Normalizing a wavefunction

1. Feb 19, 2014

fuserofworlds

I was reviewing the infinite square well, using D.J. Griffiths, and came across this small point of confusion. The time-independent solution is shown to be Asin(kx), where the constant A is determined by normalization. Then, in assembling the complete (time dependent) solution, he writes that the most general solution "is a linear combination of stationary states", where each stationary state is assigned a coefficient c_n. Griffiths then explains that |c_n|^2 is the probability of observing that state, and "the sum of all these probabilities should be 1."

My confusion is this: if c_n is the probability of observing each state, why do we use the normalization requirement to find A? Isn't this in effect normalizing twice?

2. Feb 19, 2014

atyy

You only normalize once to find A. Once that's done, c_n gives the probability without any further normalization.

3. Feb 19, 2014

fuserofworlds

Thank you, that makes sense!

4. Feb 19, 2014

Ravi Mohan

It is quiet convenient to work in orthonormal set of basis. Therefore you must normalize each element of basis set such that
$$\int_0^l f_n(x)f_m(x)\,dx = \delta_{n m}$$
where $f_n(x) = A_n\sin(n\pi x/l)$.

The total statefunction (wavefunction) can be written as a linear combination of the elements of basis set.
$$\Psi(x,t) = \sum_{n=0}^\infty c_n(t)f_n(x).$$
Now quantum mechanics says the total probability probability should always be unity, so
$$\int_0^l \Psi(x,t)\Psi(x,t)\,dx = 1,$$
which implies
$$\sum_{n=0}^\infty c_n^2 = 1.$$
(note: I have assumed real valued wave functions. For complex case, replace with complex conjugate wherever necessary)
[EDIT:]
Normalisation of the basis set just makes the mathematics easy (it is not a necessary requirement), but the normalisation of total statefunction is a constraint of quantum mechanics)

Last edited: Feb 19, 2014