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The heart of my question is below in bold. The rest is clarifying information or additional points of discussion - in case my assumptions are the heart of my misunderstanding.

In a two body attractive quantum mechanical system (namely the simplified Hydrogen atom - ignoring spin etc.), it can be shown that this reduces into two Hamiltonians. We find one Hamiltonian for the center of mass, and one for the behavior of the individual particles.

Assuming the individual particles are bound by their mutually attractive potential, the Hamiltonian has discrete Energy Eigenstates. The center of mass Hamiltonian, $H_{com}$, is a free particle solution. Does position-momentum uncertainty principle still apply to the wavefunction describing solutions to $H_{com}$ ?

In my undergrad QM class, the professor said that when we're in the center of mass reference frame, we ignore $H_{com}$. When I pressed him further on this, he said it wouldn't make sense for the atom's wavefunction to "spread out" in the center of mass frame (due to a $k$-space distribution of $\psi$). This didn't sit well with me, because being in the center of mass frame of the atom implies we know the center of mass momentum exactly (or to extreme accuracy), and therefore can't know the exact position (or only have an extremely inaccurate knowledge).

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  • $\begingroup$ You can easily apply the general uncertainty principle to any set of two observables. Have you tried this yet? $\endgroup$ Commented Aug 18, 2020 at 18:18

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Yes it holds because the corresponding operators satisfy the standard commutation relations. It can be proved in various ways. The most elementary way is to describe the system using the so-called Jacobi coordinates (in the phase space) which include the coordinates of CM and the total momentum. Assuming that the position and momentum operstors of every single particle satisfy the standard commutation rules, it arises that the position of CM and the total momentum satisfy the same relations.

However the typical time a wavefunction spends in spreading out in space, when it freely evolves, is related to the value of the mass of the system: large masses imply large times. This implies that the wavefunction of the center of mass (assuming the total wavefunction factorised) needs a typical time larger than the times of the single particles, if these are not subjected to mutual interactions.

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  • $\begingroup$ I have never used Jacobi coordinates before - is this just describing every particle in its own coordinates and following the math to its logical conclusion? Thank you for the answer $\endgroup$
    – John
    Commented Aug 18, 2020 at 18:26
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    $\begingroup$ Yes I am just saying that as soon as you assume the validity of H principle for the two particles separately, then you are logically forced to assume the validity of that principle for the center of mass and the total momentum. $\endgroup$ Commented Aug 18, 2020 at 18:29

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