Srednicki Chapter 7. QFT Problems & Solutions. A. George. May 27, Starting with equation 7.12, do the contour integral to verify.

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1 Srednck Chapter 7 QFT Problems & Solutons A. George May 7, 0 Srednck 7.. equaton 7.4 Startng wth equaton 7., do the contour ntegral to verfy Equaton 7. s: Now we can wrte ths as: G(t t ) = ) de e E(t t π E + ω G(t t ) = π dee E(t t ) E (ω ) Hence: G(t t ) = 4π ω dee E(t t ) E ω E + ω (7..) Now let s assume that t t 0. In ths case, an ntegral over a sem-crcle the lower halfplane wll go to zero, snce the exponental factor wll have a term of (-) as wrtten, a term proportonal to (-) from the crcle n the lower half-plane, and a postve term from the (t - t ). The result s e postve real number * mod(e) whch wll go to zero as E. As a result, we can tack on the ntegral over the sem-crcle (snce t equals zero), endng up wth a contour ntegral: G(t t ) = 4π ω C dee E(t t ) E ω E + ω Now the resdue theorem tells us that we can take π tmes the resdue, where the resdue s just for those sngulartes n the lower half plane (that s to say, those wth the negatve magnary part). Hence, G(t t ) = ω Res e E(t t ) E ω Now we let E = ω, whch s the sngularty. We drop the part that goes to zero (that s the beauty of takng the resdue), and we re left wth: G(t t ) = ω ω e (t t )

2 where the negatve sgn dsappeared snce we dd the ntegral n the clockwse drecton, and the defnton of the resdue s taken assumng a counterclockwse drecton. (By the way, we know t s the clockwse drecton because the real ntegral ends at postve nfntty, and we need to make our way back to negatve nfnty by gong through the lower complex plane. Ths s obvously clockwse). Now we let 0: G(t t ) = ω e ω(t t ) for t - t 0 Good. Now let s assume that t t 0. In ths case we have to add on a sem-crcle n the upper half-plane, and we need to use the other sngularty (snce that s the sem-crcle that equals zero). Everythng wll work the same as before, except for an overall mnus sgn (snce we re usng the other sngularty, whch has a mnus sgn n front of t) but even that s cancelled snce the contour ntegral s n the counterclockwse drecton ths tme. The only real dfference s n the exponental, snce the opposte sngularty was chosen: G(t t ) = Combnng these results, we have: ω eω(t t ) for t - t 0 whch s equaton 7.4. G(t t ) = ω e ω t t Srednck 7.. Startng wth equaton 7.4, verfy equaton 7.3. Equaton 7.4 s: Now we take the dervatve: Now for another dervatve: G t = G = ω e ω t t ω ( ω)sgn(t t )e ω t t G t = sgn(t t ) e ω t t G(t t ) = δ(t t )e ω t t ω sgn(t t ) t e ω t t Ths frst term s only nonzero f t = t, n whch case the exponental doesn t contrbute. Hence: G(t t ) = δ(t t ) ω sgn(t t ) e ω t t t We ll also move the last term to the other sde: G(t t ) + ω sgn(t t ) e ω t t = δ(t t ) t

3 The sgn functon squared s one everywhere except at t = t. So we ll specalze to the case where t t : G(t t ) t + ω e ω t t = δ(t t ) whch s: G(t t ) t + ω G(t t ) = δ(t t ) Ths s equaton 7.3. What about f t = t? As t turns out, ths proof and equaton 7.3 tself don t hold n that case! But that s okay because there s no need to do path ntegrals f t = t! In that case, the ampltude s just one f the ntal and fnal states are the same and zero otherwse. Srednck 7.3. (a) Use the Hesenberg equaton of moton, Ȧ = H, A, to fnd explct expressons for Q and Ṗ. Solve these to get the Hesenberg-pcture operators Q(t) and P(t) n terms of the Schrödnger-pcture operators Q and P. Usng the formula for P: P = H, P P = m P + mω Q, P P = mω Q, P P = mω (QQ, P + Q, P Q) P = mω Q (7.3.) Note that h was set to, but m was not (Srednck set them both to one). Usng the formula for Q: Q = H, Q Q = m P + mω Q, Q Q = m P, Q Q = P, QP m Q = m P (7.3.) Dfferentatng equaton (7.3.): Insertng equaton (7.3.): P = mω Q P = ω P (7.3.3) Smlarly, dfferentate equaton (7.3.): Q = m P 3

4 Insertng equaton (7.3.): Q = ω Q (7.3.4) Solvng equtatons (7.3.3) and (7.3.4), we have: P (t) = Acos(ωt) + Bsn(ωt) Q(t) = Ccos(ωt) + Dsn(ωt) Now, at tme 0, the Schrödnger and Hesenberg pctures are the same the kets haven t evolved n the Schrödnger pcture, and the operators haven t evolved n the Hesenberg Pcture. Hence, P (0) = P and Q(0) = Q, etc. So: P (t) = P cos(ωt) + Bsn(ωt) Q(t) = Qcos(ωt) + Dsn(ωt) Now we dfferentate these and take the values at t = 0. P (0) = Bω Q(0) = Dω Now we use equatons (7.3.) and (7.3.), notng that the rght hand sde of these equatons are the Schrödnger pcture operators, snce the operators haven t evolved yet at t = 0. Then, Hence, and so: P (0) = Bω = mω Q Q(0) = Dω = m P B = mωq D = mω P P (t) = P cos(ωt) mωqsn(ωt) (7.3.5) Q(t) = Qcos(ωt) + P sn(ωt) (7.3.6) mω (b) Wrte the Schrödnger-pcture operators Q and P n terms of the creaton and annhlaton operators a and a, where H = hω(a a + ). Then, usng your result from part (a), wrte the Hesenberg-pcture operators Q(t) and P(t) n terms of a and a. Ths s actually not a dervaton as much as a defnton. We use the usual defnton from quantum mechancs, see Sakura.3.4: Q = (a + a) mω P = (a a) 4

5 We know that ths defnton s okay because the consequent Hamltonan (Sakura.3.6) s equvalent to that n the problem statement. Now we use equatons (7.3.5) and (7.3.6): mω mω P (t) = (a a)cos(ωt) (a + a)sn(ωt) whch s: Hence: Q(t) = (a + a)cos(ωt) + (a a)sn(ωt) mω P (t) = a (cos(ωt) + sn(ωt)) a (cos(ωt) sn(ωt)) Q(t) = a (cos(ωt) + sn(ωt)) + a (cos(ωt) sn(ωt)) mω P (t) = a e ωt ae ωt (7.3.7) Q(t) = a e ωt + ae ωt (7.3.8) (c) Usng your result from part (b), and a 0 = 0 a = 0, verfy equatons 7.6 and 7.7. Usng equaton (7.3.8): whch s: 0 T Q(t )Q(t ) 0 = 0 T a e ωt + ae ωt a e ωt + ae ωt 0 0 T Q(t )Q(t ) 0 = 0 T a e ωt + ae ωt a e ωt + ae ωt 0 We ll assume that t t Most of these terms annhlate the vacuum. The only one that doesn t s: 0 T Q(t )Q(t ) 0 = eω(t t ) 0 aa 0 whch s: 0 T Q(t )Q(t ) 0 = e ω(t t ) What about f t t? In ths case, the tme orderng s opposte, so we smply take t t : Combnng these we have: 0 T Q(t )Q(t ) 0 = eω(t t ) 0 T Q(t )Q(t ) 0 = e ω t t 5

6 whch s: 0 T Q(t )Q(t ) 0 = e ω t t Usng equaton 7.4 (and notng that Srednck sets m = : whch s equaton T Q(t )Q(t ) 0 = G(t t ) As for 7.7, the procedure s the same. We start by assumng that t t t 3 t 4 : 0 T Q(t )Q(t )Q(t 3 )Q(t 4 ) 0 = 0 a e ωt + ae ωt a e ωt + ae ωt a e ωt + ae ωt a e ωt + ae ωt 0 ( ) 0 T Q(t )Q(t )Q(t 3 )Q(t 4 ) 0 = 0 a e ωt + ae ωt a e ωt + ae ωt a e ωt 3 + ae ωt 3 a e ωt 4 + ae ωt 4 0 Some of these terms vansh: 0 T Q(t )Q(t )Q(t 3 )Q(t 4 ) 0 = e ω(t 4 t ) whch gves: 0 T Q(t )Q(t )Q(t 3 )Q(t 4 ) 0 = e ω(t 4 t ) and then: 0 T Q(t )Q(t )Q(t 3 )Q(t 4 ) 0 = e ω(t 4 t ) ( ) 0 a a e ωt + ae ωt a e ωt 3 + ae ωt 3 a 0 ( ) 0 a a a e ω(t +t 3 ) + a ae ω(t t 3 ) +aa e ω(t 3 t ) + aae ω(t +t 3 ) a 0 ( ) 0 aa a a e ω(t +t 3 ) + aa aa e ω(t t 3 ) +aaa a e ω(t 3 t ) + aaaa e ω(t +t 3 ) 0 The frst and last terms eventually annhlate the vacuum: ( ) 0 T Q(t )Q(t )Q(t 3 )Q(t 4 ) 0 = e ω(t 4 t ) 0 aa aa e ω(t t 3 ) + aaa a e ω(t 3 t ) 0 The operators now cancel out ncely, leavng: ( ) ( 0 T Q(t )Q(t )Q(t 3 )Q(t 4 ) 0 = e ω(t 4 t ) e ω(t t 3 ) + e ω(t 4 t ) e ) ω(t 3 t ) Rewrtng ths last term, and factorng out a factor of : 0 T Q(t )Q(t )Q(t 3 )Q(t 4 ) 0 = ( ) ( e ω(t 4 t ) e ω(t t 3 ) + e ω(t 4 t ) e ω(t 3 t ) 6

7 whch gves: +e ω(t 4 t ) e ω(t 3 t ) ) 0 T Q(t )Q(t )Q(t 3 )Q(t 4 ) 0 = ( ) ( e ω(t t ) e ω(t 3 t 4 ) + e ω(t t 4 ) e ω(t t 3 ) +e ω(t t 4 ) e ω(t t 3 ) ) As before, we ve assumed that the tme orderng was as wrtten. If the tme orderng s backward, we can flp the offendng terms, and the exponental wll go postve. As a result, we can put n absolute value sgns to account for all possble orderngs: 0 T Q(t )Q(t )Q(t 3 )Q(t 4 ) 0 = ( whch gves: ) ( e ω t t e ω t 3 t 4 + e ω t t 4 e ω t t 3 +e ω t t 4 e ω t t 3 ) 0 T Q(t )Q(t )Q(t 3 )Q(t 4 ) 0 = G(t t )G(t 3 t 4 ) + G(t t 4 )G(t t 3 ) whch s equaton G(t t 4 )G(t t 3 ) Srednck 7.4. Consder a harmonc oscllator n ts ground state at t =. It s then subjected to an external force f(t). Compute the probablty 0 0 f that the oscllator s stll n ts ground state at t =. Wrte your answer as a manfestly real expresson, and n terms of the Fourer transform f(e) = dteet f(t). Your answer should not nvolve any other unevaluated ntegrals. The answer s Srednck 7.0: 0 0 f = exp de f(e) f( E) π E + ω The remander of the problem s just to smplfy ths and put t nto the form requested. Frst we note that f(e) = dte Et f(t) so and Hence f( E) = f (E). Then, 0 0 f = exp f( E) = f (E) = dte Et f(t) (4.3.) dte Et f(t) (4.3.) de f(e) f (E) π E + ω 7

8 whch s: 0 0 f = exp de f(e) π E + ω Ths s of the form: e (a+b). Now that we take the magntude of ths, we have e (a+b) e (a b) = e b. Hence, { 0 0 f de = exp Im f(e) } π E + ω Most of ths s already manfestly real, so let s move that outsde of the Imagnary Operator. Then, { } 0 0 f de = exp π f(e) Im E + ω Ths s: 0 0 f de = exp 0 0 f = exp Takng the magnary component: 0 0 f = exp Let s defne a = ω E. Then: { π f(e) Im 0 0 f = exp E + ω E + ω + E + ω + } { de π f(e) E + ω + Im ( E + ω ) + Now we re ready to let 0. We notce that: lm 0 de π f(e) de π f(e) a + = { 0 f a 0 f a = 0 ( E + ω ) + a + Ths s a delta functon by defnton. But any delta functon wll work, so: lm 0 a + = Cδ(a) If we ntegrate both sdes from to, we fnd π = C. Then, lm 0 a + = πδ(a) Hence: 0 0 f de = exp π f(e) πδ(e ω ) Usng a property of the delta functon: 0 0 f = exp de f(e) 8 ( δ(e ω) + E ) δ(e + ω) E }

9 Ths s: 0 0 f = exp Performng the ntegrals: 0 0 f = exp de δ(e ω) f(e) exp E f(ω) exp ω f( ω) ω de δ(e + ω) f(e) E Now we notced prevously (equatons (4.3.) and (4.3.)) that f( x) = f (x). Consequently, f( x) = f (x) = f(x), whch gves: 0 0 f = exp f(ω) exp ω f(ω) ω 0 0 f = exp f(ω) ω 9

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