1. Jan 28, 2010

### 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?

Last edited: Jan 29, 2010
2. Jan 29, 2010

### 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.

3. Jan 29, 2010

### bobydbcn

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?