In this case one has more rigorous results. These are due to Lieb and Simon as already said above. They published two papers about. Lieb E. A system with this distribution is a classical system. The fact that a system with a Thomas-Fermi distribution is a classical one can be seen through the following two references:. Thirring Ed. Lieb , Springer-Verlag The question can be seen from two points: rather elementary one and hot research topic.

Let me comment on both. The commutator of operators corresponds to Poisson brackets in the classical mechanics. So classical limit of this equation is:. Which are Hamilton equations of motion in classical mechanics. This is purely linear algebra: if you consider the evolution on vector space V given by this equation then operators i.

The main surprise that it does not correspond to delta functions as naively expected. The first approximation is that classical limit corresponds to Lagrangian submanifolds. Let me repeat from Coherent states vs quantization of Lagrangian submanifold. Why it is important "Lagrangian"? It is easy. Lagrangian - just restiction on the dimension - that it should be of minimal possible dimension - so after quantization we may expect finite dimensional subspace in the best case 1-dimensional. It is fruitful line of research to think about the correspondence between quantum and classical realms.

Not just the limit, but sometimes one may hope to go in opposite direction and to completely describe quantum objects in terms of classical one. Sometimes it is subject of deepest and fascinating conjectures:. We discuss a conjecture which says that the automorphism group of the Weyl algebra in characteristic zero is canonically isomorphic to the automorphism group of the corresponding Poisson algebra of classical polynomial symbols.

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### Three revolutionary principles

Chris Gerig 9, 2 2 gold badges 53 53 silver badges 94 94 bronze badges. Ehrenfest's theorem comes into play here. Good question. This means that the wave function is very localized around this position, and nearly nil anywhere else. Wave functions are defined up to a phase. This means that a rotation of the complex plane change nothing to the physical nature of the wave.

## Wave function | physics | nizudyro.tk

So a vertical localized wave would be essentially identical to the one above! So far, we have described the position of the particle. Equivalently, we can focus on its momentum. Classically, the momentum is the velocity multiplied by the mass of the particle. This has been essential to explain the photoelectric effect, which boils down to a transfer of momentum from photons to electrons. What Planck and Einstein showed is that the momentum of light depends on the frequency of its wave and its direction of motion.

What we usually mean by frequency of light is the time frequency of the wave, but since the speed of light is constant, we may equivalently consider the spatial frequency, also known as the wavenumber. A similar definition occurs in quantum physics for all particles. Basically, yes!

## Classical and Quantum Conjugate Dynamics – The Interplay Between Conjugate Variables

Now, we have to be a little more careful before actually talking about frequencies. The basis of all frequencies comes from trigonometric waves. In the complex plane, the trigonometric waves are the simplest periodic paths one can think of: Circles. In this setting, the wavenumber corresponds to the number of cycles of the trigonometric wave in 1 meter. Equivalently, it is the inverse of the spatial period, as displayed below. Note that the trigonometric function in the figure turns anti-clockwise.

This is important. It implies that the wavenumber of the trigonometric wave is positive, and that the associated wave function goes from left to right. If it went clockwise, it would be negative. This is incompatible with waves representing probabilities. We need to use the Fourier analysis. According to this, any wave is uniquely decomposable into an infinite superposition of trigonometric waves of different frequencies.

This means that, for any wave and any frequency, the wave is composed of a certain amount of the trigonometric wave of that frequency. And the decomposition is given by the Fourier transform. Basically, the Fourier transform provides a description of any wave, like the one on the left, by an infinite combination of trigonometric waves, like those on the right:. In this figure, the greater the wavenumber, the more arrows I have put on the trigonometric functions. Note that the greater the wavenumber, the more energy is associated to it.

The Fourier transform takes the wave function as an input and returns the weights of trigonometric waves on the right. Note that these weights can actually take complex numbers. The momentum is proportional to the wavenumber! From the Fourier transform that provides a mapping of every wavenumber with an amplitude, we can easily deduce a mapping of every momentum with an amplitude. We call this matching the momentum space wave function. By opposition, the wave function is sometimes called the position space wave function. A wave function has a superposition of momenta!

Classically, the position and the momentum of a particle are two distinct concepts which can be defined independently from each other. Now, the position of a wave function is distributed in space. Weirdly enough, the mere knowledge of the amplitudes of positions determine the amplitudes of momenta!

The reciprocal phrase is true too: The amplitudes of momenta determine the amplitudes of positions. Mathematically, this corresponds to saying that the Fourier transform is more or less a bijection from the Hilbert space of wave functions and the Hilbert space of momentum space wave functions. This mapping of wave functions with momentum space wave functions by the Fourier transform is displayed below. It rather aims at providing you an image of what the intimate relationship between position space and momentum space wave functions.

## Wave function

The square of the norm of the value of the momentum space wave function is the probability density function of finding the momenta in case of observation. But I believe that the word uncertainty is misleading. This is a direct consequence of their relationship through the Fourier transform. For instance, in the case of the figure above, where the momentum space wave function was localized around the pink-red momentum, while the wave function was spread all over the spectrum. Imagine that the momentum space wave function were perfectly localized at one momentum.

Recall that the wave function is obtained by summing all momentum space wave functions of all momenta. This implies that the wave function is nearly the trigonometric wave associated to the momentum at which the momentum space wave function is localized. This trigonometric wave circles over all space, and is therefore spread over all space. This is represented below:.

Conversely, were the wave function very localized, the momentum space wave function would be circling periodically, and would therefore be spread over all momentum space. Technically, the spreading of a wave function is measured by the standard deviation of the probabilistic distribution of position induced by the wave function. Indeed, the mathematical concept matches the intuitive idea of how far in space the wave is spread. This is because people often focus on the measurement part of quantum mechanics.

As a result, the wave function becomes very localized, which implies that the momentum space wave function is greatly spread. As quantum mechanics was approached with concepts of classical mechanics, this frustrating phenomenon was commented as the impossibility of knowing both position and momentum simultaneously.

But quantum mechanics rather says that when the wave is very localized, then the wave has a superposition of a large range of momenta. This has been done by Derek Muller on Veritasium in a very simple experiment! Check out this awesome video:. The momentum is still somehow the velocity of a particle multiplied by its mass. And the reason why such a reasoning holds is because of the most fundamental equation of quantum physics which describes the dynamics of wave functions. I have a lot of trouble making sense out of it. At any position of space, a wave function has a particular energy which is deduced from its global structure.

This local energy, also known as the Hamiltonian , is a composition of some potential energy which is due to external forces like electromagnetism, and some kinetic energy which depends on the superposition of momenta which makes up the wave function. This corresponds to the following formula:. This coincides with the classical setting of classical mechanics!

This should sound surprising to you. At least, it does to me.

Sorry for that. Indeed, at the scale we are looking at, distances are so short that stationary states are reached in such a small fraction of time that we can only observe these equilibria. Now, recall that the wave function is defined up to a rotation in the complex plane. Thus, stationary states involve only such rotations.

The following figure shows a wave function rotating. Arrows drawn on the wave represent the velocity at which each point of the wave moves:. Now, for a wave to be stationary, the velocity at any given point must be perpendicular to the vector from the origin to that point. Also, the farther to the origin a point is, the greater the speed must be. These two conditions correspond to the Hamiltonian at position being proportional to the wave function at this position with a real-valued proportionality coefficient independent from this position.

In mathematical terms, this means that the wave function should be an eigenvector of the Hamiltonian. The eigenvalue, which is the real proportionality coefficient, is then the energy of the wave.

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The faster the wave turns, the greater the energy. As it turns out, in any defined system, not all energies can be reached by a stationary wave. The set of all possible energies, which is called the spectrum of the Hamiltonian, is made of discrete values more often than not. In particular, this is what happens if we consider wave functions confined in a box.

No indeed. More interestingly, in an atom, the positive charges of a nucleus affect the potential energies of wave functions of electrons, which in turns modify the Hamiltonian. The eigenvalues and eigenvectors of the obtained Hamiltonian, especially the lowest eigenvalues, now correspond to possible orbits of electrons around the nucleus. The lower energies refer to stationary wave functions localized closer to the nucleus.

Fortunately, no. A quantum state is defined by the wave function and the spin of the electron, which can be up or down. Since there are only two spins, this means that there are at most two electrons with the same wave function. In particular, the energy levels cannot contain too many electrons, which implies that there are sorts of blanks around nuclei that can each contain one electron only.