1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Quantum Mechanics: Particle in Gravitational Field

  1. Oct 3, 2012 #1
    1. The problem statement, all variables and given/known data

    The Hamiltonian for a particle moving in a gravitational field and under no other influences is H = (P^2)/2m - mgZ where P is the momentum in the Z direction.

    1. Find d<Z>/dt.
    2. Solve the differential equation d<Z>/dt to obtain <Z>(t), that is, <Z> as a function of t, for the initial conditions <Z>(0) = h, <P>(0) = 0. Compare it to the classical expression Z(t) = (-gt^2)/2 + h

    2. Relevant equations


    3. The attempt at a solution

    Part 1.

    I solved the problem using the Ehrenfest theorem and expanded the commutator, but turned out the exact answer was on Wikipedia. That was disappointing but at least my final answer was right.

    http://en.wikipedia.org/wiki/Ehrenfest_theorem

    d<Z>/dt = 1/m * <P>

    Part 2.

    <P> is a definite integral and therefore is a constant. Integrate both sides.

    d<Z>/dt = <P>/m

    ∫d<Z> = ∫<P>dt/m
    <Z> - <Z(0)> = <P>/m ( t - t0), t0 = 0.

    <Z> - h = <P>t/m, <Z> = h + <P>t/m

    Dimensionally, this is correct. <Z> is meters, h is meters, <P> is meters*kilogram/second, then divide by m and times t, <P>t/m is meters.

    However, this looks nothing like the classical version and I'm wondering, does -gt/2 correspond to <P>/m?

    In addition, is <P> really, really a constant? If so then what is the use of the <P>(0) = 0 initial condition?
     
  2. jcsd
  3. Oct 3, 2012 #2
    The [itex]\frac{\langle {p} \rangle t}{m}[/itex] term you have corresponds to the linear term in the classical equation
    [itex]\Delta x = v_0 t + \frac{a t^2}{2}[/itex],
    right? You obviously can't have a time dependence in [itex]\langle p \rangle[/itex] here, since you already assumed that there isn't one, during the integration.

    Now though, why did you assume that the expected value of the momentum of a freely falling particle is constant?
     
  4. Oct 3, 2012 #3
    Yes, that was one of my main sources of confusion and I can't think of the correct way to solve this for a while already. The expectation value of the momentum should be time dependent in the gravitational field, but the definition of the expectation value is a definite integral. Also, the momentum operator has no explicit time dependence, even though it depends on Z' which is dependent on time, and this result was used for the calculation of part A.

    When the expectation value of the momentum takes on a time dependence, how do we calculate it? Since it might not be time independent what is the dependence on time?
     
  5. Oct 3, 2012 #4
    I think the most convenient method here is to use a Taylor expansion for the expectation value (around t=0). How many terms do you need?
     
  6. Oct 3, 2012 #5
    thank you that helped alot!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Quantum Mechanics: Particle in Gravitational Field
Loading...