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?

  • 3
    $\begingroup$ 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, 2015 at 15:17
  • $\begingroup$ related: physics.stackexchange.com/q/186112 $\endgroup$
    – Phoenix87
    Jun 10, 2015 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. $\endgroup$ Jun 10, 2015 at 15:25
  • 2
    $\begingroup$ 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, 2015 at 17:18

1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.