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There is some connection between classical non-interacting harmonic oscillator (OH) and the free particle in higher dimensions?

I was studying statistical mechanics and I came across the idea that we can write the Hamiltonian of $N$ tree-dimensional HO as $$\mathcal{H}_{HO}=\sum_i^{3N}\left(\frac{p^2_i}{2m}+\frac{x_i^2}{2(m\omega^2)^{-1}}\right)=\sum_i^{6N}q_i^2 $$

So it's like the harmonic oscillator DOF are equivalent to the DOF of a free particle moving in a higher dimensional space.

Is this logic valid? If it's valid can be extended to a system of QHO?

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    $\begingroup$ ...what? I don't understand what you're seeing here, you're just doing a coordinate change on phase space such that the Hamiltonian becomes a sum of squares of the coordinates. What is your question about that? (Also, this is not like a free particle, because all of your d.o.f. are squared, not just the momenta) $\endgroup$ – ACuriousMind Apr 4 '16 at 10:13
  • $\begingroup$ And with your definition you almost surely lose track of/mess with the symplectic properties of the phase space... $\endgroup$ – yuggib Apr 4 '16 at 12:18
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The answer is no: if you were to write $p_i\propto q_{i+3}$ for $i=1,2,3$, then your quantum Hamiltonian would be $$ H\sim \sum_{i=1}^6 q_i^2 $$ which looks like the Hamiltonian of a free particle in a six dimensional space, but it is not.

It is not because the commutation relations are not $[q_i,q_j]=0$ but $$ [q_1,q_2]=0\quad [q_1,q_3]=0\quad [q_1,q_4]\propto\color{red}{i\hbar}\quad [q_1,q_5]=0 \quad\cdots $$ instead. This means that $q_1,q_4$ cannot be thought of as independent position coordinates in a six dimensional space (nor $q_2,q_5$ and $q_3,q_6$). You have an oscillator, not a free particle.

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