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It seems that they are two different types of collapses within quantum mechanics. But are they somehow related, or is it two completely different collapses? Sorry if the answer is obvious, and I have not done enough research, but i cannot find a place on the internet which this is stated.

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They are the same. The wavefunction, before collapse, is considered to be in a superposition of states. Think about it like a few different instruments all playing different notes; the noise you hear is the superposition, and the sound waves are the wavefunction. Upon collapse, all the instruments except for one stop playing. The wavefunction is no longer in a superposition, and you can measure which note is being played.

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  • $\begingroup$ If i have understod things right, the propabillity distribution for a particle is caused by the uncertainty principle. In addition, the propability distribution acts like a wave. And the wave function addresses the probability of finding the particle within that wave probability $\endgroup$ – user134652 May 17 '17 at 16:15
  • $\begingroup$ ..Within that wave probability distribution. Does this mean that the uncertainty principle superposes the propability of that particle into a superposition? And that superposition is the wave distribution for that prticle, which can be described by the wave function? $\endgroup$ – user134652 May 17 '17 at 16:19
  • $\begingroup$ The uncertainty principle is actually unrelated to the probability distribution. The probability distribution is directly related to the amplitude of the wavefunction, though, so you have that right. The uncertainty principle comes from other wave mechanics, though. Think about it like this: the position of a wave is hard to define. If you have a perfect sine wave, it is impossible to say "where" it is located; it spans across all space. However, the momentum of a wave is directly related to the wavelength. The wavelength of a sine wave is easily found. If you narrow down the position,... $\endgroup$ – Pawr May 17 '17 at 16:20
  • $\begingroup$ But in a book I have read about physics. (By Stephen Hawking), it is stated that the uncertainty principle distributes the position, which smears it out like a wave. Is this wrong? $\endgroup$ – user134652 May 17 '17 at 16:23
  • $\begingroup$ ...by making the wave a single spike in space (like a delta function, if you are familiar), then you lose information about the wavelength. What is the wavelength of a single peak, when the wavelength is defined as "peak-to-peak distance"? By increasing knowledge of position, we decrease knowledge of momentum, and vice-versa. $\endgroup$ – Pawr May 17 '17 at 16:23
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In standard quantum mechanics wavefunction is said to collapse exactly because we don't observe superposition of states as a result of our measurement. For example, if we have the state $$|\Psi\rangle = \frac{1}{\sqrt{2}}|\uparrow\rangle+\frac{1}{\sqrt{2}}|\downarrow\rangle$$ where $|\uparrow\rangle, |\downarrow\rangle$ are eigenstates of pauli spin operator $\sigma_z$ then, the measurement of $\sigma_z$ will result in $|\uparrow\rangle$ with a probability $\frac{1}{2}$ and $|\downarrow\rangle$ with a probability $\frac{1}{2}$ but not both.

Hence, measurement "collapases" the superposition/wavefunction to either $|\uparrow\rangle$ or $|\downarrow\rangle$.

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