# Why fourier transforms in QM

1. Mar 3, 2007

### pivoxa15

Fourier transforms imply that the waves are added up with a continous range of frequencies hence wavelengths instead of discrete numbers.

In QM fourier transforms are used which imply the psi function employs a continous rather than discrete range of frequencies. Why is this? Is it because psi wave is not a mechanical wave so it dosen't matter wheither the frequencies are continuous or discrete?

However, I have only seen fourier transforms done on plane waves hence on free particles which can obtain a continous range of energies via E=hf. So f takes on a continous range hecne both wavenumer, k and period length lamda will all be continous. That is why the fourier transform can be done on free particles. However, if the particle is not free than fourier transforms cannot be used as the frequencies will not be continous, wavenumber and period will not be continous.

Last edited: Mar 3, 2007
2. Mar 4, 2007

### quasar987

The frequencies are continuous rather than discrete simply because momentum is not quantized.

Remember, in QM, when a measurement is made, the only possible measured values are the eigenvalues of the eigenvalue equation. In the case of momentum, the momentum operator is $\hat{p}=-i\hbar\partial_x$, and so the eigenvalue equation $$\hat{p}u_p(x)=pu_p(x)[/itex] has solutions $u_p(x)=(2\pi\hbar)^{(-1/2)}\exp(ipx/\hbar)$ and p can take any real value. And how do we know what value a measurement will yield? Well be first expand the wave function in terms of the eigenfunctions of the quantity we want to measure. In the case of the momentum, since there are an infinity of eigenfunctions, the expansion actually looks like a fourier integral [tex]\psi_(x)=\int_{-\infty}^{+\infty}\phi(p)u_p(x)dp$$

and indeed it is one because $u_p(x)$ is actually a complex exponential in px.

And $|\phi(p)|^2$ is the probability distribution for measuring p.

Last edited: Mar 4, 2007
3. Mar 4, 2007

### pivoxa15

So it has nothing to do with whether a particle is trapped or free. Momentum is always continous and never quantised for a particle in whatever situation? But momentum is related to energy and energy is quantised in for example the square well. So momentum in the square is also quantised? Hence momentum is not always continous?