Double delta function and bound states.

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Homework Statement



Given the delta function -α[δ(x+a) + δ(x-a)] where α and a are real positive constants.
How many bound states does it possess? Find allowed energies for [itex]\frac{hbar2}{ma}[/itex] and [itex]\frac{hbar2}{4ma}[/itex] and sketch the wave functions.



Homework Equations



I know there are three parts of the wave function that has the general solution

Aekx + Be-kx

I also know for -inf to -a the term cancels out leaving only Aekx and for x to inf the only term left is some Fe-kx since it can't blow up as x goes to infinity.



The Attempt at a Solution



I set up the three parts of the wave function.

1) Aekx
2) Cekx + De-kx
3) Fe-kx

I also took the derivatives and will use the fact that they will subtract to give me [itex]\frac{-2ma}{hbar2}[/itex]

I know if I solve for k I can get the energy. Honestly I am kind of lost in the understanding of this problem. Especially the [itex]\frac{hbar2}{ma}[/itex] and [itex]\frac{hbar2}{4ma}[/itex] part.

Sorry I don't know why the fraction code isn't working. It should read (hbar^2)/(ma,4ma)
 

Answers and Replies

  • #2
TSny
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Note that your potential function is symmetric about x = 0. So, you can look for solutions where the wave function is even or odd. Thus, you should be able to write the coefficients F and D in terms of A and C for the two cases of even or odd functions. This reduces the number of unknown coefficients.

I also took the derivatives and will use the fact that they will subtract to give me [itex]\frac{-2ma}{\hbar^2}[/itex]

This isn't quite correct. Did you mean to write ##\alpha## instead of ##a## (where ##\alpha## is the coefficient of the delta functions)? Even so, there is something missing. Can you show more work here?

Don't forget that you also have the condition that the wave function must be continuous.

I know if I solve for k I can get the energy. Honestly I am kind of lost in the understanding of this problem. Especially the [itex]\frac{\hbar^2}{ma}[/itex] and [itex]\frac{\hbar^2}{4ma}[/itex] part.

I'm not sure what is meant here either. Note that [itex]\frac{\hbar^2}{ma}[/itex] has the dimensions of energy times length. This is the same dimensions as ##\alpha##. So, maybe they mean to find the energy for [itex]\alpha = \frac{\hbar^2}{ma}[/itex].
 

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