Understanding Zeroth Order Approximations in Landau Level Equations

[bobydbcn]

The hamiltonian is H=\frac{1}{2}(\vec{p}+\frac{\vec{A}}{c}). We investigate this problem at a polar coordinate (r,\theta). The radial equation is
[-\frac{1}{2}\frac{d^2}{dr^2}+\frac{1}{2}(m^2-\frac{1}{4})\frac{1}{r^2}+\frac{1}{2}\omega_{L}r^2]u(r)=[E-m\omega_{L}]u(r)
in order to find the wavefunction, we must investigate the behavior at the infinity of the differential euqtion.my question is that why the approximate equation take this form at infinity.r\rightarrow\infty, the equation becomes
[-\frac{1}{2}\frac{d^2}{dr^2}+\frac{1}{2}\omega_{L}r^2]u(r)=0.
by the way, the energy is E=\omega_L(2n+|m|+m+1). we can see that when r\rightarrow\infty, then n\rightarrow\infty, so the energy above becomes infinity. then we can't omit [E-m\omega_{L}]u(r) in the above approximate equation. In my opinion, the equation should be written in the follow formulation:
[-\frac{1}{2}\frac{d^2}{dr^2}+\frac{1}{2}\omega_{L}r^2]u(r)=[E-m\omega_{L}]u(r)
Is that right? What is the reason?

 

[DrDu]

The right hand side of your equation becomes negligible compared to the second term on the left hand side so you drop it a a zeroth order approximation. The second derivative is the only term which can compensate the second term, hence you keep it.

 

[bobydbcn]

The right hand side of your equation becomes negligible compared to the second term on the left hand side so you drop it a a zeroth order approximation. The second derivative is the only term which can compensate the second term, hence you keep it.

Thank you very much. I got it. Is there experiment that varify the landau level? Can you suggest a book to me about the zeroth order approximation?