- #1
user3
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Ok here's a potential I invented and am trying to solve:
V =
-Vo in -b<x<b
and 0 in -a<x<-b , b<x<a where b<a
and ∞ everywhere elseI solved it twice and I got the same nonsensical transcendental equation for the allowed energies: [tex]\frac{-k}{\sqrt{z_0 - k^2}} \frac{e^{2kb} + e^{2ka}}{e^{2kb} - e^{2ka}} = tan(b \sqrt{z_0-k^2})[/tex].
, where [tex]k=\frac{\sqrt{-2mE}}{\hbar}[/tex] and [tex]z_0 = \frac{2mV_0}{\hbar^2}[/tex]
The problem is that when I take the limit as b→a (the ordinary infinite square well) I get only one solution, where I should get infinity many.
So is there something fundamentally wrong with trying to solve this potential? Is it wrong to have an Infinite potential and bury some of it under the 0 (negative potential) ? Note: I am solving it for negative energies(Bound bound states?).
V =
-Vo in -b<x<b
and 0 in -a<x<-b , b<x<a where b<a
and ∞ everywhere elseI solved it twice and I got the same nonsensical transcendental equation for the allowed energies: [tex]\frac{-k}{\sqrt{z_0 - k^2}} \frac{e^{2kb} + e^{2ka}}{e^{2kb} - e^{2ka}} = tan(b \sqrt{z_0-k^2})[/tex].
, where [tex]k=\frac{\sqrt{-2mE}}{\hbar}[/tex] and [tex]z_0 = \frac{2mV_0}{\hbar^2}[/tex]
The problem is that when I take the limit as b→a (the ordinary infinite square well) I get only one solution, where I should get infinity many.
So is there something fundamentally wrong with trying to solve this potential? Is it wrong to have an Infinite potential and bury some of it under the 0 (negative potential) ? Note: I am solving it for negative energies(Bound bound states?).
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