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Ehime-th-1 Kyushu-HET-64 Caustics in the Grassmann Integral Taro KASHIWA∗ Department of Physics, Ehime University, Matsuyama, 790-8577, Japan and 3 Tomohiko SAKAGUCHI† 0 0 Department of Physics, Kyushu University, Fukuoka, 812-8581, Japan 2 n February 1, 2008 a J 7 Abstract 2 v It is shown that a simple model of 2N-Grassmann variables with a four-body coupling involves 9 caustics when the integral has been converted to a bosonic form with the aid of the auxiliary field. 1 Approximationisthenperformed toassurevalidityoftheauxiliaryfieldmethod(AFM).Itturnsout 0 that even in N =2, thesmallest case in which a four-body interaction exists, AFM does work more 1 excellently if higher order effects, given by a series in terms of 1/√3N around a caustic and of 1/N 0 around a saddle point, would betaken intoaccount. 3 0 / h t - p e h : v i X r a [email protected] ∗ [email protected] † 1 1 Introduction In the path integral formalism, it is well-known that the semiclassical or the WKB approximation is a straightforward and moreover the most plausible prescription in nonpertubative treatments[1]. Under that, first we must find out solution(s) of the equation deriving from the condition that the first (functional) derivative of an action, a quantity in the exponent of path integration, should vanish. Second we should check a positivity of the second derivative of the action at that point (or trajectory), that is, convexity of the action. If a solution meets this condition, designated as the stability condition[2], it is called the saddle point. Third, expanding the action around the saddle point and regarding terms greater than the third order derivative as perturbation we have a series expansion expressed by a loop expansion parameter, such as ¯h or 1/N. Almost every cases follow these standard procedures but sometimes there happens a situation that the second derivative itself alsovanishesatthepoint,whichissocalledacaustics. Thecureforthiscasehasalreadybeengiven, for example, in the textbook by Schulman[3]: we should start with a point given on the condition that the second derivative vanishes. We then utilize the Airy integration, instead of the Gaussian integration in the standard case. A thorough study for caustics (quantum caustics) is seen in the recent works by Horie et al.[4]. Another powerful technique exclusively used in the path integral formalism is the auxiliary field method(AFM);characterizedbyanadditionaldegreeinsertedintotheoriginalexpressionintermsof aGaussianintegraltokillafour-bodyinteraction. Althoughthemethodhasalonghistory[5,6],ithas recentlybeenclarifiedthatwecanimproveresultswhengoingintohigherordersoftheloopexpansion in terms of the auxiliary field. It is important to note that even if the loop expansion parameter is not tiny butunity,results are still persistent[7]. There are many approximation schemes even under the path integral formalism, such as variational method[8], optimized perturbation theory[9], dilute gas approximation[10], valley method[11], etc., but ”Simpler is Better”. In this paper, we study a 4-fermi interaction in 0-dimension as a prototype of the Nambu–Jona- Lasinio(NJL)model[12]. AlthoughtheNJLmodelwasoriginallyhandledbythemeanfieldtechnique, itisextensivelyanalyzedwiththeaidofAFM[5]. OurmotivationistoestablishthevalidityofAFM inNJL aswas in thebosonic case[7], however,wehavefound thatthemodelitself posseses caustics. Therefore it is tempting to analyze it as a toy model to check the powerfulness of AFM even in a causticexample. Byemployingthestandardprescriptionforthecaustics,itisunderstoodthatAFM can give us a more accurate result if higher orders are taken into account. The paperis organized as follows: in 2, we propose our model and calculate its integration. We § thenintroducean auxiliary field and apply thesaddlepoint method tothat. In 3 and 4, we study § § cases of the 4-fermi coupling, λ2, being posive and negative, respectively. The negative case, in 4, § containscaustics thenwefollow thestandardprocedure,utilizingtheAiryintegration insteadofthe Gaussian one. The final section is devoted to discussions. 2 Grassmann Integral and Auxiliary Field Method Themodel considered hereis λ2 Z = dNξdNξ∗exp ω(ξ∗ ξ)+ (ξ∗ ξ)2 , (1) − · 2N · Z h i where N dNξ≡dξ1···dξN , dNξ∗ ≡dξN∗ ···dξ1∗ , (ξ∗·ξ)≡ ξi∗ξi , (2) i=1 X andthecouplingconstantλ2 issupposedreal. Wehaveintroduced2N-Grassmannvariablesandour notaion is followed from thetextbook of ref.[1]. Z in eq.(1) is exactly calculable by expandingtheexponent and using thebinomial theorem: Z = dNξdNξ∗ ∞ 1 ω(ξ∗ ξ)+ λ2 (ξ∗ ξ)2 n n! − · 2N · Z n=0 X (cid:8) (cid:9) 2 ∞ n 1 λ2 = dNξdNξ∗ ( ω)n−r( )r(ξ∗ ξ)n+r . (n r)!r! − 2N · Z n=0r=0 − XX Recalling the formula, dNξdNξ∗(ξ∗ ξ)m =( 1)NN!δmN , (3) · − Z we obtain [N2] N! λ2 Z = ωN−2r( )r , (4) (N 2r)!r! 2N r=0 − X with [α] being the Gauss’ symbol. In the following we compare this result with that of AFM with thehelp of theloop expansion, whose expansion parameter is now 1/N. Introducean auxiliary field, y, to cancel the(ξ∗ ξ)2 term in eq.(1): insert theidentity · ∞ dy 1 λ 2 1= exp y+ (ξ∗ ξ) , (5) √2π − 2 √N · Z−∞ (cid:20) (cid:8) (cid:9) (cid:21) togive Z = ∞ dy dNξdNξ∗exp y2 (ω+ λ y)(ξ∗ ξ) √2π − 2 − √N · Z−∞ Z h i = ∞ dy (ω+ λ y)Nexp y2 √2π √N − 2 Z−∞ h i y √Ny ∞ N →= dyexp[ NS(y)] , (6) 2π − Z−∞r with y2 S(y) ln(ω+λy) . (7) ≡ 2 − When N goes larger (although we will assume that N is not so large in the following), it is expected that theintegral is dominated by thesaddle point y obeying both conditions such that 0 λ 0=S(1)(y )=y , (8) 0 0− ω+λy0 and S(2)(y )>0 ; stability condition . (9) 0 Hereand hereafter we employ thenotation, dnS(y) S(n)(y) . (10) ≡ dyn Write Ω ω+λy , (11) 0 ≡ tofind that eq.(8) becomes λ2 Ω ω =0 , (12) − − Ω giving ω √ω2+4λ2 Ω= ± . (13) 2 3 ExpandingS(y) around y , 0 Z = ∞ N dyexp N S + 1S(2)(y y )2+ ∞ 1S(n)(y y )n , 2π − 0 2 0 − 0 n! 0 − 0 Z−∞r (cid:20) (cid:26) Xn=3 (cid:27)(cid:21) then makinga change of variable, y y y/√N, we obtain 0 − 7→ Z =e−NS0 ∞ dy exp S0(2)y2 ∞ N1−n/2S0(n)yn , (14) √2π − 2 − n! Z−∞ h Xn=3 i with S(n) S(n)(y ). In viewofeq.(14)thetermsofn 3becomes ineffectivebecause ofthefactor 0 ≡ 0 ≥ N1−n/2,whichenableustoregardtheseasperturbation. (Asstressedbefore,however,N willbeset in a small finitevalue in the following.) The first 6 terms are expressed as λ2 S = lnΩ, (15) 0 2Ω2 − λ2 λ2 S(2) = 1+ =1+ 0 (ω+λy )2 Ω2 0 √ω2+4λ2(√ω2+4λ2 ω) = ∓ , (16) 2λ2 2λ3 2λ3 S(3) = = , (17) 0 −(ω+λy0)3 −Ω3 6λ4 6λ4 S(4) = = , (18) 0 (ω+λy )4 Ω4 0 24λ5 24λ5 S(5) = = , (19) 0 −(ω+λy0)5 − Ω5 120λ6 120λ6 S(6) = = , (20) 0 (ω+λy )6 Ω6 0 respectively. The expansion up to the3-loop correction is therefore read as Z = e−NS0 ∞ dy exp S0(2)y2 1+ 1 S0(4)y4+ (S0(3))2y6 √2π − 2 N − 4! 2(3!)2 Z−∞ h ih n o 1 S(6) (S(4))2 S(3)S(5) 3(S(3))2S(4) (S(3))4 + 0 y6+ 0 + 0 0 y8 0 0 y10+ 0 y12 N2 − 6 2(4!)2 3!5! − (3!)34! (3!)44! n 1 (cid:16) (cid:17) o + ( ) . (21) O N3 i 3 Calculation: λ2 > 0 In thiscase, eq.(13) reads 1 √1+4λ2 Ω = ± , (22) ± 2 so that the stability condition, eq.(9), with eq.(16) gives λ2 √1+4λ2(√1+4λ2 1) S(2) =1+ = ∓ >0 , (23) 0 Ω2 2λ2 ± that is, both solutions meet the stability condition thus should be taken as the saddle points. (We haveput ω=1 hereand hereafter.) Therefore from eq.(21) the L-loop approximation for Z is given by, ZL-loop =ZL+-loop+ZL−-loop , (24) 4 lnZ tree lnZ exact 1.5 N=2 N=10 1.4 N=100 1.3 1.2 1.1 1 λ 1 2 3 4 5 6 7 8 9 10 Figure 1: The ratio; lnZ /lnZ for 1 < λ < 10. The solid, the dotted and the dashed-dotted lines tree exact desiganate N = 2,10, and 100 cases, respectively. When N becomes larger the ratio closes to unity, showing reliablity of large N approximation. where Nλ2 Zt±ree(≡Z0±-loop)≡Ω±Nexp −2Ω2 ; (25) h ±i and Nλ2 Ω2 Z3±-loop = Ω±Nexp −2Ω2 sΩ2+±λ2 h ±i ± 1 3λ4 5λ6 1+ + × N −4(Ω2 +λ2)2 6(Ω2 +λ2)3 h n ± ± o 1 5λ6 329λ8 105λ10 385λ12 + + + . (26) N2 −2(Ω2+λ2)3 32(Ω2+λ2)4 − 8(Ω2+λ2)5 72(Ω2+λ2)6 n ± ± ± ± oi with the aid of eqs.(15) (20). In eq.(26) we have the 1- or the 2-loop approximation if we neglect ∼ (1/N) or (1/N2) term respectively. O O To see how the results depend on N, we plot the ratio, lnZ to lnZ in Fig.1 for the cases tree exact of N =2,10 and 100 and 1<λ<10. (Weshould need thelogarithm since Z itself becomes huge in N =100.) FromFig.1,it isobviousthatthetreeresult becomes exactunderN ,which guaranteesthe 7→∞ large N expansion. However, as is the bosonic case[7], even when N is small, the smallest, N = 2, in thiscase, wecan get a betteranswer bytakinghigher loops intoaccount: theresultsare listed in Table 1 where the ratio Z /Z is listed for L=0,1,2,3. In spite of the fact that at λ=10 L loop exact − theerrorisreaching 50% in thetreeapproximation, it iscuredto4% in the1-loop correction and ∼ almost totheexactvalueinthe2-loop. ThesameratioisplottedinFigs.2,whichshows thathigher the loop we go the better the results are. As was discussed in the previous work[7], however, since theloop expansion is merely an asymptotic expansion, it isnot always truethat a higherloop effect can be more accurate; whose fact can be seen in Fig.3. The 2-loop result becomes better than that of the3-loop in 2 < λ < 4. ∼ ∼ 5 tree 1-loop 2-loop 3-loop λ exact tree/ ex. 1-loop/ ex. 2-loop / ex. 3-loop / ex. 1.0000 1.0000 1.0000 1.0000 10−3 1.0000 1.00 1.00 1.00 1.00 1.0001 1.0001 1.0001 1.0001 10−2 1.0001 1.00 1.00 1.00 1.00 10−1 1.0050 1.0100 1.0050 1.0050 1.0050 1.00 1.00 1.00 1.00 1.8147 1.5346 1.5039 1.4996 1 1.5000 1.21 1.02 1.00 1.00 74.679 53.050 50.866 50.902 10 51.000 1.46 1.04 0.997 0.998 Table 1: The exact value as well as that of Z and the ratio, Z /Z , are listed in N = 2 case for L loop L loop exact 10−3 λ 10. The 2-loop results almost reprod−ucethe exact value. − ≤ ≤ 1.25 1.5 tree/exact 1-loop/exact 1.45 2-loop/exact 1.2 3-loop/exact 1.4 tree/exact 1.35 1-loop/exact 1.15 2-loop/exact 1.3 3-loop/exact 1.25 1.1 1.2 1.15 1.05 1.1 1 1.05 1 0.95 λ0.95 λ 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1 2 3 4 5 6 7 8 9 10 λ<1 1<λ<10 Figure 2: The ratio; ZL loop/Zexact in N = 2 case for 10−3 λ 10. The solid, dotted, dashed-dotted and − ≤ ≤ dashed-double-dottedlines correspond to thetree, 1-, 2-, and 3-loop results, respectively. 4 Calculation: λ2 < 0 ; caustic case Next,we consider theλ2 <0 case. Write λ2 = λ2 to find Ω and S(2) are −| | 0 1 1 4λ2 Ω = ± − | | , (27) ± 2 p λ2 1 4λ2( 1 1 4λ2) S(2) = 1 | | = − | | ± − − | | , (28) 0 − Ω2 2λ2 ± p | | p respectively. From eq.(28), we find S(2) = 0 when λ = 1/2, that is, caustics. We cannot proceed 0 | | anymorearoundthisvaluewiththestandardrecipeintermsofGaussianintegration. Wethendivide the region of λ into three segments: i) λ < 1/2 , ii) λ > 1/2, and iii) λ 1/2, and designate | | | | | | | | ≈ i) and ii) as the Gaussian region and iii) as the Airy region. The reason is that in i) and ii) we can proceed as usual but in iii) we should adopt theAiry integral, as will beshown in thefollowing. 6 1.005 1.004 2-loop/exact 3-loop/exact 1.003 1.002 1.001 1 0.999 0.998 0.997 0.996 0.995 λ 1 2 3 4 5 6 7 8 9 10 Figure 3: The 2- and 3-loop results to the exact value are plotted. The 2-loop dominates the 3-loop between λ 2 and λ 4, implying that the loop expansion is merely an aymptotic expansion. ∼ ∼ 4.1 Gaussian region I: |λ| < 1/2 Let usstart with the case, λ <1/2. In view of eq.(28), only Ω is the saddle point; since Ω does + | | − not satisfy thestability condition. A general expression, eq.(21), then reads N λ2 Ω2 Z = ΩNexp | | + 1 + h 2Ω+2 isΩ+2 −|λ|2h 1 3λ4 5λ6 + | | | | N −4(Ω2 λ2)2 − 6(Ω2 λ2)3 n +−| | +−| | o 1 5λ6 329λ8 + | | + | | N2 2(Ω2 λ2)3 32(Ω2 λ2)4 n +−| | +−| | 105λ10 385λ12 + | | + | | 8(Ω2 λ2)5 72(Ω2 λ2)6 +−| | +−| | o 1 + ( ) . (29) O N3 i 4.2 Gaussian region II: |λ| > 1/2 In thiscase, eqs.(27), (28) are rewritten as 1 i 4λ2 1 Ω = ± | | − , (30) ± 2 p i 4λ2 1( 1 i 4λ2 1) S(2) = | | − ± − | | − 0 2λ2 p | | p 4λ2 1 4λ2 1 1 = | | − | | − i . (31) λ 2λ ± 2λ p | | (cid:18)p | | | |(cid:19) As S(2) is complex, the integration range should be analytically continued from the real axis to the 0 contoursatisfied with conditions, Re[S(2)y2]>0 , Im[S(2)y2]=0 . (32) 0 0 To thisend, write S0(2) = Re±iθ , (33) 7 with 4λ2 1 1 π R | | − , tanθ ; (0<θ< ) , (34) ≡ p |λ| ≡ 4|λ|2−1 2 and p y = reiϕ, (35) tofind S0(2)y2 = r2Re2iϕ±iθ. (36) Theone of theconditions (32) thusreads 0=Im[S(2)y2]=r2Rsin(2ϕ θ) , (37) 0 ± namely, θ ϕ = , (38) ∓2 and therest of it gives Re[S(2)y2]=r2Rcos(2ϕ θ)=r2R , (39) 0 ± which is apparently positive. Therefore the stability condition is established for both cases Ω and + Ω . The quantities, eqs.(30),(15), and (17) (20), read − ∼ Ω = iλ( 4|λ|2−1 1 )= λe±i(π2−θ) , (40) ± ± | | 2λ ∓ 2λ | | p | | | | S0 = −2λ2e|λ|i2(π 2θ) −ln(|λ|e±i(π2−θ)) | | ± − 1 π 1 = cos2θ ln λ i( θ sin2θ) , (41) 2 − | |∓ 2 − − 2 2iλ3 S0(3) = λ3e±i|(32|π−3θ) =∓2e±3iθ , (42) | | 6λ4 S0(4) = λ4e |i(2|π 4θ) =6e±4iθ , (43) | | ± − 24iλ5 S0(5) = − λ5e±i|(52|π−5θ) =∓24e±5iθ , (44) | | 120λ6 S0(6) = − λ6e i|(3π| 6θ) =120e±6iθ . (45) | | ± − As was in eq.(24) both solutions Ω come into play but they are complex conjugate each other, so ± that eq.(24) reads ZL-loop =ZL+-loop+ZL−-loop =2Re ZL+-loop . (46) h i Therefore Ztree 2λNe−N2 cos2θcosN π θ 1sin2θ , (47) ≡ | | 2 − − 2 (cid:18) (cid:19) and Z3-loop ≡ 2√|λR|Ne−N2 cos2θ cos N π2 −θ− 21sin2θ − θ2 (cid:20) (cid:26) (cid:18) (cid:19) (cid:27) 8 1 3 π 1 3 + cos N θ sin2θ + θ N − 4R2 2 − − 2 2 (cid:26) (cid:26) (cid:18) (cid:19) (cid:27) 5 π 1 5 + cos N θ sin2θ + θ 6R3 2 − − 2 2 (cid:26) (cid:18) (cid:19) (cid:27)(cid:27) 1 5 π 1 5 + cos N θ sin2θ + θ N2 − 2R3 2 − − 2 2 (cid:26) (cid:26) (cid:18) (cid:19) (cid:27) 329 π 1 7 + cos N θ sin2θ + θ 32R4 2 − − 2 2 (cid:26) (cid:18) (cid:19) (cid:27) 105 π 1 9 cos N θ sin2θ + θ −8R5 2 − − 2 2 (cid:26) (cid:18) (cid:19) (cid:27) 385 π 1 11 + cos N θ sin2θ + θ , (48) 72R6 2 − − 2 2 (cid:26) (cid:18) (cid:19) (cid:27)(cid:27)(cid:21) where usehas been madeof theGaussian integration formula, ∞ dy y2nexp S0(2)y2 = e∓i2n2+1θ ∞ dr r2nexp 1Rr2 √2π − 2 √2π −2 Z−∞ h i Z−∞ h i = (2Rnn−+112)!!e∓i2n2+1θ (n≥0) . (49) Hereineq.(48),Z andZ approximationsareobtainedbydroppingtermsof (1/N)and 1-loop 2-loop O (1/N2) respectively. O 4.3 Airy region: |λ| ≈ 1/2 As was mentioned several times before, the Gaussian integtration cannot be used in this region because S(2) 0; instead, we should employ a new approximation based on the Airy integral: first 0 ≈ findthesolution yõf λ2 S(2) =1 | | =0 , (50) 0 − (ω+iλy˜)2 | | instead of S(1)(y)=0 . As in eq.(11), define Ω˜ ω+iλy˜, (51) ≡ | | toobtain Ω˜ = λ . (52) ± ±| | Write S˜(n) S(n)(y˜) , (53) ≡ (again, puttingω=1) then, i iλ i S˜(1) = (1 λ) | | = (1 2λ) . (54) ± λ ∓| | − ( λ) λ ∓ | | | | ±| | | | Weshould adopt Ω˜ as asuitable solution; because which connects smoothly with thesaddle point: + Ω˜ yields S˜(1) =0. At this point + + 1 i 2 (1 λ)2 S˜= (1 λ) ln λ = −| | ln λ , (55) 2 λ −| | − | | − 2λ2 − | | (cid:26)| | (cid:27) | | 2iλ3 6λ4 S˜(3) = | | =2i , S˜(4) = | | =6 , (56) λ3 λ4 | | | | 24iλ5 S˜(5)= | | = 24i . (57) − λ5 − | | 9 Second,expand theexponent S(y) around yãnd makea change of variable, 2 1 t y y˜=i 3t= , (58) − NS˜(3) −N31 (cid:0) (cid:1) tofind Z = e−NS˜Z−∞∞ √N216πdtexphi3t3 +iat−Xn∞=4(−n!1N)nn3S˜−(n1)tni, (59) where 2 N3 a (1 2λ) . (60) ≡ λ − | | | | Note a (1) even under N since 1 2λ 0 in this region. Therefore, third we keep the ∼ O 7→ ∞ − | | ≈ first two terms in theexponent and regard others as perturbation to obtain Z = λNexp N(1−|λ|)2 2N16 | | 2λ2 π | | r ∞ h t3 i dtcos +at × 3 hZ0 h i 1 ∞dtt4cos t3 +at −4N31 Z0 h3 i + 1 1 ∞dtt8cos t3 +at +1 ∞dtt5sin t3 +at N23n32Z0 h3 i 5Z0 h3 io 1 + ( ) . (61) O N i N =2 exact 1.1 3-loop 3-loop:Airy 1 0.9 0.8 0.7 0.6 |λ| 0.2 0.4 0.6 0.8 1 Figure 4: N = 2 case for 10−3 λ 1: the solid, dotted, and dashed-dotted lines correspond to the exact, ≤ | | ≤ theGaussian 3-loop, the (Airy) 3-loop results, respectively. The Airy curvedrops down at λ 3.5, reaches the | |∼ bottom 1.88 at λ =0.238 and goes up to infinity as λ 0. ∼− | | | |7→ This expansion is not a loop expansion given in terms of 1/N but of 1/N1/3, however, we adopt the same terminology of the loop-expansion as the Gaussian case, that is, designate the first factor as a tree, the second integral of (1) as a 1-loop, and those of (1/N1/3) and (1/N2/3) as 2- and O O O 3-loop, respectively. We thus write Z as an approximation keeping terms up to (1/N2/3) in 3-loop O eq.(61). Finally, we utilize the Airy integral[13] 1 ∞ t3 Ai(a)= dtcos +at ; Im[a]=0 , (62) π 3 Z0 h i 10