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In quantum mechanics, in general, it is stated that the act of measurement changes the state of the system. For example, consider the following Stern-Gerlach setup;

A beam of silver atoms first passes through SG apparatus in $z$ direction. The ones come out from the +z direction are led to another SG apparatus that is in the $x$ direction. The ones come out from the +x direction are then led to yet another SG apparatus in the $z$ direction.

Now, if we were to perform this experiment, we would see that the silver atoms having spin in the +z direction, initially, can have a spin in the $-z$ direction at last apparatus.Therefore, as one of my professor put, the act of measuring the $x$ component of the spin erased the information about the component of the spin in the $z$ direction.

However, in the above experiment, we made a selection at the end of each apparatus; namely, we only allowed atom having spin $+z/+x$ to pass through the next stage, so consider the following;

enter image description here

In other words, in this case, we allow the atoms coming out from $-x$ direction pass to the next stage.

In this case, the act of measuring the $x$ component of the spin will not "erase" the information about the $z$ component (this experiment is explained in the QM lectures in ocw.mit.edu), so isn't the statement "the act of measurement changes the state of the system" wrong in general ?

Moreover, if we think about it, almost all of the "problematic" (weird) things in QM comes from the fact that by making a measurement, we cannot determine the state of a system priori to the measurement because by making a measurement, we are disturbing the system, hence we cannot talk about the state "priori to the measurement" by doing a measurement, but the latter experiment says "well you can actually in some cases", so I see a contradiction in here.

Edit:

Note that, in here, I'm consciously sending both output of $SG_x$ to the same $SG_z$ apparatus.

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In this case, there is no measurement of spin in the $x$-direction happening. You're just feeding the outcome of the $x$-aligned SG apparatus into the second $z$-aligned SG apparatus. But a measurement needs a "pointer" - something that indicates the result of measurement, otherwise it is not a measurement.

But there's nothing here that would force us to treat the outcome of the $x$-aligned SG apparatus as anything but an equal superposition of $\lvert -x\rangle$ and $\lvert x\rangle$, i.e. it's still just a $z$-spin eigenstate. Unless you introduce something in one of the two paths for the $x$-spins that distinguishes them, you're not forcing this superposition to resolve to a state with definite $x$-spin, and therefore you're not measuring the $x$-spin.

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  • $\begingroup$ No, we are doing a measurement with $SG_x$, at the end of the apparatus; it doesn't need to select some of them, that is not a requirement for it to be a measurement. After all, I can put a blocker to the $-x$ path way after the $SG_x$ apparatus, independent of what $SG_x$ is doing. $\endgroup$ – onurcanbektas Jan 20 at 9:57
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    $\begingroup$ @onurcanbektas Yes, and once you do that, you're doing a measurement because now the $\lvert -x\rangle$ state behaves differently than the $\lvert +x\rangle$ state. A measurement is not simply an interaction, it is an interaction of a special kind with a measurement apparatus that has pointer states allowing you to determine the outcome of the measurement. See this question and its answers for my and other people's take on "measurement vs. interaction". $\endgroup$ – ACuriousMind Jan 20 at 10:03
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Moreover, if we think about it, almost all of the "problematic" (weird) things in QM comes from the fact that by making a measurement, we cannot determine the state of a system priori to the measurement because by making a measurement, we are disturbing the system, hence we cannot talk about the state "priori to the measurement" by doing a measurement

Actually, there is what is called preparing a system and this means we fix the system to be in a definite state. This is done by measuring the system since this means that after the measurement it will be in the eigenstate associated with that measurement.

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  • $\begingroup$ Thanks for pointing out @Ullam, but this should be a comment, not an answer, IMO. $\endgroup$ – onurcanbektas Jan 20 at 10:55

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