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3

Are you asking why taking the squared modulus of a superposition of (eigen)states turns out to be considerably more complicated than the squared modulus of a single eigenstates? It this is so, I'd say because eigenstates of different energies evolve differently, and when you do the superposition and consider the square modulus you have to take into account ...


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Edit after seem Acuriousmind and glance's contributions.... they have the answer sorted out above. Interesting experimental example of this is Zewails work - for example this paper, which is not behind a 'pay wall' where evolution on femtosecond timescale of a molecular vibrational 'wavepacket' make up of 'superposition of many states was observed'.


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Why "probability zero doesn't imply that an event is impossible"? Low probability implies that an event is possible, but improbable. But probability ZERO means that the event is IMPOSSIBLE, not IMPROBABLE. The wave-function doesn't cheat. There where the wave-function is zero, the presence of the particle is FORBIDDEN. Good luck !


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This exercise is a mathematical construct and infinite potential really means the probability of finding the particle outside is zero. It will never ever be outside the well in this scenario. In a finite case, the probability of finding the particle outside of the well goes down exponentially, thus its never really zero, only very small. This is what you ...


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When dealing with the infinite square well, we must be clear that it is a limit of the finite square well case. But even though for the finite case we have as Hilbert space $L^2(\Bbb{R})$, that is, the particle can have non zero probability of being found in any region of non-zero measure, for the infinite case, the limit forces the condition of working with ...


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It seems to me your question is on the equivalence of possibility with probability, and not the infinite quantum well. Maybe I am being sloppy, but I would use them interchangeably in the following context: If probability of something is absolute zero, it means no matter how many times you try the experiment (looking for a particle outside the well), you ...


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I found it very difficult to understand your question, but here is an attempt at answering it: The whole point of embedding is to uncover the phase-space structure or, with other words, something that is topologically equivalent to the attractor. For any interesting system, this is not describable by your usual distributions (and also it is usually not ...


1

We do basic calculus and much of physics using $\mathbb{R}$. This implies that there are uncountably infinite many "points" in even the smallest region of space or smallest interval in the space where are considering. There is little reason to be certain this is true when talking about the actual universe. If position is granular on any scale -- say, the ...


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Like ACuriousMind said, the crucial thing is that you always need to integrate over some interval to get an actual probability. In that light, the probability is neither zero for $P_{x_0}$, a particle in the interval $[x_0 \pm \Delta x]$ (for $x_0 \neq L/2$), nor for $P_{L/2}$, a particle in the interval $[L/2 \pm \Delta x]$. The crucial difference ...


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If you look closely, you will find that the event "find the particle at $x$" always has zero probability of occuring: Since $\rho(x) = \lvert \psi(x) \rvert ^2$ is a probability density, it must be integrated over some subset of $\mathbb{R}$ to actually gain a probability. The probability to find the particles at one of the positions in a subset $S \subset ...


1

So let's remember that $X$ is an operator, and $\langle X \rangle$ is just a number, and we can use the definition of the expectation value $\langle O \rangle = \langle \psi | O | \psi \rangle$ to work this out. \begin{eqnarray}\Delta X^2 =& \langle X^2 - 2X\langle X \rangle + \langle X \rangle ^2 \rangle \\ =& \langle X^2 \rangle -\langle ...


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There is a very big "expectation value" bracket all around the expression, and from the 2nd to the 3rd line, you have to use its properties. So, let's write it the long way: $$ \langle (X^2 - 2X\langle X\rangle + \langle X\rangle^2) \rangle = \langle X^2\rangle - \langle 2X\langle X \rangle \rangle + \langle X \rangle^2 $$ (now use that the expectation ...


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It seems to me you are conflating micro and macrostates. Entropy is not a property defined for a specific microstate but for an ensemble. When we describe a system in terms of pressure, entropy, etc instead of the momentum and position of each particle we are giving up the possibility of discussing specific microstates. The 'states of high order' you are ...


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A very simple example of a configuration would be 0 heads and 100 tails. I. E. Coin 1 is tails, coin 2 is tails.... Coin N is tails. A configuration is the orientation (heads or tails) of each coin in the system and specifies a microstate. The total number of each state specifies a macrostate.


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I believe in this scenario, "configuration" would be synonymous with microstate. In statistical mechanics/thermodynamics, many of these terms are often synonymous in certain circumstances while still having specific meaning. "Configuration" does not have the exact meaning that microstate does, but in this case is used to mean microstate as you are more ...


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The answer would appear to be "yes", assuming every possible state exists at least once. Tegmark did a calculation of the approximate distance between "You" and the next identical instance of "You" in such a universe as 10^10^29 meters away


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I could probably go along with this assertion; except that the observable universe is not infinite. We can only see light reaching us from about 13 billion light years away. The stars/galaxies etc that formed in those distant regions are now even further away from us, but nevertheless, the observable universe is finite. Expansion of the Universe, will light ...


2

There's different levels on which you can answer this question: First, because the coin is a large enough macroscopic system that quantum-mechanical effects are negligible (i.e. we don't have to worry about the wave nature of the coin). The variables that will affect the final outcome of the coin toss are all classical in nature, and so can be measured and ...


2

At the very least you would also need the dimensions of the coin (mass alone is not enough) and the elastic properties of both the coin and of the surface on which you land... that has a big impact (since the coin can bounce or "stick", depending on the surface). It may be possible to predict for certain sizes of coin and height of drop (a carefully executed ...


1

To do it completely without error would be impossible. There are simply too many parameters to be determined to exact precision in order for there to be no error. One of the answers in John's comment links to this research that provides further details.



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