In a Quantum mechanics book, I found the following equations:
$$ \Phi(k)=\frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty \Psi(x,0)e^{-ikx}dx $$
and
$$ \Psi(x,t)=\frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty \Phi(k)e^{ikx-\frac{\hbar k^2}{2m}t}dk $$
So, with $Ψ(x,t)$ I can find $Φ(k,t)$ because the following theorem exists (Fourier transformation):
$$f(x)=\frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty F(k)e^{ikx}dk~~\Leftrightarrow~~ F(k)=\frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty f(k)e^{-ikx}dx $$
So, I suppose that $\Phi(k)$ is the probability density of the momentum. Is this true?
If so, why don't I see in books the use of the integral relation that gives us $\Phi(k)$, in order to, say, find the probability of measuring a range of values of the momentum?
Lastly, I think that this only holds if the allowable values of momentum are very close to each other (like the case of a free particle), so as to be able to make the sum of the superimposed states an integral. Am I right?
Best Answer
Almost. $\Phi(k)$ is the probability amplitude for the momentum of the particle. The probability density is obtained as usual by squaring the amplitude, giving $|\Phi(k)|^2$.
For a free particle, all values of momentum are always allowed, which enables the superposition to be expressed as an integral. The only times when this breaks down is when you have a particle confined to a finite interval or when you impose periodic boundary conditions; this does restrict the allowed momentum values to a discrete set and turns the integral into a Fourier series.