Condensed Matter Physics II-----Exercise 1
1. Derive d x 2  y 2 , d xy , d zx d yz , and d z 2 atomic wavefunctions/orbitals from
Rnl (r )Ylm ( ,  ) .
Answer:
r
x2  y2  z2 .
d x 2  y 2 orbital:
1
2
Rn 2 (r )Y2, 2 ( ,  )  Y2, 2 ( ,  ) 

1 15
Rn 2 (r ) sin 2  e 2i  e 2i
4 4



1 15
1 15
Rn 2 (r ) sin 2  cos 2 
Rn 2 (r ) sin 2  cos 2   sin 2 
2 4
2 4

1 15
x2  y2
Rn 2 (r )
2 4
r2
d xy orbital:
i
2
Rn 2 (r )Y2, 2 ( ,  )  Y2, 2 ( ,  ) 



i 15
Rn 2 (r ) sin 2  e 2i  e 2i
4 4

1 15
15
Rn 2 (r ) sin 2  sin 2 
Rn 2 (r ) sin 2  sin  cos 
2 4
4
15
xy
Rn 2 ( r ) 2
4
r
d zx orbital:
1
2
Rn 2 (r )Y2, 1 ( ,  )  Y2,1 ( ,  ) 




1 15
Rn 2 (r ) sin  cos  e i  e i
2 4
15
15
zx
Rn 2 (r ) sin  cos  cos  
Rn 2 (r ) 2
4
4
r

d yz orbital:
i
2

Rn 2 (r )Y2, 1 ( ,  )  Y2,1 ( ,  ) 
i 15
Rn 2 (r ) sin  cos  e i  e i
2 4
15
15
yz
Rn 2 (r ) sin  cos  sin  
Rn 2 (r ) 2
4
4
r

d z 2 orbital:
Rn, 2 (r )Y2,0 ( ,  ) 
 3 z2 1 
5
1
5
3
Rn 2 (r ) cos 2    
Rn 2 (r ) 2  
4
2
4
2
2
2 r
2. Explain why a pair of electrons prefers a symmetric total spin state (or parallel
spins).
Answer:
According to the Pauli Principle, the wavefunction of a pair of electrons should be
antisymmetric with respect to the permutation of these two electrons. That is



r2 ,  2 ; r1 , 1   r1 , 1 ; r2 , 2  .
If the coupling between spins and spatial motion can be ignored, the wavefunction can


 
be written as r1 ,  1 ; r2 ,  2    r1 , r2   1 , 2 ·
(1) If the spin state,   1 ,  2  , is one of the symmetric triplet states,   ,   and
  ,which can be loosely described as parallel, the spatial two-electron
 
 
wavefunction has to be antisymmetric, i.e.  r2 , r1    r1 , r2  . This
wavefunction can be described as an determinant and


1 a r1  a r2 
1
 
a r1 b r2   a r2 b r1 
 r1 , r2  

 
2 b r2  b r2 
2
It can be shown that the electrostatic repulsion energy, UA, between these two
electrons is U A U classical J , where
J
e2
40




 
r1  r2

 a* (r1 )b* (r2 ) a (r2 ) b (r1 )
d 3 r1 d 3 r2 is called the exchange integral.
J > 0 and U A  U classical . Thus, the electrostatic repulsion energy is reduced.
 


Note that if  a (r )  b (r ) ,  r1 , r2   0 ; the probability is zero.
(2) If the spin state,   1 ,  2  , is the antisymmetric singlet state,   ,in which the
two spins can be loosely said to be in opposite directions, the spatial two-electron
 
 
wavefunction has to be symmetric, i.e.  r2 , r1    r1 , r2  .
1
 
The wavefunction has the form  r1 , r2  
a r1 b r2   a r2 b r1  and the
2
electrostatic repulsion energy, US, between these two electrons is
U S U classical J  U classical . The electrostatic repulsion energy is increased.


 


However, if b (r )   a (r ) , then  r1 , r2   a (r1 ) b (r2 ) and U S U classical .
Cases (1) and (2) show that when the two electrons have a symmetric spin state or
parallel spins the electrostatic repulsion energy is the lowest, i.e. this state is the
most favorable