Question about some general potential form

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SUMMARY

The discussion centers on the general potential form introduced in the paper "Phys. Rev. B 29, 3153" published on 15 March 1984. The potential is defined as V(u,r)=A(r)(1+e cos(2πu))/[1+r²+2r cos(2πu)]^p, where A(r) serves as a normalizing amplitude function. The participants clarify that A(r) normalizes the potential to specific forms, similar to how a normalization constant C is used in quantum mechanics to ensure wavefunctions are normalized to unity. This understanding is crucial for interpreting various potential functions in quantum mechanics.

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LagrangeEuler
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In paper
Phys. Rev. B 29, 3153 – Published 15 March 1984
general potential form is introduced and from that form one can obtain different class of period potential

V(u,r)=A(r)\frac{1+e\cos (2\pi u)}{[1+r^2+2r\cos (2\pi u)]^p}
##-1<r<1## , where ##r## is real number, ##m,p## are integers, ##e=\pm 1##. In is interesting that in that paper authors call ##A(r)## normalizing amplitude function. I am not sure why?
They take for example for
##A(r)=(1-r)^4##, ## m=p=2##, ##e=1##, ##0<r<1## to obtain double well potential. Could you explain me why ##A(r)## is normalizing amplitude function?
 
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This is my understanding of your problem.

In quantum mechanics, a wavefunction ##\psi (x)## is said to be ##\textit{normalised to unity}## if ##\int |C \psi(x)|^{2} dx = 1##, for some constant ##C##. In other words, the introduction of the multiplicative factor ##C## normalised the integral of the square of the wavefunction ##\psi (x)## to unity. Here, ##C## is called the normalisation constant.

In the same way, ##A(r)## is called a normalising function (and not a normalisation constant) because, it normalises, in the same sense as before, the general form of the potential to various specific potential functions (which are not equal to unity, in general). Furthermore, ##A(r)## is called a normalising ##\textit{amplitude}## function because, hey, it appears in front of the sinusoidal factor just as in a typical wave function ##\psi (x) = A\ cos(kx- \omega t)##.

Let me know if I've made any mistakes anywhere.
 
Why is this in the calculus forum ?
 
That's a good question! :biggrin:

Why in the world did I not check that before I replied? o0)
 

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