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Let us assume we have a quantum particle in a harmonic potential with the Hamiltonian

$$H = \sum_n n \omega |n\rangle\langle n|$$

If I am not mistaken.

Now when talking about harmonic oscillators it is most of the times not mentioned that this sum $\sum_n$ should in fact go from $0$ to $\infty$.

So to my understanding the full, proper way would be to write

$$H = \sum^{\infty}_{n=0} n \omega |n\rangle\langle n|$$

if we are talking about a quantum harmonic oscillator which would have a similar form to this, with the only difference being that the energy gaps in my example are $\omega$ apart from neighbours.

So now when I would consider a finite Hamiltonian

$$H = \sum^{d}_n n\omega |n\rangle\langle n|$$

then this would be no longer be per definition a harmonic oscillator. It would be a finite-dimensional quantum system with equally spaced energy levels that can be thought of as a truncated harmonic oscillator.

Is that correct?

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    $\begingroup$ $H = \sum_{n=1}^N i \omega_n (a^\dagger_n a_n)$ does not represent a harmonic oscillator. It represents $N$ harmonic oscillators. $\endgroup$ – DanielSank Feb 23 at 0:38
  • $\begingroup$ The answer to your very last equation is "yes", provided you substitute $\omega (i+1/2)$ for your hardly meaningful $i\omega_i$. $\endgroup$ – Cosmas Zachos Feb 23 at 1:14
  • $\begingroup$ Why is $\omega(i + 1/2)$ so much better than $i \omega_i$? $\endgroup$ – CatoMaths Feb 23 at 1:33
  • $\begingroup$ @DanielSank That is good to know. I have removed that part. But the question still holds: Is it implicitly assumed when one writes $H =\sum_i i \omega i \mid i \rangle \langle i \mid$ that the sum goes from $0$ to infinity when it describes a harmonic oscillator? $\endgroup$ – CatoMaths Feb 23 at 1:36
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    $\begingroup$ Please don't use $i$ as a variable in quantum theory where it's potentially also used as $\sqrt{-1}$. It's bad discipline. Eventually people trip up doing this kind of thing. $\endgroup$ – StephenG Feb 23 at 2:12
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If $\vert n\rangle$ is an eigenstate of $\hat H$ with energy $(n+1/2)\hbar\omega$, it is easily verified that $\hat a^\dagger \vert n\rangle $ is also an eigenstate with energy $(n+3/2)\hbar\omega$. Moreover, as $\hat a^\dagger \vert n\rangle = \sqrt{n+1}\vert n+1\rangle $ it is also clear that the ladder of states of increasing $n$ never stops, i.e. $n$ is unbounded.

This is different from angular momentum, where $\hat L_\pm\vert \ell m\rangle =\sqrt{(\ell\mp m)(\ell \pm m+1)}\vert \ell,m\pm 1\rangle$. Since the factor $\sqrt{(\ell\mp m)(\ell \pm m+1)}$ becomes $0$ when $\hat L_+\vert \ell,\ell\rangle$ or when $L_-\vert \ell,-\ell\rangle$; in the case of angular momentum, the ladder of states stops and the possible values of $m$ are $-\ell \le m\le \ell$.

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  • $\begingroup$ I'd like to add a remark. You also need to check that $\langle n+1 | n+1 \rangle \neq 0$, otherwise they would be states orthogonal to everything and thus could be consistently truncated from the Hilbert space. $\endgroup$ – MannyC Feb 23 at 4:18
  • $\begingroup$ @ZeroTheHero I see but what if I assume that energies $E_n = n \hbar \omega$ are the eigenvalues to the eigenstates $\mid n\rangle$ ? This would work aswell, no ? $\endgroup$ – CatoMaths Feb 23 at 10:03
  • $\begingroup$ see this paper Atakishiyev, N. M., G. S. Pogosyan, and K. B. Wolf. "Finite models of the oscillator." Physics of Particles and Nuclei 36.3 (2005): 247-265. available here: researchgate.net/profile/Kurt_Wolf2/publication/… $\endgroup$ – ZeroTheHero Feb 23 at 13:24
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It would be a finite-dimensional quantum system with equally spaced energy levels that can be thought of as a truncated harmonic oscillator.

Is that correct?

Yes, that is correct. It's an interesting enough system (see e.g. this shameless plug), and 'truncated harmonic oscillator' is probably the best name for it.

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