A quick question - Quantum Mechanics

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For a delta potential of the form -Vδ(x-a), the wave function equations are given for regions x<a and x>a. It is not necessary to impose A'=B=0 to prevent divergence as the exponential terms represent oscillatory functions that do not diverge at infinity. The wave function remains bounded due to its representation in the complex plane. The primary boundary condition to consider is that u(a) must equal u'(a) for the wave functions on either side of the delta potential. This ensures continuity and differentiability at the point of the delta function.
atomqwerty
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Just a quick question

A delta Potential is given by -Vδ(x-a). If we wrote the equations of the wave function like this

For E>0

u(x) = Aexp(ikx) + A'exp(-ikx) for x<a
u(x) = Bexp(ikx) + B'exp(-ikx) for x>a

do i have to impose that A'=B=0 in order to u(x) not diverge in when x goes to +-infinite?

The problem is that where I read that, only impose the condition for E<0.

Thanks!
 
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but u(x) doesn't diverge at all since exp(+/-ikx) = cos(kx)+/-isin(kx) and this is the unit circle in the C plane. The only boundry condition is that u(a)=u'(a) for x <a , x>a respectivley
 
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