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For the position of a particle, there's a probability which, somewhere, there's highest probability. And as you move away from it, the probability reduces. But the particle can be anywhere. It's often found where you would expect, but sometimes it can even be in a lot less likely place.

What I don't understand is how could there be a higher probability prior to measurement when it is measured in a less likely place? What exactly causes there to be a higher probability when particle is not there? This seems bizarre and I can't think of any reason. And it's not like it's statistics. It's a single event. Could this be considered evidence for Many Worlds?

(Note that I'm very familiar with QM in plain language, but I have no knowledge of math beyond simple math.)

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  • $\begingroup$ The probabilities are derived from the state. The state evolves according to the Schrodinger equation. So the probabilities are "caused" by unitary evolution from some prior state --- just as in Newtonian mechanics, the position of a falling anvil is caused by the force of gravity and the anvil's initial position/velocity. If that's not what you mean by "cause", maybe you could clarify by explaining what you'd accept as a "cause" of the anvil's trajectory. $\endgroup$ – WillO Nov 29 '19 at 5:36
  • $\begingroup$ Even in classical mechanics, the most probable events don't always happen (unless their probability is 100%). $\endgroup$ – D. Halsey Nov 29 '19 at 22:51
  • $\begingroup$ But in classical mechanics there is only missing information. With 100% information all probabilities are always either 100% or 0%. Well a deterministic interpretation of QM I guess would also be the same but with QM we can never even hope to have all information. $\endgroup$ – Can'tThinkofaName Nov 30 '19 at 0:06
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Can'tThinkofaName,

Quantum mechanics allows you to calculate the probability of finding a particle at a certain location. For example, in a hydrogen atom, there are preferred locations that you can find graphically represented as atomic or molecular orbitals. See for example here:

https://en.wikipedia.org/wiki/Atomic_orbital

The reason some locations are more probable than others can be found in the type of interactions between the particles (in the case of atoms and molecules it's mostly electrostatic interactions between electrons and nuclei). Still, just like in classical physics it is still possible to find a particle in a less likely location. It's unlikely for a coin to land on its edge, but it can happen. No mystery here.

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  • $\begingroup$ But what determines more probable location? Earlier measurements? Imagine a free particle within a vacuum to make things more simple. Would a particle have equal probability distribution everywhere if no earlier measurement was made? Sorry if this is annoyingly ignorant. $\endgroup$ – Can'tThinkofaName Nov 29 '19 at 19:03
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The state collapses after measurement, and if you measure the precise position, it collapses to a position eigenstate (i.e. a precise location), so you no longer have a "probability somewhere else". The probability somewhere else is prior to the measurement.

If you want to learn the real ideas behind quantum mechanics, not "without the math" whatever that means, you should have a look at my partially-written blog post series here: Quantum Mechanics I - The Winding Number.

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In quantum mechanics, a system is in a superposition of all possible states.

By possible states I mean the possible outcomes of the measurement of a particular physical quantity.

For simplicity consider that the system of a single particle can exist in two states |1> and |2>. If we perform a measurement of a physical quantity, say the position of the particle, the particle in state |1> will be found at x = x1 and the particle in state |2> will be found at x = x2. In any given time the system is in a superposition of these two states. The state of the system is written as

$$ |\psi> \ =\ \alpha|1>\ +\ \beta|2> $$

And we say the system is in state $|\psi>$.

If we measure the position of the particle, the amplitude of getting the outcome x = x1 (corresponding to |1>) is $\alpha$ and the amplitude of getting the outcome x = x2 (corresponding to |2>) is $\beta$. This is same as saying the probability of getting x1 is $|\alpha|^2$ and the probability of getting x2 is $|\beta|^2$. Since the particle can only be found either at x = x1 or at x = x2, we must have

$|\alpha|^2$ + $|\beta|^2$ = 1

Now once you perform any measurement on the system you will either get x = x1 or x = x2. If you get x = x1, then we can say that the system which was in the state |$\psi$> before the measurement has collapsed into the state |1> after the measurement. If you get x = x2, then we can say that the system which was in the state |$\psi$> before the measurement has collapsed into the state |2> after the measurement. No matter how many times you perform the measurement of position you will always get either x = x1 or x = x2.

Now coming to your question if $\alpha$ = $\frac{1}{\sqrt{100}}$ and $\beta$ = $\frac{\sqrt{99}}{\sqrt{100}}$, you would expect to get x = x2 due to higher probability of getting x = x2. But there is still a probability of $|\frac{1}{\sqrt{100}}|^2=\frac{1}{100}$ to get x = x1.

The fact that the probability of getting x = x1 is $\frac{1}{100}$ means that if you perform the measurement of position in 100 independent identical systems you will get x = x1 in just one case. I think this makes it clear why "It's often found where you would expect, but sometimes it can even be in a lot less likely place."

Now to answer the part:

What I don't understand is how could there be a higher probability prior to measurement when it is measured in a less likely place? What exactly causes there to be a higher probability when particle is not there? This seems bizarre and I can't think of any reason. And it's not like it's statistics. It's a single event. Could this be considered evidence for Many Worlds?

The reason "you can't think of" is that this still is just probability. It is just like out of 99 black balls and a white ball, you can be almost certain that you will draw a black ball. But if you keep on repeating the act of drawing and replacing the ball you will get the white ball once in a while.

I hope it helps

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I think WillO's comment on the original question could also have been considered an aswer, but I'll also point out what I think could help, although I'm not sure if I fully understood the point of the question.

First of all, just to make sure you have correctly understood the underlying principles in quantum mechanics, the probability distribution of a particle is not simply not knowing where the particle is. Prior to measurement the particle does not exist in a precise location. If you have not correctly understood (or accepted) this, you might wonder what is the difference between the two cases, and how do we know the location is not definite and we just don't know it? For these matters I forward you to read on Bell inequailities (e.g. from Wikipedia).

The other point kind of relies on the first one. Since the position is not determined prior to the measurement, based on the measurement you don't get any information about the probability distribution of the particle before the measurement. (Okay, this is exaggerated, you get some information, but certainly not enough to actually determine the probability distribution or even its maximum.) To actually get information of the distribution, you need to have many identical cases and then you perform repeated measurements to get an approximation of the distribution.

Now as to what causes the particle to have a higher probability at a location where it is not found at a later measurement, the probability density depends only on the history of the particle, so it contains no information regarding the measurement. Only after the measurement the probability distribution is "updated" to correspond the result of the measurement.

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  • $\begingroup$ "Prior to measurement the particle does not exist in a precise location." - that's wrong. QM says nothing about what exists, only about what can be known. Bell's theorem does not prove that particles don't have a precise location. It proves that either physics is non-local, or distant systems are not independent. "the probability density depends only on the history of the particle" - again wrong, it depends on the interactions (like the electric interactions in an atom). Atomic/molecular orbitals have nothing to do with the history of the electrons/nuclei. $\endgroup$ – Andrei Nov 29 '19 at 9:18
  • $\begingroup$ Sorry, my bad on the second point, I formulated it poorly for a couple of reasons. Of course what really matters is the interactions, and the history as such does not even matter. My point was just that the distribution is not affected by a measurement done in the future. As for the first point, to debate this further I would need a definition for "to exist", but I guess the debate becomes then more philosophical in nature. Still, I'm not saying you're wrong, I was just trying to say things in an understandable manner without being too much wrong. $\endgroup$ – JustSaying Nov 29 '19 at 10:13
  • $\begingroup$ You used "exist" first, so, I think you should clarify your understanding of the term. From my part, I say that QM does not contradict the assumption that the particle is precisely located, at all times, just like in classical physics. Some mainstream interpretations, like consistent histories or "Copenhagen done right" maintain that the measurement actually reveals the particle's property prior to measurement. In Bohm's interpretation (non-local, non-mainstream one) the particle always has a defined position, yet the measurement does not reveal it. $\endgroup$ – Andrei Nov 29 '19 at 10:38
  • $\begingroup$ Well, I guess I stand corrected. My understanding of QM is not on the level that I could actually argue with you, and I don't actually have any explanations why I would be right, so thank you for the corrections! I'll have to keep on studying. $\endgroup$ – JustSaying Nov 29 '19 at 11:00

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