Bound states of Yukawa potential

In summary, the conversation discusses the derivation of the critical q for having bound states in a Yukawa potential. The potential, V(r) = -\frac{e^2}{r}e^{-rq}, is being used to reproduce N.F. Mott's argument about the critical spacing of a lattice of hydrogen atoms for a metal-insulator transition. The criterion for a bound state is \frac{2m}{hbar^2}\frac{|V_0|}{q^2} > 2.7, as stated in a source provided. The speaker is unsure of how to derive this criterion and suggests solving the TISE for the potential and finding the minimum value of q that allows a single bound state.
  • #1
drummerguy
3
0
Say you have a Yukawa potential (a.k.a. screened coulomb potential) [tex]V(r) = -\frac{e^2}{r}e^{-rq}[/tex] where q is the inverse screening length, how would you find the critical q for having bound states? I'm working on reproducing N.F. Mott's argument about the critical spacing of a lattice of hydrogen atoms for a metal-insulator transition.

I realize any negative potential will have bound states, and the potential just as I have written it will have at least one bound state (i.e. if q goes to infinity the potential becomes a delta function well which always has one bound state), but my professor told me to derive the condition for no bound states (he said it was an 'elementary' quantum mechanics exercise).

I found one site http://farside.ph.utexas.edu/teaching/qm/lectures/node69.html that says that the criterion for a bound state is [tex] \frac{2m}{hbar^2}\frac{|V_0|}{q^2} > 2.7[/tex] and this gives the same answer that Mott got for the lattice spacing. I just don't know how to get this criterion in the first place.
 
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  • #2
drummerguy said:
Say you have a Yukawa potential (a.k.a. screened coulomb potential) [tex]V(r) = -\frac{e^2}{r}e^{-rq}[/tex] where q is the inverse screening length, how would you find the critical q for having bound states? I'm working on reproducing N.F. Mott's argument about the critical spacing of a lattice of hydrogen atoms for a metal-insulator transition.

I realize any negative potential will have bound states, and the potential just as I have written it will have at least one bound state (i.e. if q goes to infinity the potential becomes a delta function well which always has one bound state), but my professor told me to derive the condition for no bound states (he said it was an 'elementary' quantum mechanics exercise).

I found one site http://farside.ph.utexas.edu/teaching/qm/lectures/node69.html that says that the criterion for a bound state is [tex] \frac{2m}{hbar^2}\frac{|V_0|}{q^2} > 2.7[/tex] and this gives the same answer that Mott got for the lattice spacing. I just don't know how to get this criterion in the first place.

So can't you just solve the TISE for the potential you gave, and figure out the ZPE in terms of the particle mass and the parameters of the potential? You may have to make some simplifying assumptions that make sense in the limit you want to test, but I think it is doable. Then it should be trivial to find the minimum value of q that allows a single bound state.
 

1. What is a bound state in the context of a Yukawa potential?

A bound state in the context of a Yukawa potential is a state in which the energy of a particle is lower than the potential energy of the system, causing the particle to be confined. This is due to the attractive force of the Yukawa potential, which decreases with distance.

2. How does the strength of the Yukawa potential affect the bound states?

The strength of the Yukawa potential, represented by the parameter α, directly affects the number of bound states. As the value of α increases, the potential becomes stronger and more bound states are formed.

3. Can bound states of Yukawa potential exist for all values of α?

No, bound states of Yukawa potential only exist for values of α greater than a critical value, determined by the potential energy and the mass of the particle. If the value of α is less than the critical value, there are no bound states.

4. How do the properties of the bound states change as α increases?

As α increases, the bound states become more tightly bound and the energy levels become closer together. This means that the particle is more confined and has less energy, making it more difficult to break free from the potential.

5. Are there any real-world applications of bound states of Yukawa potential?

Yes, bound states of Yukawa potential have been studied in various fields such as nuclear physics, atomic physics, and condensed matter physics. They have also been used to model the behavior of particles in potential wells, such as in the atomic nucleus or in quantum dots.

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