5. 70
CHAPTER MULTIPLE-PARTICLE SYSTEMS
This is similar to the observation in calculus that integrals of products can be factored into
separate integrals:
all r 1
all r 2
all r 1
f (r1 )g (r2 ) d3 r1 d3 r2 =
f (r1 ) d3 r1
all r 2
g ( r 2 ) d3 r 2
Answer:
| | =
1
h
Sz1 = 2
(Sz1 )(Sz1 )
1
Sz2 = 2
h
(Sz2 )(Sz2 )
and written out
1
1
| | = (+ 1 h)(+ 2 h) + ( 1 h)( 1 h) (+ 1 h)(+ 1 h) + ( 2 h)( 1 h)
2
2
2
2
2
2
and multiplying out, and reordering the second and third factor in each term, you see it is the
same the as expression obtained in the answer to the previous question,
|
1
1
1
= (+ 2 h)(+ 2 h) (+ 1 h)(+ 1 h) + (+ 1 h)( 2 h) (+ 1 h)( 1 h) +
2
2
2
2
2
1
1
( 1 h)(+ 1 h) ( 2 h)(+ 2 h) + ( 1 h)( 1 h) ( 1 h)( 1 h).
2
2
2
2
2
2
5.5.5
Example: the hydrogen molecule
5.5.5.1
Solution complexsc-a
Question: Show that the normalization requirement for gs means that
|a++ |2 + |a+ |2 + |a+ |2 + |a |2 = 1
Answer: For brevity, write
gs,0 = a [l (r1 )r (r2 ) + r (r1 )l (r2 )]
so that
gs = a++ gs,0 + a+ gs,0 + a+ gs,0 + a gs,0 .