Tuesday, 21 August 2018

quantum mechanics - What would be likely to completely stop a subatomic particle assuming it was possible?



Suppose that completely stopping a subatomic particle, such as an electron, could happen under certain conditions. What would be likely ways to get an electron to be perfectly still, or even just stop rotating the nucleus and collapse into it by electromagnetic forces? What would likely be required, below absolute-zero temperatures? Negative energy? Or could a 0 energy rest state not exist in any form, of any possible universe imaginable?


Let's say there was a magnetic field of a certain shape that we could postulate that is so intensely strong that if we put an electron in the center of it, it could not move at all in any direction. Would the energy requirement of the field be infinite? What would be the particle's recourse under this condition?


Further, suppose it were possible and one could trap an electron and stop all motion completely. What would this do to Heisenberg's Uncertainty Principle and/or Quantum Mechanics, because its position and momentum (0) would both be known? If it can be done, is Quantum Mechanics no longer an accurate model of reality under these conditions? Could we say QM is an accurate model under most conditions, except where it is possible to measure both position and momentum of a particle with zero uncertainty?



Please assume, confined in a thought experiment, that it IS possible to stop a particle so that is has 0 fixed energy. This may mean Quantum Mechanics is false, and it may also mean that under certain conditions the uncertainty commutation is 0. ASSUMING that it could physically be done, what would be likely to do it, and what would be the implications on the rest of physics?



Now here's the step I'm really after - can anyone tell me why a model in which particles can be stopped is so obviously not the reality we live in? Consider the 'corrected' model is QM everywhere else (so all it's predictions hold in the 'normal' regions of the universe), but particles can be COMPLETELY stopped {{inside black holes, between supermagnetic fields, or insert other extremely difficult/rare conditions here}}. How do we know it's the case that because the uncertainty principle has lived up to testing on earth-accessable conditions, that it holds up under ALL conditions, everywhere, for all times?



Answer



Your question is interesting, and gets specifically to the kinds of questions that quantum mechanics was intended to answer in the first place. It helps to understand the motivation behind the original Bohr model of the atom, and how that led to QM in the first place.


The problem Bohr was trying to address can be paraphrased as, "If an electron orbits a nucleus like a planet, why doesn't it gradually lose energy and spiral into the nucleus?" The answer came when Bohr realized that the orbital momentum was quantized, effectively meaning that since the electron had mass then by the relationship $p=mv$, the velocity was also quantized (note: this simple expression is more complicated when relativity is included, but the discussion can continue without including it). These quantized values of momentum/velocity are what one would call eigenvalues, or observables in quantum mechanics. Since the electron can only change orbits by given off specific quantities of energy instead of giving off energy continuously, it remained in a stable orbit relative to the nucleus, thus preventing it from spiraling in.



What is important to understand is the idea of the potential well. In the Bohr model, the electrons closest to the nucleus have higher velocity than the electrons further away. In other words, they have greater kinetic energy ($K.E. = \frac{1}{2} mass \times velocity^2$), but it had less potential energy since it was closer to the nucleus (obeying the relationship $P.E. = mass \times distance \times gravity$). However, the total energy ($K.E. + P.E.$) associated with orbits closer to the nucleus is less than those further away. So in order for the electron to move closer to the nucleus, energy must be given up. This is accomplished by the emission of a photon. Alternatively, if one wants to cause an electron to move into a more distant orbit, then one must add energy through use of a photon.


It is in contemplating how to determine the orbit of the electron that the uncertainty principle first became apparent. The only probe that we have available to determine the position of an electron in its orbit is a photon, and the photon must be of sufficient energy in order for it to be small enough to give a meaningful result, however if we use a photon small enough (in terms of wavelength), it will have enough energy to shift the electron into a different orbit, and then we would have to start the process all over again.


A free electron has sufficient energy to escape the nucleus. In other words, it has acquired sufficient energy to fill its potential energy deficit. If there were only one nucleus in the universe, the potential deficit would only be eliminated when the electron was at infinite distance from the nucleus. In that situation, the implication is that the electron would also have zero velocity. This situation is obviously unrealistic, first there are more than one nuclei in the universe, and second, to verify that a particle has zero velocity at infinite distance is clearly an impossible task.


If we move past the Bohr model and into more modern quantum mechanics, the question then is whether there are eigenstates that have eigenvalues for momentum of a particle that are equal to zero? It is important to review some basic facts about matrix operations and linear algebra



If 0 is an eigenvalue of a matrix A, then the equation A x = λ x = 0 x = 0 must have nonzero solutions, which are the eigenvectors associated with λ = 0. But if A is square and A x = 0 has nonzero solutions, then A must be singular, that is, det A must be 0. This observation establishes the following fact: Zero is an eigenvalue of a matrix if and only if the matrix is singular.



This means that the matrix in question is not of full rank. In QFT this has a very specific interpretation. The annihilation operator has the power to destroy the vacuum state and map it to zero. This situation is understood to be associated with the free field vacuum state with no particles. This state is necessary because it allows us to find vacuum state solutions for the associated quantum mechanical system.


By means of analogy, we can see that the solution to ground state problem is the solution to the homogeneous part of a differential equation.


The Schrodinger equation is a linear, homogeneous equation which governs evolution of the wave function of a particle. The solutions of the Schrodinger equation can be used to understand particle motion. The exact position and momentum of a particle can only be known if h (planck's constant) approaches zero, however, in quantum mechanics, planck's constant is fundamental to the theory, so this cannot occur in the single particle case. Because of this, momentum and position uncertainty establish an inverse relation to each other, and if the uncertainty of momentum is zero, then the uncertainty in position is infinite.



For these reason's it is not possible to talk about a particle "stopping" or being "stopped" in any meaningful or non-contrived sense.


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