So we have an infinite square well with bounds at -a and +a with a delta functii

Question

So we have an infinite square well with bounds at -a and +a with a delta functiion at x=0. Obviously we are looking for allowed energies. Once we do that we are then supposed to compare the results to the regular infinite square well (without the delta function), explain why the odd case isn't affected, and finally, comment on limiting cases as the alpha term in the delta function goes to zero and infinity.

So first we notice that since V(x) is even, psi needs to be either even or odd.

for even case, I have psi=0 on |x|>a, psi=Acos(n(pi)/(2a)) on 0<|x|<a (the evenness of function allows us to use same coefficient for -a<x<0 as we do for 0<x<a since even functions have property of being symmetric about x=0.(even with the delta function located at x=0). This can also be verified by assuring continuity at |x|=a (which leads to the two coefficients being equal on both intervals).. I'm a little confused on how to make sure of the continuity of the derivative of psi wrt x at 0. I keep coming up with E=alpha (with some constants) but that doens;t feel correct. haven;t started the odd case yet. Any advice would be appreciated. Thank you.

Explanation / Answer

YOUR ANSWER IS CORRECT E CAN MAKE AS A CONTANTANT BY DERIVATION IT WILL GET AS h2n2/8ma2

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