Does measurement of quantum system always collapses the state of system? 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?
 A: 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 $|1\rangle$).
A: 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."
