Solutions 4: Free Quantum Field Theory

Transcription

1 QFT PS4 Solutions: Free Quantum Field Theory 8//8 Solutions 4: Free Quantum Field Theory. Heisenberg icture free real scalar field We have φt, x π 3 a e iωt+i x + a e iωt i x ω i By taking an exlicit hermitian conjugation, we find our result that φ φ. You need to note that all the arameters are real: t, x, are obviously real by definition and if m is real and ositive semi-definite then ω is real for all values of. Also â â is needed. ii Since π φ t φ classically, we try this on the oerator to find d 3 ω πt, x i π 3 a e iωt+i x a e iωt i x iii In this question as in many others it is best to leave the exression in terms of commutators. That is exloit [A + B, + D [A, + [A, D + [B, + [B, D 3 as much as ossible. Note that here we do not have a sum over two terms A and B but a sum over an infinite number, an integral, but the rincile is the same. The second way to simlify notation is to write e iωt+i x e ix so that we choose 0 +ω. The simlest way however is to write  â e iωt+i x and  â e iωt+i x. You can quickly check these satisfy the same commutation relations as â and â. In fact  e iht â e iht â t for a free field, i.e. we have just alied time evolution to the free field Heisenberg icture oerators. We have the usual commutation relations for the annihilation and creation oerators with continuous labels here the three-momenta a, a q π 3 δ 3 q, [a, a q a, a q 0. 4 The first field commutation relation is then d 3 d 3 q [φt, x, πt, y π 3 ω π 3 i ωq [ a e iωt+i x + a e iωt i x, a q e iωqt+iq y a qe iωqt iq y 5 d 3 q ω i π 3 π 3 4ω q e iω ωqt+i x iq y a, a q + e iω ωqt i x+iq y a, a q 6 d 3 q ω i π 3 π 3 e iω ωqt+i x iq y π 3 δ 3 q 4ω q +e iω ωqt i x+iq y π 3 δ 3 q 7 i π 3 e +i x y + e i x y 8 i δ3 x y 9 Note that this gives the annihilation and creation oerators units of energy 3 in natural units. The usual a i, a j iδ ij is for the case of discrete labels.

2 QFT PS4 Solutions: Free Quantum Field Theory 8//8 Thus we find the canonical equal-time commutation relation [φt, x, πt, y iδ 3 x y. 0 For the commutators ˆφt, x, ˆφt, y [ˆπt, x, ˆπt, y 0. we see that they are of similar form ˆφt 0, x, ˆφt 0, y [ˆπt 0, x, ˆπt 0, y [ π 3 â e +i x + â e i x, ω d 3 q π 3 â q e +iq y + â qe iq y ωq [ d 3 π 3 i ω â e +i x â e i x, d 3 q π 3 i ωq â qe +iq y â qe iq y 3 Thus all we need to show that d 3 q [ s ω ω q s/ â e +i x + sâ e i x, â q e iq y + sâ qe +iq y, 4 is zero where s ± as the field momenta commutator is roortional to +. Using 3 and 4 we find exactly as before that s d 3 q ω ω q s/ [â, â q e +i x+iq y + s â, â q e +i x iq y +s [â, â q e i x+iq y + â, â q e i x iq y 5 d 3 q sδ qe +i x y sδ qe i x y 6 s ω s ω s e +i x y e i x y 7 If you now change integration variable in the second term to you will find it is of exactly the same form as the first term and these cancel giving zero as required. iv Given ω + m then ω ω and hence d 3 x φt 0, xe iq x d 3 x d3 π 3 ω a e i x + a e i x e iq x 8 d 3 x d3 π 3 ω a e i+q x + a e iq x 9 π 3 a π 3 δ 3 q + + a π 3 δ 3 q ω a q + a q ωq 0

3 QFT PS4 Solutions: Free Quantum Field Theory 8//8 3 d 3 x πt 0, xe iq x i d 3 x d3 ω π 3 a e i x a e i x e iq x i d 3 x d3 ω a e i+q x a e iq x 3 i i ωq π 3 π 3 ω a q a q a π 3 δ 3 q + a π 3 δ 3 q 4 5 Thus solving for a q and a q gives a q d 3 x e iq x ωq φ0, x + i π0, x ωq a q d 3 x e iq x ωq φ0, x i π0, x ωq Hence since [φx, φy [πx, πy 0 a, a q d 3 x e i x d 3 y e iq y 9 [ ω φ0, x + i ωq π0, x, ω φ0, y i π0, y 30 ωq d 3 x d 3 y e i i x+iq y ω ωq [φ0, x, π0, y + i [π0, x, φ0, y 3 4ω q 4ω d 3 x d 3 y e i x+iq y ω δ 3 x y + 4ω q d 3 x e iq x ω ωq + 4ω q 4ω ωq 4ω δ 3 y x π 3 δ 3 q v Taking the derivative we have φx i π 3 a e i x a e i x 35 ω

4 QFT PS4 Solutions: Free Quantum Field Theory 8//8 4 so P Now a a + a a d 3 x πx φx 36 ω d 3 x d3 d 3 q π 3 π 3 a e i x a e i x q a q e iq x a qe iq x 37 ωq d 3 x d3 d 3 q ω q π 3 π 3 38 ω q { } e i+q x a a q e i q x a a q e i q x a a q + e i+q x a a q 39 d 3 q ω q π 3 π 3 40 ω q { } π 3 δ 3 + q a a q + a a q π 3 δ 3 q a a q + a a q 4 a π 3 a a a a a a a 4 is an odd function under and so integrates to zero. Thus a π 3 a + a a π 3 a a + π 3 δ 3 0 since again δ 3 0 is an odd function albeit it oorly defined! π 3 a a 45 The interretation is that a a is the oerator giving the number of quanta in a small volume centred at momentum. This will indeed contribute to the total momenta. We see the same tye of term for the energy, the Hamiltonian oerator H, but we get a factor of ω not in that case. vi From we have d 3 x Π d 3 x i. i ωq ω a π 3 e iωt+i x a e iωt i x a q e iωqt+iq x a qe iωqt iq x 46 d 3 q ω ω q a π 3 e iωt+i x a 4 e iωt i x d 3 q π 3 d 3 x π 3 a q e iωqt+iq x a qe iωqt iq x 47 Using that d 3 xe i x π 3 δ 48 NOTE THIS IS FOR t 0 should add in more exonentials for non-tirvial times!

5 QFT PS4 Solutions: Free Quantum Field Theory 8//8 5 we aly the d 3 x and use the resulting delta function of δ 3 ± q to eliminate the q integral. This gives us d 3 x Π ω π 3 a a e iωt a a a a + a a e+iωt 49 From we have d 3 x φ. d 3 x π 3 ia e iωt+i x ia e iωt i x ω d 3 q π 3 iqa q e iωqt+iq x iqa qe iωqt iq x 50 ωq d 3 x d 3 π 3 d 3 q π 3.q ia e iωt+i x ia e iωt i x 4ω ω q. ia q e iωqt+iq x ia qe iωqt iq x 5 Again we aly the d 3 x and use the resulting delta function of δ 3 ±q to eliminate the q integral to find d 3 x φ π 3 a a e iωt + a a + a ω a + a a eiωt 5 Finally, in the same manner we find that d 3 x φ π 3 a a e iωt + a a + a ω a + a a eiωt 53 Putting this together we find H d 3 x Π + φ + m φ 54 ω + + m π 3 a a + a ω a ω + + m π 3 a a e iωt + a ω a eiωt 55 a a + a a with ω + + m π 3 ω d 3 π 3 ω a a + π3 δ vii [H, a k a k [ [ d 3 π 3 ω a a +, a k a k d 3 π 3 ω 57 π 3 ω a a, a k a k 58 a a, a k a k 0 59 Note that this result is not true for any interesting i.e. interacting theory. The interactions mix the modes of different momenta leading to a lack of conservation of article number. Only a continuous symmetry can guarantee conserved numbers and those are usually linked to total numbers of various articles, not the individual quanta of one article at one momentum.

6 QFT PS4 Solutions: Free Quantum Field Theory 8//8 6. Time evolution of annihilation oerator i From 56 and using the commutation relation a, a q π 3 δ q, [a, a q a, a q 0, 60 we have that d 3 [ q [H, a π 3 ω q a qa q +, a 6 d 3 q π 3 ω q a qa q a a a qa q 6 d 3 q π 3 ω q a qa q a a qa q a + π 3 δ qa 63 d 3 qω q δ qa ω a 64 as required. ii We want to rove that ith n a a [ith ω n 65 using [H, a ω a. First note that trivially the relation holds for n. Now if we assume the relation holds for n m then ith m+ a ithith m a 66 itha [ith ω m 67 ita H ω a [ith ω m 68 a [ith ω m+. 69 So the relation holds for n m + if it holds for n m. Thus, by induction, it holds for all n 0. iii Start by exanding the first exonential e iht a e iht n! ihtn a e iht 70 n0 n! ihtn a e iht 7 n n! a [ith ω n e iht 7 n a n! [ith ω n e iht 73 n a e ith ω e iht 74 a e ith ω H 75 76

7 QFT PS4 Solutions: Free Quantum Field Theory 8//8 7 Thus we see that 3 e iht a e iht a e itω 77 where we have used the obvious facts that [H, H 0 and [ω, H 0. The corresonding equation for a follows from hermitian conjugation e iht a e iht a e itω 78 e iht a e iht e itω a 79 e +iht a e iht e +itω a 80 e iht a e iht a e +iωt The Advanced Proagator We have D A x θ x 0 0 [φx, φ0 0 8 i We start by evaluating D A. We have d 3 q [φx, φ0 π 3 π 3 [a, a q e i x + [a, a q e i x 83 4ω ω q 0 ω Hence since 0 0 we have 4 d 3 q π 3 π 3 π 3 δ 3 q e i x e i x 4ω ω 0 ω 84 q π 3 ω e i x e i x 0 ω 85 D A x θ x 0 0 [φx, φ θ x 0 e i x π 3 e i x ω 0 ω 87 θ x 0 θ x 0 θ x 0 π 3 e iωx0 e i x e iωx0 e i x ω 88 π 3 e iωx0 e i x e iωx0 e i x ω in second term89 π 3 e i x ω + 0 ω e i x ω 0 ω We now need to introduce a 0 integration and rewrite the exression in terms of a contour integration. There are two standard ways to do this. In the first aroach we shift the oles of the integrand, introducing a small ositive infinitesimal ɛ into the integrand which is taken to zero from the ositive side at the end of the calculation. This is the aroach used in the lectures and it is common ractice to use this notation, esecially in the case of the time-ordered Feynman 3 Another neat way to rove this by taking the derivative of the equation with resect to t and solving the resulting oerator valued differential equation. 4 For the in second term don t forget that in changing variable in each momentum comonent + di means the range of integration changes from to + to the other way round giving another minus sign. 90

8 QFT PS4 Solutions: Free Quantum Field Theory 8//8 8 roagator. The second aroach is to make small distortions in the contour away from the real 0 axis near the oles. This is used by Tong in his derivation of the Feynman roagator sec..7. age 38 though Tong reverts to the first and standard notation later on see Tong equation Both methods are equivalent in the ɛ 0 + limit. First aroach onsider d 4 i I x e i x π 4 0 iɛ m 9 d 3 e i x π 3 πi 0 ω iɛ ω iɛ where, as usual, ω + m. The 0 integration is along the real axis with oles in the integrand as shown here -w +ie +w +ie 93 The integrand has simle oles at f 0, 0 ω iɛ 0 + ω iɛ e i x 94 0 ±ω + iɛ. 95 Near these oles the integrand looks like f R ± / 0 ω + iɛ with residues R ± given by R ± ± ω e i x 0 ±ω +iɛ 96 The idea is that we think of our exression for the advanced roagator in 90 as being of the form D A x θ x 0 π 3 R + + R. 97 In order for this to match I x of 9 we need to find a closed contour such that by using the residue theorem we can deduce that πi f0, θ x 0 R + + R 98

9 QFT PS4 Solutions: Free Quantum Field Theory 8//8 9 If x 0 > 0 then e i0 x 0 0 as I 0 i. This means that an integration of this integrand f round a large semi-circle running around the lower half lane is equal to zero πi f0, 0 ifx 0 > 0 as I 0 i. 99 We can therefore add this integration of f around the semi-circle to our 0 integration along the real axis in I without changing the result for I. So we roduce an exression for I which uses a closed contour for the 0 integration by adding this lower semi-circle. Now no oles are enclosed within this closed contour so the residue theorem tells us the result is zero + -w +ie +w +ie πi f0, 0 if x 0 > If x 0 < 0 then e i0 x 0 0 as I 0 +i. This means that an integration of this integrand f round a large semi-circle running around the uer half lane will give zero. We can therefore add this to our existing 0 integration along the real axis in I without changing the result. So we roduce a closed contour by adding the semi-circle above and now the residue theorem tells us that we ick u contributions from both oles. This gives us + -w +ie +w +ie πi f0, R + + R if x 0 < Putting the two cases together gives us the desired result πi f0, θ x 0 R + + R. 0 Alternative aroach

10 QFT PS4 Solutions: Free Quantum Field Theory 8//8 0 The second aroach to these tyes of roblem is to distort the contour away from the real 0 axis near the oles by a small amount. So now consider d 3 Ix π 3 π d 3 π 3 i m e i x 03 πi where, as usual, ω + m. The integrand f 0, 0 ω 0 + ω e i x 04 0 ω 0 + ω e i x 05 has simle oles at 0 ±ω 06 with residues f R ± / 0 ω near 0 ±ω given by Observing that we can rewrite D A x θ x 0 R ± ± ω e i x 0 ±ω 07 in order for this to match Ix we need to find a contour such that onsider the following contour π 3 R + + R 08 πi f0, θ x 0 R + + R 09 E E 0 If x 0 > 0 then e i0 x 0 0 as I 0 i and we close the contour below and ick u no oles E E πi f0, 0

11 QFT PS4 Solutions: Free Quantum Field Theory 8//8 If x 0 < 0 then e i0 x 0 oles 0 as I 0 +i and we close the contour below and icking u both E E πi f0, R + + R so that indeed πi f0, θ x 0 R + + R 3 To give a bit more detail on how we close the contour, consider x 0 > 0. We have remember 0 is comlex 0 ω e i0 x 0 eim 0 x 0 0x0 0 ω eim 0 ω 4 where the last ste comes from writing 0 0 ω + ω 0 ω + ω using the triangle inequality. losing the integral below we have Im 0 0 so that e Im 0 x 0. Thus, evaluating first at finite 0 we have, for the infinite semi-circular ath on the lower half lane 5 we have 0 ω e i0 x 0 0 ω e i0 x 0 6 eim 0 x 0 0 ω 7 0 ω 8 π 0 0 ω integrate on with 0 dθ 9 0 as 0 0 Hence we see that the contribution from the integration along is zero and we can close the contour along in order to evaluate Ix when x 0 > 0. A similar argument holds for x 0 < 0.

12 QFT PS4 Solutions: Free Quantum Field Theory 8//8 ii We have, taking the derivative of e i x + m D A x d 4 π 4 i m + m e i x d 4 π 4 ie i x iδ 4 x 3 Note the integrand has no oles on the real axis, so none of the subtleties in the ath relevant when defining D A x aear. 4. harge of a comlex scalar field We have by definition Q i d 3 x Φ Π ΠΦ 4 i We have i d 3 x ΠxΦx Now and hence i Q d 3 x d3 d 3 q ω π 3 π 3 c e i x b 4ω e i x b q e iq x + c qe iq x 5 q d 3 x d3 d 3 q ω π 3 π 3 4ωq { } e i+q x c b q + e i q x c c q e i q x b b q e i+q x b c q 6 ω d 3 q π 3 π 3 4ωq { π 3 δ 3 + q } c b q c b q + π 3 δ 3 q c c q b b q 7 π 3 c b c b + c c b b d 3 x Φ xπ x i 8 d 3 x ΠxΦx 9 π 3 c b c b c c + b b π 3 c b c b + c b c b + b b c c π 3 b b c c π 3 b b c c π 3 δ π 3 b b c c + infinite constant 34 where we have used the fact that odd functions integrate to zero.

13 QFT PS4 Solutions: Free Quantum Field Theory 8//8 3 ii Substitute in the form of the fields Φx Πx Π 3 b e i x + c e i x 35 ω Π 3 i ω c e i x b e i x 36 and hence confirm that the ETR equal time commutation relations for comlex scalar fields are [Tong.7,.34 [ [Φx, Πy x0 y 0 π 3 δ 3 x y, Φ x, Π y π 3 δ 3 x y, 37 x 0 y 0 along with the other zero commutators of the ETR: 0 [Φx, Φy x0 y 0 Φx, Φ y 38 x 0 y 0 [Πx, Πy x0 y 0 Πx, Π y 39 x 0 y 0 Φx, Π y 0, 40 x 0 y 0 and so forth. Note that the ETR for the comlex fields should match what we found for the real field case exactly. These ETR are true for ANY field. Since we are meant to treat Φ and Φ as searate fields, they should each obey exactly the same equal time commutation relations. It is trivial to check the as we can take the hermitian conjugate of [Φ, Π and should get the [Φ, Π for free and it should look the same. The factor of i is critical for that. This is really i δ 3 x y but with, the same factor as seen in the QM [q, i commutator. Then we can rove the charge-field commutator by calculating not all oerators are at equal times [ [Q, Φx i d 3 y Φ yπ y ΦyΠy, Φx 4 i d 3 y [ΦyΠy, Φx 4 i d 3 y iφyδ 3 x y 43 Φx 44 Alternatively, we can rove this as follows: d 3 q [ [ [Q, Φx π 3 π 3 c b b c, b q e iq x + c b b c, c q e iq x 45 ωq d 3 q π 3 π 3 π 3 δ 3 q b e iq x + b e iq x 46 ωq π 3 b e i x + b e i x 47 ω Φx 48

14 QFT PS4 Solutions: Free Quantum Field Theory 8//8 4 where we have used [ab, c a [b, c + [a, c b and hence [b b, b q [c, b q b π 3 δ 3 qb 49 b b, b q 0 50 c c, b q b c, b q π 3 δ 3 qb 5 [c c, b q 0 5 iii We have that [Q, ηx qηx for some oerator η with c-number charge q. onsider η η e iθq ηe iθq 53 We roceed exactly as we did for this tye of transformation for the Hamiltonian and annihilation oerator. In fact as all we used was the commutation relation, a simle substitution into the answer given above is sufficient to convince that indeed η e iθq η. However let us illustrate this with a different roof. dη dθ dη dθ iqq iqe iθq.η.e iθq + Qe iθq.η. iqe iθq 54 ie iθq.qη.e iθq ie iθq.ηq.e iθq 55 ie iθq.[q, η.e iθq ie iθq.qq.e iθq iqe iθq Q.e iθq Integrating this and using the fact that if θ 0 we have η η as a boundary condition, we see that we get our answer η e iθq η. If integration of an oerator equation worries you, note that if you substitute the solution into the differential equation, the oerators are not involved in the differentiations, just the c-number arts with which you are familiar. 5. Time evolution and roagators of a comlex scalar field i onsider two sets of annihilation and creation oerators â, â, a and a obeying â i, â jq π 3 δ 3 qδ ij, [â i, â jq â i, â jq 0, i, j,. 58 We define so that ˆb â + iâ, ĉ â iâ. 59 ˆb â iâ, ĉ â + iâ. 60 We require several commutators to be roved. The key identity here is that for any oerators A, B,, D we have [A + B, + D [A, + [A, D + [B, + [B, D 6 which you should rove if this is not familiar.

15 QFT PS4 Solutions: Free Quantum Field Theory 8//8 5 The commutators between a b or c annihilation and a ˆb or ĉ creation oerator are all of the form [ â + is â, â q + is qâ q 6 where s, s q ±. Exanding we have that [ â + is â, â q + is qâ q 63 So we find [ â, â q + is q [ â, â q + is â, â q s s q â, â q 64 δ3 q s s q δ3 q 65 s s q δ 3 q. 66 [ˆb, ĉ q 0 s +, s q + 67 [ˆb, ˆb q δ 3 q s +, s q 68 [ĉ, ĉ q δ 3 q s, s q + 69 [ĉ, ˆb q 0 s, s q 70 Note the last is the hermitian conjugate of the first so you could avoid calculating it for that reason. The commutators between a air of b or c annihilation oerators are all of the form [â + is â, â q + is q â q 7 where s, s q ±. Since annihilation oerators commute with each other even if they are of the same tye and at the same momentum, all these commutators are clearly zero. Taking the hermitian conjugate, or using a similar argument for the creation oerators, we see that all the commutators between a air of ˆb or ĉ creation oerators are also zero. ii Given the question below for Q we can save time by considering ˆb ˆb + sĉ ĉ where s ±. We then have that ˆb ˆb + sĉ ĉ â iâ â + iâ + s â + iâ â iâ 7 â â + iâ â iâ â + â â + s â â iâ â + iâ â + â â 73 + s â â + â â + i s â â â â 74 When s + we have the desired result that ˆb ˆb + ĉ ĉ â â + â â. 75 From this we see that here Z is the zero oint energy for mode Ĥ ω â iâi + Z i, ω ˆb ˆb + ĉ ĉ + Z 76 77

16 QFT PS4 Solutions: Free Quantum Field Theory 8//8 6 iii With the Hamiltonian given by 77, to show that d ˆΦ 3 H t, x π 3 ˆb e ix + ĉ e +ix 78 ω given the form at t 0 where x µ x µ and 0 +ω we can use the same method as used for the real field. That is since like all oerators 5 all we need to do is look at the behaviour of ˆΦ H t, x ex{iĥt}ˆφ H t 0, x ex{ iĥt} 79 ex{iĥt}ˆb ex{ iĥt} and ex{iĥt}ĉ ex{ iĥt}. 80 As the c and b oerators commute, the only art that matter that matters is the ˆb ˆb term in the first case and the ĉ c term in the second case. More formally we can commute one art of the Hamitonian through the ˆb or the ĉ oerator to be cancelled. The roblem reduces to that of a single tye of oerator, i.e. we are left with ex{iĥbt}ˆb ex{ iĥbt} and ex{iĥct}ĉ ex{ iĥct}, 8 where Ĥ b ω ˆb ˆb, Ĥ c ω ĉ ĉ. 8 We have already derived these so from 77 we know that ex{iĥbt}ˆb ex{ iĥbt} ex{ iω t} and ex{iĥct}ĉ ex{ iĥct} ex{+iω t}. 83 and we find the desired time evolution for the free comlex scalar field in the Heisenberg icture. iv No answer for this art so far. onsider two real 6 scalar fields ˆφ Hi t, x π 3 â i e iωt+i x + â eiωt i x i 84 ω where â i and â i are the oerators given in 58. Show that if?? is true then the field oerators obey ˆΦ H t, x ˆφH t, x + i ˆφ H t, x, ˆΦ H t, x ˆφH t, x i ˆφ H t, x. 85 v The Wightman function for a comlex scalar field is defined as Dx y : 0 ˆΦxˆΦ y Substituting 78 into 86 give us d 3 q Dx y π 3 ω π 3 0 ˆb e ix + ĉ e +ix ˆb qe +iqy + ĉ q e iqy 0 87 ωq d 3 q π 3 ω π 3 0 ˆb ˆb q 0 e ix+iqy 88 ωq 5 As this is a free Hamiltonian it is the same in Heisenberg and Scroödinger ictures and so it needs no subscrit. 6 These reresent classical real scalar fields but their quantised versions are hermitian not real.

17 QFT PS4 Solutions: Free Quantum Field Theory 8//8 7 as the c ĉ oerator annihilates the ket bra vacuum. ommuting the two b oerators gives a non-zero term containing a delta function in momentum which will kill one integral lus a second term ˆb qˆb which again annihilates the vacuum states so does not contribute. This leaves us with Dx y exactly we found before for real scalar fields. Note that from the definition we have that for any tye of field π 3 ω e ix y 89 Dx y : 0 ˆΦyˆΦ x 0 Dy x. 90 For the free field case in 89 we can see this exlicitly. vi For free comlex scalar fields by substituting 78 into 0 ˆΦxˆΦy 0 gives us 0 ˆΦxˆΦy 0 d 3 q π 3 ω π 3 0 ˆb e ix + ĉ e +ix ˆb q e iqy + ĉ qe +iqy 0 9 ωq d 3 q π 3 ω π 3 0 ˆb ĉ q 0 e ix+iqy 9 ωq as the b ĉ oerator annihilates the ket bra vacuum. However the two remaining oerators commute and again they then annihilate the vacuum so this case is zero for any time. vii The Feynman roagator for comlex scalar fields is defined as F x y : 0 TˆΦxˆΦ y 0 93 where T is the time ordering oerator. We can exress F in terms of the Wightman functions for the comlex field 86 F x y θx 0 y 0 0 ˆΦxˆΦ y 0 + θy 0 x 0 0 ˆΦ yˆφx 0 94 Following the same stes as above we can show that 0 ˆΦ d 3 q yˆφx 0 π 3 ω π 3 0 ˆb qe +iqy + ĉ q e iqy ˆb e ix + ĉ e +ix 95 0 ωq By comaring with 89 we see that d 3 q π 3 ω π 3 0 ĉ q ĉ 0 e iqy+ix 96 ωq d 3 q π 3 ω π 3 π 3 δ 3 q e iqy+ix 97 ωq π 3 e iy x 98 ω 0 ˆΦ yˆφx 0 Dy x. 99 This means that the Feynman roagator for comlex scalar fields is F x y 0 TˆΦxˆΦ y 0 θx 0 y 0 Dx y + θy 0 x 0 Dy x. 00

18 QFT PS4 Solutions: Free Quantum Field Theory 8//8 8 This is exactly the same form as we had for the real free field scalar roagator so the Feynman roagator is of exactly the same form i.e. in momentum sace it is equal to F i m + iɛ. 0 learly these are also solutions of the Klein-Gordon equation as they were in the real free scalar field case. Again the oles at m indicate that we have article-like solutions of mass m dominating the behaviour. The resences of two distinct tyes of annihliation/creation oerator in this comlex field indicates there are two indeendent degrees of freedom with the same mass, here distinct article and anti-articles. Note that 0 TˆΦ yˆφx 0 0 TˆΦxˆΦ y 0 if the times are unequal because the T oerator defines the order so how we write the fields under the T oerator has no effect.