1 Complete Set of Grassmann States

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1 Physcs 610 Homework 8 Solutons 1 Complete Set of Grassmann States For Θ, Θ, Θ, Θ each ndependent n-member sets of Grassmann varables, and usng the summaton conventon ΘΘ Θ Θ Θ Θ, prove the dentty e ΘΘ dθ d Θ e ΘΘ e Θ Θ e Θ Θ. (1 Hnt: Frst show that t s true when n 1 (brute force?. Then show that, for larger n, t factorzes nto n copes of the n 1 case. 1.1 Soluton For the case n 1, we want to show that (1+ ΘΘ Θ Θ (1+ ΘΘ (1+ Θ Θ (1 Θ Θ Θ Θ (1+ ΘΘ + Θ Θ (1 Θ Θ Θ Θ (1+ ΘΘ + Θ Θ Θ Θ ΘΘ Θ Θ Θ (Θ Θ ΘΘ Θ (1 0+ ΘΘ (2 For the general case, note that any even functon of Θ, Θ,Θ,Θ commutes wth any functon of Θj, Θ j,θ j,θ j f j. Also, exp(a + B exp(aexp(b f [ A, B ] 0. Therefore e ΘΘ e Θ Θ e Θ Θ e Θ 1 Θ 1 e Θ 1 Θ 1 e Θ 1 Θ 1 e Θ 2 Θ 2 e Θ 2 Θ 2 e Θ 2 Θ 2... (3 Further, dθ 1 d Θ 1 s even n Grassmann dervatves and commutes across any other par of Grassmann dervatves. Ths ensures that the n 1 case factorzes nto n copes of the n 1 case. 2 Movng past exponents We use the same notaton as n the prevous problem. Suppose that A s a polynomal n Grassmann varables whch s ndependent of Θ, that s, Θ A 0 for all. Suppose also 1

2 that B s a polynomal n Grassmann varables whch does not depend on Θ, so Θ B 0 for all. Frst, show that Be ΘΘ Θ A B Θ e ΘΘ A. (4 Whch condtons on A,B dd you need? [The exponent s a sum on all the Θ, but next to A stands just one specfc Θ.] Next, show that dθ Be ΘΘ A dθ B Θ e ΘΘ j A. (5 Θ j [Agan all Θ are ntegrated over, but the dervatve actng on A s just one of the Θ s.] Whch condtons on A,B dd you need n ths case? 2.1 Soluton For the frst step we must use that A 0. Then we can smply apply the dervatve: Θ ( e ΘΘ A e ΘΘ Θ Θ A+e ΘΘ Θ A Θe ΘΘ A+0 e ΘΘ ΘA. (6 For the second case we use the Θ ndependence of B to move t outsde of the ntegral; t then plays no role. Now perform ntegraton by parts: dθ e ΘΘ ( A dθ e ΘΘ A Θ j Θ j + + dθ ( Θ j ΘΘ dθ ( Θ j Θ k Θk e ΘΘ A e ΘΘ A dθ Θj e ΘΘ A. (7 B. The frst part reled on our condtons on A, the second part reled on our condtons for 3 Change of varables For ordnary (commutng numbers, f we make a change of varables x R j x j so x R 1 j x j (8 2

3 wth R j a nonsngular matrx, we usually defne partal dervatves n terms of the new varables as so that x x x j R j, x j δ j, (9 x x j R 1 j. (10 Defne changes of varables and dfferentaton wth respect to the new varables n the same way for Grassmann numbers: Θ R j Θ j so Θ R 1 j Θ j. (11 Use the fact that dθ n...dθ 1 Θ 1...Θ n 1 (12 and the fact that dθ to prove that Θ dθ n...dθ 1 Θ 1...Θ n DetR 1. (13 Argue that, as a consequence, dθ n...dθ 1 A DetR dθ n...dθ 1 A (14 where A s any polynomal n Grassmann varables. Therefore changes of varables n Grassmann varables gve rse to Jacobans whch are the nverses of those whch arse for bosonc varables. Hnt: you may fnd t easest to prove ths by dong the case of one Grassmann varable and the case of two Grassmann varables explctly, to see the pattern. Partal credt wll be gven f you can show these cases but you cannot generalze to larger numbers of varables. 3.1 Soluton Usng the expresson for Θ n terms of Θ j, we have dθ n...dθ 1 Θ 1...Θ n dθ n...dθ 1 R Θ 1...Rn 1 n Θ n. (15 Performng the dθ 1 ntegral gves n terms: n ( 1 m 1 Rm1 1 dθ n...dθ 2 Rj 1 j Θ j. (16 m1 j m For the case n 1 ths result s just R 1 DetR 1. We make an nducton assumpton that the desred relaton holds for up to n 1 Θ s. The ntegral n Eq. (16 nvolves a matrx 3

4 R 1 shorn of ts m-column and 1-row, whch s the m,1 mnor of R 1, M m1 [R 1 ]. Usng the nducton assumpton, n dθ n...dθ 1Θ 1...Θ n ( 1 m 1 Rm1 1 DetM m1 [R 1 ] (17 m1 whch s the expresson, by mnors, for the determnant of R 1. Now consder a general A a polynomal of the Θ (and perhaps of other Grassmann numbers. Only the terms contanng each Θ contrbute to Eq. (14. These terms can be wrtten as A 12...n Θ 1...Θ n where A 12...n s a coeffcent (possbly ncludng other Grassmann varables. Snce ths coeffcent s ndependent of the Θ, t can be factored out (up to some sgns and the ntegral s reduced to the one we have already performed. The ntegral over the Θ varables can be computed n the same way (wth thesame sgns when factorng A 12...n out of the ntegral, but to perform t we have to change varables back to Θ leadng to the desred determnant. 4 Shftng ntegraton varables Show explctly that, for the Grassmann varables Θ, Θ, η, and η (each a sngle varable, n ths problem we wll not deal wth n-vectors of Grassmann numbers! dθd Θ exp ( ΘΘ+ ηθ Θη exp ( ηη. (18 Hnt: expand everythng and use brute force. 4.1 Soluton dθd Θ exp ( ΘΘ+ ηθ Θη dθd Θ e ΘΘ e ηθ e Θη dθd Θ (1+ ΘΘ(1+ ηθ(1 Θη dθd Θ (1+ ΘΘ+ ηθ(1 Θη dθd Θ (1+ ΘΘ+ ηθ Θη ηθ Θη (19 Now we perform the Θ ntegral: d Θ1 0, d Θ ΘΘ Θ, d Θ ηθ 0, d Θ( Θη η, (20 and d Θ( ηθ Θη d Θ( Θ ηθη ηθη +Θ ηη. (21 4

5 Usng these expressons, dθd Θ exp ( ΘΘ+ ηθ Θη dθ ( Θ η+θ ηη 1+ ηη e ηη (22 as desred. 5 Combnng your results Consder the ntegral dθ d Θ exp ( Θ M j Θ j + η Θ Θ η. (23 Use the prevous two results to prove that ths ntegral equals DetM exp ( η M 1 j η j. (24 Hnt: consder defnng Θ M 1 Θ. Then ηθ ηm 1 Θ. Then defne η ηm Soluton Frst defne Θ M 1 Θ. In terms of Θ the exponent s ΘMΘ+ ηθ Θη ΘMM 1 Θ + ηm 1 Θ Θη ΘΘ + ηm 1 Θ Θη. (25 Now just rewrte dθ Det M dθ. Introducng η ηm 1, the exponent s e ΘΘ + η Θ Θη whch s e Θ 1 Θ 1 + η 1 Θ 1 Θ 1 η 1 tmes the same for 2,3,...n. Now we can do each ntegral for 1,2,... separately, yeldng DetM exp ( η η DetM exp ( ηm 1 η. (26 5