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Double deltafunction potential 
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#1
Mar111, 07:48 PM

P: 7

Hello, recently I had a problem in QM involving a double deltafunction potential. A bunch of qualitative questions were asked, some of which were obvious to me immediately, some of which I made an educated guess on, and others I totally guessed. I'm following Griffiths' textbook in studying QM, and luckily I found another problem analogous to the one I had before. Any guidance/explanation would be appreciated.
1. The problem statement, all variables and given/known data The potential is V(x) = α[[tex]\delta[/tex](x + a) + [tex]\delta[/tex](x  a)]. The problem specifies that we are only conserved about bound states, E < 0. 2. Relevant equations In the end, the even wave function solution yields (ħk/mα) = e^{2ka} + 1, and the odd wave function solution yields (ħk/mα) = 1  e^{2ka}. 3. The attempt at a solution Based on these solutions, I'm asked a few questions: 1) What is the approximate energies of both the even and odd bound states in the limit 2maα/ħ^{2} >> 1? 2) Show that as 2maα/ħ^{2} > [tex]\infty[/tex], both energies converge to the energy of the bound state of the single deltafunction potential. 3) Show that there is only one bound state in the limit 2maα/ħ^{2} << 1. 4) Show that the wave function behaves like the boundstate wave function of a single deltafunction potential for x >> a. Any guidance regarding these questions is appreciated. I might be able to provide more information if anything is unclear. 


#2
Mar211, 04:08 AM

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PF Gold
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[tex]\frac{2ka}{2ma\alpha/\hbar^2} = 1\pm e^{2ka}[/tex] Let [itex]c=2ma\alpha/\hbar^2[/itex] and [itex]x=2ka[/itex]. Then you get [tex]\frac{x}{c} = 1 \pm e^{x}[/tex] You might find it illuminating to plot both sides of the equation to see where the solutions are and what the effect of varying the parameter c is. 


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