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The uncertainty principle is a fundamental property of quantum systems, and is not a statement about observational success. No particle either free or in crystal can have zero momentum otherwise a nonsensical infinity is required for the standard deviation of position $\Delta x$, in the uncertainty principle $\Delta x \Delta p \geq \hbar / 2$. $0 \cdot ... 2 In Bohr's theory the smallest possible orbital angular momentum is$\hbar$. The measured value is$0$. On the other hand the picture developed by solving the (time independent) Schödinger equation reproduces the energy levels from Bohr's model and gets the minimum angular momentum and the angular momentum step size right (it also fives you the quantization ... 2 I don't think you really understood the outcome of the discussion Is uncertainty principle a technical difficulty in measurement?. This has nothing to do with our ability to perform a precise measurement. The position and momentum of the electron simply does not exist simultaneously in a definite state. It is like trying to measure the exact day that winter ... 2 The reason Bohr's theory is considered surpassed is that Heisenberg and Schroedinger developed more powerful theories, in which Bohr orbits do not play major role. Bohr's theory works nice only for few-electron systems, like hydrogen atom or ion Li$^{2+}$. For more complicated systems like the molecule of water H$_2$O it is difficult to see how to generalize ... 2 So, is there any reason for overruling the idea of fixed orbit? or is there any thing wrong in my opinion about the concept, if so please explain, so that I would not proceed with that wrong thinking? An insurmountable problem with the Bohr atom is that one has two charged particles orbiting around each other. Electromagnetism was an exact science at ... 6 The electron is stuck to the atom, which isn't going anywhere. This means that$\langle p \rangle = 0$. The uncertainty$\Delta p$measures the RMS fluctuation of the momentum: $$\Delta p^2 = \langle p^2 - \langle p \rangle^2 \rangle = \langle p^2 \rangle$$ Since$E = p^2 / 2m$, this means that $$\langle E \rangle = \frac{\Delta p^2}{2m}$$ The ... 0 Let's start off by removing the restriction of computational resources such that we're not limited by computing power and by finite precisions. Let's also use the word exact to mean absolute certainty (ie. probability is precisely 1) about a quantity. Take a real group of particles at an initial state. We may or may not be able to derive a set of governing ... 1 As Emilio pointed out, the uncertainty principle is not a limiting factor. However, as for simulating or calculating future states, this is not really generally possible for classical systems, because of chaotic behaviour. 7 The Uncertainty Principle will never, as far as we know, prevent you from simulating any physical system. The reason for this is that quantum mechanics is - except for that little problem with measurements - completely deterministic. To be more precise, say you want to simulate a given system within quantum mechanics. You begin by describing your ... 0 In the ground state, the uncertainty will always be a minimum (we know exactly which state the particle is in). So you should expect that in the ground state of any single particle quantum system,$\Delta x \Delta p \approx \frac{\hbar}{2}$. As far as linear growth, that is more a thermodynamics/statistical mechanics question. Remember that not all ... 1 It assumes a sort of democracy between$x$and$p$, and is obviously not valid everywhere. The connection between classical and quantum mechanics happens through coherent states, which are states which minimize the product$\Delta p \Delta x$, and in a sense, behave most classically (and have well-defined classical limits as$\hbar \to 0$). So if you ... 1 No, you can't do that. You can write$\lvert x\rvert = \sqrt{x^2}$, and you can then go to$\langle\lvert x\rvert\rangle = \langle\sqrt{x^2}\rangle$, but it's not valid to say$\langle\sqrt{x^2}\rangle = \sqrt{\langle x^2\rangle}$. You can pick almost any wavefunction,$\psi(x) = \pi^{-1/4}e^{-x^2/2}$for instance, and show that the two are not equal. In ... 2 My reading of the electron photon experiment is that intrinsic uncertainty enters the problem by limiting the resolving power of the photon. In other words, the electron is along for the ride, and perhaps historically it was chosen because it is such a simple system. But the recoil of the electron seems to confuse the issue. Instead of a free electron, we ... 2 Yes, the experiment is oversimplified, because the uncertainty principle is not about "disturbance through measurement". Although that's what Heisenberg said (one of the things he said), it turned out you can't interpret it that way in a very rigorous sense. Whether there is something like "disturbance through measurement" that gives rise to an uncertainty ... 4 The uncertainty principle is a mathematical consequence of wave behaviour. It is true for sound waves, electrical signals, radio waves, etc. Anywhere you might want to work with Fourier transforms. Let's say you want to send a pulse via a radio wave. Furthermore, lets say you make the amplitude of the pulse a Gaussian function with time:$f(t) \sim ...

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The answer depends on the interpretation of quantum mechanics you like the most. As they are mathematically equivalent, the question is metaphysical rather than scientific. Some of the ways to explain: There is no uncertainty in the system. The uncertainty is in the observer. By making the measurement the observer introduces the uncertainty because he ...

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This reply probably would be more suitable as comment, but i'm unable to comment yet, so: First of all we can't really describe universe as a system that even close to entropy which actually would be a lowest energy and most stable state for universe. So, following from that: universe is unstable, and as it's unstable there is no really a problem for ...

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Usually when someone asks about why a system is the way it is we answer it by saying that it is the most stable lowest energy state. That's a lousy explanation. A system doesn't have to be in the lowest energy state (in that case the Universe would be very boring). In fact, some systems don't even have a definite energy. To visualize what I've ...

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Physics isn't entitled in answering the question "why" but rather "how". The question "why" is always diverted to philosophy, and some liars use it to make religions.

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In one dimension, motion in a Coulomb-like attracting potential $U=-\frac{\alpha}{|x|}$ is quite ill-defined. I'll now try to explain this for classical particles, then will say some words about quantum mode. Let's consider two cases: 3D motion of a particle nearly free falling into the attracting center, but with nonzero angular momentum $L$. In this ...

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