So when one measures momentum in the lab, with whatever error, is one really measuring "is the particles momentum within $p\pm\Delta p$ and also in the lab?"? If so, what is the operator corresponding to this measurement?
The momentum operator is still the same old momentum operator, regardless of how well you can measure momentum. (But, see the caveats below in the update regarding "boundary conditions" and the domain of definition of the operator.)
If your measurement is not precise enough to pick out a single exact momentum value (and presumably no real measurement can be so precise) then you are projecting on to a set of states, rather than collapsing to a single state.
For example, in one dimension, let
$\psi(x)$ be your wave function in real space and let
$$
P\psi(x) = -i\hbar\frac{d\psi}{dx}(x)\;,
$$
as usual.
The wave function in momentum space is
$$
\tilde \psi(p) = \int_{-\infty}^\infty e^{-ipx}\psi(x)dx\;,
$$
and the wave function in position space can be expressed as
$$
\psi(x) = \int_{-\infty}^\infty e^{ipx}\tilde \psi(p)\frac{dp}{2\pi}\;.
$$
If you start with the state $\psi(x)$ and you only measure the momentum to be $p_0$ within $\pm \Delta p$ then the wave function is projected onto those momentum eigenstates and becomes:
$$
\psi(x)\underbrace{\to}_{measure}\psi_{after}(x) = N\int_{p_0-\Delta p}^{p_0+\Delta p} e^{ipx} \tilde\psi(p)\frac{dp}{2\pi}\;,
$$
where $N$ is a normalization constant
$$
N = \frac{1}{\sqrt{\int_{p_0-\Delta p}^{p_0+\Delta p}|\tilde \psi(p)|^2\frac{dp}{2\pi}}}
$$
Update To Address OP's Comment:
Yes, the part of projecting into a set of states I understand, and I think I mention it in the question. However, what I'm unsure about is: when we are measuring the momentum of a particle in a lab, are we not also measuring its position? After all, we gain knowledge that the particle is roughly in the lab
To some extent you do also have knowledge of the position because the system is taken to be confined to some known laboratory space. But usually the laboratory size is so much larger than any other length scale, that knowledge doesn't matter.
On the other hand, if the "lab" size really is a small "box" (like for particle-in-a-box), then the definition of the momentum operator does change (because the domain of the operator is required for its definition). (But, in this case, it also turns out that the momentum is not a physical observable, unfortunately.)
In the case where the "lab" is a "box" of size $a$, then the definition of the momentum operator can be taken to be given in two parts (one defining the action on functions in $L^2([0,a],dx)$ and one defining the domain of the operator):
- The action of $\hat P$ on $\psi(x)$ is $$
\psi(x) \to P\psi(x) = -i\hbar\frac{d\psi}{dx}
$$
- The domain of $\hat P$ is
$$
\mathcal{D}(\hat P) = \left\{\left.\psi\in L^2([0,a],dx)\;\right|\;\psi(0)=0\;, \psi(a)=0\;, \frac{d\psi}{dx}\in L^2([0,a],dx) \right\}
$$
However, it should be noted that in the case of a particle-in-a-box, the momentum is not a physical "observable," since in this case the momentum turns out to be Hermitian, but not self-adjoint, since $\mathcal{D}(\hat P)\neq \mathcal{D}(\hat P^\dagger)$.
The issue of "interpolation" between a small box where the momentum is not an observable and a very large box where the momentum is effectively an observable is interesting, but beyond the scope of this question-and-answer website, I think.
