Solving a Linear Time-Dependent Hamiltonian Problem

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SUMMARY

The discussion focuses on solving a linear time-dependent Hamiltonian problem defined by the Hamiltonian H = p²/2m - mAtq, where m is mass and A is a constant. The participant derived the equations of motion using Hamilton's canonical equations, resulting in q(dot) = p/m and p(dot) = mAt. However, the participant's solution for q did not match the expected result from their professor, indicating a misunderstanding regarding the time dependence of p. The correct approach involves integrating p as a function of time to achieve the desired expression for q.

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Homework Statement


Suppose the potential in a problem of one degree of freedom is linearly dependent on time such that the Hamiltonian has the form:

H= p^2/2m - mAtq

where m is the mass of the object and A is contant

Using Hamilton's canonical equations that are give below. Find the equations of motion and obtain the solution by integrating directly.


Homework Equations



q(dot) = ∂H/∂p
-p(dot) = ∂H/∂q


The Attempt at a Solution



Finding q(dot) = ∂H/∂p = p/m → integrating q=q(knot) + pt/m

Finding p(dot) = -∂H/∂q = -(-mAt) → integrating p= p(knot) +mAt^2/2

the initial conditions were p(knot)= p and q(knot) = q at t=0


Subbing the expression for p into the formula for q

q= q(knot) + [p(knot) + mAt^2/2]*t/m = q(knot) + p(knot)t/m + 1/2*(At^3)




The solution I obtained for my expression for q does not match the desired that was given by my professor of q(knot) + p(knot)t/m + 1/6*(At^3).



I am trying to determine if I made a mistake somewhere. It appears to me that my solution is correct. Any guidance would be greatly appreciated.
 
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I failed to realize p is also a function of time and i need to integrate it with respect to t as well. i.e. ∫ p(knot) + 1/2 mAt^2 dt
 


Yes, usually one of the 2 Hamilton's equations will help you eliminate P in favor of Q and turn the system of ODE's into a single ODE.
 

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