In QM does the observer really perturb the system to make a measurement?

Question

It is commonly taught in introductory QM courses that in order to get to know the position or momentum of a particle, be it by "sending a photon" or similar experiments, the measurement necessarily disturbs the system. What I do not like about these thoughts experiments is that they all assume that the system has some position/momentum value unknown due to some uncertainty (denoted by $\Delta x$ and $\Delta p$). They never consider the case of the system being in a superposition of eigenstates.

Anyway, let's suppose that it's true that in QM a measurement necessarily disturbs the system. I.e. $\Psi$ is necessarily modified between the moment before the measurement and the moment after the measurement. Let's say we make a measurement of a property whose corresponding operator $\hat A$ commutes with the Hamiltonian of the system $\hat H$. Let's say that we're lucky enough to get to know that $\Psi$ "collapsed" into a single non degenerate eigenstate of the operator $\hat A$. We can safely say that before the measurement $\Psi$ was in a superposition of eigenstates of $\hat A$ and that after the measurement it was equal to some eigenstate of $\hat A$, we have perturbed the system. But now if we make another measurement, we know for sure that the result will be exactly the same, because $\hat A$ commutes with $\hat H$. Does this mean that now making any new measurement does not perturb the system anymore? It doesn't make any sense!

I've read from Lubos Motl, John Reenie, Sidney Coleman, David Mermin, London, Zurek, etc. that $\Psi$ is subjective, i.e. it is a representation of the information one knows about the system. Two observers need not to agree on $\Psi$ if they do not have the same information on $\Psi$. It does make sense to me, but then there is something that I do not understand with the description in the previous paragraph. Let's imagine that the description above is from an observer X. An observer Y had made a one measurement before X, that X was unaware of. So Y had perturbed the system and according to the description of Y, $\Psi$ "collapsed" when he made his first measurement. It also means he disturbed the system at that first measurement, not on the ones from X. But from X's viewpoint, it is himself who collapsed his own $\Psi$, and thus perturbed the system in his first measurement. This contradicts Y viewpoint.

While I can buy the fact that $\Psi$ depends on the observer, I find harder to buy the fact of "disturbing the system" is also subjective.

Hmm, after all if "perturbing the system" means that $\Psi$ is changed between the moment before a measurement is performed and the moment after the measurement is performed, and if $\Psi$ is subjective, then it is no wonder that "perturbing the system" is an arbitrary statement. That is quite strange and in sharp contrast with classical mechanics. It would mean that "disturbing the system" is also subjective. I now think that this is the case, i.e. that perturbing the system is not a universal feature, but it is entirely related to the knowledge of each observer and need not a universal agreement on. It is entirely subjective.

I would love some comments, confirmation or infirmation of what I wrote. And if Lubos Motl could write some comment I would be immensely glad.

Answer 1

Perturbing the system and the collapse of the wave function are not necessarily the same thing, that is where part of your confusion lies.

Perturbing the system means that an additional interaction is introduced in the Hamiltonian: given the free Hamiltonian $H_0$, a measurement process is described by $H=H_0+H'(t)$, with $H'(t)$ vanishing before and after the interaction has occurred. From an experimental perspective this is always the case as measurements involve for the most measuring energies or momenta (or any other physical quantity for that matter) after some scatterings or the like. Usually an initial state $|\psi\rangle_i$ is given as linear combination of the eigenstates of the initial Hamiltonian and you are asked to determine what is the final state $|\psi\rangle_f$ after the interaction has occurred: notice that the final state is a linear combination of the eigenstates of the initial Hamiltonian as well, since once the interaction has occurred the additional term $H'(t)$ vanishes. In order to solve this problem there is a whole bunch of theory about scattering where states are expanded in terms of initial plus interacting eigenstates and people take limits to plus or minus infinity.

Now let us go back to the collapse of the wave function: from a theoretical point of view, given a quantum system described by a collection of self-adjoint operators, the state of the quantum system by definition collapses into any of the eigenstates of one such operator upon performing a measurement thereof. In your language you want to measure $A$ and the state collapses into $|a\rangle$ after the first observation. Provided $A$ commutes with the Hamiltonian and that the Hamiltonian itself only acts as a phase factor onto its eigenstates (which is not necessarily always true) the system remains in such eigenstate unless another interaction occurs. You want then to measure $A$ again: since (as we discussed above) performing a measurement corresponds to adding an additional term to the Hamiltonian, you have to make sure that $A$ still commutes with the full Hamiltonian again (not only with the free part). If all these conditions hold then the system may as well remain in the same eigenstate - but this does not mean that the system was not perturbed: it was indeed perturbed but everything commuted with everything else and therefore the same eigenstates are eigenstates of the initial and perturbed Hamiltonian too.