The phase states are defined usually by the finite sum

$$ |\theta \rangle = (s+1)^{-1/2}\sum_{n=0}^s \exp(i n\theta) |n\rangle, $$

where $\theta = 2\pi k/(s+1)$ and $|n\rangle$ is the $n$-th eigenstate of the harmonic oscillator. Now, there exists some subtleties when taking $s\rightarrow \infty$, but in practice my question is as follows. In programming the harmonic oscillator and the coherent states on the computer, one usually cuts the Hilbert space for a sufficiently large $s$. Does it make sense to take this same number for the phase state? (I know it may be obvious, but I want to be sure that one does have two cut again a coherent state when working with the phase states.)


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The difference with the oscillator or the coherent state is that, for most states, the probability density is concentrated below some state $\vert n_0\rangle$, i.e. there is $n_0$ so that $\sum_{n=0}^{n_0} \vert\langle n\vert \psi\rangle\vert^2=1-\epsilon$ for some arbitrarily small $\epsilon$. Two examples of $\vert\langle n\vert \psi\rangle\vert^2$ are given in the figures below, for coherent states with $\alpha=\frac{1}{2}(\sqrt{3}+4i)$ and $\alpha=\sqrt{2}$ respectively.

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You can see that there isn't much of an overlap for any state past $n_0=15$, so in such a case it would make sense to truncate your Hilbert space at $n_0=15$ (or beyond if you want greater accuracy). Basically, the physics is well captured by the first $n_0=15$ states.

This is fundamentally different for your $\vert\theta\rangle$ since by definition the amplitude is constant over all the states $\vert s\rangle$. Thus, there is no guarantee that, extending the sum from $s$ to $s+1$, you will get results that don't change much, i.e. there no reason to believe the physics is captured by the first $s$ states of the sum.


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