What does Solve for the time dependence of mean?

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Homework Help Overview

The discussion revolves around interpreting the phrase "Solve for the time dependence of x(t)" in the context of quantum mechanics, specifically using the Heisenberg equation of motion and a given Hamiltonian.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore what is meant by "time dependence" and whether it refers to the derivative of x(t) or the function itself over time. There are questions about the necessity of the Hamiltonian in this context and the interpretation of the position operator in relation to its eigenfunctions.

Discussion Status

The discussion is active, with participants sharing their interpretations and seeking clarification on the role of the Hamiltonian and the meaning of time dependence in quantum mechanics. Some participants express confusion over the definitions and implications of the terms used.

Contextual Notes

There are indications of differing interpretations regarding the relationship between the position operator and its eigenfunctions, as well as the necessity of the Hamiltonian for solving the problem.

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.
 


That's how I would interpret it too.
 


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


wotanub said:
Please explain. I'm sure x(t) = x in real space.
?? What do you mean by "x(t)= x"??
 


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)
 

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