Here is a follow-up question.

For one of the N - H bonds in NH_{3} the wave function can be written as psi = 1s_{Ha}(1)sp_{1}^{3}(2) + 1s_{Ha}(2)sp_{1}^{3}(1). This is for two electrons in one MO. What is the rationale for taking products of wave functions and then adding them? Is each of the products the same as allowing an electron in each AO and the sum is to allow them to exchange places?

Imagine that ##H_A## is the Hamiltonian of system A and ##H_B## the Hamiltonian of system B. The time-independent SchrÃ¶dinger equations are

$$ H_A \Psi(x_A) = E_A \Psi(x_A) $$

$$ H_B \Psi(x_B) = E_B \Psi(x_B) $$

with ## E_A ## and ## E_B ## the respective energies. Multiplying the first equation by a function of coordinates ##x_B## and the second by a function of coordinates ##x_A## does not change anything because ## H_A ## only works on ## x_A ## and ## H_B ## only does on ## x_B ##. I.e.,

$$ H_i \Psi(x_i)\Psi(x_j) = \Psi(x_j) H_i \Psi(x_i)$$

Now consider a total Hamiltonian ## H_{AB} = H_A + H_B ## and use the above equations, you obtain

$$ H_{AB} \Psi(x_A)\Psi(x_B) = ( E_A + E_B ) \Psi(x_A)\Psi(x_B) $$

I.e. the function for the total system is a product ## \Psi(x_A,x_B) = \Psi(x_A)\Psi(x_B) ##

Mathematically, this method is named separation of variables.

You can repeat this for the NH

_{3} and obtain a product ## \Psi(x_1,x_2) = \Psi(x_1)\Psi(x_2) ## for the MO. Notice that up to this point we are considering a product of functions but not saying what functions.

One possibility is ## \Psi'(x_1,x_2) = 1\mathrm{s}(x_1)\mathrm{sp}^3(x_2) ## and other ## \Psi''(x_1,x_2) = \mathrm{sp}^3(x_1)1\mathrm{s}(x_2) ##. Mathematically, a linear combination of two special solutions of a differential equation give a general solution of the equation. That is the reason which you sum both to obtain a more general solution.

Physically, you can interpret this sum as that a given electron (e.g. 1) is neither in atom N nor in atom H but in a superposition.