The motivation of this question is the following experiment:
Assume you have quantum mechanical oscillator, e.g. a particle in a potential $V(q_x)\propto q_x^2$. Now the position of the particle shall be measured by having photons scattered from the particle and then detect $k$ vector and phase of the photons (in fact an indirect quantum measurement). Heisenberg's uncertainty principle now tells us that we would, by measuring the scattered photon, introduce a back action on the particles momentum. The strength of the back action is given by $\Delta q_x\cdot\Delta p_x\geq\hbar/2$.
The question is: Do the scattered photons, that I do not detect with my measurement device (and for example get absorbed at some wall or in some optics on the way) contribute to the perturbation introduced to the particles momentum?
There are two arguments that I heard in this discussion:
As every scattered photon interacts with the particle, every scattered photon will perform some kind of measurement on the particle position and therefore introduce a back action following Heisenberg's uncertainty principle. Thus I have to consider the position uncertainty $\Delta q_x$ that I could reach by using the information from all scattered photons (including the ones that I do not actively detect).
It only depends, which observable you actually measured. As for the photons that are not detected by the measurement device, the observable $Q_x$ is not measured, there is no perturbation on the momentum $p_x$ introduced to the particle through this particular photon. Thus I only have to consider the photons that actually reach my measurement device (e.g. my photo detector) for the estimation of the perturbation.
This distinction can become important, if for example the potential $V(q_x)$ is realized by an optical tweezer that relies on (a large number of) photons being scattered from the particle. If I use a different weak measurement beam (e.g. at a different wavelength), it would be nice if only this measurement beam would introduce perturbation.
EDIT: Hamiltonian of the System and Derivation of the Back Action
The Hamiltonian of the particle in the optical tweezer potential is given by the electric-dipole Hamiltonian
$$ \hat{\mathcal{H}}=\hat{\mathcal{H}}_\text{obj} + \hat{\mathcal{H}}_\text{int} + \hat{\mathcal{H}}_\text{prb} = \frac{\hat{p}^2}{2m}-\alpha\hat{E}\cdot\hat{E}+\frac{1}{2}\int d\vec{r}\,\left(\varepsilon_0\hat{E}^2+\mu_0^{-1}\hat{B}^2\right) $$
with $\alpha$ being the polarizability of the particle. I am only interested in the first two terms for the next steps, as the object does not interact with the Hamiltonian of the probe (any observable of the object and $\hat{\mathcal{H}}_\text{prb}$ commute). The second term also describes the potential, because the electric field is focused and $\hat{E}$ is a function of $\hat{q}$.
A fraction of the electric field (after interacting with the particle) gets collected and is used for detection of the particle position $q$ (see also here). The information on the position of the particle is stored in the phase of the scattered field. The back action on the particle position $\hat{q}$ is then given by:
$$ i\hbar\frac{d}{dt}\hat{q}(t)=-\left[\hat{\mathcal{H}},\hat{q}\right]=-\left[\hat{\mathcal{H}}_\text{obj},\hat{q}\right]-\left[\hat{\mathcal{H}}_\text{int},\hat{q}\right]-\left[\hat{\mathcal{H}}_\text{prb},\hat{q}\right]=\frac{i\hbar}{m}\hat{p}+0+0 $$
Because of the uncertainty relation the measurement of $q$ will perturb $p$ which in turn gives a random back action on $q$.
$$ \frac{d}{dt}q(t)=\frac{p+\Delta p}{m}\geq \frac{p}{m}+\frac{\hbar}{2m \Delta x} $$
Now this was for the fraction of the electromagnetic field that I used to measure the position. What kind of back action do I get from the other fraction of the field, on which I do not explicitly perform the position measurement? Obviously the scattered photons are also correlated with the state of the particle, if I don't detect its position ...
