What does Solve for the time dependence of mean?

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

Homework Statement


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

[itex]H = \frac{p^{2}(t)}{2m} + mgx(t)[/itex]


Homework Equations



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

The Attempt at a Solution



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

I already have a value for that. [itex]\frac{d x(t)}{dt} = \frac{\hbar}{im}\frac{d}{dx}[/itex]
 
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I think it means x(t) for all t.
 


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


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

[itex]\hat{x}(t)\left|ψ\right\rangle = x\left|ψ\right\rangle[/itex]

As in, the time dependent position operator's eigenfunction is [itex]x[/itex]
 


wotanub said:
What do I need a Hamiltonian and such for?
You clearly won't be able to use Heisenberg's equation without it.

wotanub said:
[itex]\hat{x}(t)\left|ψ\right\rangle = x\left|ψ\right\rangle[/itex]

As in, the time dependent position operator's eigenfunction is [itex]x[/itex]

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.
 
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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)