Задачи по квантовой механике. Часть 2. Алмалиев А.Н - 28 стр.

UptoLike

ˆ
T
a
hp|
ˆ
T
a
|p
0
i =
1
(2π})
3
Z
exp
µ
i
}
pr
ˆ
T
a
exp
µ
i
}
p
0
r
d
3
r =
=
1
(2π})
3
exp
µ
i
}
p
0
a
Z
exp
½
i
}
r(p
0
p)
¾
d
3
r
| {z }
(2π})
3
δ(p
0
p)
= δ(p
0
p) exp
µ
i
}
p
0
a
.
δ
ˆ
T
a
= exp
µ
i
}
pa
= exp
µ
i
}
ap
.
ˆ
T
a
= exp
µ
i
}
a
ˆ
p
ˆ
p = i}
r
ˆ
T
a
ˆ
T
a
p ¤
{Φ
n
(r)}
hn
0
|
ˆ
A
ˆ
B |ni hn
0
|
ˆ
A |ni hn
0
|
ˆ
B |ni
P
n
Φ
n
(r
n
(r
0
) = δ(r
0
r)
hn
0
|
ˆ
A
ˆ
B |ni =
Z
Φ
n
0
(r
0
)
ˆ
A
ˆ
BΦ
n
(r
0
) d
3
r
0
=
=
ZZ
Φ
n
0
(r
0
)
ˆ
A δ(r
0
r)
ˆ
BΦ
n
(r
0
) d
3
r
0
d
3
r =
=
X
n
00
ZZ
Φ
n
0
(r
0
)
ˆ
AΦ
n
00
(r
0
n
00
(r)
ˆ
BΦ
n
(r) d
3
r
0
d
3
r =
=
X
n
00
Z
Φ
n
0
(r
0
)
ˆ
AΦ
n
00
(r
0
) d
3
r
0
·
Z
Φ
n
00
(r)
ˆ
BΦ
n
(r) d
3
r =
=
X
n
00
hn
0
|
ˆ
A |n
00
ihn
00
|
ˆ
B |ni.
hn
0
|
ˆ
A
ˆ
B |ni =
X
n
00
hn
0
|
ˆ
A |n
00
ihn
00
|
ˆ
B |ni,
îïåðàòîðà T̂a â èìïóëüñíîì ïðåäñòàâëåíèè:
                        Z     µ          ¶   µ       ¶
            0       1              i             i 0
  hp| T̂a |p i =      3
                          exp − pr T̂a exp        p r d3 r =
                 (2π})             }            }
                µ      ¶Z      ½           ¾                   µ     ¶
      1           i 0            i      0    3         0         i 0
 =          exp     pa     exp     r(p − p) d r = δ(p − p) exp     pa .
   (2π})3         }              }                               }
                        |            {z        }
                                     (2π})3 δ(p0 −p)

Ìíîæèòåëü, íàõîäÿùèéñÿ ïîñëå δ -ôóíêöèè, êàê ðàç è áóäåò îïåðàòîðîì
ñäâèãà â èìïóëüñíîì ïðåäñòàâëåíèè:
                                µ     ¶      µ     ¶
                                  i            i
                      T̂a = exp     pa = exp     ap .                  (2.21)
                                  }            }
                                                    µ      ¶
                                                      i
 êîîðäèíàòíîì ïðåäñòàâëåíèè îïåðàòîð T̂a = exp         ap̂ , ò.å. ïî ñòðóê-
                                                      }
òóðå ôîðìàëüíî ñîâïàäàåò ñ (2.21). Îäíàêî â ýòîì ñëó÷àå p̂ = −i}∇r ,
ò.å. îïåðàòîð, è T̂a âûðàæàåòñÿ ÷åðåç îïåðàòîðíóþ ýêñïîíåíòó, à â èì-
ïóëüñíîì ïðåäñòàâëåíèè (2.21) T̂a  îáû÷íûé ìíîæèòåëü, ÿâëÿþùèéñÿ
ôóíêöèåé p.                                                                ¤

Ïðèìåð 2.7. Ôóíêöèè {Φn (r)} îáðàçóþò ïîëíóþ îðòîíîðìèðîâàííóþ
ñèñòåìó. Âûðàçèòü hn0 | ÂB̂ |ni ÷åðåç hn0 |  |ni è hn0 | B̂ |ni.
                                               P ∗
Ðåøåíèå. Âñïîìèíàÿ óñëîâèå ïîëíîòû                  Φn (r)Φn (r0 ) = δ(r0 − r), ïîëó÷à-
                                                n
åì:
                 Z
  hn | ÂB̂ |ni = Φ∗n0 (r0 )ÂB̂Φn (r0 ) d3 r0 =
    0

                    ZZ
                  =     Φ∗n0 (r0 )Â δ(r0 − r)B̂Φn (r0 ) d3 r0 d3 r =
                 X ZZ
              =        Φ∗n0 (r0 )ÂΦn00 (r0 )Φ∗n00 (r)B̂Φn (r) d3 r0 d3 r =
                 n00
                XZ                                         Z
            =           Φ∗n0 (r0 )ÂΦn00 (r0 ) d3 r0   ·       Φ∗n00 (r)B̂Φn (r) d3 r =
                n00
                                                                     X
                                                                 =         hn0 | Â |n00 i hn00 | B̂ |ni .
                                                                     n00

Èòàê,
                                          X
                       hn0 | ÂB̂ |ni =         hn0 | Â |n00 i hn00 | B̂ |ni ,                     (2.22)
                                          n00


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