The phase operator \(\hat{\phi}\) described in Section 6.3.1 Laflamme
[1] follows the equation \(e^{\pm i\hat{\phi}}|n\rangle=|n\mp 1\rangle\).
Assuming \(e^{i\hat{\phi}}\) is unitary \(e^{i\hat{\phi}}(e^{i\hat{\phi}})^{\dagger}=1\), and combined with the fact \(e^{i\hat{\phi}}e^{-i\hat{\phi}}=1\), we get \((e^{i\hat{\phi}})^{\dagger}=e^{-i\hat{\phi}}\).
This means
$$\displaylines{
\begin{aligned}
\langle n|e^{i\hat{\phi}} = & \space \left( (e^{i\hat{\phi}})^{\dagger} \langle n|^{\dagger} \right)^{\dagger} \\
= & \space\left( e^{-i\hat{\phi}}|n\rangle \right)^{\dagger} \\
= & \space \left( |n+1\rangle \right)^{\dagger} \\
= & \space \langle n+1| \\
\end{aligned}
}$$
With \(\langle n|e^{i\hat{\phi}}=\langle n+1|\), we have
$$\displaylines{
\begin{aligned}
\langle m|\space [e^{i\hat{\phi}}, \hat{n}] \space |n\rangle = & \space \langle m|\space e^{i\hat{\phi}}\hat{n} \space |n\rangle - \langle m|\space \hat{n}e^{i\hat{\phi}} \space |n\rangle \\
= & \space n\langle m+1|n\rangle - m\langle m|n-1\rangle \\
\end{aligned}
}$$
which has non-zero entries only when \(m=n-1\), \(\langle m|\space [e^{i\hat{\phi}}, \hat{n}] \space |n\rangle = \delta_{m+1, n}\).
At the same time, we have \(\langle m|\space e^{i\hat{\phi}} \space |n\rangle = \langle m|n-1\rangle = \delta_{m+1, n}\).
Thus,
$$
[e^{i\hat{\phi}}, \hat{n}] = e^{i\hat{\phi}}
$$
We show below that \([\hat{n}, \phi]=i\) is the solution to above equation \([e^{i\hat{\phi}}, \hat{n}] = e^{i\hat{\phi}}\):
$$\displaylines{
\begin{aligned}
[\hat{n}, \hat{\phi}^{n}] = & \space \hat{n}\hat{\phi}^{n} - \hat{\phi}^{n}\hat{n} \\
= & \space \hat{n}\hat{\phi}^{n} - \hat{\phi}^{n-1}(\hat{n}\hat{\phi}-i) \\
= & \space \hat{n}\hat{\phi}^{n} + i\hat{\phi}^{n-1} - \hat{\phi}^{n-1}\hat{n}\hat{\phi} \\
= & \space \hat{n}\hat{\phi}^{n} + i\hat{\phi}^{n-1} - \hat{\phi}^{n-2}(\hat{n}\hat{\phi}-i)\hat{\phi} \\
= & \space \hat{n}\hat{\phi}^{n} + 2i\hat{\phi}^{n-1} - \hat{\phi}^{n-2}\hat{n}\hat{\phi}^{2} \\
= & \space \hat{n}\hat{\phi}^{n} + 3i\hat{\phi}^{n-1} - \hat{\phi}^{n-3}\hat{n}\hat{\phi}^{3} \\
= & \space \vdots \\
= & \space \space \hat{n}\hat{\phi}^{n} + n\cdot i\hat{\phi}^{n-1} - \hat{n}\hat{\phi}^{n} \\
= & \space i\cdot n\hat{\phi}^{n-1} \\
\end{aligned}
}$$
Thus,
$$\displaylines{
\begin{aligned}
[e^{i\hat{\phi}}, \hat{n}] = & \space [\sum_{n=0}^{\infty }\frac{(i\hat{\phi})^{n}}{n!}, \hat{n}] \\
= & \space [\frac{1}{0!}, \hat{n}] + \sum_{n=1}^{\infty }\frac{i^{n}}{n!}[\hat{\phi}^{n}, \hat{n}] \\
= & \space 0 + \sum_{n=1}^{\infty }\frac{i^{n}}{n!}(-i\cdot n\hat{\phi}^{n-1}) \\
= & \space \sum_{n=1}^{\infty }\frac{i^{n-1}}{(n-1)!}\hat{\phi}^{n-1} \\
= & \space e^{i\hat{\phi}} \\
\end{aligned}
}$$
However, \([\hat{n}, \phi]=i\) results in a mathematical conundrum:
$$\displaylines{
\begin{aligned}
\langle m|\space [\hat{n}, \hat{\phi}] \space |n\rangle = & \space \langle m|\space \hat{n}\hat{\phi} - \hat{\phi}\hat{n} \space |n\rangle \\
= & \space (m-n)\langle m| \hat{\phi} |n\rangle \\
\end{aligned}
}$$
$$\displaylines{
\begin{aligned}
\langle m|\space i \space |n\rangle = & \space i\langle m|n\rangle \\
= & \space i\delta_{m, n} \\
\end{aligned}
}$$
Thus, \(\langle m| \hat{\phi} |n\rangle = \frac{i}{m-n}\delta_{m, n}\), which is undefined when \(m=n\).
This suggests \(e^{i\hat{\phi}}\) is not unitary, which implies \(\hat{\phi}\) is not Hermitian.
There are proposed solutions under the names Susskind-Glogower and Pegg-Barnett, but none are satisfactory
[2] as of date.