 ) = (g + ) 4 [ ; ] = = 2 (g g ) [ ; ] = 2(g g ) (2) = 2 [ ; ] satsfy Eq(2). It s not hard to see that for the nte Lorentz transformaton, we have Note that 0 (x 0 ) = S (x); wth S = exp[ y = 0 0 and S y = 0 S 0 4 ] (3)
Thus S s not untary. From 0 (x 0 ) = S _ we get y 0 (x 0 ) = y S y = y 0 S 0 ; or 0 (x 0 ) = _ (x)s Ths shows that has smpe transformaton property. Fermon bnears Even though the Drac wave functon has rather compcate transformaton under the Lorentz transformaton as shown n Eq(3). The fermon b-nears (x) (x) has rather smpe transformaton. For exampe, 0 (x 0 ) 0 (x 0 ) = (x)s S (x) = (x) (x) Ths means that the combnaton (x) (x) s Lorentz nvarant. Smary, we can work out the other combnaton to get the foowng resuts. 5 5 4-vector axa vector 2nd rank antsymmetrc ensor pseudo scaar where 5 = 0 2 3 Hoe Theory ( Drac 9 ) To sove the probem wth negatve energy states, Drac proposed that the vaccum s the one n whch E < 0 states are a ed and E > 0 states are empty. Then Pau excuson prncpe w prevent an eectron from movng nto E<0 states. In ths pcture hoe n the negatve sea,.e. absence of an eectron wth charge jej wth negatve energy jej s equvaent to a presence of a partce wth energy jej and charge + jej. Ths new partce s caed "postron" and sometme aso caed ant partce. Ths correspondence of partce and ant-partce s caed charge conjugaton: