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Momentum probability Distribution

  1. Dec 6, 2004 #1
    To get at a particle's postion probability distribution we take the modulus of the wave function and square it etc. , but my lecturer said next year we will learn how to get at the particle's momentum probability distribution from the wavefunction, and mentioned something about taking the Fourier Transform of the wavefunction. I can't wait til next year, any one care to explain how the transform gives a momentum distribution?

  2. jcsd
  3. Dec 6, 2004 #2


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    Hopefully i'm not mistaking,i haven't used this ormula in decades (damn QFT has other formula for changing representation) so here goes my say:
    [tex] <\vec{r}>=\int_{R^3} d^{3}\vec{r}\psi^{*}(\vec{r}) \vec{r} \psi(\vec{r}) [/tex]
    [tex] <\vec{p}>=\int_{R^3} d^{3}\vec{p}\psi^{*}(\vec{p}) \vec{p} \psi(\vec{p}) [/tex]
    [tex]\psi(\vec{p})=\frac{1}{(2 \pi\hbar)^{\frac{3}{2}}}\int_{R^3} d^{3}\vec{r}\psi(\vec{r})\exp(-\frac{i}{\hbar}{\vec{r}\cdot\vec{p}}) [/tex]
    is called the Fourier transform of [itex] \psi(\vec{r}) [/itex].

    I sincerely hope u know how to calculate the average of the momentum operator in the coordinate representation...
    Last edited: Dec 6, 2004
  4. Dec 6, 2004 #3
    Thanks i understand how that works, and i know how to compute Fourier Transforms, but i thought Fourier Transforms transform a function from the time domain to the frequancy domain...is there any underlying reason why transforming the position domain goes to momentum space?

  5. Dec 6, 2004 #4


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    To calculate the probability distrubtion of any observable, you need to write your wavefunction as a linear combination of eigenfunctions of that observable.
    Reasoning in one-dimension:
    Since [tex]\hat p = \frac{\hbar}{i}\frac{\partial}{\partial x}[/tex]
    We can find the eigenfunctions of [itex]\hat p[/itex] by the following DE:
    [tex]\hat p \psi(x) = p \psi(x) \iff \frac{\hbar}{i}\frac{\partial \psi (x)}{\partial x}=p \psi(x)[/tex]
    The solutions are clearly: [tex]\psi(x)=C\exp\left(\frac{ipx}{\hbar}\right)[/tex], where [itex]p[/itex] can take any real value.
    So you have to write your wavefunction as a linear combination of these eigenfunctions of [itex]\hat p[/itex]. Since you have a continuous spectrum, the combination takes the form of an integral:

    [tex]\psi(x)=C\int_{-\infty}^{+\infty}\bar \psi(p)\exp\left(\frac{ipx}{\hbar}\right) dp[/tex]
    As you can see, apart from normalization factors, it is the Fourier transform of [itex]\bar \psi(p)[/itex].
    EDIT: The above didn't really answer the question. The probability of measuring a momentum between p and p+dp is:
    [tex]|\langle \exp\left(\frac{ipx}{\hbar}\right)| \psi(x) \rangle|^2 dp = C\int_{-\infty}^{+\infty}\psi(x)\exp\left(\frac{-ipx}{\hbar}\right) dx = \bar \psi(p)[/tex]
    which is the fourier transform of [itex]\psi(x)[/itex].
    Last edited: Dec 6, 2004
  6. Dec 6, 2004 #5
    I see that's very clever, thanks a lot!
  7. Dec 6, 2004 #6


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    Fourier trasforms apply in the case of any two canonically conjugated observables.Either momentum-position,energy-time,or any other pair.In the last case,u can say the energy is proportional to the (quantum physics used)frequency "omega" (by means of Planck formula),and u find frequancy-time.
    This is put together in QFT where energy/frequency are the same thing (hbar is set to one) and one uses 4-vectors for momentum and coordinates.
    I edited the post above because it didn't give the reason why the normalization constant in the Fourier transform included hbar.Hbar sould have appeared (and now does) in the exponential to make its argument dimentionless.
    As i said,i worked with these only in QFT lately,where hbar is set to one and i don't have to worry about it.

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