An Introduction to Quantum Field Theory (Peskin and Schroeder) Solutions

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1 An Introduction to Quantum Field Theory Pekin and Schroeder Solution Andrzej Pokraka December, 7 Content The Dirac Equation. Lorentz grou.... Gordon Identity Sinor roduct Majorana fermion Suerymmetry Fierz tranformation....7 Dicrete ymmetrie of the Dirac field Bound tate The Dirac Equation. Lorentz grou The Lorentz commutation relation are J µν,j ρσ ]ig νρ J µσ g µρ J νσ g νσ J µρ + g µσ J νρ. a Define the generator of rotation and boot a L i ϵijk J jk and K i J i, where ijk i a ermutation of 3. AninfiniteimalLorentztranformationcanthenbewritten Φ iθ L iβ KΦ. Write the commutation relation of thee vector oerator exlicitly. Show that the combination J + L + ik and J L ik

2 commute with one another and earately atify the combination relation of angular momentum. Proof: The L oerator commutation relation are L i,l j] L i L j L j L i 4 ϵ ikl J kl,ϵ jmn J mn] 4 ϵikl ϵ jmn J kl,j mn] i 4 ϵikl ϵ jmn g lm J kn g km J ln g ln J km + g kn J lm] i ϵ ikl ϵ jmn g lm J kn ϵ ikl ϵ jmn g km J ln ϵ ikl ϵ jmn g ln J km + ϵ ikl ϵ jmn g kn J lm] 4 for the econd and lat term k l i ϵ ikl ϵ jmn g lm J kn ϵ ilk ϵ jmn g lm J kn ϵ ikl ϵ jmn g ln J km + ϵ ilk ϵ jmn g ln J km] 4 i ϵ ikl ϵ jmn g lm J kn + ϵ ikl ϵ jmn g lm J kn ϵ ikl ϵ jmn g ln J km ϵ ikl ϵ jmn g ln J km] 4 for the third and lat term m n i ϵ ikl ϵ jmn g lm J kn + ϵ ikl ϵ jmn g lm J kn ϵ ikl ϵ jnm g lm J kn ϵ ikl ϵ jnm g lm J kn] 4 i ϵ ikl ϵ jmn g lm J kn + ϵ ikl ϵ jmn g lm J kn + ϵ ikl ϵ jmn g lm J kn + ϵ ikl ϵ jmn g lm J kn] 4 iϵ ikl ϵ jmn g lm J kn iϵ ikl ϵ jmn δ lm J kn iϵ ikl ϵ jln J kn iϵ ikl ϵ jnl J kn i δ ij δ kn δ in δ kj J kn i δ ij J kk J ji ij ij iϵ ijk L k where Jαβ kk i δα k δk β δk β δk α and ϵ ijk L k ϵijk ϵ klm J lm ϵkij ϵ klm J lm δ il δ jm δ im δ jl J lm J ij J ji J ij.

3 The K commutation relation are K i,k j] K i K j K j K i J i J j J j J i i g i J j g J ij g ij J + g j J i. Thi i imlified uing roertie of the metric g i, g, g ij and the generator J K i,k j] ij ij Next, we need iϵ ijk L k. L i,k j] ϵikl J kl,j j] i ϵikl g lj J k + g kj J l i ϵikl δ lj J k δ kj J l i ϵ ikj J k ϵ ijl J l i ϵ ijk J k + ϵ ijk J k iϵ ijl K l. Latly we comute the angular momentum commutator J +, J ] L + ik, ] L ik L, L] i L, K]+i K, L]+K, K]} 4 i K, L] i K i,l j] ê i ê j i K i,l i] ] J± i,jj ± L i ± ik i, L j ± ik j] L i,l j] ± i L i,k j] ± i K i,l j] K i,k j]} 4 iϵ ijk L k ± i iϵ ijl K l i iϵ jil K l + iϵ ijk L k} 4 iϵ ijk L k ϵ ijl K l ± ϵ ijl K l + iϵ ijk L k} iϵ ijk L k ϵ ijl K l} i ϵijk L k ± ik k} i ϵijk J k ± 3

4 b The finite-dimenional rereentation of the rotation grou correond reciely the to the allowed value for angular momentum: integer or half-integer. The reult of art a imlie that all finite-dimenional rereentation of the Lorentz grou correond to air of integer or half integer, j +,j,correondingtoairofrereentationofthe rotation grou. Uing the fact that J σ/ in the in-/ rereentation of angular momentum, write exlicitly the tranformation law of the -comonent object tranforming according to the, and, rereentation of the Lorentz grou. Show that thee correond reciely to the tranformation of ψ L and ψ R giving in Proof: The rereentation of the Lorentz grou are denoted by m, n π m,n where m, n are either half-integer or integer. The irreducible rereentation are given by π m,n L i I m+ J n i + J m i I n+ π m,n K i i I m+ J n i J m i I n+. With J σ/ the, rereentation i found to be π m,n L i I I + σi I I + σi I σi I σi π m,n K i i I I σi I i I σi I i σi I i σi. Thu, with L i σ i / and K i iσ i / the tranformation law become Φ, e i ωµν J µν Φ, i ω µνj µν Φ, i ω νj ν i ω iνj iν Φ, i ω J i ω ij i i ω ij i i ω ijj ij i ω ij i i ω ij i i ω ijj ij Φ, iω i K i i ω ijϵ ijk L k Φ, iβ K iθ L Φ, β σ i θ σ Φ, where we have defined β i ω i ω i and θ k ω ij ϵ ijk. Now for the, rereentation we have Φ, π m,n L i I σi σ i + I I I + I σi π m,n K i i I σi σ i I I i I I i σi I. 4

5 Thu, with L i σ i / and K i iσ i / the tranformation law become Φ, Uon comarion with equation 3.37 we identify e i ωµνj µν Φ, iβ K iθ L Φ, + β σ i θ σ Φ,. Φ, ψ L and Φ, ψ R. Thu, the left- and right-handed inor tranform according toearaterereentationofthelorentzgrou. c The identity σ T σ σσ allow u to rewrite the ψ L tranformation in the unitarily equivalent form ψ ψ + iθ σ/+β σ/, where ψ ψl T σ.uingthilaw,wecanrereenttheobjectthattranforma, a a matrix that ha the ψr tranformation law on the left and imultaneouly, the tranoed ψ L tranform on the right. Parametrize thi matrix a V + V 3 V iv V + iv V V 3. Show that the object V µ tranform a a 4-vector. Proof: Left-handed inor tranform according to ψ L β σ i θ σ ψ L. With ψ ψ T L σ we verify the tranformation law ψ β σ i θ σ ψ L T σ ψ T L σ σ β σ i θ σ T σ ψ σ β σt i θ σt σ ψ + β σ + i θ σ. Now we are intereted in the tranformation roertie of a, object. We arameterize the, object a the matrix V Φ, + V 3 V iv V + iv V V 3 V µ σ µ 5

6 where it will be hown that V µ i a 4-vector. Alying the tranformation to Φ, we get Φ, + β σ i θ σ Φ +, β σ + i θ σ + β iθ σ V µ σ µ + β σ + iθ σ V µ σ µ + β iθ σv µ σ µ + V µ σ µ β + iθ σ + O θ,β V µ σ µ + V µ β σσ µ + σ µ σ V µ iθ σσ µ σ µ σ V µ σ µ + V µ βi σ i,σ µ } V µ iθi σ i,σ µ ] V µ σ µ + V βi σ i,σ } + V j βi σ i,σ j } V iθi σ i,σ ] V j iθi σ i,σ j ] V µ σ µ + V βi σ i, I} + V j βi σ i,σ j } V iθi σ i, I] V j iθi σ i,σ j ] V µ σ µ + V βi σ i + V j βi δ ij V j iθi iϵ ijk σ k V µ σ µ + V β i σ i V i β i + V j θ i ϵ ijk σ k where σ i, I} σ i, σ i, I], σ i,σ j } δ ij, σ i, I] iϵ ijk σ k. Alo note that β i V j δ ij β i V i becaue g ij δ ij. Recall that we have defined the anti-ymmetric tenor ω i β i and ω ij ϵ ijk θ k. Inerting thee exreion into the above, we have V µ σ µ + V β i σ i V i β i + V j θ i ϵ ijk σ k V µ σ µ + V ω i σ i V i ω i + V j ω jk σ k V µ σ µ + V ω i σ i + V i ω i σ + ω ij V i σ j V µ σ µ + ω µν σ ν V µ δ ν µ + ω ν µ Vν σ µ. We would like to how that thi i identical to equation 3.9inP&S.P&Saertthata4-vectorV µ tranform a follow V α δβ α i ω µν J µν α β V β for J µν αβ δ µ αδ ν β δν αδ µ β.withthidefinitionofj µν α β the tranformation condition become δ α β i ω µν J µν α β δ α β i ω µν J µν γβ g γα δ α β i ω µν δ γ µ δβ ν δγδ ν µ β g γα δ α β i ω γβ ω βγ g γα δ α β iω γβg γα δ α β iω α β. Thi i the identical reult obtained from tranforming V µ σ µ.thu,weeethatv µ i indeed a 4-vector. 6

7 . Gordon Identity Derive the Gordon identity, where q. Proof: The comutation i traightforward: ū γ µ u ū µ + µ ] m + iσµν q ν u, m σ µν q ν i γµ,γ ν ] ν ν i γµ γ µ i γµ γµ i gµν ν γ µ γ µ i γµ gµν ν + γ µ i µ γ µ i γ µ µ ū µ + µ ] m + iσµν q ν u ū µ + µ µ γ µ +γ µ µ m m ū γ µ + γ µ ] u m mγ ū µ + γ µ ] m u m ū γ µ u where we have ued the Fourier tranformed Dirac equation and it adjointequation m u and ū m. ] u.3 Sinor roduct Together i Problem 5.3 and 5.6 we develo an efficient comutational method for rocee involving male article. Let k µ, kµ be fixed 4-vector atifying k, k, k k. Define baic inor in the following way: Let u L be the left-handed inor for for a fermion with momentum k.letu R k u L.Then,foranyuch that i lightlike define u L k u R and u R k u L. Thi et of convention define the hae of inor unambiguouly excet when i arallel to k. a Show that k u R.Showthat,foranylightlike, u R. Proof: k u R k k u L k µ kν γ µ γ ν u L k µ kν g µν γ ν γ µ u L k k u L k k u L 7

8 where we have ued the Dirac equation for a male article k u L and the dot roduct k k.nowforanylightlike 4-momentum the definition of the inor above atify the male Dirac equation u R k u L k u L. b For the choice k E,,, E, k,,,, contructu L, u R, u L, andu R exlicitly. Proof: For thi roblem we will need the oerator k which in the Chiral or Weyl bai i given by k k k σ k + k σ k k 3 k ik k + ik k + k 3 k + k 3 k ik. k + ik k k 3 From equation 3.5 in P&S the Dirac inor for are given by k σξ u k k σξ E σ + σ 3 ξ E σ σ 3 ξ ξ E ξ ξ E T } T where ξ, and σ σ σ.takingξ ξ u L E E T the left-handed inor i while the right-handed inor i given by taking ξ u R E T E. 8

9 The chiral inor for any momentum are then given by u L k u R E E i + i i + i 3 + i + 3 u R k u L E E i 3 i + i i + i 3 c Define the inor roduct, and t,for, lightlike, by, ū R u L and t, ū L u R. Uing the exlicit form for the u λ given in art b, comute the inor roduct exlicitly and howthatt,, and,,.inaddition,howthat,. Thu the inor roduct are the quare root of 4-vector dot roduct. Proof: 9

10 , ū R u L i i i + i i i i + i 3 + i + 3 i i ,, Since and are lightlike i i i + i i 3 i i + 3 i i 3 i we have,

11 .4 Majorana fermion Recall form Eq. 3.4 that one can write a relativitic equation for male -comonent fermion field that tranform a the uer two comonent of a Dirac inor ψ L.Callucha-comonentfieldχ a x, a,. a Show that it i oible to write an equation for χ x a a maive field in the following way: i σ χ imσ χ. That i, how, that thi equation i relativitically invariant and, econd, that it imlie the Klein-Gordon equation, + m χ.thiformofthefermionmaicalledmajoranamaterm. Proof: The unitary matrix, Λ,whichLorentztranformafermionfieldinorigivenbyequation3.3 where Λ θ, β ex i ω µνs µν S µν i 4 γµ,γ ν ] Λ, Λ, i the Lorentz tranformation generator for inor, β i ω i ω i and θ k ω ij ϵ ijk. Thi tranformation matrix i block diagonal. Thi i een from the block diagonal form of the generator of boot 3.6 and 3.7 S i i σ i σ i where the in oerator of Dirac theory i S ij ϵijk Σ k Σ k σ i σ i. The block Λ, and Λ are the left and right handed rereentation of the Lorentz grou. A inor which, tranform according to Λ i called a Dirac inor. Becaue of the block diagonal form left and right inor tranform in different rereentation of the Lorentz grou Λ Λ ψ, ψl Λ,ψ L Λ, ψ R Λ, ψ. R Under infiniteimal rotation by θ and boot β the tranformation law for the left and right handed inor are β σ i θ σ Λ, Λ, β σ + i θ σ. Now equation 3.4 tell u that we can write the relativitic field equation for a male fermion which tranform under Λ, a i σ χ.furthermore,weknowhowγµ tranform. Thi i given by equation 3.9 Λ γ µ Λ Λµ ν γν. On the left the Lorentz tranformation are acting on the inor indice of γ µ while on the right the Lorentz tranformation

12 act on the ace time index µ. Wecanworkoutthetranformationfor σ becaue σ µ i jut the off-diagonal block of γ µ Λ µ ν σν Λ µ ν σ ν Λ µ νγ ν Λ γ µ Λ Λ,σµ Λ, Λ, σµ Λ, From the above we conclude that σ µ Λ ν µ Λ, σν Λ,. Wewillaloneedthefactthatifχtranform according to Λ, then σ χ tranform under Λ, i.e., χ i left-handed and σ χ i right-handed. We rove thi here. Alying Λ, to σ χ yield σ χ σ β σ i θ σ χ σ β σ + i θ σ + β σ i θ σ χ Λ, χ. χ. Lorentz tranform; review of a calar field χ no atial orientation.. χ: UnderLorentztranformationthecalarχ tranform a χ x χ x χ Λ x where x x Λx.. µ χ:underlorentztranformation µ χ tranform a µ χ x µ χ x xν Λ x µ x ν χ β α x α Λ ν ρ x ρ Λ x µ x ν χ β Λ Λ νρ gρ µ x ν χ β Λ ν µ νχ Λ x α x α α x α where x x Λx. Noticethatthiiooitetohowthevectorx µ tranform x µ x µ Λ µ νx ν.thihow that while x µ i a contravariant vector, µ i a covariant vector. Alying the Lorentz tranformation to the field equation we have i σ χx imσ χ x i σ µ Λ νµ νλ,χ Λ x imλ, σ χ Λ x where we know how µ χ tranform. Uing σ µ Λ ν µ Λ, σν Λ, we conclude i σ χx imσ χ x iλ, σν Λ, νλ,χ Λ x imλ, σ χ Λ x Λ, i σ ν ν χ Λ x imσ χ Λ x }. Thu, we ee that the field equation i σ χx imσ χ x i relativitically invariant. Next, we mut how that i σ χx imσ χ x imlie the Klein-Gordon equation. The equation i σ χ x+ imσ χ x imlie χ x m σ σ χx. Subtitutionintothecomlexconjugateofi σ χx imσ χ x

13 we have a required. i σ µ µ χ x imσ χ x i σ µ µ m σ σ ν ν χ x imσ χ x σ σ σ µ σ σ ν µ ν χ x+m σ χ x σ σ µ σ ν µ ν χ x+m σ χ x σ µ σ ν µ ν χ x m χ x g µν µ ν χ x m χ x χ x+m χ x b Doe the Majorana equation follow from a Lagrangian? The ma term would eem to be the variation of σ ab χ a χ b ; however, ince σ i antiymmetric, thi exreion would vanih if χ x were an ordinary c-number field. When we go to quantum field theory, we know that χ x will become an anti-commuting quantum field. Therefore, it make ene to develo it claical theory by conidering χ x a a claical anti-commuting field, that i, a a field that take a value Gramann number which atify αβ βα for any α, β. Note that thi relation imlie that α.agramannfieldξ x can be exanded in a bai of function a ξ x n α n φ n x, where the φ n x are orthogonal c-number function and the α n are a et of indeendent Gramann number. Define the comlex conjugate of a roduct of Gramann number to revere the order: αβ β α α β. Thi rule imitate the Hermitian conjugation of quantum field. Show that the claical action, S d 4 x χ i σ χ + im χ T σ χ χ σ χ ], where χ χ T ireals S, andthatvaryingthis with reect to χ and χ yield the Majorana equation. Proof: The trick to thi roblem i realizing that S C and o S S.Thecomlexconjugateoftheactionitherefore χ S d 4 x i σ µ µ χ im ] + χ T σ χ χ σ χ d 4 x µ χ i σ µ χ im χ σ χ T χ σ χ ] d 4 x µ χ i σ µ χ im χ σ χ χ T σ χ ] ] d 4 x µ χ im i σ µ χ + d 4 x χ T σ χ χ σ χ ] IBP ds µ iχ σ µ χ ] d 4 x χ i σ µ µ χ ] im + d 4 x χ T σ χ χ σ χ ] d 4 x χ i σ µ µ χ ] im + d 4 x χ T σ χ χ σ χ ] S. 3

14 Notice that we have aumed that the field χ vanihe at the boundary of the integration region. Recall that the action i the integral of the Lagrange denity L. Inourcaewehave L χ i σ χ + im χ ai σ µ ab µχ b + im χ T σ χ χ σ χ σ ab χ a χ b σ abχ aχ b. The Euler-Lagrange equation are derived from variation oftheaction,requiringthattofirtorderδχ that δs. However, we cannot jut ue the Euler-Lagrange equation becaue χ i a Gramann field. Varying the action with reect to χ yield δs S χ + δχ ] S χ ] d 4 xl χ + δχ, µ χ + µ δχ d 4 xl χ, µ χ d 4 x χ a + δχ a i σµab µχ b + im ] σab χ aχ b χ a + δχ a χ b + δχ b d 4 x χ ai σ µ ab µχ b + im ] σ ab χ a χ b χ aχ b d 4 x δχ a i σµab µχ b + im ] σab χ a δχ b δχ a χ b d 4 x δχ ai σ µ ab µχ b + im σ ab δχ aχ b σabδχ aχ ] b d 4 xδχ a i σ µ ab µχ b imσabχ ] b Since thi hold for all integration volume i σ µ µ χ imσ χ which i the Majorana equation. Similarly, with the action S S d 4 x µ χ i σ µ χ im χ σ χ χ T σ χ ] and varying with reect to χ yield the comlex conjugate equation i σ µ µ χ imσ χ. c Let u write a 4-comonent Dirac field a ψ x ψl ψ R, and recall that the lower comonent of ψ tranform in a way equivalent by a unitary tranformation to the comlex conjugate of the rereentation ψ L.Inthiwaywecanrewritethe4-comonentDiracfieldintermoftwo-comonent inor: ψ L x χ x and ψ R x iσ χ x. Rewrite the Dirac Lagrangian in term of χ and χ and note the form of the ma term. 4

15 Proof: The Dirac Lagrange denity i L D ψ iγ µ µ m ψ. TheDiracinoriψ T ψ L ψ R χ iσ χ T. where we have ued σ σ µ σ σ µ. L D ψ iγ µ µ m ψ χ iχ T m iσ µ µ χ σ i σ µ µ m iσ χ iχ T σ χ mχ + iσ µ µ iσ χ i σ µ µ χ miσ χ iχ T σ mχ + iσ µ µ iσ χ + χ i σ µ µ χ miσ χ iχ T σ σ σ χ + iχ σ χ + imχ T σ χ miχ σ χ iχ σ χ + iχ T σ σσ χ χ + im T σ χ χ σ χ iχ σ χ + iχ T σ χ χ + im T σ χ χ σ χ iχ σ χ + iχ T σ χ + im χ T σ χ χ σ χ Since the econd term, iχ T σ χ,ijutanumberwecanimlyitfurtheruingthefactthat σµ T σ µ iχ T σµ µ χ iχ T σµ µ χ T i µ χ σµ χ i µ χ σµ χ. Thu, the Lagrange denity i L D iχ σ χ + iχ σ χ + im χ T σ χ χ σ χ. Notice that if χ χ we have twice the Majorana Lagrange denity from art b: L M iχ σ χ + iχ σ χ + im χ T σ χ χ σ χ iχ σ χ + im χ T σ χ χ σ χ }. d Show that the action of art c ha a global ymmetry. Comute the divergence of the current J µ χ σ µ χ and J µ χ σµ χ χ σµ χ, for the theorie of art b and c, reectively, and relate your reult to the ymmetrie of thee theorie. Contruct atheoryofnfreemaive-comonentfermionfieldwitho N ymmetry that i, the ymmetry of rotation in an N-dimenional ace. Proof: The action S i invariant under the U ymmetry ψ e iθ ψ. ] d 4 x iχ σ χ + iχ σ χ + im χ T σ χ χ σ χ 5

16 To verify thi we write χ iσ χ ψ e iθ ψ e iθ χ iσ e iθ χ. Thu, in term of the χ the tranformation ψ e iθ ψ i χ e iθ χ and χ e iθ χ. Becaue in each term of the action χ i artnered with either it comlex conjugate or χ and vice vera for χ the action i invariant under the U ymmetry χ e iθ χ and χ e iθ χ. We now comute the divergence of the current, J µ χ σ µ χ J µ χ σµ χ χ σµ χ, for the theorie of art b and c. To derive thee current we aly equation.. Part c: Under the U tranformation the Lagrangian of art c i invariant. To comute the current we need to know the equation of motion. The equation of motion i obtained from the Euler-Lagrange equation L µ L. µ χ i χ i With χ i χ the EL equation yield µ iχ σµ imχ T σ µ χ σ µ mχ T σ σ µ µ χ mσ χ T σ µ µ χ mσ χ. Similarly, for χ i χ µ iχ σµ + im µ χ χ χ T σ χ σ µ m χ T χ σ χ µ χ σ µ mχ T σ σ µ µ χ mσ χ. Now the divergence of the current can be comuted µ J µ µ χ σµ χ χ σµ χ µ χ σµ χ µ χ σµ χ + χ σµ µ χ χ σµ µ χ σ µ µ χ χ σ µ µ χ χ + χ σµ µ χ χ σµ µ χ mσ χ χ mσ χ χ + χ mσ χ χ mσ χ mχ T σ χ mχ T σ χ + mχ σ χ mχ σ χ. 6

17 Since the divergence of the current vanihe, the current i conerved under U tranformation. Part c: Under the U tranformation the Lagrangian of art b i NOT invariant χ i σ χ + im χ T σ χ χ σ χ χe iφ i σ χe iφ + im χe iφ T σ χe iφ χe iφ σ χe iφ χ i σ χ + im χ T σ χe iφ χ σ χ e iφ L. However, if m then the Lagrangian i ymmetric under U hae rotation. The equation of motion are σ µ µ χ mσ χ µ χ σ µ mχ T σ. Thu the divergence of the current i Like the Lagrangian the current i only conerved if m. µ χ σ µ χ µ χ σ µ χ + χ σ µ µ χ mχ T σ χ + χ mσ χ mχ T σ χ + mχ σ χ. For the lat art of thi quetion we contruct a theory of N free maive -comonent fermion field with ON ymmetry the ymmetry of rotation in an N-dimenional ace. Each free maive article i decribed by the Lagrangian of art b L a χ ai σ χ a + im χ T a σ χ a χ aσ χ a where a,,..., N}. ThetotalLagrangianitheumoftheindividualLagrangian L a χ a i σ χ a + im χ T a σ χ a χ a σ χ a. Each fermion field atifie the equation of motion σ µ µ χ a mσ χ a µ χ a σ µ mχ T a σ. To have ON ymmetry the Lagrangian mut be invariant under rotation in N-dimenional ace. Let the rotation oerator be denoted by R ab where a, b are the comonent of the oerator. Alied to the b th fermion field the rotation oerator take the b th fermion field to the a th fermion field R ab χ b χ a. Since R i a rotation, it i an orthogonal matrix i.e., R R T and R ab R. Alication of R to the Lagrangian yield L R ab χ b i σ R ab χ b + im R ab χ b T σ R ab χ b R ab χ b σ R ab χ b ab ab ab b R ab R abχ b i σ χ b + im R ba R ab χ b i σ χ b + im χ b i σ χ b + im RabR T ab χ T b σ χ b R ab R abχ b σ χ b R ba R ab χ T b σ χ b R ba R ab χ b σ χ b χ T b σ χ b χ b σ χ b L. 7

18 Thu, we ee that the Lagrangian for maive two-comonent fermion i invariant under ON tranformation. e Quantize the Majorana theory of art a and b. That i, romoteχ x to a quantum field atifying the canonical anti-commutation relation χ a x,χ b y } δ ab δ 3 x y, contruct a Hermitian Hamiltonian, and find a rereentation ofthecanonicalcommutationrelationthatdiagonalize the Hamiltonian in term of a et of creation and annihilation oerator. Hint: Comareχ x to the to two comonent of the quantized Dirac field. Proof: The book ugget comaring χ to the uer two comonent of the quantized Dirac field d 3 ψ a π 3 u e i x + b v e i x E d 3 π 3 E σξ a σξ ση e i x + b ση e i x. Setting ψ R iσ ψ L in the Dirac Lagrangian yield the Majorana Lagrangian. Thu, we exect the Dirac field to be a olution to the Majorana equation under certain retriction i.e., η iσ ξ. Setting χ equal to ψ L,thequantized Majorana field become d 3 σ χ a π 3 E ξ e i x + b η e i x. The condition of charge conjugation u iγ v and v iγ u lace a retriction on the inor, namely η iσ ξ.thu,themajoranafieldbecome d 3 σ χ a π 3 E iσ η e i x + b η e i x. We will now tet our olution in the Majorana equation of motion i σ µ µ χ imσ χ.therhsbecome i σ µ µ χ i σ µ d 3 σ µ a π 3 E iσ η e i x + b η e i x i σ µ d 3 σ a π 3 E iσ η i µ e i x +i µ b η e i x d 3 i σ π 3 d 3 im π 3 σ E σ σ E where we have ued the fact that σ σ m.thelhsi imσ χ imσ imσ d 3 σ π 3 E σ d 3 π 3 E a σ η e i x + ib η e i x a σ η e i x + ib η e i x 8 a iσ η e i x + b η e i x ia σ η e i x + b η e i x.

19 To deal with the quare root of σ we note that the quare root i defined a a taylor erie. Inerting unity in the form σ σ between each conecutive term of σ, σ n σ σσ and the taylor erie become one in σ intead of σ.withthi in mind the LHS become imσ χ imσ im d 3 π 3 σ d 3 π 3 σ E σ E σ Comaring the LHS and RHS we ee that they are equal only if ia σ η e i x + b η e i x. a iη e i x + b σ η e i x. b a. Thu, the quantized Majorana field i χ d 3 σ a π 3 E iσ η e i x + a η e i x. From thi mode exanion of the Majorana field we can ee that the Majorana article i it own anti-article. To determine the commutation relation of the article oerator a we imlify the commutator } χ a x,χ b y δ ab δ 3 x y in term of a commutator. Simlifying, we have } δ ad δ 3 x y χ a x,χ d y d 3 d3 ] / σ σ π 3 π 3 E r ab E ia σbcη c e i x + a ηbe i x, d 3 d3 ] / σ σ π 3 π 3 E r ab E σbcη c σ feη r f iη bσ feη r f ] / ed ia r σ feη r f e i y + a r ηr ] / ed } a,a r e i x+i y + iσbcη c ηe r } a,a r e i x+i y + η bη r e } i y e e a,a r } e i x i y a,a r } e i x i y ]. Thi i very comlicated and it i not obviou what commutation relation for a yield the RHS. However, ince we are uing a modified Dirac field we would exect that a atifie the uual relation a,a r } a } a,a r },ar π 3 δ r δ. 9

20 Let tet if thee commutation relation are indeed the one we require χ a x,χ d y } When taken at equal time the commutator become χ a x,χ d y } ] / d 3 d3 σ σ π 3 π 3 E r ab E ed } σbc η c σ fe ηr f a,ar e i x+i y + η } b ηr e a,ar e i x i y ] d 3 π 3 d 3 ] / ] / σ σ E r ab E ed ] δ r δ σbc η c σ fe ηr f e i x+i y + ηb ηr e ei x i y d 3 ] / ] / σ σ π 3 E ab E ed σbc η c σ fe η x+ie y y f e iex + η b ηe x ie y y ] eiex d 3 σ π 3 E d 3 π 3 σ E ] / ab ] / ab σ E ] / ed σ bc η c η f σ fe ] / σ bc η c σ fe ηr f ei x i y + η b η e e i x+i y] σ E To imlify thi further we um over and ue the identity, ηa ηb δ ab, to get χ a x,χ d y } d 3 π 3 d 3 π 3 d 3 π 3 σ E ] σ E ] σ E ] / ab σ bc δ cfσ fe e i x y + ad e i x y + ad σ ] / E ] σ ] / ed E ] σ E ed σ e i x y + E σ e i x y + E ] e i x y ad ] e i x y. ad To imlify further we need to know σ ad.here,weworkoutthematrixelementof σ, ] / ] / ab ab η b η e σ E ] / ed e i x y ] ] ] / σ δ be e i x y E ed Subtitution into the commutation relation yield σ ad E σ ad E δ ad i σad. i χ a x,χ d y } d 3 E δ ad i σad i π 3 E d 3 π 3 δ ad δx y δad ei x y + δ ad e i x y e i x y + E δ ad i σad i ] e i x y E ] d 3 i σad i π 3 e i x y + i σ i ] ad e i x y E E

21 where the econd integral i odd in. We now turn to calculating the Hamiltonian for Majorana field. The Hamiltonian i the Legendre tranform of the Lagrangian, H d 3 x π χ L d 3 x iχ χ iχ σ χ + im χ T σ χ χ σ χ where The imlification i traight forward and mey. We jut note the reult here Which i exactly / the Dirac Hamiltonian, H D with the retriction b a. L iχ σ χ + im π L χ iχ. H d 3 π 3 E d 3 π 3 E We imlify the Hamiltonian term by term. The firt term i iχ χ i i χ T σ χ χ σ χ, a a. a a + b b, d 3 d3 ia π 3 π 3 η T σ e i x + a η e i x σ E r d 3 d3 π 3 π 3 E ia η T σ e i x + a η e i x σ E r σ a r E iσ η r e i x + a r ηr e i x σ a r E σ η r e i x + ia r ηr e i x.5 Suerymmetry It i oible to write field theorie with continuou ymmetrie linking fermion and boon; uch tranformation are called uer-ymmetrie. a The imlet examle of a uerymmetric field theory i the theory of a free comlex boon and a free Weyl fermion, written in the form L µ φ µ φ + χ i σ χ + F F. Here F i an auxiliary comlex calar field whoe field equation i F.ShowthatthiLagrangianiinvariantutoa total divergence under the infiniteimal tranformation δφ iϵ T σ χ, δχ ϵf σ φσ ϵ, δf iϵ σ χ,

22 where the arameter ϵ a i a -comonent inor of Gramann number. b Show that the term L mφf + ] imχt σ χ +comlex conjugate i alo left invariant by the tranformation given in art a. Eliminate F from the comlete Lagrangian L+ L by olving it field equation, and how that the fermion and boon field φ and χare given the ame ma. c It i oible to write uerymmetric nonlinear field equation by adding cubic and higher-order term to the Lagrangian. Show that the following rather general field theory, containing the field φ i,χ i, i,..., n, iuerymmetric: L µ φ i µ φ i + χ i i σ χ i + Fi F i W φ] + F i + i W φ] χ T i φ i φ i φ σ χ j + c.c, j where W φ] i an arbitrary function of the φ i,calledtheuer-otential. Fortheimlecaen and W gφ 3 /3, write out the field equation for φ and χ after elimination of F..6 Fierz tranformation Let u i, i,...4, befour4-comonentdiracinor.inthetext,werovedthefierzrearrangementformulae3.78and The firt of thee formulae can be written in 4-comonent notation a +γ ū γ µ 5 +γ 5 +γ u ū 3 γ µ u 4 ū γ µ 5 +γ 5 u 4 ū 3 γ µ u. The above identity lay an imortant art in the weak interaction. Thu, we derive thi equation from 3.78 and 3.79 ū R σ µ u R ū 3R σ µ u 4R ū R σ µ u 4R ū 3R σ µ u R ū L σ µ u L ū 3L σ µ u 4L ū 3L σ µ u 4L ū L σ µ u L. Denoting the handedne oerator by P R +γ5 and P L γ5 we have the following: ū γ µ P R u ū 3 γ µ P R u 4 ū R γ µ u R ū 3R γ µ u 4R ū R σ µ u R ū 3R σ µ u 4R ū R σ µ u 4R ū 3R σ µ u R ū γ µ P R u 4 ū 3 γ µ P R u ū γ µ P L u ū 3 γ µ P L u 4 ū 3L γ µ P L u 4L ū L γ µ P L u L. In fact, there are imilar rearrangement formulae for any roduct ū Γ A u ū3 Γ B u 4, where Γ A, Γ B are any of the 6 combination of Dirac matrice lited in Section 3.4. a To begin, normalize the 6 matrice Γ A to the convention Tr Γ A Γ B] 4δ AB. Thi give Γ A,γ,iγ j,... } ;writeall6elementofthiet. Proof: We already know that the 6 Dirac matrice are orthogonal ubject to the inner roduct Tr AB]. Thu,weonly have to find the normalization contant.

23 . Scalar Γ A : Tr ] 4 thu, the identity i roerly normalized.. Vector Γ A γ µ : Tr γ ] Tr ] 4 Tr γ i ] Tr σ i ] σ i Tr ] 4 Thu, the normalized vector matrice are: γ and iγ i. 3. Tenor Γ A σ µν : Tr σ µν ] i Tr γ µ,γ ν ] ] 4 Tr γµ γ ν γ µ γ ν γ ν γ µ γ µ γ ν γ ν γ µ γ µ γ ν + γ ν γ µ γ ν γ µ ] 4 Tr γµ γ ν γ µ γ ν ] Tr γ ν γ µ γ µ γ ν ]} 4 Tr γµ γ ν γ µ γ ν ]+Tr γ µ γ ν γ µ γ ν ] 4g µν Tr γ µ γ ν ]} 4 4Tr γµ γ ν γ µ γ ν ] 6g µν g µν } 4 6 gµν g µν g µµ g νν + g µν g µν ] 6g µν g µν } 4 6 gµν g µν g µµ g νν ] 4 g µν g µν g µµ g νν } µ ν 4g µµ g νν µ ν µ ν 4 µ ν and both µ, ν are acelike or timelike 4 µ ν and µ i acelike timelike and ν i timelike acelike Thu, the normalized tenor matrice are: σ i, σ i,σ,σ ii. 4. Peudo-vector Γ A γ µ γ 5 : Tr γ µ γ 5] Tr γ µ γ 5 γ µ γ 5] Tr γ µ γ 5 γ 5 γ µ] Tr γ µ γ µ ] 4g µµ 4 µ 4 µ i Thu, the normalized eudo-vector matrice are: γ γ 5,γ i γ Peudo-calar Γ A γ 5 : Thu, γ 5 i already roerly normalized. Tr γ 5 γ 5] 3

24 b Write the general Fierz identity a an equation ū Γ A u ū3 Γ B u 4 C AB CD ū Γ C u 4 ū3 Γ D u with unknown coefficient C AB CD.Uingthecomleteneofthe6ΓA matrice, how that Proof: C,D C AB CD 6 Tr Γ C Γ A Γ D Γ B]. Uing the comletene of the Gamma matrice we write Γ A abγ B cd C AB C Γ C ad C AB D Γ D bc C AB CDΓ C adγ D bc C D CD from which it follow ū Γ A u ū3 Γ B u 4 We now calculate C AB C D.MultilyingbyΓC da ΓD bc we have C,D C AB CD ū Γ C u 4 ū3 Γ D u. Γ C da ΓD bc ΓA ab ΓB cd C,D C AB CD ΓC da ΓD bc ΓC ad ΓD bc which imlie Tr Γ C Γ A Γ D Γ B] C AB CDTr Γ C Γ C] Tr Γ D Γ D] C,D Thu, we ee that C AB CD 6 Tr Γ C Γ A Γ D Γ B]. 6 C,D 6C AB C D. C AB CD δcc δ DD c Work out exlicitly the Fierz tranformation law for the roductū u ū 3 u 4 and ū γ µ u ū 3 γ µ u 4. Proof: Firt, the roduct ū u ū 3 u 4 ū Γ A u ū3 Γ B u 4 where Γ A,B.Thu,weneedallnonzerotraceoftheform C CD 6 Tr Γ C Γ D ] 6 Tr Γ C Γ D] 4 δ CD. Thu, the roduct rule for Γ A,B become ū u ū 3 u 4 C AB CD ū Γ C u 4 ū3 Γ D u C,D 4 δ CD ū Γ C u 4 ū3 Γ D u C,D ū Γ C u 4 ū3 Γ C u 4 C 4

25 where the index C cycle through all 6 Γ C. Next, we examine the roduct ū γ µ u ū 3 γ µ u 4.ThecoefficientC AB CD are determined a uual C µµ CD 6 Tr Γ C γ µ Γ D γ µ ]. Uing the matrice from our normalized bai of art a we define Γ µ γ,iγ.intermofγ µ we have C µµ CD Tr Γ C Γ Γ D Γ ] 6 Tr Γ C Γ Γ D Γ ] Tr Γ C iγ i Γ D iγ i]} i 3 Tr Γ C Γ i Γ D Γ i]}. i Suoe that Γ C γ then C µµ D Tr Γ Γ Γ D Γ ] 3 + Tr Γ Γ i Γ D Γ i]} 6 i Tr Γ D Γ ] 3 Tr Γ Γ D]} 6 δ D. i Similarly, if Γ C γ j then C µµ jd Tr Γ j Γ Γ D Γ ] + 6 Tr Γ j Γ D] δ Dj i 3 Tr Γ j Γ i Γ D Γ i]} i Tr g ij Γ i Γ j Γ D Γ i]} 3 g ij Tr Γ D Γ i] i 4δ Dj Tr Γ D Γ j] 4δ Dj 8δ Dj δ Dj} δdj 3 Tr Γ i Γ j Γ D Γ i]} i 3 Tr Γ j Γ D]} i 3 δdj Thu, we ee that for Γ C γ ν we mut have Γ D γ ν and vice vera. For Γ C,D we have C µµ CD. The remaining matrice are the anti-ymmetric matrice form theetγσ µν,γ µ γ 5,γ 5 }we have 5

26 for: Γ C γ 5 for: Γ C γ γ 5 C µµ 5D Tr Γ 5 Γ Γ D Γ ] + 6 Tr Γ 5 Γ D] 3 6 δ D5 i 3 Tr Γ 5 Γ i Γ D Γ i]} i Tr Γ 5 Γ D]} C µµ ν5,d Tr Γ 5 Γ Γ D Γ ] 3 + Tr Γ 5 Γ i Γ D Γ i]} 6 i Tr γ γ 5 γ Γ D γ ] 3 + Tr γ γ 5 iγ i Γ D Γ i]} 6 i Tr γ 5 γ Γ D] 3 + Tr iγ i γ γ 5 Γ D Γ i]} 6 6 Tr Γ 5 Γ D] + i 3 Tr Γ 5 Γ D]} i 6 6δ5,D δ 5,D for: Γ C γ i γ 5 C µµ j5,d Tr Γ j5 Γ Γ D Γ ] + 6 Tr Γ Γ j5 Γ D Γ ] δj5,d Tr Γ j5 Γ D] 4δ j5,d 4δ j5,d + 3 Tr Γ j5 Γ i Γ D Γ i]} i 3 Tr iγ j γ 5 iγ i Γ D Γ i]} i 3 Tr iγ j iγ i γ 5 Γ D Γ i]} i 3 Tr i g ij iγ i iγ j γ 5 Γ D Γ i]} i 3 Tr g ij γ 5 Γ D Γ i] + i 4δ j5,d +Tr γ 5 Γ D Γ j] + 4δ j5,d +Tr Γ D Γ j5] + 3 Tr iγ i iγ j γ 5 Γ D Γ i]} i 3 Tr Γ j5 Γ D]} i 3 Tr Γ j5 Γ D]} i 3 δj5,d 6

27 for: Γ C Γ µν σ i, σ i,σ,σ ii σ µν,γ α } i γµ γ ν γ ν γ µ,γ α } i γµ γ ν,γ α } i γν γ µ,γ α } i γµ,γ α } γ ν + i γµ γ ν,γ α ] i γν,γ α } γ µ i γν γ µ,γ α ] ig µα γ ν + iγ µ γ ν γ α g να ig να γ µ iγ ν γ µ γ α g µα ig µα γ ν + iγ µ γ ν γ α iγ µ g να ig να γ µ iγ ν γ µ γ α + iγ ν g µα ig µα γ ν ig να γ µ + iγ µ γ ν γ α iγ ν γ µ γ α ig µα γ ν ig να γ µ + iγ µ γ ν γ α iγ ν γ µ γ α.7 Dicrete ymmetrie of the Dirac field Thi roblem concern the dicrete ymmetrie P, C, andt. a Comute the tranformation roertie under P, C, and T of the antiymmetric tenor fermion bilinear, ψσ µν ψ,with σ µν i γµ,γ ν ].Thicomletethetableofthetranformationroertieof bilinear at the end of thi chater. Proof: Starting with the arity tranformation we have P ψσ µν ψp P ψpp i γµ,γ ν ] PPψP η aη a ψ t, x γ i γµ,γ ν ] γ ψ t, x ψ t, x γ i γµ γ ν γ ν γ µ γ ψ t, x. Notice that µ ν otherwie the commutator vanihe. There are two cae:. µ ν where µ, ν,, 3. P ψσ ij ψp i ψ t, x γ i γ j γ j γ i γ γ ψ t, x ψ t, x σ ij ψ t, x.. µ ν and either µ or ν.forν the arity tranformation i Similarly, for µ the arity tranformation i P ψσ i ψp ψ t, x γ i γ i γ γ γ i γ ψ t, x ψ t, x i γ i γ γ γ i γ γ ψ t, x ψ t, x σ i ψ t, x. P ψσ j ψp ψ t, x σ j ψ t, x. 7

28 The arity tranformation of the bilinear tenor can be ummarized by the following formula P ψσ µν ψp µ ν ψ t, x σ µν ψ t, x where µ µ µ,, 3. Next we look at the time reveral tranformation Since γ γ, T ψσ µν ψt T ψtt i γµ,γ ν ] TTψT ψ t, x γ γ 3 i γµ,γ ν ] γ γ 3 ψ t, x ψ t, x γ γ 3 i γµ γ ν γ ν γ µ γ γ 3 ψ t, x γ µ γ γ 3 γ γ 3 γ µ µ γ γ 3 γ µ µ,, 3 and we can imlify the tranformation law. Like the arity tranformation there are two cae:. µ ν and µ, ν,, 3. T ψσ ij ψt ψ t, x γ γ 3 γ γ 3 i γ i γ j γ j γ i ψ t, x ψ t, x σ ij ψ t, x.. µ ν and either µ or ν.forν the time reveral tranformation i T ψσ i ψt ψ t, x γ γ 3 i γ i γ γ γ i γ γ 3 ψ t, x ψ t, x γ γ 3 γ γ 3 i γ i γ γ γ i ψ t, x ψ t, x σ µ ψ t, x. Similarly, for µ the arity tranformation i T ψσ j ψt ψ t, x σ j ψ t, x. The time reveral tranformation of the bilinear tenor can be ummarized by the following formula T ψσ µν ψt µ ν ψ t, x σ µν ψ t, x. Latly, we work out how the bilinear tenor tranform under charge conjugation C ψσ µν ψc C ψcσ µν CψC iγ γ ψ T σ µν i ψγ γ T ψγ γ σ µν T γ γ ψ ψγ γ σ µν T γ γ ψ ] T 8

29 where in the lat line we have ued the fact that iγ γ ψ T σ µν i ψγ γ T i jut a number o the tranoe doe not change anything. The tranoe of the gamma matrice are γ µ T σ µ T σ σ ν µ T σ ν T. Since, σ T σ T σ T σ 3 T σ σ σ σ 3 we ee that γ µ T γ µ µ, γ µ µ, 3. Therefore, γ γ γ µ T γ µ γ γ and we can how that γ γ σ µν T σ µν γ γ.thu,thebilineartenorunderchargeconjugationbecome C ψσ µν ψc ψσ µν γ γ γ γ ψ ψσ µν ψ. The charge conjugation tranformation of the bilinear tenor can be ummarized by the following formula C ψσ µν ψc ψ t, x σ µν ψ t, x. b Let φ x be a comlex-valued Klein-Gordon field, uch a we conidered inproblem.. Findunitaryoerator P, C and an anti-unitary oerator T all defined in term of their action on the annihilation oerator a and b for the Klein-Gordon article and antiarticle that give the following tranformation of the Klein-Gordon field: Pφt, x P φ t, x; Tφt, x T φ t, x; Cφt, x C φ t, x. Find the tranformation roertie of the comonent of the current J µ i φ µ φ µ φ φ under P, C and T. Proof: The quantized Klein-Gordon field i φ t, x φ t, x d 3 π 3 a e i x + b ei x E d 3 π 3 a e i x + b e i x E 9

30 The natural definition for the arity oerator i Pa P a Pb P b. Let u tet thi definition on the KG field. Alication of P yield d 3 PφP π 3 Pa Pe i x +Pb P e i x E d 3 π 3 a e iet+i x + b eiet i x. E Defining E, E,, theaboveequationmaybewrittena PφP d3 π 3 a e ie t i x + b eie t+i x E d 3 π 3 a e i t, x + b ei t, x E φt, x. Thu, P take φt, x to φt, x a required. We note here that Pφ P φ t, x. Thi will be ued later to find the tranformation roertie of the current. Define the time reveral oerator to be anti-linear o that when commuted at a c-number it conjugate the c-number. The time reveral oerator acting on the creation oerator i defined to be: Ta T a Tb T b. Let u tet thi definition on the KG field. Alication of T yield d 3 TφT π 3 Ta Te i x +Tb T e i x E d 3 π 3 a e i x + b e i x E d 3 π 3 a e i t, x + b e i t, x E d 3 π 3 a e i t,x + b e i t,x E a required. We note here that φ t, x Tφ T φ t, x. Thi will be ued later to find the tranformation roertie of the current. Define the time reveral oerator to be anti-linear o that when commuted at a c-number it conjugate the c-number. The time reveral oerator acting on the creation oerator i defined to be: Ca C b Cb C a. 3

31 Let u tet thi definition on the KG field. Alication of T yield d 3 CφC π 3 Ca Ce i x +Cb C e i x E d 3 π 3 b e i x + a e i x E d 3 π 3 b e i x + a ei x E a required. We note here that φ t, x Cφ C φt, x. Thi will be ued later to find the tranformation roertie of the current. Now, we can find the tranformation roertie of the current. Inadditiontoknowinghowφ and φ tranform under P, T,andC, weneedthetranformationruleforthederivative µ.paritytakex x and t t and therefore, and.timereveraltakex x and t t and therefore, and.chargeconjugationdoenot effect the derivative. With g µµ defined uch that there i no um we can write the tranformation of the derivative in a comact way, Thu, under P, T,andC, thecurrentbecome P µ P g µµ µ T µ T g µµ µ C µ C µ. PJ µ t, xp i Pφ t, xp µ Pφt, xp µ Pφ t, xppφt, xp i φ t, xg µµ µ φt, x g µµ µ φ t, x φt, x µ J µ t, x TJ µ t, xt i Tφ t, xt µ Tφt, xt T µ Tφ t, xttφt, xt i φ t, x g µµ µ φ t, x g µµ µ φ t, x φ t, x µ J µ t, x CJ µ t, xc i Cφ t, xc µ Cφt, xc µ Cφ t, xccφt, xc i φt, x µ φ t, x µ φt, x φ t, x i φt, x µ φ t, x µ φt, x φ t, x J µ t, x. c Show that any Hermitian Lorentz-calar local oerator built from ψ x, φ x, and their conjugate ha CPT +. Proof: A Hermitian, Lorentz invariant oerator mut have φ aired with φ and ψ aired with ψ. Additionally, the Lorentz indice mut be contracted to form a Lorentz calar. Looking at table in P&S we can ee that any combination of thee bilinear which yield a Hermitian, Lorentz invariant will have CPT. 3

32 .8 Bound tate Two in-/ article can combine to a tate of total in either or. The wavefunction for thee tate are odd and even, reectively, under the interchange of the two in. a Ue thi information to comute the quantum number under P and C of all electron-oitron bound tate with S, P, or D wavefunction. The P Schrodinger equation i the Schrodinger equation for aarticleofelectricchargee and reduced ma µ m e / in the Coulomb otential V x α/ x where e < itheelectroncharge,m e i the electron ma, α /37 i the fine tructure contant and x i the earation ditance between the electron and oitron in P. Intead of directly olving the P Schrodinger equation, the oition ace wave functionofpcanbeobtainedbytakingadvantageofthe imilarity between the Schrodinger equation for P and that for hydrogen. Secifically, the oition ace wave function can be obtained by relacing the hydrogen Bohr radiu with the P Bohr radiu in the hydrogen oition ace wave function. Thee wave function are characterized by the rincial quantum number or energy quantum number, n, theorbitalangularmomentumquantumnumber, l, andtheorbitalangular momentum rojection quantum number, m l.theoitionacewavefunctionare ψ nlml x 3 n l! na n n +!] 3 e x /na l x L l+ n l na where a /m e α i the Bohr radiu of P, L m n m are aociated Laguerre olynomial and Y m l l x Yl m θ, φ,. na are herical harmonic. Now that we have the oition ace wave function of P, we can contructthepboundtate. Inthecentreofma frame, the P bound tate can be exreed by Ψ nlml ;,m d 3 k m P π 3 ψ nlm l k k; k;, m,. me me where i the total in of the P bound tate and m i the in rojection along the z-axi. The momentum ace wavefunction, ψ nlml k d 3 xe ik r ψ nlml x,.3 give the amlitude for finding P in a given configuration where the electron ha momentum k. Thefreetateinequation. i k; k;, a k b k a k b k /, for -P m and k; k;,m for o-p where i the vacuum tate. Alying the arity oerator, we obtain P Ψ nlml ; me me a k b η a me k for m, a k b k + a k b k / for m, a k b k for m, d 3 k π 3 ψ nlm l kp k; k; d 3 k π 3 ψ nlm l kp d 3 k π 3 ψ nlm l kp a k b k a k b k P a k b k a k b k P 3

33 where we have aumed that the vacuum tate i even under arity. To imlify the above further, note from equation.3, that the arity of ψ nlml k i the ame a ψ lnml x. Therefore, we can invert the atial coordinate in. to obtain the arity of the momentum ace wavefunction. The only art of. thatienitivetouchaninverioni the herical harmonic, Y lm θ, φ l Y lm θ, φ. Thu,-Ptranforma under arity. Similarity, o-p tranform a P Ψ nlml ; l+ Ψ nlml ; P Ψ nlml ;m l+ Ψ nlml ;m under arity and ha an P eigenvalue of l+. Notice that the relative hae difference between the fermion and anti-fermion inverion hae, ηb η a,contributeafactorof to the arity of P -P and o-p in addition to the arity of the wave function; thi extra factor i called the intrinic arity of a fermion--anti-fermion ytem. The remaining C and T eigenvalue of P are obtained in a imilar manner and are lited below in Table. Dicrete Tranform P Eigenvalue P l+ C l+ T + Table : P, C and T eigenvalue of the P tate Ψ nlml ;m.here,l i the orbital angular momentum quantum number and i the in angular momentum quantum number. b Since the electron-hoton couling i given by the Hamiltonian H d 3 xea µ j µ, where j µ i the electric current, electrodynamic i invariant to and C if the comonent of the vector otential have the ame P and C arity a the correonding comonent of j µ. Show that thi imlie the following urriing fact: The in- ground tate of oitronium can decay to hoton, but the in- tate mut decay to 3 hoton. Find the election rule for the annihilation of higher oitronium tate, and for -hoton tranition between oitronium level. 33