Calculate the normalization constant for an oscillator with v = 3?

Question

Help :( We've been looking at this problem all over the weekend, and we don't seem to be closer to cracking what it is... 

We know that the normalization constant multiplied with the wavefunction must give the integral of 1 (right?)

We know the Hermite polynomial for v=3 is 8y^3-12y

In our book, Quanta, Matter and Change, we see that the integral from -infinity to infinity, of the normalized wavefunction times the wavefunction is απ^(1/2)*2^v*v!, which in our case is 48α*π^(1/2).

(1/(48α*π^(1/2)))^1/2 should then equal N? (α=x/y) N=y/(96x*π^(1/2))

But in that case, shouldn't the intergral from -infinity to infinity of y/(96x*π^(1/2)) * 8y^3-12y * e^(y^2/2)  = 1 ? Because it doesn't, it gives us all sorts of mess :(

Please, if someone can give us some clues!! We have 2 more problems over the same theme, but I'm hoping that if we figure this one out, we will be able to do the others...

Steve4Physics · Accepted Answer

There are too many mistakes/ambiguities in your question to make it answerable.

If ψ(y) is an un-normalised wave function and its normalisation constant is N, 
then taking the simple case that ψ is real:
∫(Nψ)(Nψ)dy = 1 (±infinity limits taken as read)
∫ψ²dy = 1/N²

So N = 1/√ [∫ψ²dy]

Note that it is the square of the wave function that gets integrated.  For a complex wave function you integrate the product of the wave function and its complex conjugate: ∫(Nψ*)(Nψ)dy = 1
______

You said:
“We know that the normalization constant multiplied with the wavefunction must give the integral of 1 (right?)”

No. I think you are saying  ∫Nψdy = 1, which is wrong. See above.
______

You said:
“we see that the integral from -infinity to infinity, of the normalized wavefunction times the wavefunction is...”
Do you really mean that? You have described:
∫(Nψ)ψdy 
which, in fact equals 1/N and is probably not what you mean.
______

You refer to the integral of y/(96x*π^(1/2)) * 8y^3-12y * e^(y^2/2)

Where has ‘x’ come from? Do you mean α?

Shouldn’t there be a minus sign in the exponential? – or the wave amplitude tends to infinity as y increases!

The expression is ambiguously written and you may mean:
[y/(96 α *π^(1/2))] * [8y^3-12y] * e^((-y^2)/2) (I’ve guessed the brackets)
_____

So in its present form the question is garbled and unanswerable. Sorry.

However, I strongly recommend looking through the link below.

Trending News

Calculate the normalization constant for an oscillator with v = 3?

1 Answer