Groenewold in his book On the Principles of Elementary Quantum Mechanics (1946, Springer Netherlands) page 45, maps the canonical momentum $p^2$ in the classical phase space to a general canonical operator ${\mathbf{x}_1}$:

\begin{equation}p^{2} \rightarrow {\mathbf{x}_1}. \tag{4.06} \end{equation}

Hence, the Poisson brackets are mapped to the canonical commutation relations as the following:

\begin{equation}\begin{array}{l} \frac{1}{2}\left(p^{2}, q\right)=p \rightarrow \frac{1}{2}\left[\mathbf{x}_{1}, \mathbf{q}\right]=\mathbf{p} \\ \frac{1}{2}\left(p^{2}, p\right)=0 \rightarrow \frac{1}{2}\left[\mathbf{x}_{1}, \mathbf{p}\right]=0 \end{array} \tag{4.07} \end{equation}

and from those relations we find that $p^{2} \rightarrow \mathbf{p}^{2}+c_{1}$ where $c_1$ is some constant.

My question is why did we add this constant ($c_1$) to the momentum operator?

Groenewold in his book On the Principles of Elementary Quantum Mechanics (1946, Springer Netherlands) page 45, maps the canonical momentum $p^2$ in the classical phase space to a general canonical operator ${\mathbf{x}_1}$:

\begin{equation}p^{2} \rightarrow {\mathbf{x}_1}. \tag{4.06} \end{equation}

Hence, the Poisson brackets are mapped to the canonical commutation relations as the following:

\begin{equation}\begin{array}{l} \frac{1}{2}\left(p^{2}, q\right)=p \rightarrow \frac{1}{2}\left[\mathbf{x}_{1}, \mathbf{q}\right]=\mathbf{p} \\ \frac{1}{2}\left(p^{2}, p\right)=0 \rightarrow \frac{1}{2}\left[\mathbf{x}_{1}, \mathbf{p}\right]=0 \end{array} \tag{4.07} \end{equation}

and from those relations we find that $p^{2} \rightarrow \mathbf{p}^{2}+c_{1}$ where $c_1$ is some constant.

My question is why did we add this constant ($c_1$) to the momentum operator?

In Groenewold's Theorem, Groenewold maps the canonical momentum $p^2$ in the classical phase space to a general canonical operator ${\mathbf{x}_1}$:

\begin{equation}p^{2} \rightarrow {\mathbf{x}_1}.\end{equation}

Hence, the Poisson brackets are mapped to the canonical commutation relations as the following: \begin{equation}\begin{array}{l} \frac{1}{2}\left(p^{2}, q\right)=p \rightarrow \frac{1}{2}\left[\mathbf{x}_{1}, \mathbf{q}\right]=\mathbf{p} \\ \frac{1}{2}\left(p^{2}, p\right)=0 \rightarrow \frac{1}{2}\left[\mathbf{x}_{1}, \mathbf{p}\right]=0 \end{array}\end{equation}

and from those relations we find that $p^{2} \rightarrow \mathbf{p}^{2}+c_{1}$ where $c_1$ is some constant.

My question is why did we add this constant ($c_1$) to the momentum operator?

Groenewold in his book On the Principles of Elementary Quantum Mechanics (1946, Springer Netherlands) page 45, maps the canonical momentum $p^2$ in the classical phase space to a general canonical operator ${\mathbf{x}_1}$:

\begin{equation}p^{2} \rightarrow {\mathbf{x}_1}. \tag{4.06} \end{equation}

Hence, the Poisson brackets are mapped to the canonical commutation relations as the following:

\begin{equation}\begin{array}{l} \frac{1}{2}\left(p^{2}, q\right)=p \rightarrow \frac{1}{2}\left[\mathbf{x}_{1}, \mathbf{q}\right]=\mathbf{p} \\ \frac{1}{2}\left(p^{2}, p\right)=0 \rightarrow \frac{1}{2}\left[\mathbf{x}_{1}, \mathbf{p}\right]=0 \end{array} \tag{4.07} \end{equation}

and from those relations we find that $p^{2} \rightarrow \mathbf{p}^{2}+c_{1}$ where $c_1$ is some constant.

My question is why did we add this constant ($c_1$) to the momentum operator?