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The physical meaning of the sum of two independent Hamiltonians is well known. But... what about the difference between the Hamiltonians of two independent (i.e. non interacting) systems?

As an example: Consider the difference of the Hamiltonians of two independent oscillators: $$ \left[\frac{P_1^2}{2m} +m\omega^2 \frac{x_1^2}{2}\right]-\left[\frac{P_2^2}{2m} +m\omega^2 \frac{x_2^2}{2}\right] $$ Is it still an Hamiltonian? Just formally? Does it have a physical interpretation?

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  • $\begingroup$ When we sum the Hamiltonians of two independent single particle systems, the state space of the two particle system is the tensor product of the two single particle state spaces. What would a difference of Hamiltonians imply about the resulting state space? $\endgroup$ – Alfred Centauri Nov 25 '16 at 11:40
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No: if both Hamiltonians are bounded from below, then their difference is in general unbounded, and therefore the system is unstable. In your example, the kinetic term of the second part, $-\frac{p_2^2}{2m_2}$ can be arbitrarily negative.

Formally speaking, you have a function on phase space and therefore "it works" as a Hamiltonian, but it makes little sense as a physical system.

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  • $\begingroup$ @NorbertSchuch , could you please explain better your point? $\endgroup$ – AndreaPaco Nov 27 '16 at 1:50
  • $\begingroup$ @NorbertSchuch I have the feeling that OP is asking about classical mechanics, not QM :/ $\endgroup$ – AccidentalFourierTransform Nov 27 '16 at 12:46
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    $\begingroup$ @NorbertSchuch yeah, thanks for noticing it. I changed the tags back into classical-mechanics. Though at this point I'm not sure if OP is asking about CM or QM. Why would he/she write $P_i$ instead of $p_i$ for the momenta? $\endgroup$ – AccidentalFourierTransform Nov 27 '16 at 12:53
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    $\begingroup$ I'm referring to CM indeed! $\endgroup$ – AndreaPaco Nov 27 '16 at 13:48

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