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B Position representation

  1. Dec 22, 2016 #1
    I understand that the fourier transform to obtain the representation of a wavefunction in k space is

    $$ \phi(k) =\frac{1}{2\pi}\int{dx \psi(x)e^{-ikx} } $$
    and that $$p=\bar{h} k$$

    But why then is $$\phi(p) =\frac{\phi(k)}{\sqrt{\bar{h}}} $$

    Many thanks in advance :)
  2. jcsd
  3. Dec 22, 2016 #2


    Staff: Mentor

    Where are you getting this equation from?
  4. Dec 22, 2016 #3


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    In QM of non-specially relativistic particles, the Fourier transform is usually defined with a ##\frac{1}{\sqrt{2\pi\hbar}}## factor in front of the integral. Thus the OP should be amended by the user, so his (intended) question should automatically find an answer.
  5. Dec 22, 2016 #4
    Well, I think you would agree that we must have

    dp |ø(p)|2 = dk |ø(k)|2 ,

    for the corresponding intervals [p, p + dp] and [k, k + dk], where p=hbark.

    But dp = hbar dk; so,

    hbar dk |ø(p)|2 = dk |ø(k)|2 ,

    and therefore,

    |ø(p)|2 = |ø(k)|2/hbar .
  6. Dec 24, 2016 #5


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    That's simply, because ##|\phi(p)|^2## is a distribution function (namely the probability distribution for momentum). In this sense we can write
    $$|\phi(p)|^2 = \frac{\mathrm{d} N}{\mathrm{d} p},$$
    where ##N## is the number of particles. Since the wave number and momentum are related by ##p=\hbar k## you get
    $$\frac{\mathrm{d} N}{\mathrm{d} p} = \frac{\mathrm{d} N}{\mathrm{d} k} \frac{\mathrm{d} k}{\mathrm{d} p} = \frac{1}{\hbar} \frac{\mathrm{d} N}{\mathrm{d} k}.$$
    Now the phase of the wave function is arbitrary, and thus you can conclude from this that
    $$\phi(p)=\frac{1}{\sqrt{\hbar}} \tilde{\phi}(k) = \frac{1}{\sqrt{\hbar}} \tilde{\phi}(p/\hbar).$$
    Note that you should use a different function symbol for the momentum and the wave-number distribution!
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