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## Main Question or Discussion Point

The hamiltonian is [tex]H=\frac{1}{2}(\vec{p}+\frac{\vec{A}}{c})[/tex]. We investigate this problem at a polar coordinate [tex](r,\theta)[/tex]. The radial equation is

[tex][-\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)[/tex]

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.[tex]r\rightarrow\infty[/tex], the equation becomes

[tex][-\frac{1}{2}\frac{d^2}{dr^2}+\frac{1}{2}\omega_{L}r^2]u(r)=0[/tex].

by the way, the energy is [tex]E=\omega_L(2n+|m|+m+1)[/tex]. we can see that when [tex]r\rightarrow\infty[/tex], then [tex]n\rightarrow\infty[/tex], so the energy above becomes infinity. then we can't omit [tex][E-m\omega_{L}]u(r)[/tex] in the above approximate equation. In my opinion, the equation should be written in the follow formulation:

[tex][-\frac{1}{2}\frac{d^2}{dr^2}+\frac{1}{2}\omega_{L}r^2]u(r)=[E-m\omega_{L}]u(r)[/tex]

Is that right? What is the reason?

[tex][-\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)[/tex]

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.[tex]r\rightarrow\infty[/tex], the equation becomes

[tex][-\frac{1}{2}\frac{d^2}{dr^2}+\frac{1}{2}\omega_{L}r^2]u(r)=0[/tex].

by the way, the energy is [tex]E=\omega_L(2n+|m|+m+1)[/tex]. we can see that when [tex]r\rightarrow\infty[/tex], then [tex]n\rightarrow\infty[/tex], so the energy above becomes infinity. then we can't omit [tex][E-m\omega_{L}]u(r)[/tex] in the above approximate equation. In my opinion, the equation should be written in the follow formulation:

[tex][-\frac{1}{2}\frac{d^2}{dr^2}+\frac{1}{2}\omega_{L}r^2]u(r)=[E-m\omega_{L}]u(r)[/tex]

Is that right? What is the reason?

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