Consider the following functions of the vector coordinates
of three particles: **a**, **b**, and **c**.

*f*(**a**,**b**,**c**)=|**c**| |**a**-**b**|

*g*(**a**,**b**,**c**)=|**c**| (|**a**|-|**b**|)

*h*(**a**,**b**,**c**)=|**c**| |**a**| |**b**|

*f* is symmetric under exchange of **a** and **b**,
because:

*f*(**a**,**b**,**c**)=*f*(**b**,**a**,**c**)

*g* is antisymmetric under exchange of **a** and **b**,
because:

*g*(**a**,**b**,**c**)=-*g*(**b**,**a**,**c**)

*h* is fully symmetric under exchange,
because:

*h*(**a**,**b**,**c**)=*h*(**b**,**c**,**a**)=
*h*(**c**,**a**,**b**)=*h*(**a**,**c**,**b**)= etc

*f* is neither symmetric nor antisymmetric under exchange of
**a** and **c**,
because:

*f*(**a**,**b**,**c**)=|**c**| |**a**-**b**|

and

*f*(**c**,**b**,**a**)=|**a**| |**c**-**b**|

have no particular relationship.

Thus if we want to have one particle in the 1*s* and the other
in the 2*p* state there are two possible ways:

symmetric: 1*s*(**r**_{1})2*p*(**r**_{2})+
2*p*(**r**_{1})1*s*(**r**_{2})

antisymmetric: 1*s*(**r**_{1})2*p*(**r**_{2})-
2*p*(**r**_{1})1*s*(**r**_{2})

The sort of wavefunction we were considering:

1*s*(**r**_{1})2*p*(**r**_{2})

has no particular symmetry and is ruled out.

Note that if we wanted both particles in the 1*s* groundstate
our only choice would be to make a symmetric function:

1*s*(**r**_{1})1*s*(**r**_{2})

Let us then construct symmetric and antisymmetric versions of the wavefunction we were considering on the previous page and calculate the energy correction for those states.

Note if we perform a change of variables in the second integral:
renaming **r**_{1} **r**_{2}
and **r**_{2} **r**_{1},
|**r**_{2}-**r**_{1}| is invariant and the second
integral is seen to be exactly the same as the first. We call this
common integral the *direct interaction integral* and name
it *J*. Note that *J* is the integral calculated
on the previous page. The same game on the fourth integral shows
it to be the same as the third. We call this
common integral the *exchange interaction integral* and name
it *K*.

We proceed to evaluate *K* much as we did *J*:
by plugging in our expansion for 1/|**r**_{2}-**r**_{1}|
in terms of the *Y*_{lm}:

Thus the integral *K* becomes a sum of integrals with a typical
term like this:

The *Y*_{00} factors are in fact constant and can be pulled
outside the integrals (where they can cancel the overall 4
factor); the resulting 2-*Y*_{lm} integrals
are either 0 or 1 by the usual orthonormality of the *Y*_{lm}.
Thus the sum of terms reduces to one term; the *lm* in the sum must match
the *kn* in the wavefunction.

We are left doing the radial integral using *Mathematica*:

The energy of the 1*s*2*s* symmetric state
(called parahelium or singlet) is
approximately:

-½(1+¼)+½(0.209877+0.0219479)= -0.509

The energy of the 1*s*2*s* antisymmetric state
(called orthohelium or triplet) is approximately:

-½(1+¼)+½(0.209877-0.0219479)= -0.531

The energy of the 1*s*2*p* symmetric state is
approximately:

-½(1+¼)+½(0.242798+0.0170706)= -0.495

The energy of the 1*s*2*p* antisymmetric state is
approximately:

-½(1+¼)+½(0.242798-0.0170706)= -0.512

i.e., the orthohelium (antisymmetric, triplet) 2*p* state lies
.019×(*Z*^{2}27.2 eV)=2 eV
above the orthohelium 2*s* state;
the parahelium (symmetric, singlet) 2*p* state lies
.014×(*Z*^{2}27.2 eV)=1.5 eV
above the parahelium helium 2*s* state. Again, the splittings
are about a factor of two too large, but the relative placement of
states agrees with experiment.

The main point here is that the antisymmetric states [i.e., the states that
will turn out to have the largest net electron spin] lie below
the equivalent symmetric states. The reason for this is the
antisymmetric wavefunction must be zero if evaluated where
**r**_{1}=**r**_{2}. Thus there is no probability
that the electrons are on top of each other, and via wavefunction continuity
a generally small probability that the electrons are near each other.
The repulsive electron-electron interaction is minimized by
having the electrons relatively far apart.