Конспект лекций по статистической физике. Коренблит С.Э - 79 стр.

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ᮣ« á­® (8.10), (8.11), ª®­áâ ­âã ­®à¬¨à®¢ª¨ (8.12), ¨ ᢮¤¨âáï ª 㦥
§­ ª®¬®© ª®¬¡¨­ â®à­®© § ¤ ç¥ ®¡ ®¯à¥¤¥«¥­¨¨ ç¨á« ¬ˆªà®á®áâ®ï­¨©
á¨á⥬ë á § ¤ ­­®© ¯®«­®© í­¥à£¨¥© ¨ ç¨á«®¬ ç áâ¨æ (8.7), ¯à¨ 䨪á¨-
஢ ­­ëå § ᥫ¥­­®áâïå [nf ] ®¤­®ç áâ¨ç­ëå á®áâ®ï­¨© jf i. ¥ à¥è¥­¨¥
(6.16), (6.17), á â®ç­®áâìî ¤® ¯¥à¥®¡®§­ 祭¨©, ᮢ¯ ¤ ¥â á (5.15){(5.16),
çâ® ¢á¢ï§¨ á ¢ë¢®¤®¬ (6.14){(6.18) ®¡á㦤 «®áì ¢ëè¥ ¢ ‡ ¬¥ç ­¨¨ II.
    à¨ § ¤ ­­®¬ ­ ¡®à¥ [nf ] ¢®«­®¢ ï äã­ªæ¨ï E (8.8) ¨¤¥ «ì­®© á¨-
áâ¥¬ë ­¥á¥â ¨­ä®à¬ æ¨î ⮫쪮 ® ¥¥ ᨬ¬¥âਨ, ª®®à¤¨­ â­ ï äã­ª-
æ¨ï 'E (8.5){(8.8) ä¥à¬¨- á¨áâ¥¬ë ¢ ®¡®§­ 祭¨ïå (8.1) ¡ã¤¥â ¯à®áâ® ­®à-
¬¨à®¢ ­­ë¬ ®¯à¥¤¥«¨â¥«¥¬ det jj fk (j )jj, ª®â®àë© ¯à¨ N = 2 ¨ ¥áâì (8.3):
        'EN (1; : : : ; N ) = p1
                                         X
                                                   
                               N ! f1fN 2ffkgN1 f1fN f1
                                                              (1)         N) 
                                                                          fN (

         p1N ! X [nf ]( 1)P c f1 (1)    fN (N ) ;                              (8.13)
                 fb f g
£¤¥: f1fN - ¥¤¨­¨ç­ë© ¯®«­®áâìî ­â¨á¨¬¬¥âà¨ç­ë© ⥭§®à à ­£ N ,
c { ¯® ¯à¥¦­¥¬ã, ®¯¥à â®à ¯¥à¥áâ ­®¢ª¨ ç áâ¨æ (á¬. (8.54) ¨ (16.4)).
,

3   ®«ìè ï áâ âá㬬 ª¢ ­â®¢®© á¨á⥬ë
    Š ­®­¨ç¥áª ï áâ âá㬬 ï¥âáï á«¥¤®¬ áâ â¨áâ¨ç¥áª®£® ®¯¥à â®à
(4.23), ­¥§ ¢¨áï騬 ®â ¢ë¡®à ¯à¥¤áâ ¢«¥­¨ï. ’®£¤ , ¢ ¯à¥¤áâ ¢«¥­¨¨
(8.8), á ãç¥â®¬ (8.7), (8.12), á«¥¤ãï (8.9){(8.11), ¨¬¥¥¬:
                     n             o
            ZN = Tr exp Hc                    
                                                                    (8.14)
                 X
                      h(f1 : : : fN )j exp Hc j(f1 : : : fN )i =
              f1 fN 2ff g
                                   0               1
                                               "fj A 1
                   X                      N
                                          X
          =
              f1 fN 2ff g
                               exp @
                                      C [nf ] h[nf ]j[nf ]i =
                                         j=1
                                                                                    (8.15)

          = exp ( E [nf ]) N; N [nf ] 1
            X                                   X
           fnf g                      C [nf ] fb f g [nf ]1 =                      (8.16)
                          0                  1
              X                   X
          =           exp @            "f nf A N; P nf = ZN ;                      (8.17)
              fnf g                f              f

£¤¥ ¢ëà ¦¥­¨¥ (8.15) ᮢ¯ ¤ ¥â á (6.18), â.ª., ­ ¯à¨¬¥à, ᮣ« á­® (8.13),
®â«¨ç­ë¥ ®â ­ã«ï ¬ âà¨ç­ë¥ í«¥¬¥­âë h[nf ]j[nf ]i = 1 (áà. (8.54)), ¨