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There are two ways to interpret the boundary conditions you are imposing. The first case is that of a system which is infinite in extent, but has a periodic regularity. This is like an electron in an idealised 1D crystal, where the periodic boundary condition is imposed by the presence of nuclei regularly spaced. In this case, the plane wave solution has ...

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This is what happens if one cares not for the subtlety that quantum mechanical operators are typically only defined on subspaces of the full Hilbert space. Let's set $a=1$ for convenience. The operator $p =-\mathrm{i}\hbar\partial_x$ acting on wavefunctions with periodic boundary conditions defined on $D(p) = \{\psi\in L^2([0,1])\mid \psi(0)=\psi(1)\land ... 8 Notice that$\psi(x)$is defined on a circle of circumference$a$. Multiplying$x$on this circle is really multiplying a periodic extension of$x$, i.e., the sawtooth function$x - a\lfloor x/a\rfloor$, where$\lfloor y\rfloor$means the largest integer not greater than$y$. So, the commutator of the position and momentum operators involves the derivative ... 4 Your question or confusion is mostly based on several misconceptions of the premises: Chaos theory is not a theory in the scientific sense like, e.g., the theories of relativity, evolution or quantum mechanics. It does not make predictions about the laws of nature. You do cannot make statements about reality like: “According to chaos theory, …”, or: “This ... 5 It is part of the postulates of quantum mechanics that the expectation value of the observable corresponding to the hermitian operator$A$in the normalized state$|\psi\rangle$is given by$\langle A\rangle_\psi =\langle\psi|A|\psi\rangle$. Alternatively, you can postulate that the expectation value is given by$\langle A \rangle_\psi = ...

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You say: Let us just focus on a single electron. I know the exact velocity (magnitude and direction) to which this electron is accelerated to (energy conservation). but this isn't true. You know the electron energy has increased by $E$ eV, where $E$ is the potential difference you're using but you don't know what its energy was initially i.e. when it ...

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The uncertainity principle does indicate indeterminism in quantum mechanics, in the sense that uncertainity in position measurement necessitates a corresponding uncertainity in momentum measurement. There is another aspect of indeterminacy in quantum mechanics in the context of measurements, which states that the exact outcome cannot be predicted. On the ...

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In particle physics we continually go to the center of mass system for the interactions under study, because the mathematical expressions are simpler and Lorenz invariance assures that the results will be the same in whatever system one studies the interaction. Suppose we have an electron in a bubble chamber. This is a pion decaying into a muon and an ...

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Your question has several flaws. First, you say the electron is at rest at the origin. As John Rennie noted, this implies that the position and momentum are both sharp, which contradicts the uncertainty principle. There is no such thing as an electron at rest at a particular point. An electron is described by a wave function spread over an extended region ...

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If you say: According to me the electron is at rest. that means you have measured the electron momentum to be zero, in which case the electron position is completely uncertain. So you can't be sitting on the electron. If you say: Let us say I sit on an electron. that means you have measured its position precisely so you have no idea what its ...

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An intuitive explanation would require the situation to be translated to a non-quantum scale, away from the subatomic scale and into something that most people would understand. Imagine a child is holding a balloon on a windy day. Suddenly, the wind pulls the balloon from their hand. The balloon is moved around unpredictably due to the wind blowing on it. ...

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Depending on your level, the wave packet might be intuitive enough. But let me add a metaphysical justification that's intuitive on a very satisfying level: The uncertainty principle exists because the universe has a smallest scale. If you zoom in on a digital image you get a grid. Reality doesn't have a regular grid in the same sense, but you can think ...

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1) It is a product of operators. And they are not so much decomposed into real and imaginary parts, but rather into self-adjoint and antiself-adjoint parts. If we take self-adjoint $A,B$ linear operators on some suitable Hilbert space, it is clear that $$AB=\frac{1}{2}(AB+BA)+\frac{1}{2}(AB-BA),$$ since  ...

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The proof here shows that Hermitian operators $A,\,B$ satisfy $\sigma_A\sigma_B\geq\frac{1}{2}\left|\left\langle\left[A,\,B\right]\right\rangle\right|$. (It's a consequence of the Cauchy-Schwarz inequality on the Hilbert space.) Since $\left[x,\,p\right]=\text{i}\hbar$, $\sigma_x\sigma_p\geq\frac{\hbar}{2}$.

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The uncertainty principle states that product of the standard deviations of two observables is bounded below by a multiple of the absolute value of the expectation value of the commutator of the two operators. In detail, if $\Delta A=\sqrt{\langle \Psi|\left(A^2-\langle \Psi|A|\Psi\rangle^2\right)|\Psi\rangle}$ and $\Delta B=\sqrt{\langle ... 0 The velocity$\hat v$is a well-defined operator for particles, equal to$\hat p/m$in non-relativistic physics. The non-relativistic, non-linear correction factors could be added but that would create a can of worms because the correct way to describe relativistic particles requires quantum field theory where$\hat x,\hat v,\hat p$aren't quite well-defined ... 0 They are consistent with each other because quantum mechanical momentum is not the change in position. There is no quantum notion of velocity. Classically, velocity is the time derivative of$x(t)$along a particular trajectory. The quantum theory has no notion of a real-valued trajectory$x(t)$. Quantum states, in general, are not position eigenstates, ... 0 The Heisenberg uncertainty principle establishes a lower bound on the uncertainty (hence the$\lt$sign). It follows that if you have additional error (due to your measurement instrument) it can only increase the uncertainty over this quantum mechanical limit. 1 Imagine that the information that describes position and momentum is digital and of limited precision. There is a constant total precision for both of them, but you can slice it up differently. If you dedicate more bits to momentum, you get fewer bits for position and vice versa. 1 Non-scientific joking answer may be like that: The Product Release Uncertainty Principle says that you may know what your product will do or when it will be released - but not both things together. Quick explanation: the company will never have enough "resources" to do the full testing in a certain amount of time. In one situation you can fix the release ... 2 Yes, indeed any measurement where you learn something will disturb the state. This is also known as "no information without disturbance". Recently, there has been a lot of flurry of activity trying to pin this down quantitatively. As you said, this is not what the Heisenberg relation you write down is about, but the effect still certainly exists. Here is a ... 4 Actually while measuring, the Probability Distribution function of a particle also changes, Does this means that the measuring instrument has some effect ? The measuring process may change the boundary conditions of the solutions of the quantum mechanical equations of the system under measurement, so the complex conjugate square of the wave function ( ... 0 These are standard deviations of a probability distribution indeed. The probability distribution is that of getting a particular value while the system is in the state prior to measurement. So we imagine measuring a state - it collapses to some value, and then somehow resetting time back before measurement and measuring again. If we repeat this process, ... 2 maybe not the kind of answer you are looking for, but from a theological perspective it is necessary so that electrons don't collapse into protons thus destroying the universe. this is what would happen without it http://www.feynmanlectures.caltech.edu/II_01.html#Ch1-S1 2 The wikipedia article for Particle in a Box neatly explains how it obeys the uncertainty principle. Basically, a smaller box gives the particle a wider distribution of momentum, or more uncertainty in momentum. 4 I find it to be difficult to explain the Heisenberg uncertainty principle intuitively in one step. I find it helpful to split it into two halves. The first half explains why uncertainty like behaviors appear in wave mechanics. The second half argues why you have to consider wave mechanics when dealing with small particles. For wave mechanics, I like to ... 8 Here's a maths-free explanation which, i think, can be understood intuitively. A lot of people think (incorrectly) that they understand Heisenberg's Uncertainty Principle: they think that it's a measurement problem: ie, that we can't measure the properties without interacting with it, and this interaction changes the property, so we don't actually know what ... 1 Another way to explain it to lay persons is to first consider why we have effective laws of physics valid at the macroscopic scale in the first place. So, stripped of all its details, one should consider that there are mathematical laws that apply to some small microscopic scale. But this seems to preclude the possibility of there being simple mathematical ... 2 There has been some disagreement above about the appropriate way to explain the HUP. I think that the more abstract explanation is the correct way to explain it, and that it can be illustrated with examples to make the abstraction clearer. The classical way of thinking about the world runs something like this. There are particles and waves and fields and ... -1 The Heisenberg uncertainty principle can be understood intuitively. In order to make a complicated thing simple, you first have to get a good understanding about the accessory conditions: -The principle of complementarity (momentum-position, energy-time etc.) with the delta-function -The multidimensionality: Complex spaces may be doubling the number of ... 2 The uncertainty principle is a mathematical effect related to Fourier duals. In normal maths everything disappears down to infinetesimals, so is (was) rarely mentioned. (IIRC its the point where Newton's difference between two points 'only just' disappeared) Heisenberg identified that in QM, with it's fixed wave speed (radio, EM, Light, Gravity waves), ... 96 The best intuitive analogy I've heard is with classical sound waves. Consider a musical instrument playing a pure sine wave of frequency$\nu$and amplitude$A$, and no other harmonic frequencies at all. Graphing this in frequency-amplitude space ($x$-axis=frequency,$y$=amplitude) gives you a$\delta$-function-like point function with value$y=A\$ at ...

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I think: Yes there's an intuitive explanation for the Uncertainty principle. The explanation is the following: The most important thing to convince the non-scientist listener is that particles in Quantum Mechanics ARE NOT OBJECTS! This is observed in interference experiments and is a fact that we are very certain about. So they are waves. Once they wrap ...

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The explanation you have heard, extended, goes as follows: suppose I want to find the position of a particle in a box. To do so, I shine light on it, and, in a very similar way to what happens in the macroscopic world, by how the light bounces off I understand where the object is. However, the particle is so small that the momentum of a photon can push it ...

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No, there is no intuitive explanation for the Heisenberg Uncertainty Principle, or most of other QM. Feynman was rumored to say Anyone who claims to understand quantum theory is either lying or crazy. To answer your second question, the HUP states the product of the uncertainties of two measurements on a system has a lower bound, provided those ...

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In my experience non scientists tend to turn quantum mechanics into metaphysics. A non scientist would not know even what a measurement error is, which is inherent in all data. For mathematically inclined people the Fourier transform uncertainties are directly related to the HUP. Heisenberg identified h_bar as the lowest limit for pairs of conjugate ...

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