Nucleon Potential: Why 3D Harmonic Oscillator?

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Why is the potential energy of a nucleon within the nucleus approximated by 3D harmonic oscillator potential?
Is the 3D harmonic oscillator potential approximately equal to the yukawa potential experienced by the nucleon as a result of all the other nuclei(ie collectively) ?
 
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A harmonic oscillator has a potential that goes as ~x2. The Taylor series of any potential looks like x2 near the minimum. So you can always approximate with a harmonic oscillator. That's just mathematics.

How good this approximation is, that's physics.
 
The SHO is used because the many body wave functions are easier to work with.
The actual effective potential is more like a somewhat rounded square well.
The Yukawa potential does not act because the Pauli principle tends to let the nucleons pass freely through the nucleus until they near an edge.
 
The Hultén potential is an analytical potential for the deuteron, i.e the N-P force.

Remember also that we here only discuss the RADIAL part of the potential, there are SS parts, LS parts, etc. as well
 
malawi_glenn said:
The Hultén potential is an analytical potential for the deuteron, i.e the N-P force.

Remember also that we here only discuss the RADIAL part of the potential, there are SS parts, LS parts, etc. as well
The Hulthen potential is an approximate two-nucleon potential, it is not the potential seen by a nucleon in a nucleus.
 
clem said:
The Hulthen potential is an approximate two-nucleon potential, it is not the potential seen by a nucleon in a nucleus.

Yes, I have not stated anything else.

Another used potential for nucleons inside a nucleus which has not been mentioned is the wood saxon, which is the one you are referring to as "a rounded square well".

I am not familiar with the Yukawa potential when talking about the potential energy of a nucleon inside a nucleus, only when discussing the nucleon-nucleon force, since the yukawa potential arises due to scalar meson exchange.. that's why I mentioned the hulten potential.
 
The YP is due to pseudoscalar exchange.
The YP and HP are both used in the N-N interaction, but not in larger nuclei.
You are not "not familiar with the Yukawa potential when talking about the potential energy of a nucleon inside a nucleus", because as I said in my post, it does not act there.
You are right in identifying my simplification as the Woods-Saxon potential.
 
clem said:
The Yukawa potential does not act because the Pauli principle tends to let the nucleons pass freely through the nucleus until they near an edge.


What PEP has got to do?
Plz make it more clear.
What do you mean by edge?
 
The nucleons in a nucleus are all in quantized bound states.
In order for a nucleon to change its state due to interaction, it must be raised to a higher unoccupied state. Most of the interactions do not provide enough energy for that.
The nucleus is like a sphere of radius 1.2A^(1/3)fm. A nucleon does not get affected by the forces until it tries to leave the nucleus at its edge.
You should read a nuclear physics book instead of this forum.
 
  • #10
clem said:
You should read a nuclear physics book instead of this forum.

Which authors do you recommend??
BTW what potential is used to model the interaction between neutron and proton in deuterium nucleus?
should'nt it be the Yukawa potential.
If so does the equation admit solutions that are similar to that of electron in hydrogen atom for the theta-phi part .The radial part depending on the potential
 
  • #11
The Hulthen potential is a fairly simple for the deuteron.
The N-N potential has a hard core at small distance, which is lacking in the Yukawa potential.
The angular part for any spherically symmetric potential is the same (spherical harmonics).
However the N-N system has only one bound state, and it is sphericallly symmetric with no angular dependence.
I am not familiar with new books on nuclear physics, but Henley and Fraunfelder wrote a good book some time ago. You should know QM to read it.
 
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