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Free particle in One Dimension

  1. Jun 24, 2008 #1
    Consider a free particle moving in one dimension. The state functions for this particle are all elements of [tex]L^2[/tex]. Show that the expectation of the momentum [tex]\langle p_x \rangle[/tex] vanishes in any state that is purely real. Does this property hold for [tex]\langle H \rangle[/tex]? Does it hold for [tex]\langle H \rangle[/tex]?

    For [tex]\langle p_x \rangle [/tex], we have

    [tex] \langle p_x \rangle = \int_{-\infty}^{\infty} \phi^* \hat{p_x} \phi dx[/tex]
    [tex] = - i \hbar \int_{-\infty}^{\infty} (Ae^{ikx} + Be^{-ikx})(-Aik \cdot e^{-ikx} + Bik \cdot e^{ikx}) dx[/tex]
    [tex] = -\hbar k \int_{-\infty}^{\infty} [A^2 - B^2][/tex],

    since we need to have [tex] kx = n\pi[/tex] to satisfy the condition that the wavefunction must be real. But, the above integral diverges to infinity (assuming that [tex] A \neq B[/tex]).

    I'll post the second and third parts a bit later, but have I correctly shown that the expectation of the momentum vanishes?
  2. jcsd
  3. Jun 24, 2008 #2
    I would do this:
    First try to get purely real functions that are elements of L2; squaring wavefunction gives you something like (A^2 +B^2)* exp(2ikx) + 2AB -> 0, when x-> inf. Considering purely real functions gets you A=-B... put that in your equation in the end and you get desired result.
  4. Jun 24, 2008 #3


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    kx=n*pi has nothing to do with the wavefunction being real. Nor does A=(-B). The condition you want is B=conjugate(A). That must be true for every momentum component of the wave function. exp(ikx) and exp(-ikx) represent opposite direction momentum components. What does that tell you?
  5. Jun 24, 2008 #4
    Ah, yes, I understand why this must be true now... because we need [tex]\phi^* = \phi[/tex]... that was a careless error on my part heh. Also, what exactly does it mean for it to 'vanish'? Does this mean go to 0 or go to infinity? Infinity would be nicer...
  6. Jun 24, 2008 #5


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    I've never heard of 'vanish' meaning 'go to infinity'. In your example the contribution of the exp(ikx) part is equal and opposite to the contribution of the exp(-ikx) part. They cancel. They are also orthogonal, so there are no cross terms. A general wavefunction is a superposition of such functions where A is a function of time and the wavenumber k, A(k,t). But the conclusion still holds.
  7. Jun 24, 2008 #6
    Hmm... okay i think I understand now. Thanks =)
  8. Oct 12, 2008 #7
    what is the uncertainity of a free particle moving in one dimension
  9. Oct 12, 2008 #8
    Your proof is incomplete: you focus on a very particular type of real wavefunction, namely with a fixed value for [itex] p_x^2 [/itex]. However, what you have to prove is that for *any* real wavefunction [itex] \langle p_x\rangle =0. [/itex] There is no need to expand your wavefunction in plane waves, although, as Dick said, it gives a simple intuitive picture for why the momentum is zero on average.
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