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jmtome2
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Homework Statement
Show that the conditions for a bound state, Eqn1 and Eqn2, can be obtained by requiring the vanquishing of the denominators in Eqn3 at k=i[tex]\kappa[/tex]. Can you give the argument for why this is not an accident?
Homework Equations
Eqn1: [tex]\alpha[/tex]=q*tan(qa)
Eqn2: [tex]\alpha[/tex]=-q*cot(qa)
Eqn3:
R= i*e[tex]^{-2ika}[/tex][tex]\frac{(q^{2}-k^{2})*sin(2qa)}{2kq*cos(2qa)-i(q^{2}+k^{2})*sin2qa}[/tex]
T=e[tex]^{-2ika}[/tex][tex]\frac{2kq}{2kq*cos(2qa)-i(q^{2}+k^{2})*sin2qa}[/tex]
The problem concerns a potential well of V[tex]_{0}[/tex] from -a to a.
I believe that k[tex]^{2}[/tex]=[tex]\frac{2mE}{h(bar)^{2}}[/tex] and that [tex]\kappa[/tex][tex]^{2}[/tex]=-[tex]\frac{2m(E-V_{0})}{h(bar)^{2}}[/tex]
I also know that in order to reach these boundaries through another method in the book, they let -[tex]\alpha[/tex][tex]^{2}[/tex]=[tex]\frac{2mE}{h(bar)^{2}}[/tex]...
The Attempt at a Solution
I removed the demoninators from Eqn3 and tried to substitute k=[tex]\sqrt{\frac{2mE}{h(bar)^{2}}}[/tex] into Eqn3... but not only was this result messy but I couldn't get the boundary conditions, Eqn1 and Eqn2 to come out. Help?
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