# Correction to the quantum phase operator for photons

## TL;DR

This paper refines the quantum phase operator for photons by proposing a unitary operator that closely resembles the Susskind-Glogower operator, with a correction factor that converges to 1 for large photon numbers.

## Contribution

It introduces a modified phase operator for photons that ensures unitarity and improves upon the Susskind-Glogower operator by including a correction factor.

## Key findings

- The new phase operator is unitary.
- The correction factor approaches 1 as photon number increases.
- The operator reduces to the SG operator with a small correction.

## Abstract

The vector potential operator, $\hat{\boldsymbol A}$, is transformed and rewritten in terms of cosine and sine functions in order to get a clear picture of how the photon states relate to the $\boldsymbol A$ field. The phase operator, defined by $\hat E = \exp(-i \hat \phi)$, is derived from this picture. The result has a close resemblance with the known Susskind-Glogower (SG) operator, which is given by $\hat E_{SG}=(\hat a_{\boldsymbol k} \hat a_{\boldsymbol k}^\dagger)^{-1/2} \hat a_{\boldsymbol k}$. It will be shown that $\hat a_{\boldsymbol k}$ should be replaced by $(\hat a_{\boldsymbol k} + \hat a_{-\boldsymbol k}^\dagger)$ instead to yield $\hat E = ((\hat a_{\boldsymbol k} + \hat a_{-\boldsymbol k}^\dagger ) (\hat a_{\boldsymbol k}^\dagger + \hat a_{-\boldsymbol k}))^{-1/2} (\hat a_{\boldsymbol k} + \hat a_{-\boldsymbol k}^\dagger)$, which makes the operator unitary. $\hat E$ will also be analyzed when restricted to the space of only forward moving photons with wave vector $\boldsymbol k$. The resulting phase operator, $\hat E_+$, will turn out to resemble the SG operator as well, but with a small correction: Whereas $E_{SG}$ can be equivalently written as $\hat E_{SG} = \sum_{n=0}^{\infty} |n\rangle \langle n+1 |$, the operator, $\hat E_+$, is instead given by $\hat E_+ = \sum_{n=0}^{\infty} a_n |n \rangle \langle n+1|$, where $a_n = (n+1/2)!/(n! \sqrt{n+1})$. The sequence, $(a_n)_{n \in \lbrace 0, 1, 2, \ldots \rbrace}$, converges to $1$ from below for $n$ going to infinity.