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I have a question regarding the delta function potential well.

Given the following potential:

V(x) = -αδ(x) for -a/2 < x < a/2 (α- positive constant) and V(x) = 0 elsewhere, how would one show that the ground state is the only eigenstate with E <0. One could of course solve the problem explicitly and show that this is the case. However, I was trying to come up with a reasonable physical argument without actually solving the problem.

So, here are my thoughts:

1.) V(x) is even, so the eigenfunctions are either even or odd.

2.) By the node theorem, all odd eigenfunctions have x=0 as a node. In particular, the first excited state has x=0 as a node. So, the probability of finding the particle in the first excited state at x=0 is zero (strictly speaking the probabilty density is zero). So, the particle then doesn't "feel" the delta potential. So, it must be unbounded, i.e. the nergy must be E>0.

3.) Since any excited state has energy larger than the first excited state, it follows that all excited states have E>0.

4.) But, every 1D potential has at least one bound state. So, it must be the ground state.

Does this sound like a reasonable argument?

Thank you.

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# 1D delta funtion potential well

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