Confusion about $1/|\vec{p}|$ in helicity operator The helicity operator is defined as
$$
h = \frac{1}{|\vec{p}|} \vec{\sigma} \cdot \hat{\vec{p}}
$$
One of the first exercises into QED is to check whether this commutes with the Dirac Hamiltonian. Should the question arise, everyone will say (or the books will hint), that $1/|\vec{p}|$ is "just a number" (hence no hat on that bit). Years ago when I took the course of quantum electrodynamics I didn't pay this much attention, but now I revisited this concept and I'm confused.
We know, that $\hat{\vec{p}} = - i \hbar \vec{\nabla}$, so to find an operator that corresponds to $|\hat{\vec{p}}|$ would mean to find an operator that, when applied twice, gives $- \hbar^2 \Delta$. People usually shrug this off as unnecessary and that $|\vec{p}|$ is to be understood as just a real number.
$1/|\vec{p}|$ can be "just a number" only if the state we're applying this operator to is an eigenstate of the momentum operator (i.e. a plane wave). It can be then shown (in p-representation, where any operator composed of $\vec{p}$ is interpreted as multiplication), that $1/|\hat{\vec{p}}|$ gives $1/|\vec{p}|$.
Moreover, if it's so easy to just shrug off this whole issue about $1/|\vec{p}|$ as it being just a number, why did Dirac obsess so much about finding a square root of the Klein-Gordon equation? He could just say "well, anything here is just a number, so we plug in the corresponding numbers and then take the square root."
Does anyone have any satisfactory answer to clear my confusion? Please, if you think you do, I indulge you to calculate the following commutator
$$
\left[ 1/|\hat{\vec{p}}|, x_i \right]
$$
 A: I think I figured it out.
As for $1/|\vec{p}|$ being just a number this is allowed, since the Hamiltonian for free particles does not contain position, just momentum ($H = \vec{\alpha} \cdot \hat{\vec{p}} + \beta m$), so operator-ness of $1/|\vec{p}|$ is not important.
However, it is still possible (and potentially useful) to discuss this as an operator. I tried to derive a few commutation relations as an exercise and it is possible
$$
\left[ x_i, | \hat{\vec{p}} | \right] = \frac{i \hbar \hat{p}_i}{\left| \hat{\vec{p}} \right|}
$$
$$
\left[ x_i, 1/| \hat{\vec{p}} | \right] = - \frac{i \hbar \hat{p}_i}{\left| \hat{\vec{p}} \right|^3}
$$
Example how the first one can be shown: we use the fact, that no matter how complicated an operator is, it always yields the appropriate eigenvalue on plane waves $e^{i \vec{k} \cdot \vec{x}}$.
$$
\left[ x_i, | \hat{\vec{p}} | \right] f (\vec{x}) = \left( x_i |\hat{\vec{p}}| - |\hat{\vec{p}}| x_i \right) \int \mathrm{d}^3 k \, f (\vec{k}) \: e^{i \vec{k} \cdot \vec{x}} = \\
= x_i \int \mathrm{d}^3 k \, f(\vec{k}) \left( | \hat{\vec{p}} | e^{i \vec{k} \cdot \vec{x}} \right) - \int \mathrm{d}^3 k \, \left( - i \frac{\partial f}{\partial k^i} \right) \left( | \hat{\vec{p}} | e^{i \vec{k} \cdot \vec{x}} \right) = \\
= x_i \int \mathrm{d}^3 k \, (\hbar k) \, f (\vec{k}) \: e^{i \vec{k} \cdot \vec{x}} - i \int \mathrm{d}^3 k \, (\hbar k) \, \frac{\partial f}{\partial k^i} \: e^{i \vec{k} \cdot \vec{x}} = \\
= x_i \int \mathrm{d}^3 k \, (\hbar k) \, f (\vec{k}) \: e^{i \vec{k} \cdot \vec{x}} + i \int \mathrm{d}^3 k \, \left( \frac{\hbar k_i}{k} + i \hbar k x \right) f (\vec{k}) \: e^{i \vec{k} \cdot \vec{x}} = \\
= i \hbar \int \mathrm{d}^3 k \, \left( \frac{k_i}{k} \right) f (\vec{k}) \: e^{i \vec{k} \cdot \vec{x}} = \frac{i \hbar \hat{p}_i}{\left| \hat{\vec{p}} \right|} f (x)
$$
In the first line I just wrote out the commutator and expanded the function in plane waves, on the second line I used that $\left\langle k \right| \hat{x} \left| \psi \right\rangle = i \partial_k \, \psi (k)$, on the fourth line I integrated by parts (throwing out the boundary term, since $f$ is as quickly going to zero as needed) and finally I interpreted the integral on the last line as the result. Any $p$-related operator can be treated like that:
$$
\left[ x_i, f(| \hat{\vec{p}} |) \right] = i \hbar \, f^\prime (| \hat{\vec{p}} |) \frac{\hat{p}_i}{\left| \hat{\vec{p}} \right|}
$$
