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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?

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    $\begingroup$ Yes you can think of suitably normalized $|\phi(k)|^2$ as the probability density for the momentum. I'm pretty sure most quantum mechanics textbooks do explicitly point this out. $\endgroup$ – octonion Jun 10 '15 at 15:17
  • $\begingroup$ related: physics.stackexchange.com/q/186112 $\endgroup$ – Phoenix87 Jun 10 '15 at 15:24
  • $\begingroup$ Note that you can (and should) use LaTeX notation to represent your maths. Thus $\Psi(x,t)$ is written $\Psi(x,t)$. Double $$ signs center the equations, and \int_{a}^{b} makes an integral. $\endgroup$ – Emilio Pisanty Jun 10 '15 at 15:25
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    $\begingroup$ Your images are barely readible. I changed them to TeX, but please make sure that I didn't introduce any mistakes. $\endgroup$ – Martin Jun 10 '15 at 17:18
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So, I suppose that $Φ(k)$ is the probability density of the momentum. Is this true?

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.

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