(6-13) Let’s take just two of these eigenfunctions, ψ1 and ψ2, corresponding to the solutions where n=1 and n=2. They need not necessarily be mixed in equal amounts, so let’s choose an example where our superposition state is:
ψs = N(c1ψ1 + c2ψ2)
Here, c1 and c2 are just numbers and N is a normalisation factor. ψ1 and ψ2 are individually normalised but this does not mean the superposition will be, so we will have to begin by working out N.
Normalisation means that:
So in this case:
① To evaluate the integral, begin by expanding the bracket, to get a sum of three integrals. You should then be deal easily with two of them, based on the normalisation of the original eigenfunctions, which we have already established.
② What about the other term? This includes the integral:
This one is not so easily dealt with. Before trying to work out the integral, see if you can deduce / guess what the answer will be by having a look at the case where c1 and c2 are equal: in other words, just adding up the n=1 and n=2 solutions. Here are the separate ψ1 and ψ2 curves.
We are thinking about the product, ψ1ψ2. The definite integral of this function will be the area between this product function curve and the x-axis. Remember that in integration the area will positive if it's above the axis and negative below it.
Can you see how this will work out?
③ Now let's see if your result from ② applies generally, when we allow c1 and c2 to vary, by working out the integral. This looks non-trivial but, once again, it'll be trigonometric identities to the rescue.
First, write out the explicit forms of ψ1 and ψ2 in this integral.
Then recall your trigonometric identities again. This time, the key relationship is:
sin(a)sin(b) = ½ (cos(a-b) – cos(a+b)).
This should enable you to evaluate this integral reasonably straightforwardly.
④
So this term (in which this is just multiplied by just disappears. This enables you easily to go back to your answer from ① and find an expression for N. Go ahead and do that.
⑤ Now we can test this superposition wavefunction to see if it is an eigenfunction of the Schrödinger equation. This would require that:
Ĥ ψs = E ψs where E is an energy eigenvalue.
We have the explicit equation (equation VI-6) ready to go:
So, try plugging in your expression for ψs and see what happens.
