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Expectation Value of Momentum in H-Atom

  1. May 31, 2007 #1

    Tom Mattson

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    Here's a silly question. I'm sure I should know the answer, but alas my most recent QM course was 9 years ago.

    I sat down to calculate the expectation value of momentum in the H-atom today, because some kid on another forum wanted to know how fast an electron in an atom is. I was going to calculate [itex]<p>[/itex] and divide it by [itex]m[/itex] for him.

    So I wrote down [itex]<p>=\int_V\psi_{100}^*(-i\hbar\nabla)\psi_{100} d^3x[/itex] ([itex]\psi_{100}[/itex] being the ground state H-atom wavefunction). Upon noticing that I've got a solitary [itex]i[/itex] sandwiched between two purely real functions, I didn't even bother to continue. It's going to come out imaginary.

    What am I doing wrong? :confused:
     
  2. jcsd
  3. May 31, 2007 #2
    Not evaluating the integral. Just compute it and see.

    Then, calculate <p^2> and take its root mean square and tell the kid that.
     
  4. May 31, 2007 #3

    Meir Achuz

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    The vector momentum =0.
     
  5. Jun 1, 2007 #4

    vanesch

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    It should come out 0, and in fact you just proved this:
    the p operator being hermitean, its expectation value must be real for any vector. You showed that it must be a purely imaginary number. Hence it can only be 0.

    And that's normal too: the expectation of momentum for any stationary bound state must be 0: given that a stationary state will not have the momentum change in time, and given that the fact that it is bound will confine (most of) the wavefunction to a bounded region, the C.O.G. cannot leave that bounded region, which it should, if there was a non-zero expectation value of momentum, after some time.
     
  6. Jun 1, 2007 #5

    Tom Mattson

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    The integral is not zero. The ground state wavefunction is:

    [tex]\psi_{100}(r)=\frac{1}{\sqrt{\pi}}\frac{1}{a_0^{3/2}}\exp(-r/a_0)[/tex]

    If we integrate [itex]\psi^*_{100}(-i\hbar\nabla)\psi_{100}[/itex] over all space we get:

    [tex]\frac{-i\hbar}{\pi a_0^3}\int_0^{2\pi}\int_{-1}^1\int_0^{\infty}\exp(-r/a_0)\frac{d}{dr}\exp(-r/a_0)r^2drd(\cos(\theta))d\phi[/tex]

    Since the wavefunction doesn't depend on the angular coordinates, this integral is proportional to the following:

    [tex]i\int_0^{\infty}r^2\exp(-2r/a_0)dr[/tex]

    The integrand is not identically zero, and is nonnegative everywhere. So it's integral cannot vanish. So, it seems this result is purely imaginary and not zero.

    There's got to be a mistake somewhere, but I don't see where. :confused:
     
    Last edited: Jun 1, 2007
  7. Jun 1, 2007 #6
    The nabla is a vector operator. It produces something proportional to r_vector. Integrating that over spherically symmetrical space gives zero vector.
     
  8. Jun 1, 2007 #7

    nrqed

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    I agree with Smallphi.
    You forgot to include [itex] \hat{r}[/itex]. Integrating this over all angles gives zero.
     
  9. Jun 12, 2007 #8

    olgranpappy

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    One simple way to get an appropriate estimate of the velocity is to divide 13.6eV by the electron mass, multiply by 2 and then take the square-root thus finding that the electron travels at a bit less than one hundredth of the speed of light. This is correct because the expectation value of the kinetic energy is equal to the negative of the expectation value of the Hamiltonian for a Coulomb potential. I.e. -(-13.6eV).

    Also, you can see that <\vec p> must be zero for any real wave-function (btw, eigenfunctions of this Hamiltonian can always be *chosen* as real) by just integrating by parts... you end up with the equation <\vec p>=-<\vec p> which implies that <\vec p>=0
     
  10. Jun 12, 2007 #9

    Hans de Vries

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    To get rid of the " i " you might want to use following momentum
    density operator:

    [tex]{ i\hbar \over 2} \left( \frac{\partial \psi^*}{\partial x_i}\psi\ -\ \psi^*\frac{\partial \psi}{\partial x_i }\right)[/tex]

    Where the psi is the relativistic wave function related to the Schrödinger
    wave function as follows:

    [tex]\psi\ =\ \Psi \ e^{-imc^2t/\hbar}, \qquad \mbox{$\Psi$ is the Schroedinger wave function}[/tex]

    This gives the same result without the imaginary part. It's always real.
    This is the Klein-Gordon momentum density operator, as well as the Dirac
    momentum density operator, (after the spin contributions are split off
    via the Gordon decomposition).


    Regards, Hans
     
    Last edited: Jun 12, 2007
  11. Jun 12, 2007 #10

    Hans de Vries

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    A little bit less than one hundredth of the speed of light is 1/137.036 to be
    a bit more exact, a value also known as alpha. :wink:


    Regards, Hans
     
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