# In practice, how does one work with the phase states?

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.)

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.
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.