Конспект лекций по статистической физике. Коренблит С.Э - 23 стр.

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¢á¥£¤ ¯®¤à §ã¬¥¢ ¥â ç¨áâë¥ á®áâ®ï­¨ï ¨ ¥¥ ¯®¤á¨á⥬. ’®«ìª® ¥á«¨
¢á¥, ªà®¬¥ ®¤­®£®, wk à ¢­ë ­ã«î, ¯®¤á¨á⥬ 1 â ª¦¥ ®ª ¦¥âáï ¢ ç¨-
á⮬ á®áâ®ï­¨¨ ji, ¯à®¥ªâ®à®¬ ­ ª®â®à®¥ ¨ ï¥âáï ⮣¤ ¬ âà¨æ
¯«®â­®áâ¨, çâ® ¤ ¥â ç¨áâ® ¤¨­ ¬¨ç¥áª®¥, ­¥ ¢¥à®ïâ­®áâ­®¥ ®¯¨á ­¨¥:
                           %bc = jihj; (%bc)2 = %bc :                     (2.11)
ˆá¯®«ì§ãï ­¥§ ¢¨á¨¬®áâì á«¥¤ ®¯¥à â®à ®â ¢ë¡®à ¯à¥¤áâ ¢«¥­¨ï, § -
¯¨è¥¬ ¢ëà ¦¥­¨¥ ¤«ï á।­¥£® §­ 祭¨ï ®¯¥à â®à bb ¢ ᮡá⢥­­®¬ ¯à¥¤-
áâ ¢«¥­¨¨ ¤«ï bb, á ãç¥â®¬ (2.7), â ª¦¥, ¢ ¯à¥¤áâ ¢«¥­¨¨ (2.9), (2.10):
                                      X
         bbj'ii = bij'i i; â:¥:; bb = j'iibi h'i j; wi = h'i j%bj'ii;       (2.12)
                                      i
            b > = Tr(%bbb) = X wi bi = X wkhk jbbjki  X wk b(k);
         <                                                               (2.13)
                                i        k                  k
         £¤¥: b(k)  hk jbbjk i = Xhk j'i ibih'i jki  X bi jh'ijkij2: (2.14)
                                     i                      i
                 b >, ª ª à ¢¥­á⢮ (2.13) ¥áâì áâ â¨áâ¨ç¥áª®¥ á।­¥¥
’ ª¨¬ ®¡à §®¬, <
®â ª¢ ­â®¢®¬¥å ­¨ç¥áª¨å á।­¨å b(k) ®¯¥à â®à bb ¯® ­á ¬¡«î ç¨áâëå
á®áâ®ï­¨© jki ­ 襩 ¯®¤á¨á⥬ë 1, ॠ«¨§ã¥¬ëå ¢ í⮬ ­á ¬¡«¥ á ¢¥-
à®ïâ­®áâﬨ wk . ˆ «¨èì ¢ á«ãç ¥ ç¨á⮣® á®áâ®ï­¨ï, w1 = 1; wk6=1 = 0,
®­® ᢮¤¨âáï ª ®¡ëç­®¬ã ª¢ ­â®¢®¬¥å ­¨ç¥áª®¬ã á।­¥¬ã, ª ª १ã«ì-
â âã ç¨áâ® ¤¨­ ¬¨ç¥áª®£® ª¢ ­â®¢®£® ®¯¨á ­¨ï:
                 b > =) b(1) = h1jbbj1i = X bi jh1j'iij2:
               <                                              (2.15)
                                                 i

2   “à ¢­¥­¨¥ ä®­ ¥©¬ ­
   ‡ ¢¨á¨¬®áâì ¢¥ªâ®à á®áâ®ï­¨ï ¢ ¯à¥¤áâ ¢«¥­¨¨ ˜à¥¤¨­£¥à ®â ¢à¥-
¬¥­¨ t § ¤ ¥âáï ¢ëà ¦¥­¨¥¬: jk(t)i = exp iHt= c h jk(0)i, íª¢¨¢ «¥­â-
­ë¬ ãà ¢­¥­¨î ˜à¥¤¨­£¥à , ih @tjk(t)i = Hcjk(t)i. ˆá¯®«ì§ãï ¯à¥¤áâ -
¢«¥­¨¥ (2.10), ¯®«ã稬 ¬ âà¨æã ¯«®â­®á⨠%b(t), ª ª äã­ªæ¨î t, ¢ ¢¨¤¥:
           X                         c              c 
    %b(t) = jk (t)iwkhk (t)j = exp iHt= h %b(0) exp iHt=h :    (2.16)
             k
„¨ää¥à¥­æ¨à®¢ ­¨¥¬ ¯® t, á ãç¥â®¬ ¯¥à¥áâ ­®¢®ç­®á⨠Hc ¨ exp(iHt=       c h ),
   ­ 室¨¬: @ %b(t)=@t = (iH=  c h ) exp( iHt=
                                             c h )%b(0) exp(iHt=
                                                              c h ) +
           c h )%b(0)(iH=
   + exp( iHt=          c h ) exp(iHt=
                                     c h ) = (i=h )[Hc%b(t) %b(t)Hc];