1 (1 + ( )) = 1 8 ( ) = (c) Carrying out the Taylor expansion, in this case, the series truncates at second order:

Transcription

1 68A Solutons to Exercses March 05 (a) Usng a Taylor expanson, and notng that n 0 for all n >, ( + ) ( + ( ) + ) We can t nvert / because there s no Taylor expanson around 0 Lets try to calculate the nverse of alpha by lookng at ( + ) ( + ( )) Unfortunately, we can t take the lmt 0, due to the sngular term There s thus no Grassman number correspondng to (b) Usng a bnomal expanson, notng that the expanson truncates at lnear order, we obtan: p ( + ) + 8 ( ) + + (c) Carryng out the Taylor expanson, n ths case, the seres truncates at second order: cos[ + ] ( + ) + 6 ( + ) 4 ( )( ) () (d) Smlarly, carryng out a Taylor expanson of the matrx exponental, (e) If f (, ) f + exp B@ + CA + g s a Grassmanan functon of two varables, f (, f ( g ) g, Snce d erentaton over Grassman varables s equvalent to ntegraton d d f (, f g

2 (f) " # d d d d exp, det 5 In ths queston, I wanted to check that you had understood how to set up a fermonc path ntegral (a) The frst step n settng up the path ntegral s to wrte the Trace usng coherent states For a general operator Â, we may wrte h c  c A oo h c 0h0 c + A 0 h c 0h c + A 0 h c h0 c + A h c h c A oo + A 0 c + A 0 c + A cc, () so that the trace may be wrtten Tr[A] A 00 + A d cdce cc h c  c (3) Applyng ths to the partton functon, Tr[e H ] d c 3 dc e c 3c 0 0 h c 3 e H c 0 d c 3 dc 3 e c 3c 3 h c 3 e H c 0, (4) where we have used the defnton, c 3 c 0 We now use the completeness relaton d cdce cc chc (5) to ntroduce two tme-slces nto the matrx element h c 3 e H c 3, whch we wrte as h c 3 e H c 3 d c dc d c dc h c 3 e H c h c e H c h c e H c 0 e c c c c (6) Fnally, usng the expanson of the matrx element n terms of coherent states, h c j+ e H c j e c j+c j + O( ), (7) where ( ), we obtan 3 d c 3 dc 3 d c dc d c dc e [ c 3c + c c + c c 0 ] [ c 3 c 3 + c c + c c ] >< d c 3 dc 3 d c dc d c dc exp >: ( c 3, c, c ) 0 B@ CA B@ 0 c 3 c c 9 > CA >;, (8) where we have set c 0 c 3 n the last step

3 (b) Snce ths ntegral s Gaussan, the ntegral s gven by the determnant of the matrx: det 0 B@ CA + 3 (9) 0 (c) Generalzng ths result to N tme-slces, we obtan In the lmt N, N det 0 [M] M B@ CA det[m] + N (by nspecton) (0) so that Lt N N e, () N N + e () 3 We need to evaluate D[ f, f ]e S, (3) where S d ~ ~B, where ~B Bẑ s the appled feld Snce ths s a Gaussan ntegral, ~ ~B f, (4) F T ln TTr ~ ~B (5) To evaluate the trace, we go nto the frequency n F T Tr ln n ~ ~B (6) n 3

4 Formally, to carry out the Matsubara sum, we convert the summaton to a contour ntegral around the poles of the Ferm functon f (z) We then dstort the ntegral around the branch-cuts n the logarthm and carry out the ntegral to obtan: F T ln ( n B) n, branch cut d B B dz f (z) ln( z B) f ()Im ln( B + ) d f () T ln( + e B ) T ln[ cosh B ] (7) The partton functon s cosh B cosh B + spn + (8) Notce how ths d ers from the partton functon of a spn because the remander term derved from the the empty and doubly occuped states One way of removng these states s usng the Popov Fedatov trck, n whch one adds an magnary chemcal potental to the Hamltonan as follows: H f ~ ~B T f + ( f f ) The addtonal complex chemcal potental term has the e ect of cancellng out the unwanted empty state wth the unwanted doubly occuped state, to gve the correct partton functon 4 Let us evaluate I d cdch c  ce cc, (9) where h c  c (A 0 A cc)e cc (0) Carryng out the ntegral, we have I d cdc(a 0 A cc)e cc 4

5 as expected d cdc(a 0 A cc)( cc) d cdc( cca 0 A cc) (A 0 + A ), () 5 I m sorry, ths was a lot more challengng that I had n mnd at frst The hnt was only half-way helpful, because I had made a mstake n my orgnal soluton We had M e P, j A j c c j, where A j s an N N antsymmetrc matrx, and the c j are a set of N canoncal Ferm creaton operators Our task s to use coherent states to calculate t Tr[MM ], where the trace s over the N dmensonal Hlbert space of fermons As we shall see, the answer for A real, s smply t det[ + A] Convertng ths nto Grassmanan calculus, we have Y t d c dc h c MM ce P c c () Now snce MM s already normal ordered, we know that 0 3 h c MM c exp 64 B@ A j c c j + (Hc) CA 75 e P c c,, j so that the full trace can be wrtten 0 Y t d c dc exp 64 B@ A j c c j + (Hc) CA c c 375, j Y " # d c dc exp A A, (3) where s twce the N dmensonal unt matrx and ( c, c N, c c N ), 0 c c N, c B@ CA c N ) (4) 5

6 defnes a N dmensonal spnor and ts conjugate Now (3) s a Gaussan ntegral, so we can defntely do t, but you ve got to be a bt careful, because you can t just take the determnant of the matrx, because the c occur n both and, so and aren t ndependent You can do the ntegral n varous ways One way to do t would be to frst dagonalze 0 A A B@ N N CA, (5) where the are the egenvalues of the matrx The doublng of the egenvalues s guaranteed by the partcle-hole symmetry of the matrx Now when we do the ntegral over the, we have to notce that the frst N and second N terms n the exponental of the ntegrand are dentcal and should be grouped together, so that the ntegral over the varables s then, s Y t d d e P Y A ( ) N det A, (6) where the ( ) N s a result of makng the dentfcaton det A Y A ( ) N 0B@ CA (7) Another way to do the ntegral s to square t, wrtng the square as the ntegral as two separate ntegrals, then combnng them nto a sngle ntegral n whch and are ndependent (The ( ) N s pcked up n the combnaton process) The fnal answer s then the square root of the determnant Anyway, we can check our answer for A 0, and ths gves N, whch s the rght answer You can also check the result for the N case where A 0 a, whch gves t 4 a by drect a 0 ntegraton, and from the above formula We can actually smplfy the result a bt more If we square the matrx, t becomes dagonal, so that A det A det 4 + AA 4 + A det(4 + A A) det(4 + AA ) det(4 + A A) (8) A In the above expresson, we are able to dentty det(4 + AA ) det(4 + AA ) T det(4 + A A T ) det(4 + ( A )( A)) det(4 + A A), where the mddle step reled on the antsymmetry of A Here 6

7 we ve been a bt sloppy, wrtng 4 where we really should wrte 4 Thus, A det A ( ) N det(4 + A A) (9) Combnng everythng, we then have t Tr[MM ] h det(4 + A A) / [det(4 A A)] /, (30) where the last step follows from the antsymmetry of A For the specal case where A s real, so that A A, ths can be further smplfed nto t p det[ A] det[ + A] p det[ A] det[ A] T det[ + A] (3) 7