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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:

  1. 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).

  2. 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 ...

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Assume you have quantum mechanical oscillator, e.g. a particle in a potential V(qx)∝q2x. 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).

Note that in a quantum mechanical potential the particle is in a stable energy level. The scattering will not be against the particle but against the whole system. The experiment has been done in measuring the orbitals of the hydrogen atom recently.

Note the quantum mechanical language:

demonstrate how photoionization microscopy directly maps out the nodal structure of an electronic orbital of a hydrogen atom placed in a dc electric field . This experiment—initially proposed more than 30 years ago—provides a unique look at one of the few atomic systems that has an analytical solution to the Schrödinger equation. To visualize the orbital structure directly, the researchers utilized an electrostatic lens that magnifies the outgoing electron wave without disrupting its quantum coherence.

hydrogen orbitals

A photoionization microscope provides direct observation of the electron orbital of a hydrogen atom. The atom is placed in an electric field E and excited by laser pulses (shown in blue). The ionized electron can escape from the atom along direct and indirect trajectories with respect to the detector (shown on the far right). The phase difference between these trajectories leads to an interference pattern, which is magnified by an electrostatic lens.

So in the quantum mechanical system there is no direct "action reaction" just probabilities . To learn about the wavefunction structure in space the experiment is destructive, the photon energy kicks off an electron and that atom is no longer described by the original wavefunction. The history is recorded cumulatively

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 Δqx⋅Δpx≥ℏ/2.

No, at most if we scatter a photon off an atom non destructively, it would tell us something about the position of the atom, and the whole atom would interact to the scatter not the particle in the well, the whole atomic system.

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?

The particle is either happy in the potential well, or kicked to a higher level, or the particle has been completely kicked out and the QM problem has changed. It is the WHOLE atom (particle in potential well) that will have a perturbation of a photon is scattered off it in the non destructive case.

In any case the heisenberg uncertainty principle is taken care of in the probabilistic form of the solution of quantum mechanical equations.

Again 1, 2 have no meaning withing auantum mechanics the way written. It is the atom, not the particle.

My answer involves known solutions of the Schrodinger equation with a given potential. It still holds for any potential that a particle in a potential well is not "free" to run around and interact, the whole system absorbs photons and raises particles to a higher state.

I am wondering if you have misunderstood and are talking about virtual photons

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  • $\begingroup$ I think I understand now that I have to treat not only the particle but the particle in the potential as a whole. But I'm not sure yet what this means. As in my case the potential is shaped by an optical trap (strongly focused laser beam causing optical forces), the "whole system" is the particle plus the optical trap. Does this imply that the photons forming the optical trap are not to be treated as a measurement of the particle position? Does it also mean that the measurement photons are interacting with the entire trap-particle system and does it change the nature of the perturbation? $\endgroup$ – erik Dec 1 '14 at 12:43
  • $\begingroup$ I am not familiar with this type of potentials. Do you have a link I could look up? I expect the answer to be"yes", since it is a quantum mechanical problem, on the other hand there could exist simplifying approximations I am not aware of. later. $\endgroup$ – anna v Dec 1 '14 at 13:43
  • $\begingroup$ Thanks to your answer I figured some things out myself, especially how the Hamiltonian of the system looks like and that I have to consider the "entire system" as one. I updated my question with some additional information. I also linked to a publication that describes the physical system (although there the oscillator is not yet cooled to a quantum state). $\endgroup$ – erik Dec 6 '14 at 20:31
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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?

Yes, scattering events in general changes the state of the quantum object/particle. If you know the exact initial state of the light/photon, and you measure the scattered state, then you can infer the change in state of the object it scattered off of (and thus update your knowledge of the particle's quantum state). But just because you don't look (or measure), doesn't mean that the interaction (and subsequent state change) didn't happen.

If you do not measure the scattered beam/particle, then you must average over your ignorance of all possible scattering possibilities (weighted by the correct probabilities etc). The result of this process is to create or add (classical) uncertainty or mixing of your quantum state (this process is also known as decoherence).

In your example of light scattering off of a trapped particle there are two primary (related) physical effects. The first is light scattering of each photon happens in a random direction, which gives the particle a momentum kick in a random direction. The result is a form of heating of the center of mass degree of freedom of the particle which is known as recoil heating. A very complete quantum derivation can be found here.

The second physical effect from this scattering is the suppression or decoherence of macroscopic quantum superpositions (or so called Schrödinger cat states). A real nice/readable introduction can be found here (section IVa explicitly addresses decoherence due to scattering).

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The tentative answer to your question is "No" - not all quantum interactions produce a back action. There is an entire field of study under the heading of "Quantum Non Demolition Measurement". A simple example of this is illustrated by the Bomb Detector thought experiment.

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    $\begingroup$ I am aware of QND measurements, which don't work in this case where the position is not a constant of motion. Furthermore, as I understand this, QND measurements are still introducing perturbations on the object. The specialty is just that this perturbation does not influence the measured observable. E.g. a QND measurement of a resonators energy will still perturb its phase, but if carefully designed this perturbation will not change the energy. I probably should be more careful: I mean perturbation when I say back action. So is there any perturbation if I don't actively measure my photons? $\endgroup$ – erik Dec 1 '14 at 12:30

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