ÎÌ»ÒÎϳعֵÁÏ¿2005ǯÂè12²ó


¢«Âè11²ó¤Ø¡¡2005ǯ ½éÅùÎÌ»ÒÎϳء¿ÎÌ»ÒÎϳؤιֵÁÏ¿Ìܼ¡¤ËÌá¤ë¡¡Âè13²ó¤Ø¢ª

15.3  ±é»»»Ò¤Ë¤è¤ë²òË¡

¡¡¤¹¤Ç¤Ë½Ò¤Ù¤¿¤è¤¦¤Ë¡¢¥¨¥Í¥ë¥®¡¼¸ÇÍ­ÃͤÏ1/2hbar.png¦Ø¤òºÇÄãÃͤȤ·¤Æ¡¢hbar.png¦Ø¤º¤ÄÂ礭¤¯ ¤Ê¤Ã¤Æ¤¤¤¯¤À¤í¤¦¤È¿ä¬¤¹¤ë¤³¤È¤¬¤Ç¤­¤ë¡£¤½¤³¤Ç°Ê²¼¤Ç¤Ï¡¢¡Ö¥¨¥Í¥ë¥®¡¼¸ÇÍ­Ãͤòhbar.png¦Ø¤À¤±¾å¾º¤µ¤»¤ë±é»»»Ò¡×¤òºî¤Ã¤Æ¤¤¤¯¤³¤È¤Ë¤¹¤ë¡£¤â¤·¤½¤ó¤Ê±é»»»Ò¤òºî¤ë¤³ ¤È¤¬¤Ç¤­¤ì¤Ð¡¢¤½¤Î±é»»»Ò¡Êa¢÷¤È̾ÉÕ¤±¤ë¡Ë¤ò¤É¤ó¤É¤ó¤«¤±¤Æ¤¤¤¯ ¤³¤È¤Ç¡¢¤¤¤í¤ó¤Ê¥¨¥Í¥ë¥®¡¼¤ò»ý¤Ã¤¿¾õÂÖ¤òµá¤á¤ë¤³¤È¤¬¤Ç¤­¤ë¡£
  ¥¨¥Í¥ë¥®¡¼¤Ï±é»»»ÒH¤Î¸ÇÍ­ÃͤǤ¢¤ë¤«¤é¡¢¥¨¥Í¥ë¥®¡¼E¤ò»ý¤Ä¾õÂÖ¡ÊÇÈÆ°´Ø¿ô¤ò¦×E¤È½ñ¤³¤¦¡Ë¤¬¤¢¤Ã¤¿¤È¤·¤Æ¡¢¤³¤Î¾õÂÖ¤ÎÇÈÆ°´Ø¿ô ¤Ëa¢÷¤ò¤«¤±¤ë¤È¡¢¥¨¥Í¥ë¥®¡¼E+¦Å¤ò»ý¤Ã¤¿¾õÂÖ¡ÊÇÈÆ°´Ø¿ô¤òa¢÷ ¦×E¤È½ñ¤³¤¦¡Ë¤Ë¤Ê¤ë¤ï¤±¤Ç¤¢¤ë¡£¦Å¤Ï¸å¤Ç·è¤Þ¤ëÄê¿ô¤Ç¤¢¤ë ¡Ê¼Â¤Ïhbar.png¦Ø ¤Ë¤Ê¤ë¤È¤¤¤¦¤³¤È¤Ï¤¹¤Ç¤Ë½Ò¤Ù¤¿¤È¤ª¤ê¡Ë¡£¤Ä¤Þ¤ê¡Ö¦×E¤ËH¤ò¤«¤±¤ë¤È¸ÇÍ­ÃÍE¤À¤¬¡¢a¢÷¤ò¤«¤±¤Æa¢÷ ¦×E¤Ë¤·¤Æ¤«¤éH¤ò¤«¤±¤ë¤È¸ÇÍ­ÃͤÏE+¦Å¤Ë¤Ê¤ë¡×¤è¤¦¤Ê±é»»»Ò¤òºî¤ë¤ï¤±¤Ç¤¢¤ë¡£¤µ¤é¤ËÊ̤θÀ¤¤Êý¤ò¤¹¤ë¤È¡ÖH¤ò¤«¤±¤Æ¤«¤éa¢÷¤ò¤«¤±¤¿¤Î¤È¡¢a¢÷¤ò ¤«¤±¤Æ¤«¤éH¤ò¤«¤±¤¿¤Î¤Ç¤Ï¡¢¦Å¤À¤±¸ÇÍ­Ãͤ¬°ã¤¦¡×¤È¤¤¤¦¤³¤È¤Ë¤Ê¤ë¡£¤³¤ì¤ò¼°¤Çɽ¤»¤Ð¡¢
H a¢÷ ¦×E ¡Ý a¢÷ H ¦×E = ¦Åa¢÷¦×E
(15.29)
¤Ç¤¢¤ë¡£¤æ¤¨¤Ë¡¢
Ha¢÷¡Ýa¢÷ H=¦Åa¢÷  ¤¹¤Ê¤ï¤Á  [H,a¢÷]=¦Åa¢÷
(15.30)
¤È¤Ê¤ë¤è¤¦¤Ê±é»»»Ò¤¬ºî¤ì¤ì¤Ð¤è¤¤¤È¤¤¤¦¤³¤È¤Ë¤Ê¤ë¡£
  ¤³¤Îa¢÷¤ò¡Ö¥¨¥Í¥ë¥®¡¼¤ò¤¢¤²¤ë±é»»»Ò¡×¤È¤¤¤¦¤³¤È¤Ç¡Ö¾å¾º±é»» »Ò¡×¤È¸Æ¤Ö¡£¾å¾º±é»»»Ò¤ò¤ï¤¶¤ï¤¶a¢÷¤È¢÷¤Ä¤­¤Ç½ñ¤¤¤Æ¤¤¤ë¤Î¤Ï¡¢ÀΤ«¤é²¼¹ß±é»»»Ò¡Ê¥¨¥Í¥ë¥®¡¼¤ò²¼¤²¤ëÊý¤Î±é»»»Ò¡Ë¤ÎÊý¤òa¤È½ñ¤¯¤Î¤¬½¬´· ¤À¤«¤é¤Ç¤¢¤ë¡£¤Ä¤Þ¤ê¡¢¾å¾º±é»»»Ò¤È²¼¹ß±é»»»Ò¤Ï¸ß¤¤¤Ë¥¨¥ë¥ß¡¼¥È¶¦Ìò¤Ç¤¢¤ë¡£
 [H,a¢÷]=¦Åa¢÷¤Î¼°¤ÎξÊդΥ¨¥ë¥ß¡¼¥È¶¦Ìò¤ò¤È¤ë¤È¡¢[a,H]=¦Åa¤È¤Ê¤ë(¸ò´¹´Ø·¸¤Î¥¨¥ë¥ß¡¼¥È ¶¦Ìò¤ò¼è¤ë¤È½çÈÖ¤¬¤Ò¤Ã¤¯¤ê¤«¤¨¤ë¤³¤È¤ËÃí°Õ)¡£¤³¤ì¤Ï¤Ä¤Þ¤ê[H,a]=¡Ý¦Åa¤À¤È¤¤¤¦¤³¤È¤Ç¤¢¤ë¡£¤Ä¤Þ¤ê¡¢a¤Ï¥¨¥Í¥ë¥®¡¼¤ò¡Ý¦Å¤¢¤²¤ë¡Ê¦Å²¼¤²¤ë¡Ë±é »»»Ò¤Ç¤¢¤ë¡£°Ê¾å¤«¤é¡¢a¢÷¤¬¾å¾º±é»»»Ò¤Ê¤é¤Ða¤¬²¼¹ß±é»»»Ò¤Ê¤é ¤Ç¤¢¤ë¤³¤È¤¬¤ï¤«¤Ã¤¿¡£
  a¢÷¤òµá¤á¤ë¤Î¤Ï¤½¤ó¤Ê¤ËÆñ¤·¤¯¤Ê¤¤¡£¤Þ¤º¡¢
a¢÷ = C(x+ a p)
(15.31)
¤Î¤è¤¦¤Ëx,p¤Î°ì¼¡¼°¤Ç½ñ¤±¤ë¤³¤È¤ò²¾Äꤹ¤ë(C,a¤Ï¸å¤Ç·è¤á¤ë¤¬¡¢Ê£ÁÇ¿ô¤ÎÄê¿ô)¡£·¸¿ô¤¬´Êñ¤Ë¤Ê¤ë¤è¤¦¤Ë¤³¤³¤Ç¤â̵¼¡¸µ²½¤·¤¿É½µ­¤ò»È¤¦¤³¤È¤Ë¤· ¤Æ¡¢
a¢÷ = A(¦Î+¦Á ¢ß
¢ß¦Î
)
(15.32)
¤È¤·¤è¤¦¡ÊA,¦Á¤â¤Þ¤¿¡¢Ê£ÁÇ¿ô¤ÎÄê¿ô¡Ë¡£ º£H=¡Ý1/2[(¢ß2)/(¢ß¦Î2)]+1/2¦Î2¤Ç ¤¢¤ë¤«¤é¡¢¸ò´¹´Ø·¸¤ò¤È¤Ã¤Æ¤ä¤ë¤È¡¢

[H,a¢÷] =
A [ ¡Ý 1
2

¢ß2
¢ß¦Î2
+ 1
2
¦Î2, ¦Î+¦Á ¢ß
¢ß¦Î
]


(15.33)
¤Ç¤¢¤ë¡£¸ò´¹´Ø·¸¤Î¸å¤í¤Î¦Î¤Ë´Ø·¸¤¹¤ëÉôʬ¤ò·×»»¤¹¤ë¡£¤Þ¤º¦Î¤Ï¼«Ê¬¼«¿È¤È¤Ï¸ò´¹¤¹¤ë¤Î¤Ç¡¢¼«Ê¬¼«¿È¤Î¼«¾è¤È¤â¸ò´¹([¦Î2,¦Î]= 0)¤·¡¢

[ ¡Ý 1
2

¢ß2
¢ß¦Î2
+ 1
2
¦Î2, ¦Î ] =¡Ý 1
2

[ ¢ß2
¢ß¦Î2
, ¦Î ]
(15.34)
¤È¤Ê¤ë¡£¼¡¤Ë¸ø¼°[AB,C]=[A,C]B+A[B,C]¤ò»È¤¦¤È¡¢
¡Ý 1
2

[ ¢ß2
¢ß¦Î2
, ¦Î ] = ¡Ý 1
2

[ ¢ß
¢ß¦Î
,¦Î ]
¢ß
¢ß¦Î
¡Ý 1
2

¢ß
¢ß¦Î

[ ¢ß
¢ß¦Î
,¦Î ]
(15.35)
¤Ç¤¢¤ë¤¬¡¢[[¢ß/¢ß¦Î],¦Î]=1¤Ç¤¢¤ë¤«¤é¡¢·ë²Ì¤Ï

[ ¡Ý 1
2

¢ß2
¢ß¦Î2
+ 1
2
¦Î2, ¦Î ] =¡Ý ¢ß
¢ß¦Î

(15.36)
¤Ç¤¢¤ë¡£
¡¡Ç°¤Î¤¿¤á¤Ë¤¯¤É¤¤¤è¤¦¤À¤¬¤â¤¦°ìÅÙ½ñ¤¤¤Æ¤ª¤¯¤È¡¢¤³¤¦¤¤¤¦·×»»¤Î»þ¤Ë¤Ï¤µ¤é¤Ë¸å¤í¤ËǤ°Õ¤Î´Ø¿ô f(¦Î)¤¬¤¤¤ë¤È¹Í¤¨¤ë¡£¤Ä¤Þ¤ê¡¢[[¢ß/¢ß¦Î],¦Î]=1¤È¤Ï
¢ß
¢ß¦Î
¡Ê¦Î£æ¡Ë¡¡¡Ý¡¡¦Î ¢ß£æ
¢ß¦Î
=¡¡¡¡£æ
¤È¤¤¤¦¤³¤È¡£

¡¡Æ±Íͤˡ¢

[ ¡Ý 1
2

¢ß2
¢ß¦Î2
+ 1
2
¦Î2, ¢ß
¢ß¦Î
] =¡Ý¦Î
(15.37)
¤È¤Ê¤ë¤Î¤Ç¡¢
[H,a¢÷] = [ H,A (
¦Î+¦Á ¢ß
¢ß¦Î
) ] = ¡ÝA ( ¢ß
¢ß¦Î
+¦Á¦Î )
(15.38)
¤È¤¤¤¦·ë²Ì¤¬½Ð¤ë¡£
¡¡¤³¤Î¼°¤Î±¦ÊÕ¤¬¸µ¤Î±é»»»Ò¤Î¦Å ÇܤÀ¤È¤¤¤¦¾ò·ï¤òÃÖ¤¯¤È¡¢¡ÝA([¢ß/¢ß¦Î]+¦Á¦Î)=¦ÅA(¦Î+¦Á[¢ß/¢ß¦Î])¤«¤é¡¢

¡Ý1
=
¦Å¦Á

(15.39)

¡Ý¦Á
=
¦Å

(15.40)
¤È¤¤¤¦¼°¤¬½Ð¤ë¡£
¡¡¤Þ¤º(5.39)¤Ë(5.40)¤òÂåÆþ¤¹ ¤ë¤È
1=¦Á2
(15.41)
¤È¤Ê¤ê¡¢·ë²Ì¦Á = ¡Þ1¤È¤ï¤«¤ë¡£¤³¤ì¤ò(5.40)¤ËÂåÆþ¤·¤Æ ¡Þ1 = ¦Å¤È¤Ê¤ë¤¬¡¢¦Å¤¬Àµ¤Î¿ô¤Ë¤Ê¤ëÊý¤ò¤È¤Ã¤Æ¡¢¦Á = ¡Ý1¤È¤ë¤³¤È¤Ë¤¹¤ë¡£·ë¶É¡¢
a¢÷ =A  ( ¦Î¡Ý ¢ß
¢ß¦Î
)
(15.42)
¤È¤Ê¤ë¡Ê¡Ê¢ß/¢ß¦Î¡Ë¢÷¡á -¢ß/¢ß¦Î¤ËÃí°Õ¡Ë¡£a¤ÎÊý¤Ï
a=A* ( ¦Î+ ¢ß
¢ß¦Î
)
(15.43)
¤È¤Ê¤ë¡£
¡¡¥Æ¥­¥¹¥È¤Ç¤Ï¤³¤³¤ÎA¤ÈA*¤¬Æþ¤ìÂؤï¤Ã¤Æ¤¤¤¿¡£
¡¡¤â¤·¡¢¾å¤Ç¦Å¤¬Éé¤Ë¤Ê¤ë²ò¤ò¤È¤Ã¤¿¤È¤¹¤ë¤È¡¢a¤Èa¢÷¤¬Æþ¤ìÂØ¤ï ¤ë¤è¤¦¤ÊÅú¤¨¤Ë¤Ê¤Ã¤Æ¤·¤Þ¤Ã¤¿¤³¤È¤Ë¤Ê¤ë¡£a¢÷¤¬¥¨¥Í¥ë¥®¡¼¤ò¾å¾º ¤µ¤»¤ë±é»»»Ò¤Ë¤Ê¤ë¤è¤¦¤Ë¤·¤¿¤«¤Ã¤¿¤Î¤À¤«¤é¡¢¦Å¤òÀµ¤È¼è¤Ã¤¿¤Î¤ÏÀµ¤·¤«¤Ã¤¿¡£
¤³¤³¤Ça¤Èa¢÷¤Î¸ò´¹´Ø·¸¤ò¤È¤Ã¤Æ¤ß¤ë¤È¡¢

[a,a¢÷] =
AA* [ ¦Î+ ¢ß
¢ß¦Î
,¦Î¡Ý ¢ß
¢ß¦Î
]
=
AA* ( [ ¢ß
¢ß¦Î
,¦Î ] ¡Ý [ ¦Î, ¢ß
¢ß¦Î
] )
=
2 AA*

(15.44)
¤È¤Ê¤ë¡£A=[1/¢å2]ei¦Â¤È¤¹¤ì¤Ð±¦ÊÕ¤Ï1¤È¤Ê¤Ã¤Æ¸å¡¹³Ú¤Ê¤Î¤Ç¡¢¤½¤¦¤¹¤ë¤³¤È¤Ë¤¹¤ë¡£A¤Î°ÌÁê¦Â¤Ï·è¤Þ¤é¤Ê¤¤¤Î¤Ç¡¢¤³¤ì¤Þ ¤¿´Êñ¤Î¤¿¤á¦Â = 0¤ÈÁª¤ó¤Ç¤ª¤¯¡£a¤Èa¢÷¤Î¸ò´¹´Ø·¸¤Ï
[a,a¢÷]=1
(15.45)
¤È¤Ê¤Ã¤¿¡£¤Þ¤È¤á¤ë¤È¡¢
a= 1
¢å2

( ¦Î+ ¢ß
¢ß¦Î
) ,    a¢÷= 1
¢å2

( ¦Î¡Ý ¢ß
¢ß¦Î
)
(15.46)
¤Ç¤¢¤ë¡£¤³¤ì¤òµÕ¤Ë²ò¤¯¤È¡¢
¦Î = 1
¢å2
(a+a¢÷),    ¢ß
¢ß¦Î
= 1
¢å2
(a¡Ýa¢÷)
(15.47)
¤È¤Ê¤ë¡£¥Ï¥ß¥ë¥È¥Ë¥¢¥ó¤Ë¤³¤Î¼°¤òÂåÆþ¤¹¤ì¤Ð¡¢

H =
¡Ý 1
2


1
¢å2
(a¢÷¡Ýa)
2

 
+ 1
2


1
¢å2
(a+a¢÷)
2

 

=
¡Ý 1
4
(a¢÷¡Ýa)+ 1
4
(a+a¢÷)2
=
¡Ý 1
4
((a¢÷)2¡Ýaa¢÷ ¡Ýa¢÷ a + a2)+ 1
4
(a2 + aa¢÷ + a¢÷ a+(a¢÷)2)
=

1
2
(aa¢÷ + a¢÷ a) = a¢÷ a + 1
2


(15.48)
¤È¤Ê¤ë¡£ºÇ¸å¤Ï¸ò´¹´Ø·¸(5.45)¤«¤éa a¢÷=a¢÷ a+1¤È¤Ê¤ë¤³¤È¤ò»È¤Ã¤Æ¤¤¤ë¡£
¤³¤³¤Ç¡¢¼¡¸µ¤òÉü³è¤µ¤»¤è¤¦¡£a,a¢÷¤Ï̵¼¡¸µ¤Î¤Þ¤Þ¤È¤¹¤ë¡£ ¦Î¤òx¤Ëľ¤¹¤Ë¤Ï¡¢Ä¹¤µ¤Î¼¡¸µ¤ò»ý¤ÄÎÌ¢å{[hbar.png/m¦Ø]}¤ò¤«¤±¤ì¤Ð¤è¤¤¡£±¿Æ°Î̤ϡÝihbar.png[¢ß/¢ßx] ¤Ç¤¢¤ë¤«¤é¡¢¡Ýihbar.png¤ò¤«¤±¤¿¸å¡¢Ä¹¤µ¤Î¼¡¸µ¤ò»ý¤ÄÎÌ¢å{[((h/2p))/m¦Ø]}¤Ç³ä¤ì¤Ð¤è¤¤¡£¤æ¤¨¤Ë¡¢
x=   ¢å

hbar.png
2m¦Ø
 
(a+a¢÷),    p=i   ¢å

hbar.png m¦Ø
2
 
(a¢÷¡Ýa)
(15.49)
¤Ç¤¢¤ë¡£¤³¤ì¤«¤éa,a¢÷¤ò½Ð¤¹¤È¡¢
a=   
¢å

m¦Ø
2hbar.png
 

( x+i 1
m¦Ø
p ) ,    a¢÷=   
¢å

m¦Ø
2hbar.png
 

( x¡Ýi 1
m¦Ø
p )
(15.50)
¤È¤Ê¤ë¡£¥Ï¥ß¥ë¥È¥Ë¥¢¥ó¤Ï

H =hbar.png¦Ø ( a¢÷ a + 1
2
)

(15.51)
¤È¤Þ¤È¤Þ¤ë¡ÊH¤Ï¥¨¥Í¥ë¥®¡¼¤Î¼¡¸µ¤ò»ý¤Ã¤Æ¤¤¤ë¤«¤é¡¢hbar.png¦Ø¤ò¤«¤±¤Æ¤½¤Î¼¡¸µ¤òÉü³è¤µ¤»¤ë¡Ë¡£
¤³¤Î·Á¤Ë¤Ê¤ë¤È¡¢[H,a¢÷]=hbar.png¦Øa¢÷¤Ï¼«ÌÀ¤Ç¤¢¤ë¡Ê[a¢÷ a, a¢÷]=a¢÷[a,a¢÷]=a¢÷¡Ë¡£
¥¨¥Í¥ë¥®¡¼¸ÇÍ­Ãͤˤϲ¼¸Â¤¬¤Ê¤¯¤Æ¤Ï¤Ê¤é¤Ê¤¤¡£¤½¤¦¤Ç¤Ê¤¤(Äì̵¤·)¾ì¹ç¡¢¡Öʪ»ö¤Ï¥¨¥Í¥ë¥®¡¼¤ÎÄ㤤Êý¤ËÍî¤Á¤Æ¤¤¤¯¡×¤È¤¤¤¦Ë¡Â§¤Ë¤·¤¿¤¬¤Ã¤Æ¤É¤ó¤É¤ó¥¨¥Í ¥ë¥®¡¼¤¬Äã ¤¯¤Ê¤Ã¤Æ¤¤¤Ã¤Æ¤·¤Þ¤¦¡£

ºÇÄã¾õÂÖ¤¬¤¢¤ë¤È¤¹¤ë¤È¡¢¤½¤ì¤Ëa¤ò¤«¤±¤Æ¿·¤·¤¤¾õÂÖ¤òºî¤ë¤³¤È¤Ï¤Ç¤­¤Ê¤¤(¤â¤·¤Ç¤­¤¿¤é¡¢¤½¤Î¾õÂ֤ϡֺÇÄã¾õÂÖ¤è¤ê¤âÄ㤤¾õÂÖ(¡©)¡×¤Ë¤Ê¤Ã¤Æ¤·¤Þ ¤¦)¡£¤½¤³¤Ç¡¢ºÇÄã¾õÂÖ¤ÎÇÈÆ°´Ø¿ô¦×0¤Ï a¦×0=0 ¤ò¤ß¤¿¤¹¤È¤·¤è¤¦¡£
¤¹¤Ê¤ï¤Á¡¢

( ¦Î+ ¢ß
¢ß¦Î
) ¦×0=0
(15.52)
¤Ç¤¢¤ë¡£¤³¤¦¤Ê¤ë¤è¤¦¤Ê´Ø¿ô¤Ï
¦×0 = e¡Ý1/2¦Î2
(15.53)
¤Ç¤¢¤ë¡£¤³¤Î¦×0¤Çɽ¤µ¤ì¤ë¾õÂÖ¤ò¡Ö´ðÄì¾õÂ֡פȸƤ֡£
¡¡
¡¡¢¬¤Î¥°¥é¥Õ¤Ï´ðÄì¾õÂ֤Υ°¥é¥Õ¤Ç¤¢¤ë¡£¥°¥é¥Õ¤ÎÌÜÀ¹¤ê¤Ï̵¼¡¸µ²½¤·¤¿ºÂɸ¦Î¤Ç½ñ¤«¤ì¤Æ¤¤¤ë¡£´ðÄì¾õÂ֤ϥ¨¥Í¥ë¥®¡¼¤¬1/2hbar.png¦Ø¡¢Ìµ¼¡¸µ²½ ¤·¤Æ¹Í¤¨¤¿¾ì¹ç¤Ç1/2¤Ç¤¢¤ë¡£°ÌÃÖ¥¨¥Í¥ë¥®¡¼¤¬Ìµ¼¡¸µ²½¤·¤Æ½ñ¤¯¤È1/2¦Î2¤Ç ¤¢¤ë¤³¤È¤ò¹Í¤¨¤ë¤È¡¢¸ÅŵŪ¤Ë¤Ï¡Ý1 < ¦Î < 1¤Î¤È¤³¤í¤Þ¤Ç¤·¤«Î³»Ò¤Ï¸ºß¤Ç¤­¤Ê¤¤¤Ï¤º¤Ç¤¢¤ë¡Ê¿¶¤ê»Ò¤È¤·¤Æ¹Í¤¨¤¿¤Ê¤é¤Ð¡¢¦Î = ¡Þ1¤¬¿¶¤ê»Ò¤¬Ìá¤ë°ÌÃÖ¡Ë¡£ÇÈÆ°´Ø¿ô¤Ï¤½¤Î³°Â¦¤Ë¤â¸ºß¤¹¤ë¡£¤¿¤À¤·¡¢¤³¤ì¤Þ¤Ç¤Î¡Ö±¿Æ°¥¨¥Í¥ë¥®¡¼¤¬Éé¤Ë¤Ê¤ëÎΰè¡×¤ÈƱÍÍ¡¢ÇÈÆ°´Ø¿ô¤Î¶Ê¤¬¤ê¶ñ¹ç¤¬ÊѤï¤Ã ¤Æ¤¤¤ë¡Ê¶ñÂÎŪ¤Ë¸À¤¦¤Ê¤é¤Ð¡¢£²³¬Èùʬ¤ÎÉä¹æ¤¬°ã¤¦¡Ë¡£¿Þ¤Ç¤Ï¿å¿§¤Ç¡Ö¸ÅŵŪ¤Ëµö¤µ¤ì¤ëÎΰè¡×¤ò¼¨¤·¤¿¡£

¡¡¸ÅŵŪ¤Ëµö¤µ¤ì¤Ê¤¤Îΰè¤Ï¡¢¥È¥ó¥Í¥ë¸ú²Ì¤Î¾ì¹ç¤ÈƱÍÍ¡¢ÇÈÆ°´Ø¿ô¤¬µÞ®¤Ë ¸º¿ê¤·¤Æ¤¤¤¯¡£¤³¤Î¥°¥é¥Õ¤Ç¤Ïµö¤µ¤ì¤Ê¤¤Îΰè¤Ë¤â¤±¤Ã¤³¤¦Â礭¤ÊÃͤÇÇÈÆ°´Ø¿ô¤¬¤¢¤ë¤¬¡¢¼ÂºÝ¤Ë¸ÅŵÎϳؤǰ·¤¦¤è¤¦¤ÊÌäÂê¤Ç¤Ï¡¢¤¢¤Ã¤È¤¤¤¦¤Þ¤Ë¸º¿ê¤¬µ¯¤³ ¤ë¡£



¡¡Îã¤Ë¤è¤Ã¤Æ¤³¤ì¤é¤ò¥¢¥Ë¥á¡¼¥·¥ç¥ó¤Ç¸«¤ë¥¢¥×¥ì¥Ã¥È¤¬¤¢¤ë¤Î¤Ç¸«¤Æ¤¯¤À¤µ¤¤¡£

¡¡´ðÄì¾õÂÖ¤ÎÇÈÆ°´Ø¿ô¤Ëa¢÷¤ò¤«¤±¤Æ¤¤¤¯¤³¤È¤Ç¤½¤ì¤è¤ê¤âhbar.png¦Ø¤À¤±¥¨¥Í¥ë ¥®¡¼¤¬¹â¤¤¾õÂÖ¤ò¼¡¡¹¤È¤Ä¤¯¤ê½Ð¤·¤Æ¤¤¤¯¤³¤È¤¬¤Ç¤­¤ë¡£º¸¤Î¥°¥é¥Õ¤Ï´ðÄì¾õÂÖ¤«¤éÂ裳Î嵯¾õÂ֤ޤǤÎÇÈÆ°´Ø¿ô¡Ê¤Î¼Â¿ôÉô¡Ë¤òɽ¤·¤¿¤â¤Î¤Ç¤¢¤ë¡£´ðÄì¾õÂÖ¤« ¤é¡¢¥¨¥Í¥ë¥®¡¼¤¬¹â¤¯¤Ê¤ë¤´¤È¤Ë¡Ên¤¬Áý²Ã¤¹¤ë¤´¤È¤Ë¡ËÇÈÆ°´Ø¿ô¤ÎÇȤοô¤¬Áý¤¨¤Æ¤¤¤¯¤ÈƱ»þ¤Ë¡¢¶õ´ÖŪ¹­¤¬¤ê¤âÂ礭¤¯¤Ê¤Ã¤Æ¤¤¤ë¡£
  ¶ñÂÎŪ¤Ëa¢÷¤ò¤«¤±¤ë¤È¤¤¤¦·×»»¤ò¼Â¹Ô¤¹¤ë¤È¡¢

a¢÷ e¡Ý1/2¦Î2
=
¢å2¦Îe¡Ý1/2¦Î2

(15.54)

a¢÷ (¢å2¦Îe¡Ý1/2¦Î2)
=

( ¦Î¡Ý ¢ß
¢ß¦Î
) (¦Îe¡Ý1/2¦Î2)



=
(2¦Î2¡Ý1)e¡Ý1/2¦Î2

(15.55)
¤È¤Ê¤ë¡£°Ê²¼Æ±Íͤ˥¨¥Í¥ë¥®¡¼¸ÇÍ­ÃͤȸÇÍ­´Ø¿ô¤òµá¤á¤Æ¤¤¤¯¤³¤È¤¬¤Ç¤­¤ë¡£ÅöÁ³¡¢¤³¤ÎÅú¤Ï¤Þ¤¸¤á¤ËÈùʬÊýÄø¼°¤òµé¿ôŸ³«¤ò»È¤Ã¤Æ²ò¤¤¤¿·ë²Ì¤È°ìÃפ¹¤ë¡£
¡¡¶õ´ÖŪ¹­¤¬¤ê¤¬Â礭¤¯¤Ê¤ë¤³¤È¤Ï¡¢Ã±¿¶Æ°¤Î¿¶Éý¤¬Â礭¤¯¤Ê¤ë¤³¤È¤ËÂбþ¤¹¤ë¡£¤Þ¤¿¡¢ÇȤοô¤¬Â¿¤¯¤Ê¤ë¤È¤¤¤¦¤³¤È¤ÏÇÈŤ¬Ã»¤¯¤Ê¤ë¤È¤¤¤¦¤³¤È¤Ç¡¢Ê¿¹ÕÅÀ¤ò Ä̲᤹¤ë»þ¤Î±¿Æ°¥¨¥Í¥ë¥®¡¼¤¬Â礭¤¯¤Ê¤Ã¤Æ¤¤¤ë¤³¤È¤ËÂбþ¤·¤Æ¤¤¤ë¡£
¡¡¤Ê¤ª¡¢¿Þ¤Ë½ñ¤«¤ì¤¿ÇÈÆ°´Ø¿ô¤Ï¤¹¤Ù¤ÆÄê¾ï¾õÂ֤Ǥ¢¤Ã¤Æ¡¢¤³¤Î¾õÂ֤ǤϸÅŵŪ¤Ê°ÕÌ£¤Ç¤Î¡Ö±¿Æ°¡×¤Ï¸«¤¨¤Ê¤¤¡£Æó¤Ä°Ê¾å¤Î¥¨¥Í¥ë¥®¡¼¸ÇÍ­¾õÂÖ¤¬½Å¤Í¹ç¤ï¤µ¤ì ¤ë¤È¡¢ÇÈÆ°´Ø¿ô¤ÏÄê¾ï¾õÂ֤ǤϤʤ¯¤Ê¤ê¡¢Î³»Ò¤Î¸ºß³ÎΨ¤¬±¦¤Øº¸¤Ø¤ÈÆ°¤¯ÍͻҤ¬¸«¤¨¤Æ¤¯¤ë¡£¤³¤Î¤¢¤¿¤ê¤ÏÊɤËÊĤ¸¤³¤á¤é¤ì¤¿Î³»Ò¤Î¾ì¹ç¤ÈƱ¤¸¤Ç¤¢¤ë¡£  

¡¡Â裱Î嵯¾õÂ֤Ǥϥ¨¥Í¥ë¥®¡¼¤¬3/2hbar.png¦Ø¤Ç¤¢¤ê¡¢´ðÄì¾õÂ֤Σ³Çܤ¢¤ë¤Î¤Ç¡¢¸ÅŵŪ¤Ë ¸ºß¤Ç¤­¤ë¾ì½ê¤â¢å3Çܹ­¤¬¤ê¡¢¡Ý¢å3 < ¦Î < ¢å3¤È¤Ê¤ë¡£¤ä¤Ï¤ê¤³¤Î¾ì½ê¤ÇÇÈÆ°´Ø¿ô¤Î£²³¬Èùʬ¤¬Éä¹æ¤òÊѤ¨¤Æ¤¤¤ë¤³¤È¤¬¸«¤Æ¼è¤ì¤ë¡£
¸ÅŵŪ¤Ëµö¤µ¤ì¤ëÎΰè¤Ç¤ÏÇÈÆ°´Ø¿ô¤Ï¤Þ¤µ¤Ë¡ÖÇȡפȤʤäƿ¶Æ°¤¹¤ë¤è¤¦¤Ë¿¶¤ëÉñ¤¤¡¢¤½¤Î³°¤Ç¤ÏµÞ®¤Ë¸º¿ê¤¹¤ë´Ø¿ô¤Ë¤Ê¤Ã¤Æ¤¤¤ë¤Î¤Ç¤¢¤ë¡£
¡¡´ðÄì¾õÂÖ¤ÏE=1/2hbar.png¦Ø¤À¤±¤Î¥¨¥Í¥ë¥®¡¼¤ò»ý¤Ä¡£¤³¤ÎºÇÄ㥨¥Í¥ë¥®¡¼¤Î¤³¤È¤ò¡ÖÎíÅÀ¿¶Æ°¤Î¥¨¥Í¥ë ¥®¡¼¡×¤È¸Æ¤Ö¡£¸ÅŵÎϳØŪ¤Ë¤ÏºÇÄ㥨¥Í¥ë¥®¡¼¤È¤Ïγ»Ò¤¬¸¶ÅÀ¤ËÀŻߤ·¤¿¾õÂ֤Ǥ¢¤ê¡¢¥¨¥Í¥ë¥®¡¼¤Ï0¤Ç¤¢¤ë¡£¤·¤«¤·ÎÌ»ÒÎϳØŪ¤Ë¤Ï¸¶ÅÀ¤ÇÀŻߤ·¤Æ¤¤¤ë(¤Ä¤Þ ¤ê±¿Æ°Î̤â°ÌÃÖ¤â0¤È¤¤¤¦Ãͤ˳ÎÄꤷ¤Æ¤¤¤ë)¤È¤¤¤¦¤³¤È¤ÏÍ­¤êÆÀ¤Ê¤¤¡£¤³¤ì¤ÏÉÔ³ÎÄê´Ø·¸¤Î¤ª¤«¤²¤Ç¤¢¤ë¡£
²æ¡¹¤Ë´Ñ¬¤Ç¤­¤ë¤Î¤Ï¾ï¤Ë¥¨¥Í¥ë¥®¡¼¤ÎÊѲ½Î̤ʤΤǡ¢¥¨¥Í¥ë¥®¡¼¤Î¸¶ÅÀ¤Ï¤É¤³¤ËÁª¤ó¤Ç¤â¤è¤¤¡£¤è¤Ã¤Æ¡¢ÎíÅÀ¿¶Æ°¤Î¥¨¥Í¥ë¥®¡¼¤ò0¤ÈÃÖ¤¤¤Æ¤â¤è¤¤¡£¤¿¤À¤·¤½ ¤Î¾ì¹ç¤Ï¸ÅŵŪ¤ÊÄ´Ï¿¶Æ°»Ò¤Î¥¨¥Í¥ë¥®¡¼H=[1/2m]p2+1/2m¦Ø2 x2¤ËÈæ¤Ù¡¢1/2hbar.png¦Ø¤À¤±¾®¤µ¤¤Î̤ˤʤäƤ¤¤ë¡£ÎÌ»ÒÎϳؤǤϥ¨ ¥Í¥ë¥®¡¼¤Ïh¦Í¤Çɽ¤»¤ë¤Î¤Ç¡¢¡Ö¥¨¥Í¥ë¥®¡¼¤Î¸¶ÅÀ¤òÊѤ¨¤ë¤È¤¤¤¦¤³¤È¤ÏÇÈÆ°´Ø¿ô¤Î¿¶Æ°¿ô¤òÊѤ¨¤ë¤³¤È¤Ë¤Ê¤Ã¤Æ¤·¤Þ¤¦¤¬¡¢¤½¤ì¤Ï¤¤¤¤¤Î¤«¡©¡×¤È¤¤¤¦¿´ÇÛ¤¬µ¯ ¤­¤ë¤«¤â¤·¤ì¤Ê¤¤¤¬¡¢ÎÌ»ÒÎϳؤˤª¤¤¤Æ¡¢¥¨¥Í¥ë¥®¡¼¤Î¸¶ÅÀ¤òE0¤º¤é¤¹¤È¤¤¤¦¤³¤È¤ÏÇÈÆ°´Ø¿ô¤Î¦×¢ª e¡Ý[i/hbar.png]E0t¤È ¤¤¤¦ÃÖ¤­´¹¤¨¤ËÂбþ¤¹¤ë¡£¤³¤ÎÊѲ½¤Ï¡Ê¥¨¥Í¥ë¥®¡¼¤Î´üÂÔÃͤθ¶ÅÀ¤¬¤º¤ì¤¿¤È¤¤¤¦¤³¤È°Ê³°¤Ë¤Ï¡Ë´Ñ¬·ë²Ì¤Ë±Æ¶Á¤òÍ¿¤¨¤Ê¤¤¡£
¤³¤³¤Þ¤Ç¡¢ÇÈÆ°´Ø¿ô¤Îµ¬³Ê²½¤Ï¹Í¤¨¤Æ¤¤¤Ê¤«¤Ã¤¿¡£´ðÄì¾õÂÖ¤«¤é½ç¤Ëµ¬³Ê²½¤ò¹Í¤¨¤Æ¤¤¤³¤¦¡£¸ø¼°¢é¡Ý¡ç¡ç dx e¡Ýx2=¢å{¦Ð}¤¬¤¢¤ë¤Î¤Ç¡¢´ðÄì¾õÂÖ¤Ï
¦×0 = ¦Ð1/4e¡Ý1/2¦Î2
(15.56)
¤Èµ¬³Ê²½¤µ¤ì¤ë¡£¤¿¤À¤·¡¢¤³¤Î¾õÂÖ¤¬µ¬³Ê²½¤µ¤ì¤ë¤Î¤Ï¢é¡Ý¡ç¡ç d¦Î¦×*(¦Î)¦×(¦Î) ¤È¤¤¤¦ÀÑʬ¤Ë´Ø¤·¤Æ¤Ç¤¢¤ë¡£ÀÑʬ¤òx¤Ç¹Ô¤¦¤Ê¤é¤Ð¡¢d¦Î = ¢å{[m¦Ø/hbar.png]}dx¤È¤¤¤¦°ã¤¤¤ÎÉôʬ¤òµÛ¼ý¤¹¤ë¤¿¤á¤Ë¡¢
¦×0 = ( ¦Ðm¦Ø
hbar.png
) 1/4

 
exp[¡Ý[m¦Ø/(2hbar.png)]x2]
(15.57)
¤È¤·¤Æ¤ª¤¯É¬Íפ¬¤¢¤ë¡£¦Î¤Ï̵¼¡¸µ¤Ê¤Î¤Ç¡¢¦×(¦Î)¤Ë¤â¼¡¸µ¤Ï¤Ê¤¤¡£¤·¤«¤·x¤ÏŤµ¤Î¼¡¸µ¤ò»ý¤Ä¤Î¤Ç¡¢¢édx ¦×*¦× = 1¤È¤Ê¤ë¤¿¤á¤Ë¤Ï¦×(x)¤Ï(Ťµ)¡Ý1/2¤Î¼¡¸µ¤ò»ý¤¿¤Ê¤¯¤Æ¤Ï¤¤¤±¤Ê¤¤¡£(5.56)¤È(5.57)¤Î°ã¤¤¤Î °ø»Ò¤Ï ¤Á¤ç¤¦¤É¼¡¸µ¤Îʬ¤Ç¤¢¤ë¡£
¤³¤ì¤Ç¦×0¤Ïµ¬³Ê²½¤µ¤ì¤¿¤Î¤Ç¡¢¼¡¤ËÂ裱Î嵯¾õÂÖ(¦×1=a¢÷¦×0)¤ò¹Í¤¨¤Æ¤ß¤ë¡£¶ñÂÎŪ¤ËÀÑʬ¤¹¤ë¤³¤È¤Ç¤â·×»»¤Ç¤­¤ë¤¬¡¢ a,a¢÷¤Î¸ò´¹´Ø·¸¤ò»È¤Ã¤¿Êý¤¬´Êñ¤Ç¤¢¤ë¡£



¡ç

¡Ý¡ç 
d¦Î(a¢÷ ¦×0)* a¢÷ ¦×0 =


¡ç

¡Ý¡ç 
d¦Î¦×0* a a¢÷ ¦×0
=


¡ç

¡Ý¡ç 
d¦Î¦×0* (a¢÷ a +1) ¦×0
=


¡ç

¡Ý¡ç 
d¦Î¦×0* ¦×0=1

(15.58)
¤È¤Ê¤Ã¤Æ¡¢¤Á¤ç¤¦¤Éµ¬³Ê²½¤µ¤ì¤Æ¤¤¤ë¤³¤È¤¬¤ï¤«¤ë¡£¤³¤³¤Ç¤Ï¡¢¸ò´¹´Ø·¸¤«¤éaa¢÷ = a¢÷ a+1¤È¤Ê¤ë¤³¤È¤È¡¢a¦×0=0 ¤Ç¤¢¤ë¤³¤È¤ò»È¤Ã¤Æ¤¤¤ë¡£
Â裱Î嵯¾õÂ֤ϵ¬³Ê²½¤·¤Ê¤¯¤Æ¤â¤è¤«¤Ã¤¿¤¬¡¢Â裲Î嵯¾õÂÖ¤«¤éÀè¤Ï¤½¤¦¤Ï¤¤¤«¤Ê¤¤¡£¾å¤Î·×»»¤È¤ÎÂ礭¤Êº¹¤Ïa¦×0=0¤Ç¤¢¤Ã¤¿¤¬ a¦×1 ¡â 0¤À¤È¤¤¤¦¤³¤È¤Ç¤¢¤ë¡£ Â裲Î嵯¾õÂÖ¤ò¤È¤ê¤¢¤¨¤º¦×2=(a¢÷)2¦×0¤È ¤ª¤¯¤È¡¢


¡ç

¡Ý¡ç 
d¦Î((a¢÷)2 ¦×0)* (a¢÷)2 ¦×0 =
¡ç

¡Ý¡ç 
d¦Î¦×0* a a a¢÷ a¢÷ ¦×0
(15.59)
¤Ç¤¢¤ë¤«¤é¡¢±é»»»Òaaa¢÷ a¢÷¤ÎÉôʬ¤ò¹Í¤¨¤Æ¡¢¡Öa¤ò±¦¤Ë´ó¤»¤Æ¤¤¤¯¡×¤È¤¤¤¦·×»»¤ò¤¹¤ë¡£

a

aa¢÷
= a¢÷ a+1 
a¢÷ = aa¢÷

a a¢÷
= a¢÷ a+1 
+ a a¢÷ = aa¢÷ a¢÷ a + 2aa¢÷

(15.60)
¤È¤Ê¤ë¡£Â裱¹à¤Ï¦×0¤Ë¤«¤«¤ë¤È0¤È¤Ê¤ê¡¢Â裲¹à¤Ï¦×1¤ÈƱ¤¸ÊýË¡¤Ç·×»»¤Ç¤­¤ë¤¬¡¢·¸¿ô2¤Îʬ¤À¤±¡¢Åú¤¨¤¬ Â礭¤¯¤Ê¤ê¡¢¢éd¦Î¦×*2¦×2=2¤Ç¤¢¤ë¤³¤È¤¬¤ï¤«¤ë¡£¤æ¤¨¤Ëµ¬³Ê²½¤µ¤ì¤¿¦×2¤Ï[1/¢å2](a¢÷)2 ¦×0¤Ç¤¢¤ë¤³¤È¤¬¤ï¤«¤ë¡£
ƱÍͤη׻»¤ò·«¤êÊÖ¤¹¤È¡¢µ¬³Ê²½¤µ¤ì¤¿ÂènÎ嵯¾õÂÖ¤¬[1/¢å[n!]](a¢÷)n ¦×0¤Ç¤¢¤ë¤³¤È¤¬³Îǧ¤Ç¤­¤ë¡£¤è¤Ã¤Æ¡¢¦×n¡Ý1=[1/¢å[(n¡Ý1)!]](a¢÷)n¡Ý1¦×0¤Ëa¢÷¤ò¤«¤±¤Æ¢ån¤Ç³ä¤Ã¤¿¤â¤Î¤¬¦×n¤Ç¤¢¤ë¡£
¦×n = 1
¢ån
a¢÷ ¦×n¡Ý1
(15.61)



[Ì䤤15-0] ¿ô³ØŪµ¢Ç¼Ë¡¤ò»È¤Ã¤Æ¦×n=[1/¢å[n!]](a¢÷)n¦×0¤¬µ¬³Ê²½¤µ¤ì¤Æ¤¤¤ë¤³¤È¤ò¾ÚÌÀ ¤»¤è¡£



n=5¤Þ¤Ç¤Îµ¬³Ê²½¤µ¤ì¤¿ÇÈÆ°´Ø¿ô¤Ï°Ê²¼¤ÎÄ̤ꡣ

¦×0=
¦Ð1/4 e¡Ý1/2¦Î2

(15.62)

¦×1=
¦Ð1/4¢å2¦Îe¡Ý1/2¦Î2

(15.63)

¦×2=

¦Ð1/4
¢å2
(2¦Î2¡Ý1) e¡Ý1/2¦Î2

(15.64)

¦×3=

¦Ð1/4
¢å3
(2¦Î3¡Ý3¦Î)e¡Ý1/2¦Î2

(15.65)

¦×4=

¦Ð1/4
2¢å6
(4¦Î4¡Ý12¦Î2+3)e¡Ý1/2¦Î2

(15.66)

¦×5=

¦Ð1/4
2
  __
¢å15
 
(4¦Î5¡Ý20¦Î3+15¦Î) e¡Ý1/2¦Î2

(15.67)
¤³¤ì°Ê¹ß¤âƱÍͤ˷׻»¤µ¤ì¤ë¡£
¤Ê¤ª¡¢¸ÅŵÎϳØŪ¤Ë¤³¤ÎÄ´Ï¿¶Æ°»Ò¤ò¹Í¤¨¤ë¤È³Ñ¿¶Æ°¿ô¤Ï¦Ø¤Ç¤¢¤ë¡£ÎÌ»ÒÎϳØŪ¤Ë¹Í¤¨¤¿»þ¤ÎÇÈÆ°´Ø¿ô¤Î³Ñ¿¶Æ°¿ô¤Ï(n+1/2)¦Ø ¤Ç¤¢¤ê¡¢°ìÃפ·¤Ê¤¤¡£¤½¤ì¤ÏÅöÁ³¤Ç¤¢¤ë¡£º£¹Í¤¨¤¿ÎÌ»ÒÎϳØŪ¾õÂÖ¤ÏÄê¾ï¾õÂÖ¡Ê¥¨¥Í¥ë¥®¡¼¤Î¸ÇÍ­¾õÂ֡ˤǤ¢¤Ã¤Æ¡¢¸ÅŵÎϳØŪ¤Ê±¿Æ°¤Ï¡ÖÄê¾ï¾õÂÖ¤ÎÇÈÆ°´Ø¿ô¡× ¤Ç¤Ïɽ¤»¤Ê¤¤¡£¸ÅŵÎϳØŪ±¿Æ°¤ËÂбþ¤¹¤ë¤è¤¦¤Ê¾õÂÖ¤ÏÊ£¿ô¤Î¥¨¥Í¥ë¥®¡¼¸ÇÍ­¾õÂ֤νŤ͹ç¤ï¤»¤¬É¬ÍפʤΤǤ¢¤ë¡£¤¿¤È¤¨¤Ðn=0¤Î¾õÂÖ¡Ê´ðÄì¾õÂÖ¡£³Ñ¿¶Æ°¿ô1/2¦Ø¡Ë ¤Èn=1¤Î¾õÂÖ¡ÊÂ裱Î嵯¾õÂÖ¡£³Ñ¿¶Æ°¿ô3/2¦Ø¡Ë¤ò½Å¤Í¹ç¤ï¤»¤ì¤Ð¡¢¤½¤Î»þ¤Îx¤Î´üÂÔÃÍ < x > ¤Ï¡¢Æó¤Ä¤Î¾õÂ֤γѿ¶Æ°¿ô¤Îº¹¤Ç¤¢¤ë¤È¤³¤í¤Î3/2¦Ø¡Ý1/2¦Ø = ¦Ø¤Î³Ñ¿¶Æ°¿ô¤Ç¿¶Æ°¤¹¤ë¡£

15.4  Åż§ÇȤΥ¨¥Í¥ë¥®¡¼¤¬h¦Í¤Ç¤¢¤ë¤³¤È

ºÇ¸å¤ËÎÌ»ÒÎϳؤλϤޤê¤È¤Ê¤Ã¤¿»ö¼Â¡¢¡Ö¸÷¤Î¥¨¥Í¥ë¥®¡¼¤¬h¦Í¤ÎÀ°¿ôÇܡפȤ¤¤¦¤³¤È¤ò³Îǧ¤·¤Æ¤ª¤³¤¦¡£zÊý¸þ¤Ë¿Ê¹Ô¤¹¤ëÅż§ÇȤξì¹ç¡¢Åžì¤ÎxÀ®Ê¬Ex¤Î ½¼¤¿¤¹ÊýÄø¼°¤Ï
(
¢ß2
¢ßz2
¡Ý 1
c2

¢ß2
¢ßt2
) Ex=0
(15.68)
¤³¤ÎEx¤ò¥Õ¡¼¥ê¥¨ÊÑ´¹¤·¤Æ
Ex(x,y,z,t) = 1
(2¦Ð)3/2


dkx dky dkz Ex(kx,ky,kz,t)ei(kxx+kyy+kzz)
(15.69)
¤È¤·¤è¤¦¡£¤³¤ì¤òÊýÄø¼°¤ËÂåÆþ¤¹¤ë¤È¡¢


( ¢ß2
¢ßz2
¡Ý 1
c2

¢ß2
¢ßt2
) (
1
(2¦Ð)3/2


dkx dky dkz Ex(kx,ky,kz,t)ei(kxx+kyy+kzz) ) =
0

1
(2¦Ð)3/2


dkx dky dkz ( ¡Ý(kz)2 ¡Ý 1
c2

¢ß2
¢ßt2
)
Ex(kx,ky,kz,t)ei(kxx+kyy+kzz) =
0

(15.70)
¤³¤ì¤Ï¤Ä¤Þ¤ê¡¢

( ¡Ý(kz)2 ¡Ý 1
c2

¢ß2
¢ßt2
) Ex(kx,ky,kz,t)=0
(15.71)
¤¢¤ë¤¤¤Ï¡¢

¢ß2
¢ßt2
Ex(kx,ky,kz,t) = ¡Ýc2(kz)2 Ex(kx,ky,kz,t)
(15.72)
¤È¤¤¤¦¤³¤È¤Ç¤¢¤ê¡¢Ä´Ï¿¶Æ°»Ò¤Î±¿Æ°ÊýÄø¼°

d2
dt2
x(t) = ¡Ý¦Ø2 x(t)
(15.73)
¤ÈÈæ¤Ù¤ë¤È¡¢¦Ø = ckz¤È¤¹¤ì¤Ð¡¢Ex(kx,ky,kz,t) ¤¬x(t)¤ËÂбþ¤·¤Æ¤¤¤ë¤³¤È¤Ë¤Ê¤ë¡£¤Ä¤Þ¤ê¡¢Åžì¤ò¥Õ¡¼¥ê¥¨Å¸³«¤·¤¿³ÆÀ®Ê¬¤¬°ì¸Ä°ì¸Ä¡¢Ä´Ï¿¶Æ°»Ò¤ËÂбþ¤·¤Æ¤¤¤ë¤³¤È¤Ë¤Ê¤ë¤Î¤Ç¤¢¤ë¡£
γ»Ò¤ÎÎÌ»ÒÎϳؤǡֺÂɸx¡¢±¿Æ°ÎÌp¤ò±é»»»Ò¤È¹Í¤¨¤ë¡×¤È¤¤¤¦ÊýË¡¤Ç¥·¥å¥ì¡¼¥Ç¥£¥ó¥¬¡¼ÊýÄø¼°¤òºî¤Ã¤¿¤è¤¦¤Ë¡¢Åż§¾ì¤Ë¤¿¤¤¤·¤Æ¤â¡ÖÅž좪E¡¢¼§¾ì¢ªH¤ò±é»»»Ò¤È¹Í¤¨¤ë¡×¡Ê¼Â ºÝ¤Ë¤ÏÅžì¤ä¼§¾ì¤è¤ê¡¢¥Ù¥¯¥È¥ë¥Ý¥Æ¥ó¥·¥ã¥ëA¤ÈÀÅÅťݥƥ󥷥ã¥ë¦Õ¤ò´ðËܤ˹ͤ¨¤ë¤³¤È¤¬Â¿¤¤¡£¡Ê¤³¤ì¤Ï¤â¤Á¤í¤ó¡¢Î̻Ҳ½¤¹¤ë¾ì¹ç¤Î¤³¤È¡Ë¡Ë¤È¤¤¤¦ÊýË¡¤Ç¡ÖÅż§¾ì¤ÎÎÌ »ÒÏÀ¡×¤òºî¤ë¤³¤È¤¬¤Ç¤­¤ë¤¬¡¢¾å¤Ë½Ò¤Ù¤¿¤è¤¦¤ËÊýÄø¼°¤¬Æ±¤¸·Á¤ò¤·¤Æ¤¤¤ë¤Î¤Ç¡¢·ë²Ì¤âƱÍͤˤʤ롣¤¿¤ÀÅż§¾ì¤ÎÊý¤¬(kx,ky,kz) ¤Î´Ø¿ô¤Ç¤¢¤ëʬ¤À¤±¡Ö¿ô¤¬Â¿¤¤¡×¤À¤±¤Î¤³¤È¤Ç¤¢¤ë¡£
¸÷¤Î¥¨¥Í¥ë¥®¡¼¤¬hbar.png¦Ø¤òñ°Ì¤È¤·¤ÆÎ̻Ҳ½¤µ¤ì¤¿¤Î¤Ï¡¢¸÷(Åż§¾ì)¤¬Ìµ¸Â¸Ä¤ÎÄ´Ï¿¶Æ°»Ò¤Î¤¢¤Ä¤Þ ¤ê¤Ç¤Ç¤­¤Æ¤¤¤ë¤«¤é¤Ç¤¢¤ë¤È¹Í¤¨¤ë¤³¤È¤¬¤Ç¤­¤ë¡£¸÷¤Ë¸Â¤é¤º¡¢ÅŻҤʤɤ½¤Î¾¤Îʪ¼Á¤Ë¤Ä¤¤¤Æ¤â¡¢¶õ´Ö¤ËʬÉÛ¤·¤¿Êª¼Á¾ì¤òÄ´Ï¿¶Æ°»Ò¤Î½¸¤Þ¤ê¤È¹Í¤¨¤ÆÎ̻Ҳ½ ¤¹¤ë¤³¤È¤¬¤Ç¤­¤ë¡£¤³¤ì¤ò¡ÖÎ̻Ҿì¤ÎÍýÏÀ¡×¤È¸Æ¤Ó¡¢¸½Âå¤ÎÁÇγ»ÒÏÀ¡¢ÊªÀ­ÍýÏÀ¤Ê¤É¤Î´ðÁäȤʤë¹Í¤¨Êý¤Ç¤¢¤ë¡£

15.5  ±é½¬ÌäÂê

[±é½¬ÌäÂê15-1] Ä´Ï¿¶Æ°»Ò¤ÎÆó¤Ä¤Î¥¨¥Í¥ë¥®¡¼¸ÇÍ­¾õÂ֤νŤ͹ç¤ï¤»¦× = Cm ¦×m(¦Î,t)+Cn¦×n(¦Î, t)¤ò¹Í¤¨¤è¤¦¡Ê|Cm|2+|Cn|2=1¤È¤¹¤ë¡Ë¡£
¦×k(¦Î,t) = ¦Õk(¦Î)e¡Ýi(k+1/2)¦Øt
(15.74)
¤È½ñ¤±¤ë¤³¤È¡¢


d¦Î¦Õm*(¦Î) ¦Õn(¦Î)=¦Ämn
(15.75)
¤È¤¤¤¦Ä¾¸ò´Ø·¸¤¬À®Î©¤¹¤ë¤³¤È¤ò»È¤Ã¤Æ¡¢ < ¦Î > ¤ò·×»»¤»¤è¡£
[±é½¬ÌäÂê15-2] ¥¨¥ë¥ß¡¼¥È¿¹à¼°¤Ï
exp(2¦Ît¡Ýt2) = ¡ç
­ô
n=0 

Hn(¦Î)tn
n!

(15.76)
¤È¤¤¤¦¼°¤òËþ¤¿¤¹¡£º¸Êդδؿô¤òt¤Î´Ø¿ô¤È¤·¤Æ¤ß¤Æ¥Æ¥¤¥é¡¼Å¸³«¤·¤¿»þ¤În ¼¡¤Î·¸¿ô¤¬Hn(¦Î)¤À¤È¤¤¤¦¤³¤È¤Ë¤Ê¤ë¡£¤³¤Î¼°¤« ¤é¡¢
Hn(¦Î) = (¡Ý1)nexp(¦Î2) ¢ßn
¢ß¦În
( exp(¡Ý¦Î2))
(15.77)
¤òƳ¤±¡£
[±é½¬ÌäÂê15-3] ¥¨¥ë¥ß¡¼¥È¿¹à¼°¤Î´Ö¤Ë¡¢°Ê²¼¤Î´Ø·¸¼°¤¬À®Î©¤¹¤ë¤³¤È¤ò¾ÚÌÀ¤»¤è¡£
  1. [d/d¦Î]Hn(¦Î) = 2nHn¡Ý1(¦Î)
  2. Hn+1(¦Î)=2¦ÎHn(¦Î)¡Ý2nHn¡Ý1(¦Î)
[±é½¬ÌäÂê15-4] Ä´Ï¿¶Æ°»Ò¤ÎÂènÎ嵯¾õÂ֤ξì¹ç¤Ç±¿Æ°¥¨¥Í¥ë¥®¡¼¤Î´üÂÔÃͤȰÌÃÖ¥¨¥Í¥ë¥®¡¼¤Î´üÂÔÃͤò·×»»¤»¤è¡£


³ØÀ¸¤Î´¶ÁÛ¡¦¥³¥á¥ó¥È¤«¤é

  ÎÌ»ÒÎϳر齬¤Ç¤Ïµé¿ôŸ³«¤ò»È¤Ã¤Æ²ò¤¤¤¿¤¬¡¢±é»»»Ò¤ÎÊý¤¬´Êñ¤À¤Ã¤¿¡Ê¿ô ¿Í¡Ë¡£
¡¡·×»»¤Ï±é»»»ÒË¡¤ÎÊý¤¬³Ú¤Ç¤¹¡£¤¿¤À¡¢±é»»»ÒË¡¤Ï¤¤¤Ä¤Ç¤â¤Ç¤­¤ë¤ï¤±¤Ç¤Ï¤Ê ¤¤¤Î¤Ç¡¢ÌäÂê¤Ë¤è¤Ã¤Æ¤Ï¤É¤¦¤·¤Æ¤âµé¿ôŸ³«¤ò»È¤ï¤Ê¤¯¤Æ¤Ï¤¤¤±¤Ê¤¯¤Ê¤ê¤Þ¤¹¡£

¡¡Â¾¤Ë±é»»»ÒË¡¤ò»È¤Ã¤Æ²ò¤±¤ë¤Î¤Ï¤É¤ó¤Ê¤â¤Î¤¬¤¢¤ë¤ó¤Ç¤¹¤«¡©
¡¡¤¹¤°¸å¤Ç¤ä¤ë³Ñ±¿Æ°Î̤ʤ󤫤Ǥâ»È¤¨¤Þ¤¹¡£

¡¡Â¾¤Î¥Ï¥ß¥ë¥È¥Ë¥¢¥ó¤Î»þ¤Ï¡¢¤Þ¤¿¾å¾º±é»»»Ò¤òºî¤êľ¤¹¤Î¤Ç¤¹¤«¡©
¡¡¤½¤¦¤Ç¤¹¡£º£Æü¤ä¤Ã¤¿¤Î¤ÏÄ´Ï¿¶Æ°»ÒÀìÍѤǤ¹¡£

¡¡ÇȤò½Å¤Í¹ç¤ï ¤»¤ë¤³¤È¤Ç¸ÅŵŪ¤Ê¿¶¤ê»Ò¤ÎÆ°¤­¤¬¸«¤¨¤ë¤È¤¤¤¦¤Î¤ÏÎÌ»ÒÎϳؤ餷¤¤¡£
¡¡¤¤¤Ä¤â¸ÅŵÎϳØŪ¤Ë¸«¤Æ¤¤¤ë¿¶Æ°¤Ï¡¢¼Â¤Ï¤¿¤¯¤µ¤ó¤ÎÎÌ»ÒÎϳØŪ¾õÂÖ¤Î½Å¤Í ¹ç¤ï¤»¤Ç¤Ç¤­¤Æ¤¤¤ë¤Î¤Ç¤¹¡£

¡¡±¿Æ°¥¨¥Í¥ë¥®¡¼¤¬¥Þ¥¤¥Ê¥¹¤Ë¤Ê¤Ã¤Æ¤¤¤ëÎΰ褬Âç»ö¤Ë¤Ê¤ë¤è¤¦¤ÊʪÍý¸½¾Ý¤Ï¤¢¤ê¤Þ¤¹¤«¡©¡©
¡¡¤¦¡¼¤ó¡¢¥È¥ó¥Í¥ë¸ú²Ì¤Î¤è¤¦¤Ê¸½¾Ý¤Ç¤¹¤Í¡£Ä´Ï¿¶Æ°»ÒŪ¤Ê¥Ý¥Æ¥ó¥·¥ã¥ë¤¬ ¼þ´üŪ¤Ë·«¤êÊÖ¤µ¤ì¤Æ¤¤¤ë¤è¤¦¤Ê»þ¡¢¥Ý¥Æ¥ó¥·¥ã¥ë¤Î¶Ë¾®Ãͤ«¤é¤È¤Ê¤ê¤Î¶Ë¾®ÃͤؤÈÈô¤Ó°Ü¤ë¤è¤¦¤Ê¸½¾Ý¤¬µ¯¤³¤Ã¤¿¤ê¤·¤Þ¤¹¡£

¡¡ÇÈÆ°´Ø¿ô¤Î¥° ¥é¥Õ¤Ç¡¢¿¶Æ°¤·¤Æ¤Ê¤¤¤è¤¦¤Ë¸«¤¨¤ëÅÀ¤Ç¤Ï¡¢µõ¿ôÉôʬ¤Ï¿¶Æ°¤·¤Æ¤¤¤ë¤ó¤Ç¤¹¤«¡©
¡¡¤¤¤¤¤¨¡¢Äê¾ïÇȤÎÀá¤Î¤è¤¦¤Ê¤â¤Î¤Ç¡¢¿¶Æ°¤·¤Æ¤¤¤Ê¤¤ÅÀ¤¬²¿¥«½ê¤«¤¢¤ê¤Þ ¤¹¡£

¡¡ÇÈÆ°´Ø¿ô¤ÇÇȤβô¤¬¤¢¤ë¤È¡¢Î³»Ò¤Ï¤½¤Î¤É¤³¤«¤Ë¤¤¤ë¤Î¤Ç¤¹¤«¡©
¡¡¤¤¤Þ¤¹¡£´Ñ¬¤·¤¿¤é¤½¤ÎÊÕ¤ê¤Ç¸«¤Ä¤«¤ê¤Þ¤¹¡£´Ñ¬¤·¤Ê¤¤´Ö¤Ï¡¢Á´ÂΤ˹­ ¤¬¤Ã¤Æ¸ºß¤·¤Æ¤Þ¤¹¡£

¡¡´ðÄì¾õÂ֤Ǥ⡢¿¶Æ°¤Ïµ¯¤³¤Ã¤Æ¤¤¤ë¤ó¤Ç¤¹¤«¡©
¡¡¿¶Æ°¤Ë¤â¡Ê£±¡Ë¦×¤Î¿¶Æ°¤È¡¢¡Ê£²¡Ë´üÂÔÃÍ<x>¤Î¿¶Æ°¤¬¤¢¤ê ¤Þ¤¹¡£´ðÄì¾õÂ֤Ǥϡ¢¦×¤Ï¿¶Æ°¤·¤Æ¤¤¤Þ¤¹¤¬¡¢<x>¤Ï¿¶Æ°¤·¤Æ¤¤¤Þ¤»¤ó¡£

¡¡¸ÅŵŪ¤Ê¿¶Æ° ¤Ï¥¨¥Í¥ë¥®¡¼¸ÇÍ­ÃͤΰۤʤëÇÈÆ°´Ø¿ô¤Î­¤·¤¢¤ï¤»¤¿¤â¤Î¤À¤½¤¦¤Ç¤¹¤¬¡¢¸ÇÍ­¾õÂÖ¤ËÂбþ¤¹¤ëÇÈÆ°´Ø¿ô¤Ï¸ºß¤·¤Ê¤¤¤ó¤Ç¤¹¤«¡©
¡¡Ä㤤²¹ÅÙ¡ÊÄ㤤¥¨¥Í¥ë¥®¡¼¡Ë¤Ç¤Ê¤é¡¢¼Â¸½¤·¤Þ¤¹¡£Æü¾ï¥ì¥Ù¥ë¤ÎÏäǤϡ¢¤¢ ¤ëÄøÅÙ¤ÎÉý¤Î¥¨¥Í¥ë¥®¡¼¤ò»ý¤Ã¤¿¾õÂ֤νŤ͹ç¤ï¤»¤¬¼Â¸½¤·¤Æ¤Þ¤¹¡£

¡¡¥°¥é¥Õ¤Î±¿Æ°¤Î¿¶¤ê»Ò±¿Æ°¤ÎÏäϡ¢·²Â®Å٤ȰÌÁê®Å٤δط¸¤ÈƱ¤¸¤È¹Í¤¨¤Æ¤¤¤¤¤Î¤Ç¤¹¤«¡©
¡¡¤Ï¤¤¡£¤½¤ÎÄ̤ꡣƱ¤¸¹Í¤¨Êý¤Ç¤¹¡£

¡¡±é»»»Ò¤Ç¤Î²òË¡¤âÆñ¤·¤«¤Ã¤¿¤Ç¤¹¡£
¡¡Âç³Ø3ǯ¤Ç¡¢¤â¤¦Âç³Ø¤Ç¤ÎÊÙ¶¯¤âÂçµÍ¤á¤Ç¤¹¤«¤é¡¢Æñ¤·¤¤Ï䬽ФƤ¯¤ë¤Î¤â ÅöÁ³¤Ç¤¹¡£Ä¥¤ê¹ç¤¤¤ò¤â¤Ã¤Æ¤¬¤ó¤Ð¤ê¤Þ¤·¤ç¤¦¡£

¡¡²¼¹ß±é»»»Ò¤ò¤É¤ó¤É¤ó¤«¤±¤Æ¤¤¤¯¤È¥¨¥Í¥ë¥®¡¼¤Ï¤É¤ó¤É¤ó¥Þ¥¤¥Ê¥¹¤Ë¤Ê¤ê¤Þ¤¹¤«¡©
¡¡¤¤¤¤¤¨¡£¾å¤Ç¤âÀâÌÀ¤·¤Æ¤Þ¤¹¤¬¡¢´ðÄì¾õÂ֤˲¼¹ß±é»»»Ò¤«¤±¤ë¤ÈÇÈÆ°´Ø¿ô¼« ÂΤ¬£°¤Ë¤Ê¤ê¤Þ¤¹¡£