Calculating Fourier Transform in Circular Wells

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SUMMARY

The discussion focuses on calculating the Fourier transform for an infinitely deep circular well, specifically using the radial wave function R=N_m J_m (k r), where k is defined as k=α_{mn}/R. Participants confirm that the k in J_{m}(k r) is distinct from the wave number k in momentum p=ħk, clarifying that the former is a scalar while the latter is a vector. Additionally, the Schrödinger equation is referenced, highlighting its reformulation in one dimension as d²/dx²ψ(x)=-k², with implications for three-dimensional systems using the Laplacian operator.

PREREQUISITES
  • Understanding of Fourier transforms in quantum mechanics
  • Familiarity with Bessel functions, specifically J_m
  • Knowledge of the Schrödinger equation and its applications
  • Basic concepts of wave functions and quantum confinement
NEXT STEPS
  • Study the properties and applications of Bessel functions in quantum mechanics
  • Learn about the implications of the Schrödinger equation in three-dimensional systems
  • Research the concept of quantum confinement in circular wells
  • Explore the relationship between wave numbers and momentum in quantum systems
USEFUL FOR

Quantum physicists, students of quantum mechanics, and researchers focusing on wave functions and quantum confinement in circular geometries will benefit from this discussion.

dongsh2
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Hi everyone,

do you know how to calculate the Fourier transform for the infinitely deep circular well (confined system)? The radial wave function is given by R=N_m J_m (k r). k=\alpha_{mn}/R. R is the radius of the circular well. R(k R)=0. Thanks.

Another question is that The k in J_{m}(k r) is same as the wave number k in momentum p=\hbar k?
 
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Another question is that The k in J_{m}(k r) is same as the wave number k in momentum p=\hbar
k?

Simly said:yes k is the same,thus the root of 2mE/h(bar)^2
But p= d^2/dx^2+...*-h(bar)^2/2m in three dimensions(I don't know if this was just accidental but anyway?!), what you probably ment was that the Schrödinger eqaution can be rewriten as

d^2/dx^2psi(x)=-k^2, that is in one dimensions with respect to x in three dimensions you would just use the laplacian.
 
Last edited:
I think the k in J_{m}(k r) is different from the wave number k in p=\hbar k. The wave number k is a vector. The k in J_{m}(kr) is a scalar, k=\alpha_{mn}/R, in which R is the radius of the circular well.
 

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