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In quantum mechanics, you usually hear the phrase "the act of observing the state modifies it", but they never really define what they mean by "observing".

For example, suppose we have Stern-Gerlach analyzers aligned in the $z$-axis and we shoot atomic beams into the analyzer thus we don't know if the state is spin up or down in the $z$-axis where both have a 50 percent probability. We won't know the state until we "observe" it. By observing, do they mean that we have a measuring device and the act of measuring the state disturbs the system? Does the act of measuring it use photons which interacts with the measurement? Or is it just the act of looking? What exactly is "disturbing the system"? Is the disturbance the actual photons in the measuring device?

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'Observe' means 'making a measurement' and hence interacting with the system. The result of this interaction is the wavefunction collapse, a type of time-evolution that is not accounted for within the current quantum mechanical formalism (which is, otherwise, deterministic in that the initial state evolves according to the Schrödinger equation).

Detection of particles is a specific kind of interaction, one that would often involve photons. But you don't necessarily need to do that.

Now to the Stern-Gerlach experiment.

Before the SG magnet, your state is in a superposition state $$|\psi\rangle = \alpha |\uparrow\rangle + \beta|\downarrow\rangle,$$ but you know that the SG apparatus makes a measurement and divides the incoming stream into two sets, each with a specific spin orientation.

But if you don't make another measurement to know which stream is up and which is down, you need to encode this uncertainty by writing your state as a mixed state in the density matrix formalism: $$ \rho = \alpha_{\uparrow}|\Psi_\uparrow\rangle\langle\Psi_\uparrow| + \alpha_{\downarrow}|\Psi_\downarrow\rangle\langle\Psi_\downarrow|, $$ where the $\alpha$'s are probabilities so $|\alpha_{\uparrow}|^2 + |\alpha_{\downarrow}|^2 = 1$.

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  • $\begingroup$ Thank you this is very helpful and clarifies my misconception. $\endgroup$ – Robben Jul 25 at 5:04
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We observe objects if a photon is emitted on object.In quantum mechanics,observation of quantum particle is done by shooting a photon at it. But for e.g. when a photon is emitted at an electron (which behaves both wave and particle before emitting), electrons behaves like a particle.It disturbs the system. The electron doesn't behave like a wave anymore. Note:(1)Observe means taking a measurement 2)I am not a quantum mechanics student. But this things are in my syllabus of J.E.E. in 11 th grade.So do not expect some complex answers.)

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  • $\begingroup$ Electron does show wave nature like diffraction. Is it not true? $\endgroup$ – user268766 Jul 25 at 5:15
  • $\begingroup$ It does show both wave properties and particle properties .It is called dual nature of matter. Not just quantum particles ,all matter do show both wave and particle properties. But the wavelength generated is inversely proportional to its Momentum (mass and velocity). And macroscopic objects have insignificant wavelengths. The above answer(by Superciocia) is perfect in nature ,if you are a beginner in quantum mechanics and are learning it.I assume you are just curious about quantum mechanics and are reading a popular book.Sorry if I assumed wrong. $\endgroup$ – user95251 Jul 26 at 4:53

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