Does measurement in quantum mechanics always disturb the system? The measurement postulate states that "when we do a measurement on a quantum system, the state of the system is collapsed to one of the eigenstates of the system and successive measurements of the same observable yields the same result". Is this always true? Or is there any type of measurement which doesn't affect the state of the system but yield the desired result?
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$\begingroup$ It's not possible for a measurement not to collapse the wavefunction. If your measuring instruments measure a specific number, you can no longer have a probability distribution for values at the time of measurement, but rather a sharp peak at the value you just measured $\endgroup$– John KCommented Sep 23, 2019 at 19:13
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$\begingroup$ You can do a "weak" measurement, where you only acquire a little bit of information about the system - this will correspondingly lead to a smaller perturbation. $\endgroup$– Norbert SchuchCommented Sep 25, 2019 at 18:23
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$\begingroup$ @NorbertSchuch but what kind of “weak information” will I get? $\endgroup$– FearlessVirgoCommented Sep 27, 2019 at 4:40
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$\begingroup$ Something like "it is a little bit more likely that the spin is up than down". $\endgroup$– Norbert SchuchCommented Sep 27, 2019 at 11:38
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2 Answers
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Yes, except for the specific case in which the state is in an eigenstate of the measurement operator.
Measurements with deterministic outcomes
More specifically, for any quantum (pure) state $|\psi\rangle$, there is just one class of measurements that can be performed without changing the state, and these are the measurements which ask questions of the form "is the state $|\psi\rangle$ or something else?". More formally, this means that if the measurement basis is of the form $\{|\psi\rangle,|\psi^\perp_1\rangle,...,|\psi^\perp_n\rangle\}$ for any orthonormal set of vectors $\{|\psi^\perp_i\rangle\}_i$ orthogonal to $|\psi\rangle$, then measuring the state in this basis will give a deterministic outcome (corresponding to the answer "yes, the state $|\psi\rangle$ is indeed $|\psi\rangle$"), and no collapse will be occurring.
It should be noted that even the above comes with caveats. In most realistic scenarios, any measurement will result in the destruction of the state, and "measurements" like the one described above will have to be understood in the sense of post-selection. By this, I mean scenarios such as the following one: suppose you have a photon with some polarisation, and you send it through a polariser beamsplitter (which sends the photon in one direction or the other conditionally on its polarisation state). If you only look and operate on the photon on one of the two PBS's output modes, and then measure its state afterwards, then the result of the computation will be identical to what you would have had if you had somehow measured the polarisation at the PBS without absorbing the photon, even though you actually didn't.
The more information you gain, the more the state is changed
This caveat aside, one can say even more about how measurements collapse quantum states. The generalisation of the fact that measurements with deterministic outcomes to not disturb the state, is the fact that, roughly speaking, the greater the amount of information learnt from a measurement, the more the state of the system is changed in the process. And note that this is totally independent of how exactly the measurements are performed: it is a fundamental aspect of how quantum mechanics works.
The extreme case was already discussed above: if the outcome is deterministic (which corresponds to zero knowledge learnt from the distribution, in the sense that the Shannon entropy of the output probability distribution is zero), then the state is unperturbed. On the other hand, a measurement which results in the maximum gain of information is one corresponding to output probabilities $p_0=p_1=1/2$ (which has maximal Shannon entropy). Such a measurement will always result in the maximal disruption of the system's state (which will change from $\sim|0\rangle+|1\rangle$ to either $|0\rangle$ or $|1\rangle$).
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The idea of the collapse of the state is not a fundamental part of quantum mechanics. It's a feature of the Copenhagen interpretation (CI). The CI is not the only way to think about quantum mechanics.
Even within CI, it's not necessarily true that measurement must disturb the system, depending on what you mean by "measurement" and "disturb." In the Stern-Gerlach experiment, you can split a beam of electrons into spin-up and spin-down beams, and then you can recombine them. After recombining, the beam is the same as before. There has been no disturbance. Whether you consider this to be a "measurement" will depend on your definition of that word. Note that if you look at presentations of CI, they generally never define "measurement."
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$\begingroup$ A corollary of your first paragraph is that "collapse" is not a physical change in the system, so you cannot meaningfully talk about anything "causing" it. $\endgroup$ Commented Sep 24, 2019 at 1:55
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$\begingroup$ I strongly disagree (especially with the second paragraph). When observing a quantum mechanical system, you cannot gather any information about it without leading to some kind of disturbance. $\endgroup$ Commented Sep 25, 2019 at 18:23
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1$\begingroup$ I get what you are trying to say, but I think that this wording is misleading. When one talks of "measurement" in QM one specifically refers to acquiring classical information from a quantum system, which cannot be done without collapsing the system (unless the state is in an eigenstate of measurement operator). CI and other interpretations concern how you choose to interpret/understand the transition between quantum and classical information, but they do not change how it happens $\endgroup$– glSCommented Sep 28, 2019 at 19:32
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1$\begingroup$ If you split a beam of electrons into spin-up and spin-down, and then recombine them to obtain their original state, you don't actually get any information about their spins, so I don't see how this could qualify as an actual measurement. $\endgroup$ Commented Sep 28, 2019 at 21:14