Question about Landau level

  • Thread starter bobydbcn
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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?
 
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Answers and Replies

  • #2
DrDu
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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
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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?
 

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