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Old May 14th 2016, 12:04 AM   #1
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Quantum Field Theory

Dear physicists
I have a question about the plots of wave functions of quarkonia. Why their plots are Gaussian like? For example the ground state figure for charmonia is like:

Please give me a brief explanation.
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Old Nov 28th 2016, 08:07 PM   #2
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Quantum field theory (QFT) is the theoretical framework for constructing quantum mechanical models of subatomic particles in particle physics and quasiparticles in condensed matter physics. QFT treats particles as excited states of the underlying physical field, so these are called field quanta.
Reference-http://www.ribbonfarm.com/2015/08/20/qft/
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Old Dec 2nd 2016, 09:14 AM   #3
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Some time ago I was browsing the net and came across an article discussing the shape of wave functions.
The basic indication was that there is no compelling reason to choose any particular shape,
although there will obviously be reasons for discarding obviously incorrect shapes.
Basically the Gaussian seems reasonable and sensible and is not obviously incorrect.
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Old Dec 2nd 2016, 11:56 AM   #4
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Originally Posted by Woody View Post
Some time ago I was browsing the net and came across an article discussing the shape of wave functions.
The basic indication was that there is no compelling reason to choose any particular shape,
although there will obviously be reasons for discarding obviously incorrect shapes.
Basically the Gaussian seems reasonable and sensible and is not obviously incorrect.
Even so I don't really see why they would be Gaussian, either. Gaussian wave functions arise in the harmonic oscillator problem and the quark-quark problem is anything but. At a bad approximation the quark potential is closer to a Yukawa potential than anything else and those get really messy.

-Dan
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Old Dec 5th 2016, 05:32 PM   #5
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The "Central Limit Theorem" is, in my opinion, one of the most remarkable theorems in mathematics. It says that if a number of samples are taken from a population with any probability distribution, as long as the mean is $\displaystyle \mu$ and the standard deviation is $\displaystyle \sigma$, then the average of all the samples will be, at least approximately, normally distributed with mean $\displaystyle mu$ and $\displaystyle \sigma$. The sum of n such samples will be, at least approximately, approximately normal with mean $\displaystyle n\mu$ and standard deviation $\displaystyle \sigma\sqrt{n}$.

Any time we have a measurement of a physical quantity we can think of that as combining many "sub-quantities" and so expect that it will have, at least approximately, the normal distribution.

("At least approximately": if the population from which the samples are drawn is normal, both sum and average will have, exactly, the normal distribution. If the population is not normally distributed, the distribution of average and sum will approach normal as n goes to infinity.)
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