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Expected Value of Momentum

  1. Jun 26, 2008 #1
    This is more of a general question... let's say we have a wavefunction [tex]\psi[/tex], and we want to find the expected momentum, [tex]\langle p \rangle [/tex], of the state. Isn't this just

    [tex]\langle p \rangle = \int_{-\infty}^{\infty} \psi^* i\hbar \frac{\partial \psi}{\partial x}[/tex]​

    Now, if the wavefunction happened to be purely real, wouldn't the momentum yield an imaginary value? How do we remedy this? I know that we can expand out [tex]\psi[/tex] in terms of the momentum eigenstates and go from there, but is there any way to just use the above?
  2. jcsd
  3. Jun 26, 2008 #2
    Yes it would.

    I think we set <p>=0 , since p is observable (it must be real)
  4. Jun 26, 2008 #3
    So whenever [tex]\psi[/tex] is a real function the expected value of the momentum is almost always 0? Would this imply that the probability distribution of the momentum is symmetric about the y axis?
  5. Jun 26, 2008 #4


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    Can you think of any situations where the wavefunction would be purely real? I don't think there can be, because if Psi is purely real then the LHS of the schrodinger equation is imaginary, but the RHS is real. (Where the SE is in the form here http://en.wikipedia.org/wiki/Schrödinger_equation)
  6. Jun 26, 2008 #5
    We could have a wavefunction independent of time, so the left hand side of the time-dependent Schroedinger equation goes to zero.

    Edit: Either this, or we can consider the wavefunction at a particular time, say t = 0. Here, the left hand side of the Schroedinger equation goes to zero.
  7. Jun 26, 2008 #6


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    True. But we know
    [tex]<p>=\frac{\partial <x>}{\partial t} = -\int_{-\infty}^{\infty} \psi^* i\hbar \frac{\partial \psi}{\partial x}dx[/tex]

    Since [itex]\psi[/tex] is time indepdent, so too must be <x>, so the LHS is zero. That means the RHS must be 0, or [itex]\psi[/tex] must be position independent as well. In other words, [itex]\psi[/itex] must be constant. But constants are not normalizable, so unless I'm missing something, purely real functions cannot solve the SE.

    Edit: Unless there are non constants functions f such that f(x)f'(x) = -f(-x)f'(-x), then the RHS could be zero that way.... Might have to think about this.
    Last edited: Jun 26, 2008
  8. Jun 26, 2008 #7
    I'm extremely confused :frown: So you're saying that there is no such thing as a pure real wavefunction? What about the eigenstates of a particle in box? These are purely real...
  9. Jun 26, 2008 #8


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    That is for the time-independent SE though, right?. The time-dependent part is still complex. And the expected value of the momentum is found using the full solution (time independent * time dependent).
  10. Jun 26, 2008 #9
    Okay, so what I meant was that if we have a particular state at t = 0 and the momentum is uncertain, at t = 0, what would the expected momentum be? I don't have any experience with working with the time dependent Schroedinger equation.
  11. Jun 26, 2008 #10


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    If the time-independent wavefunction is purely real, then [itex]\left<p\right>[/itex] will be zero. To prove this, integrate by parts:

    [tex]\int_{-\infty}^{\infty}dx~\psi\frac{d\psi}{dx} = \left.\psi^2\right|^{\infty}_{-\infty} - \int_{-\infty}^{\infty}dx~\frac{d\psi}{dx}\psi[/tex]

    The boundary term evaluates to zero, and what you're left with is an expression like A = -A, which of course means A = 0.

    Note that the time-dependent part just contributes phase factors which will cancel out (as long as the Hamiltonian is time-independent):

    [tex]\left<\hat{p}(t)\right> = \left<\psi\right|e^{i\mathcal{H}t/\hbar}\hat{p}e^{-i\mathcal{H}t/\hbar}\left|\psi\right> = \left<\psi\right|e^{iEt/\hbar}\hat{p}e^{-iEt/\hbar}\left|\psi\right> = \left<\psi\right|\hat{p}\left|\psi\right>[/tex]

    However, this isn't the whole story. If your system is in a particular state, then that's true, but in general,

    [tex]\left|\psi\right> = \sum_{n}c_ne^{iE_nt/\hbar}\left|\phi_n\right>[/tex]

    that is, the system is really in a superposition of states. In this case, what I wrote above doesn't work, because that assumed [itex]\psi = \phi e^{iEt/\hbar}[/itex]. With this superposition of states the time dependence will not cancel out (except at t = 0), and so the wavefunctions are complex.
    Last edited: Jun 26, 2008
  12. Jun 26, 2008 #11
    Okay, so let me sort of see if I understand this now... let's say we have a wavefunction [tex]\psi = \sqrt{2} \sin (10 \pi x)[/tex]. This is clearly not an eigenfunction of the momentum operator, but is an eigenfunction for the energy operator (for the particle in a box scenario... here, we have a = 1, n = 10). Now, if we find the expected momentum from this, we get zero:

    [tex]\langle p \rangle = \int \psi^* \hat{p} \psi dx = 0[/tex]​

    Okay, so now let's try to do this another way... let's express [tex]\psi[/tex] in terms of the eigenstates of the momentum operator,

    [tex]\phi_n = \frac{1}{\sqrt{2\pi}} e^{ikx}[/tex]​

    We have that

    [tex]\psi = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} b(k) e^{ikx} dx[/tex]​

    for complex valued [tex]b(k)[/tex]. We can invert this and get (note that the bounds for the integral change to [0,a] since the wavefunction is zero everywhere else excepting for from x = 0 to a...

    [tex]b(k) = \frac{1}{\sqrt{2\pi}} \int_{0}^a \psi e^{-ikx} dx[/tex]​

    After we substitute in [tex]\psi[/tex], we find that (after calculations...),

    [tex]|b(k)|^2 = \frac{200\pi (1- \cos k)}{(k-10\pi)^2 (k+10\pi)^2}[/tex]​

    Now, here's my argument: does this mean that

    [tex] \int_{-\infty}^{\infty} k|b(k)|^2 dk = 0[/tex]​


    Edit: YES! It works!! This is simply because

    [tex]\frac{200\pi (1- \cos k)}{(k-10\pi)^2 (k+10\pi)^2}[/tex]​

    is an even function! Thanks very much guys... I really appreciate it. My understanding of QM just went up a bunch! :smile: :smile: :smile: Just so you know... if you wanted to know why the evenness of this function caused the integral to be 0, if you multiply x by f(x) where f(x) is even, x*f(x) becomes odd. If you take the integral from -t to t of an odd function, you always get 0.
    Last edited: Jun 26, 2008
  13. Jun 26, 2008 #12


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    Careful there - that's only strictly true if the integrand tends to zero at [itex]\pm \infty[/itex], otherwise the integral is an indeterminate form. As k gets large your integrand goes as (1-cosk)/k^3, so it does indeed vanish at the boundary and you get zero for the integral.
  14. Jun 26, 2008 #13
    Well, yes, heheh I forgot to mention that :biggrin: But yea... it goes to zero since the denominator grows arbitarily large while the numerator stays fixed within 0 and 400 pi.
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