What does Solve for the time dependence of mean?

[wotanub]

What does "Solve for the time dependence of" mean?

Homework Statement

Use the Heisenberg equation of motion to solve for the time dependence of x(t) given the Hamiltonian

H = \frac{p^{2}(t)}{2m} + mgx(t)

Homework Equations

The Heisenberg equation of motion is
\frac{dA(t)}{dt} = \frac{i}{\hbar}\left[H,A(t)\right]

The Attempt at a Solution

As I said, I'm not sure what is meant by "Solve for the time dependence of x(t)". Do they just want \frac{d x(t)}{dt}?

I already have a value for that. \frac{d x(t)}{dt} = \frac{\hbar}{im}\frac{d}{dx}

 

[sunjin09]

I think it means x(t) for all t.

 

[Fredrik]

That's how I would interpret it too.

 

[wotanub]

Please explain. I'm sure x(t) = x in real space. What do I need a Hamiltonian and such for?

 

[HallsofIvy]

Please explain. I'm sure x(t) = x in real space.

?? What do you mean by "x(t)= x"??

 

[wotanub]

?? What do you mean by "x(t)= x"??

\hat{x}(t)\left|ψ\right\rangle = x\left|ψ\right\rangle

As in, the time dependent position operator's eigenfunction is x

 

[Fredrik]

What do I need a Hamiltonian and such for?

You clearly won't be able to use Heisenberg's equation without it.

\hat{x}(t)\left|ψ\right\rangle = x\left|ψ\right\rangle

As in, the time dependent position operator's eigenfunction is x

This still doesn't make sense. If anything in this equation should be called "eigenfunction", it's |ψ>, not x. (The terms "eigenvector" and "eigenket" are more common when the ket notation is used). And if this equality would hold, then $\hat x$ would be a constant function, since the right-hand side is independent of t.

 

[sunjin09]

In the Heisenberg picture, it is the operators that evolve with time. The equation of motion, which is an operator equation, can be solved for the time evolution of the operator x as a function of t. (Hope this helps, it's been years since I last looked at QM problems)