Cyclic variables for Hamiltonian

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SUMMARY

The discussion centers on the Hamiltonian for a single particle described by the equation ##H=\frac{m\dot{x}^{2}}{2}+\frac{m\dot{y}^{2}}{2}+\frac{x^{2}+y^{2}}{2}##, which can also be expressed in terms of momenta. It concludes that the system does not exhibit cyclic coordinates, as all relevant coordinates are explicitly present in the Hamiltonian. However, it suggests that alternative coordinate systems, such as polar coordinates, may reveal cyclic properties. The Hamiltonian represents a 2-dimensional harmonic oscillator with a frequency of ##\frac{1}{\sqrt{m}}##.

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digogalvao
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A single particle Hamitonian ##H=\frac{m\dot{x}^{2}}{2}+\frac{m\dot{y}^{2}}{2}+\frac{x^{2}+y^{2}}{2}## can be expressed as: ##H=\frac{p_{x}^{2}}{2m}+\frac{p_{y}^{2}}{2m}+\frac{x^{2}+y^{2}}{2}## or even: ##H=\frac{p_{x}^{2}}{2m}+\frac{p_{y}^{2}}{2m}+\frac{\dot{p_{x}}^{2}+\dot{p_{x}}^{2}}{4}##. Does it mean that the system has no cyclic coordinates? Since all relevant coordinates appear explicitly in ##H##. In that case, there are no constants of motion?
 
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First of all, the Hamiltonian is a function of the generalised coordinates and the corresponding canonical momenta (or more generally, of coordinates on phase space). You should not write the Hamiltonian as a function of derivatives of the momenta. Second, it is unclear to me what "it" that is supposed to mean that the system has no cyclic coordinates is. What property of the expression are you referring to? Clearly, none of the coordinates that you have are cyclic, but that does not mean that there are necessarily other coordinates you could use where one of them would be cyclic (as in fact there is in this case).
 
Orodruin said:
First of all, the Hamiltonian is a function of the generalised coordinates and the corresponding canonical momenta (or more generally, of coordinates on phase space). You should not write the Hamiltonian as a function of derivatives of the momenta. Second, it is unclear to me what "it" that is supposed to mean that the system has no cyclic coordinates is. What property of the expression are you referring to? Clearly, none of the coordinates that you have are cyclic, but that does not mean that there are necessarily other coordinates you could use where one of them would be cyclic (as in fact there is in this case).
What other coordinate is that?
 
digogalvao said:
What other coordinate is that?
What symmetries does the Hamiltonian have? In what coordinates would the Hamiltonian therefore not depend on one of the coordinates?
Hint: What you have is quite clearly a 2-dimensional harmonic oscillator.
 
Orodruin said:
What symmetries does the Hamiltonian have? In what coordinates would the Hamiltonian therefore not depend on one of the coordinates?
Hint: What you have is quite clearly a 2-dimensional harmonic oscillator.
Yes, the frequency is ##\frac{1}{\sqrt{m}}##. Should I put everything in polar coordinates?
 
What happens if you do?
 

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