Quantum Mechanics: Particle in Gravitational Field

In summary, 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. <P> is a definite integral and therefore is a constant. Integrate both sides to find d<Z>/dt which is meters*kilogram/second. Dimensionally, this is correct. However, this looks nothing like the classical version and I'm wondering, does -gt/2 correspond to <P>/m?
  • #1
chill_factor
903
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?
 
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  • #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?
 
  • #3
Hypersphere said:
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?

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?
 
  • #4
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?
 
  • #5
thank you that helped alot!
 

FAQ: Quantum Mechanics: Particle in Gravitational Field

What is a particle in a gravitational field?

A particle in a gravitational field is a theoretical concept in quantum mechanics that describes the behavior of a particle in the presence of a gravitational field. It takes into account both the principles of quantum mechanics and the theory of general relativity to explain how particles interact with gravity.

How does quantum mechanics explain the behavior of particles in a gravitational field?

In quantum mechanics, particles are described as both particles and waves, and their behavior is determined by the Schrodinger equation. This equation takes into account the effects of gravity on the particle's wavefunction, which determines the probability of the particle's location and momentum. Therefore, quantum mechanics can explain how particles are affected by a gravitational field.

What is the role of gravity in quantum mechanics?

Gravity plays a crucial role in quantum mechanics by affecting the behavior of particles in a gravitational field. It is one of the four fundamental forces in nature and is responsible for the attraction of bodies with mass. In the context of quantum mechanics, gravity is described by the curvature of spacetime, which is determined by the distribution of mass and energy.

Can quantum mechanics and the theory of general relativity be reconciled?

At present, there is no unified theory that combines quantum mechanics and general relativity. Both theories work well in their respective domains, but they break down when trying to explain phenomena at the quantum level and the gravitational level. Many physicists are currently working on theories that aim to reconcile these two fundamental theories.

How is the concept of a particle in a gravitational field used in practical applications?

The concept of a particle in a gravitational field is used in many practical applications, such as in the development of quantum computers and in the field of quantum cryptography. It is also essential in understanding the behavior of particles in extreme environments, such as black holes, where the effects of gravity are significant. Additionally, the study of particles in a gravitational field allows us to better understand the fundamental laws of nature and the behavior of the universe.

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