# Quick Fourier-Like Question

1. Jun 17, 2008

### Domnu

http://electron6.phys.utk.edu/QM1/modules/m1/free_particle.htm

please look at the part where it says "We may write the Fourier transform of...". I am unable to understand how the next following steps work. Could someone explain this please? :)

2. Jun 17, 2008

### Domnu

Particularly, I don't understand why the following two equations (on the same line) happen to work out. Where does the h-bar as the constant come from in the 1/sqrt(2*pi*hbar)? In addition, in the earlier steps, where it says: "let us first concentrate on one-dimensional systems", where does the constant 1/sqrt(2pi) come from? I'm guessing it has something to do with the integral e^(-x^2)...

3. Jun 17, 2008

### lbrits

The $$\frac{1}{\sqrt{2\pi}}$$ per dimension comes from one way to define the Fourier transform. With this normalization convention, the Fourier transform is effectively its own inverse. The $$\hbar$$ comes from the fact that $$p$$ is momentum, and not a wave-vector, which is what the Fourier transform is normally expressed in.

As for your first question, the first equation may be taken as an indirect definition of $$\psi(p)$$. One then simply inverts the Fourier transform to solve for $$\psi(p)$$. It's important that you understand the Fourier transform process, though. It follows from $$\int\!dx\, e^{i k x} = (2\pi) \delta(k)$$, but I suggest you get comfortable with Fourier transforms first.

4. Jun 17, 2008

### Fredrik

Staff Emeritus
You can define the Fourier transform g of a function f by

$$g(p)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty f(x)e^{-ipx}dx$$

(The Fourier transform is the map $f\mapsto g$). Now there's a theorem that says that if f is "nice enough" (I don't remember the exact conditions), then

$$f(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty g(p)e^{ipx}dp$$.

We could have chosen to omit the $1/\sqrt{2\pi}$ in the first equation, but then a factor of $1/2\pi$ would have shown up in the second equation. A lot of people think the equations look prettier when the constant in front of the integrals are the same. (If this had been a Wikipedia article, someone would insert "citation needed" here ). That's really the only reason why the constant was chosen that way in the first equation.

What they're doing on that web page is to use the second equation on the the wave function at a fixed time. They are also using the interpretation of $|\psi(x)|^2$ as a probability density. You have to multiply it by a length (a volume when we're considering 3 spatial dimensions) to get the probability that a measurement will localize the particle in a region of that size near x.

A similar thing holds for momentum. The wave function for momentum is just the Fourier transform of the regular wave function.

Last edited: Jun 17, 2008