Canonical transformation for Harmonic oscillator

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SUMMARY

The discussion focuses on the conditions under which the transformation from the phase space variables (x, p) to the new variables (Q, P) is canonical, specifically using the equations Q = ap/x and P = bx². The participant determined that a = -1/2b and is attempting to apply this transformation to the harmonic oscillator's Hamiltonian, H = 1/2m(p² + (mωx)²), where ω² = k/m. The transformed Hamiltonian is expressed as H = P(Q² + ω²). The participant seeks guidance on solving the resulting differential equations derived from Hamilton's equations in the new coordinate basis.

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  • Understanding of canonical transformations in Hamiltonian mechanics
  • Familiarity with Hamilton's equations
  • Knowledge of harmonic oscillators and their Hamiltonians
  • Basic skills in solving differential equations
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  • Explore techniques for solving differential equations, including trigonometric substitutions
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aaaa202
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Find under what conditions the transformation from (x,p) to (Q,P) is canonical when the transformation equations are:
Q = ap/x , P=bx2
And apply the transformation to the harmonic oscillator.
I did the first part and found a = -1/2b
I am unsure about the next part tho:
We have the hamiltonian in (x,p):
H = 1/2m(p2+(m\omegax)2 , \omega2=k/m
So transforming to (Q,P) whilst setting am=-1 you get the nice equation:
H = P(Q2+\omega2)
Should I now use hamiltons equation in the new coordinate basis, i.e.;
dH/dP = Q2+\omega2 = dQ/dt
dH/dQ = 2QP = -dP/dt
And solve these differential equations for Q,P and transform back? I think so but these equations are just not very easu. I mean the second one is easy to do by means of substitution but really the first one is just a mess. What is the smartest way to do this? I can't really get the first one solved.
 
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aaaa202 said:
Find under what conditions the transformation from (x,p) to (Q,P) is canonical when the transformation equations are:
Q = ap/x , P=bx2
And apply the transformation to the harmonic oscillator.
I did the first part and found a = -1/2b

Do you mean a=-\frac{1}{2b}, or a=-\frac{1}{2}b? Brackets are important! :wink:

I am unsure about the next part tho:
We have the hamiltonian in (x,p):
H = 1/2m(p2+(m\omegax)2 , \omega2=k/m

Again, brackets are needed if you meant H=\frac{1}{2m}\left(p^2+(m\omega x)^2\right)

So transforming to (Q,P) whilst setting am=-1 you get the nice equation:
H = P(Q2+\omega2)

(1) Why are you setting am=-1?
(2) You don't get that when you set am=-1

Without setting am=-1, you should get H=-\frac{1}{4m}PQ^2 -am\omega^2P

Should I now use hamiltons equation in the new coordinate basis, i.e.;
dH/dP = Q2+\omega2 = dQ/dt
dH/dQ = 2QP = -dP/dt
And solve these differential equations for Q,P and transform back? I think so but these equations are just not very easu. I mean the second one is easy to do by means of substitution but really the first one is just a mess. What is the smartest way to do this? I can't really get the first one solved.

The second DE is only easy to solve once you know Q. The first one is seperable, so al that is needed is some striaghtforward integration... try a trig substitution if you are stuck on the integration part
 

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