Srednc Chapter 4 QFT Problems & Solutons A. George September, Srednc 4.. Derve a generalzaton of Feynman s formula, = Γ ( α ) A αa... A αn n Γ(α df xα n ) (n )! ( x A ) Hnt: start wth: Γ(α) A α = dt t α e At whch defnes the gamma functon. Put an ndex on A, α, and t, and tae the product. Then multply on the rght hand sde by ( = ds δ s ) t Mae the change of varable t = sx, and carry out the ntegral over s. Followng the hnt, we start wth: Γ(α) A α = dt t α e At Puttng an ndex on A, α and t, and tang the product of both sdes, we have: Γ(α ) A α = dt t α e A t α Now we multply on the rght by equaton 4.5: Γ(α ) A α = dt ds t α e A t δ ( s t ) Next we mae the change of varable t = sx : Γ(α ) A α = dx ds (sx ) α e A sx sδ ( s sx )
Recall that the delta functon has the property of f(cx) = f(x). Snce s s always postve, c we have: Γ(α ) = ( dx A α ds (sx ) α e A sx δ ) x Now let s add n an dentty: Γ(α ) A α = ( dx ds (sx ) α e A sx (n )! (n )! δ Next, note that all the x terms must be postve (thans to our ntegral), but can t be greater than one (thans to the delta functon). So, we can rewrte the x ntegral: Γ(α ) = ( dx A α ds (sx ) α e A sx (n )! (n )! δ ) x Usng equaton 4.: Γ(α ) A α = df n ds (sx ) α e A sx (n )! x ) Separatng out the s-ntegral: Γ(α ) A α = df n (x ) α (n )! dss α e A sx Ths s qucly gong to get confusng wth the product. To avod ths, let s have the product act on each term separately. Ths, of course, s not always allowed, but t s fne n ths case snce everythng s multpled together. ] [ Γ(α ) A α = df n [ (x ) α (n )! ds s α ] [ ] e A sx These last two terms gve products of exponentals, whch of course s a sum n the exponent. So: [ ] Γ(α ) = df A α n (x ) α ds [ s ] [ α e s A x ] (n )! We can solve ths s-ntegral on Mathematca or the equvalent, the result s: Γ(α ) A α = df n [ (x ) α ] [ ] ( (n )! Γ α Dstrbutng the product on the left hand sde, we have: Γ(α ) A α... A αn n = df n [ (x ) α ] [ ] ( (n )! Γ α A x A x ) α ) α
Then: A α... A αn n = Γ ( α ) Γ(α ) (n )! df n (x ) α ( A x) α Srednc 4.. Verfy equaton 4.3. Srednc gave us the openng here when he suggested tang a Gaussan ntegral n cartesan and sphercal coordnates. Let s do cartesan coordnates frst (we ll also use x as our dummy varable): d d xe ( x ) = ( π) d Next let s do t n sphercal coordnates: dω drr d e r = Ω Γ(d/) Equatng these results, we have: Srednc 4.3. (a) Show that: Ω = πd/ Γ(d/) d d qq µ f(q ) = d d qq µ q ν f(q ) = C g µν d d qq f(q ) and evaluate the constant C n terms of d. Hnt: use Lorentz symmetry to argue for the general structure, and evaluate C by contractng wth g µν. The frst equaton s odd, and so ntegrates to zero. The second equaton has two ndces, but cannot depend on any four-vectors. Hence, the general structure s, by Lorentz Symmetry: d d q q µ q ν f(q ) = Ag µν (4.3.) Contractng wth g µν, we fnd: d d qq f(q ) = Ad and so: A = d d d qq f(q ) Pluggng ths nto equaton (4.3.) gves: d d qq µ q ν f(q ) = d gµν d d qq f(q ) 3
Comparng ths to equaton 4.53, we fnd that C = /d. (b) Smlarly evaluate d d q q µ q ν q ρ q σ f(q ). As before, Lorentz Symmetry tells us that we must use the g. Ths tme there are three ways to combne the ndces: d d q q µ q ν q ρ q σ f(q ) = Ag µν g ρσ + Bg µρ g νσ + Cg µσ g ρν (4.3.) To evaluate A, B, and C, let s start by contractng wth g µν g ρσ d d q q 4 f(q ) = Ad + Bd + Cd And now let s contract wth g µρ g νσ : d d q q 4 f(q ) = Ad + Bd + Cd And now let s contract wth g µσ g ρν : d d q q 4 f(q ) = Ad + Bd + Cd Equatng the three rght hand sdes, we fnd that A = B = C. Then, we have: d d q q 4 f(q ) = A(d + d ) Hence, we have: A = B = C = d + d d d q q 4 f(q ) Puttng ths nto equaton (4.3.), we have: d d q q µ q ν q ρ q σ f(q ) = gµν g ρσ + g µρ g νσ + g µσ g ρν d + d d d q q 4 f(q ) Srednc 4.4. Compute the values of κ A and κ B. Start wth Srednc 4.39: Π( ) = α Imposng the frst boundary condton, we have: ( ) dxdln(d/m ) + α 6 κ A + κ B m + O(α ) Π( m ) = α m dx( x + x )ln( x + x ) α 6 κ Am + ακ B m = 4
Evaluatng the ntegral: ( 3 ) 3π 9 36 6 κ A + κ B = (4.4.) Now let s tae the dervatve of equaton 4.39 so that we can mpose the second boundary condton: Π ( ) = α x( x) + m dxx( x) + ln + α m 6 κ A Imposng the second boundary condton: Solvng ths for κ A : Π ( m ) = dxx( x) [ + ln( x + x ) ] + 3 κ A = κ A = 7 3π 3 =.374 Usng ths n equaton (4.4.), we have: κ B = 3π =.9 6 Srednc 4.5. Compute the O(λ) correctons to the propagator n φ 4 theory (see problem 9.), and compute the O(λ) terms n A and B. We essentally have to repeat the entre analyss of ths chapter for the φ 4 theory. Let s start by drawng the O(λ) correctons to the vertex. They are: l x Note that ths second dagram s necessary because t may contrbute at O(λ). Srednc told us (and proved n equatons 4.37, 4.3) that ths dagram s of O(g ) n φ 3 theory. We mght assume that ths dagram s smlarly of O(λ ) n φ 4 theory, but there s no way to now for sure we ll have to assume for the moment that t does correct the propagator at O(λ). We wll have calculated A and B by the end of the problem, so then we ll now whether t contrbuted anythng or not. If not, then no harm s done, we ll just state that the dagram doesn t contrbute and gnore all hgher-order correctons, ncludng those from ths dagram. Our frst step s to wrte down the self-energy. Ths wll be the values of these Feynman dagrams, rememberng to gnore the external propagators. Followng the Feynman rules for φ 4 theory, the frst dagram contrbutes the followng terms to the self-energy: -λ for the vertex /(l + m ɛ) for the nternal propagator of the loop. 5
d d l/(π) d to ntegrate over the loop momentum. / for the symmetry factor of the loop because the stated Feynman Rules are for τ (or Π, as the case may be) The second dagram contrbutes only the counterterm vertex factor, whch s the same as n φ 3 theory, snce the counterterm Lagrangan s the same. We also put n the factor of, for the same reason as before. Puttng all ths together, we have: Π( ) = ( ) ( λ) Wrtng ths more neatly, we have: Π( ) = λ It s convenent to wrte ths as: Π( ) = λ d d l (π) d l + m ɛ + ( )( )(A + Bm ) d d l (π) d l + m ɛ + (A + Bm ) d d l (π) d l + m ɛ A Bm The man challenge s to calculate ths ntegral. Srednc presented a bag of trcs, but the only one we need n ths problem s the Wc Rotaton. We defne l = l d and l j = l j. Then l = l, but d d l = d d l. The pont of ths, of course, s that we can get rd of the ɛ n the denomnator, per equaton 4.6. Then, Π( ) = λ d d l (π) d l + m A Bm We now how to solve ths ntegral, but t s worth pausng to consder the dmensonalty. Ths ntegral appears to be well defned only for d = (or lower). However, usng Srednc s trc #3, we could tae two dervatves wth respect to m. Ths would be allowed because we have two degrees of freedom, so we could then ntegrate twce, solve the boundary condtons, and return to the orgnal problem. The advantage of ths s that after dfferentaton, we would have an l 6 n the denomnator, and so the ntegral wll converge so long as d < 6. As t turns out, we get a dmensonless couplng constant for d = 4, so ths theory s renormalzable n the case of nterest. Just as n the text, t s not necessary to actually tae the two dervatves, but t s necessary to now that the theory s renormalzable, and that t s therefore vald to analytcally contnue our results to d = 4. Now we can use equaton 4.7: ) Π( ) = λ Γ ( d (4π) d/ (m ) +d/ A Bm Now t s tme to defne ε = 4 d (4 because we get a dmensonless couplng constant when d = 4). Let s also set λ µ ε λ, such that all mass dmensonalty s projected onto the µ. Ths gves: Π( ) = λ µε Γ ( ) + ε (4π) ε/ (m )(m ) ε/ A Bm (4π) 6
whch s: Π( ) = m λ ( (4π) Γ + ε ) ( ) 4π µ ε/ A Bm m Usng equaton 4.6 and 4.33, we smplfy: [ ] Π( ) = m λ [ (4π) ε γ + + ε ( )] 4π µ ln A Bm m Now multply together, and eep only those terms of O() n ε, snce we want to wor n four dmensons. Then: Π( ) = m λ 4π µ (4π) ε γ + + ln A Bm m Next, let s solve equaton 4.35, to fnd that 4π µ = e γ µ. Then: Π( ) = m λ e γ (4π) ε γ + + ln µ A Bm m whch s: and so: Π( ) = m λ (4π) Π( ) = µ ε γ + + ln(eγ ) + ln A Bm m m λ (4π) µ ε + + ln A Bm m By the way, ths s equaton 3.5, so we now that we dd well. The problem s that we need Π( ) to be fnte and ndependent of µ. In terms of, everythng s fne at ths order, so we ll tae A = κ A λ + O(λ ), where κ A s purely numerc. Snce we re gnorng O(λ ) correctons, we have: Π( ) = m λ µ (4π) ε + + ln κ m A λ Bm Now let s mpose the boundary condton Π ( m ) = where the prme represents a dervatve wth respect to. Ths gves κ A λ =. Hence, κ A =, and so A = O(λ ). Ths s because A clearly does not contrbute at lowest order (otherwse the free propagator would have ths counterterm vertex n t), and we ve just shown that t does not contrbute at frst order. The self-energy s now: Π( ) = m λ µ (4π) ε + + ln Bm m Now for B. Ths s nether fnte nor ndependent of µ, so we wll choose: B = λ µ (4π) ε + + ln + λκ m B + O(λ ) 7
whch s Now we have: B = λ [ (4π) ε + ( µ ) ] + ln + λκ B + O(λ ) m Π( ) = λκ B m + O(λ ) Imposng the other boundary condton, Π( m ) =, we fnd that κ B =. Then, and B = λ [ (4π) ε + ( µ ) ] + ln + O(λ ) m Π( ) = O(λ ) So the propagator s uncorrected to frst order n λ. Further, we found that the counterterm vertex does contrbute at O(λ). Srednc 4.6. Repeat problem 4.5 for the theory of problem 9.3. Fortunately, ths theory s nearly dentcal to that of the prevous problem, the only dfference s the symmetry factor. Thans to the complex feld n problem 9.3, S =, and so there s an extra factor of n the frst dagram that propagates through the problem. The results are: A = O(λ ) B = λ µ (4π) ε + + ln + O(λ ) m Π( ) = O(λ ) Srednc 4.7. Renormalzaton of the anharmonc oscllator. Consder an anharmonc oscllator, specfed by the Lagrangan L = Z q Z ωω q Z λ λω 3 q 4 We set h = and m = ; λ s then dmensonless. (a) Fnd the Hamltonan H correspondng to L. Wrte t as H = H + H, where H = P + ω Q, and [Q, P] =. We tae the conjugate momentum: Then, the Hamltonan s gven by: P = L q = Z q H = P q L
H = Z q + Z ωω q + Z λ λω 3 q 4 whch s, tang Q = q and P of course s the conjugate momentum as defned above, H = Z P + Z ωω Q + Z λ λω 3 Q 4 Then: H = P + ω Q + (Z )P + (Z ω )ω Q + Z λ λω 3 Q 4 where the frst two terms are H and the remander are H. Choosng P to be the conjugate momentum of Q means that the commutaton relaton requred by the problem s automatcally satsfed. (b) Let and be the ground and frst excted states of H, and let Ω and I be the ground and frst excted states of H. We tae all these egenstates to have unt norm. We defne ω to be the exctaton energy of H, ω = E I E Ω. We normalze the poston operator Q by settng I Q Ω = Q = (ω) /. Fnally, to mae thngs mathematcally smpler, we set Z λ equal to one, rather than usng a more physcally motvated defnton. Wrte Z = + A and Z ω = + B, where A = κ A λ + O(λ ) and B = κ B λ + O(λ ). Use Raylegh-Schrödnger perturbaton theory to compute the O(λ) correctons to the unperturbed energy egenvalues and egenstates. Ths s just quantum mechancs. The energy of the unperturbed th state s gven by E = (+)ω. The frst-order correcton to ths s, by Shanar 7..7: E = H whch s: E = (Z )P + (Z ω )ω Q + Z λ λω 3 Q 4 Next we set Z λ =, Z = + A, and Z ω = + B. Then, E = AP + Bω Q + λω 3 Q 4 Splttng ths up, and usng the expanson of A and B, we have: E = κ Aλ P + κ Bλω Q + λω 3 Q 4 Now we quantze the operators n the usual way, see Shanar 7.4. and 7.4.9, whch state that: Q = (a + a) ω ω P = (a a) 9
Ths scheme s great because the poston operator s normalzed out of the box to (ω) / (between and ) as Srednc prescrbed. Thus our correcton to the energy egenvalues are: E = ωλ 4 κ A (a a) + ωλ 4 κ B (a + a ) + λω 4 (a + a ) 4 whch s: E = ωλ 4 κ A aa + a a + ωλ 4 κ B aa + a a + λω 4 aaa a + aa aa + aa a a + a aaa Now let s smplfy to the case where = : +a aa a + a a aa E = ωλ 4 κ A + ωλ 4 κ B + 3λω 4 Hence, the ground state energy of the perturbed system s: Now for the case where = : E Ω = ω + λω 4 ( κ A + κ B + 3) + O(λ ) E = 3ωλ 4 κ A + 3ωλ 4 κ B + λω 4 (6 + 4 + + + + ) Hence, the frst excted state energy of the perturbed system s E I = 3ω + 3ωλ 4 ( κ A + κ B + 5) Now for the correctons to the egenstates, gven by Shanar 7..4: Thus: Ω = + n m whch smplfes to: Ω = + n m ω (m+)ω n = n + n m m H m E Em [ κ Aλ m P + ] κ Bλω m Q + λω 3 m Q 4 m λ [ κa m (a a) κ B m (a + a) m (a + a) 4 ] m 4m In the frst two bra-ets, (a ± a) = a a ± aa ± a a + aa. These last two terms annhlate the vacuum, gvng a zero. The second term gves a zero also snce m n. Then: Ω = + n m λ [ κa m a a κ B m a a m (a + a) 4 ] m 4m
These frst two terms are only defned f m =. Then, Smplfyng: Ω = + λ [ κa a a κ B a a ] m n m λ 4m m (a + a) 4 m λ(κa + κ B ) Ω = λ 4m m (a + a) 4 m n m Dong the expansons as before, we fnd that the last bra-et gves 4 f m = 4, and 6 f m =. Then, λ(κa + κ B ) Ω = 3λ 6λ 4 4 Ths gves: λ(κa + κ B + 6) Ω = 6λ 4 + O(λ ) We do the same procedure for I, I ll sp the gory detals. The result s: I = λ 6(κ A + κ B + ) 3 λ 3 5 + O(λ ) (c) Fnd the numercal values of κ A and κ B that yeld ω = E I E Ω and I Q Ω = (ω) /. Imposng the frst condton, we have: Ths mples that: Imposng the second condton, we have: [ I Q Ω = ω E I E Ω = ω + ωλ ( κ A + κ B + 6) = ω κ A κ B = 6 (4.7.) λ 6(κ A + κ B + ) ] 3 λ 3 5 ( a + a ) [ ] λ(κa + κ B + 6) 6λ 4 = ω Usng the operators and equaton (4.7.), we have: [ ] I Q Ω = λ 6(κ A + ) 3 λ 3 5 ω 4 [ 6λκA 3 λκ ] A 3λ 6λ 4 5 4 3
Ths gves: I Q Ω = [ λκ ] A ω + O(λ ) Hence, κ A =. Usng equaton (4.7.), we have then that κ B = 6. (d) Now thn of the Lagrangan n equaton 4.54 as specfyng a quantum feld theory n d = dmensons. Compute the O(λ) correcton to the propagator. Fx κ A and κ B by requrng the propagator to have a pole at = ω wth resdue one. Do your results agree wth those of part (c)? Should they? Let s start by wrtng the Lagrangan agan: Now we ll wrte ths n the usual way: L = Z q Z ωω q Z λ λω 3 q 4 L = q ω q + (Z ) q (Z ω )ω q Z λ λω 3 q 4 where the frst two terms are the free feld terms, the second two terms are the counterterms, and the fnal term s the nteracton. Next, let s draw the terms that would correct the propagator at O(λ): l x Ths s a new theory, so we have to use new Feynman rules. Fortunately these are easy to derve from the Lagrangan. For the frst term, we have the vertex factor, the nternal propagator, the symmetry factor, and the ntegral over l. Then, Π ( ) = 4λω 3 dl l + ω ε π where the symmetry factor s snce exchangng the two ends of the propagator s equvalent to exchangng those two vertces, the vertex factor ncludes a factor of 4 snce the Qs can be pared wth the vertces n any of 4! dfferent ways, and the ntegral s one dmensonal snce the problem tells us to wor on one dmenson. By the way, f the factor of 4 seems unusual, that s because most Lagrangans have cleverly-chosen coeffcents to cancel these factors, see for example the Lagrangan of problem 9.. Note also that we have ω rather than m, snce ω has taen the place of mass n our problem. Next we do a Wc rotaton, the result s: Π ( ) = 6λω3 π l + ω dl
Ths ntegral s smple to evaluate, usng equaton 4.7 or other methods. We fnd: Π ( ) = 6λω That seems suffcently smplfed! Let s turn to the contrbuton from the second dagram. The only thng we have to worry about s the vertex factor, whch we read off from the counterterm Lagrangan. The result s: Π ( ) = ( (Z ) (Z ω )ω ) Ths should not be unclear, but f t s, t can also be obtaned by comparng our Lagrangan wth equaton 9.9, and mang the necessary substtutons to modfy Feynman Rule #9 on page 77. At any rate, we now use the nformaton n part (b), Z = κ A λ and Z ω = κ B λ. Then, Π ( ) = λ ( κ A κ B ω ) The entre self-energy (at O(λ)) s then: Π( ) = 6λω λ ( κ A κ B ω ) The propagator wll have a pole at = m wth resdue one f the boundary condton 4.7 and 4. are met. Imposng 4., we fnd: Hence κ A =. Imposng 4.7, we have: Π ( ) = λκ A = Π( m ) = 6λω + λκ B ω = whch mples that κ B = 6, n agreement wth the result from part (c). Fnally, we wrte the correcton to the propagator: ( ) = + ω ε + O(λ ) Fnally, we have the queston of whether the agreement between (c) and (d) s expected. Our method of correctng the propagator n part (d) nvolves usng our Feynman Rules, whch we derved from the LSZ formula, whch depends on the Klen-Gordon Equaton. It may seem as though ths s therefore ncompatble wth quantum mechancs. On the other hand, perturbaton theory s not strctly lmted to quantum mechancs: a close loo at the dervaton of perturbaton theory (see for example Grffths 6.) shows that there s no dependence on the Schrödnger equaton or any other tenets that volate feld theory. In fact, the bra-ets that we evaluated n part (b) are dependent only on the defnton of ω and Q, and both of those are defned n a way whch s consstent wth feld theory. Hence, we do expect these to agree. Note: In Srednc s solutons, hs equaton 4.6 has the order of the terms n the denomnator reversed. Ths s easly verfed, see for example Shanar 7..3, Saura 5..44, or Grffths 6.3. Fortunately, ths just results n a negatve sgn that cancels out n the mddle, so hs conclusons are correct, though some terms n hs corrected wave-functons n part (b) are off by a sgn. 3