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In finite-dimensional quantum mechanics, we are free to assume that our Hamiltonian is traceless. We can define $H' = H - \mathrm{tr}(H)\cdot I$, and since \begin{equation} \exp(-iH't) = \exp(-iHt)\cdot\exp(it\cdot\mathrm{tr}(H)), \end{equation} we see that the dynamics driven by $H$ and $H'$ only differ by a global phase that we can ignore. I interpret this as meaning that we are free to rescale the energies of a finite-dimensional system such that they average to zero, without any measurable consequences.

If we want to play the same game for an infinite-dimensional Hamiltonian $H$ it needs to have a well-defined trace; I'm not much of a functional analyst, but I understand that this is effectively equivalent to the eigenvalues of $H$ forming a convergent series. For example, the hydrogen atom Hamiltonian has eigenvalues that scale as $\frac{1}{n^2}$, which will yield a converging sum, while the QHO eigenvalues diverge.

People often talk about the zero-point energy of the QHO but I've never heard reference to the zero-point energy of a hydrogen atom wavefunction - is this related to the traces (or lack thereof) of these Hamiltonians? And if not, is there any physical meaning associated with a Hamiltonian (or other observable) having a well-defined trace?

Tobias Fünke
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    Some comments: You might want to check the notion of trace-class operators. You can rescale any (physically sensible, i.e. bounded from below) Hamiltonian by an additive constant, which merely shifts the energy levels. And last but not least, note that the eigenstates of the hydrogen atom you refer to do not form a complete orthonormal basis, as these are only the states corresponding to the "discrete" spectrum; check e.g. this – Tobias Fünke Jan 02 '23 at 07:48
  • I think "zero point energy" is generally only used in the context of ground states that can also be interpreted as vacuum states (as in the QSHO or in QFTs, but not the hydrogen atom). The ZPE is generally inconsequential for the reasons you state–except, e.g., in the Casimir effect. Also, the Hamiltonian always has a trace, since it must be diagonalizable, though the trace certainly may diverge. – just a phase Jan 02 '23 at 19:16
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    You've crucially failed to normalize the identity matrix by the dimension of your space. Only this will make H' traceless. Do you now see the limit for large Hilbert spaces? – Cosmas Zachos Jan 02 '23 at 23:30

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