Quantum Mechanics: Particle in Gravitational Field

chill_factor
Messages
898
Reaction score
5

Homework Statement



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

Homework Equations

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?
 
Physics news on Phys.org
The \frac{\langle {p} \rangle t}{m} term you have corresponds to the linear term in the classical equation
\Delta x = v_0 t + \frac{a t^2}{2},
right? You obviously can't have a time dependence in \langle p \rangle 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?
 
Hypersphere said:
The \frac{\langle {p} \rangle t}{m} term you have corresponds to the linear term in the classical equation
\Delta x = v_0 t + \frac{a t^2}{2},
right? You obviously can't have a time dependence in \langle p \rangle 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?

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?
 
chill_factor said:
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?

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?
 
thank you that helped alot!
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top