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Quantum Mechanics: Delta Potential Sudden Change

  1. Mar 6, 2016 #1
    1. The problem statement, all variables and given/known data
    I have a particle which is initially in a bound state for a given voltage in the form of a delta function at the origin,

    V = -αδ(x)

    initial state is ψα = (√αm)/h2*exp(-m*α*|x|/(h2)

    At t=0, voltage is changed to V = -βδ(x)

    both α and β are greater than zero. Right now I'm just struggling to find ψβ
    2. Relevant Equations

    All equations I have are included above.

    3. The attempt at a solution

    I realize the new bound state is in the same form as it was earlier, as all bound states for attractive delta functions have only one bound state and if I solve schrodinger's equation for this delta function, it comes out to the same equation except replaced with β. However, I realize there's also a probability of the particle radiating away. My attempt at solving this was treating it as a free particle radiating away to the left or right and looking into what I get , thinking that if I added them together then I would get the answer for ψβ, but it blew up and made no sense (and seemed unsolvable, at least for how I tried to do it).

    I'm a bit lost as to how to do this problem. I know there are three possibilities: It stays in the bound state, it moves off to the left, and it moves off to the right. However, I have no clue as to find the function for the unbound particle state.
     
  2. jcsd
  3. Mar 6, 2016 #2
    i could not follow the term radiating away - is it tunneling or charge particle radiation- in a bound state the particles do not radiate - what is the width of the potential being considered?
    how is the time involved? well i am asking as a novice pl. bear with me.

    /
     
  4. Mar 6, 2016 #3
    It's a delta function. There is no width. While it is in a bound state, you are right. However, if the potential suddenly changes to some other value, then there is a chance that it moves off.

    I am not sure how time is involved. When I tried to include time while treating it as a free particle, the equations turned into messes that couldn't be evaluated. That is my issue.
     
  5. Mar 6, 2016 #4
    i think one should write down the equations and get solutions which are continuous at the boundary -if it has a bound state some width is necessary -perhaps your alpha and beta is related to the width -as the integral of delta function goes to unity -your potential will go to perhaps alpha.... i guess.
     
  6. Mar 7, 2016 #5

    blue_leaf77

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    What is ##\psi_\beta##? Is it just the bound state of the new delta potential? If yes, then simply interchange ##\alpha## with ##\beta## in ##\psi_\alpha##.
    Do you mean, the free particle solutions? Yes, they are one of the states the initial state can collapse to. But here, it depends on what you mean by ##\psi_\beta##.
     
  7. Mar 7, 2016 #6
    I was referring to ψβ as the overall superposition of all the possible states. Ergo, the bound state added to the free particle states. Notice that the problem itself doesn't specify exactly what ψβ is; it only tells me what the new potential function is.

    It's also possible that I'm a bit confused on the problem. The actual problem itself asks me to find |∫ψα(x)ψβ(x)dx|2, where x goes from negative infinity to positive infinity and then prove that this quantity is less than 1... but if I include the free particle solutions with the bound state then all possibilities should be covered so I would think this quantity would be 1. However, I can't figure out how to find the free particle solutions in order to prove that. I tried to find them using the solution to Schrodinger's equation for a free particle, Ψ(x,0) =(1/√2π) ∫Φ(k)exp(ik)dk where k goes from -infinity to infinity (k=(√2mE)/ħ) where Ψ(x,0) = ψα, but I got gibberish and an unsolveable integral. It didn't really make sense to me anyway since I'm looking for ψβ(x) and not looking for any time dependence and this approach leads to a time dependent solution. I realize that it probably didn't make sense to do it that way but I'm not really sure how to find this at all. All I know is the general solution to the free particle is

    ψ(x,0) = Aexp(-ikx) + Bexp(ikx)

    and I'm not really sure where to go from there. We haven't covered perturbation theory or transmission/reflection yet which is included in the textbook so I don't think the solution would involve using either of those topics.
     
    Last edited: Mar 7, 2016
  8. Mar 7, 2016 #7
    Asked professor today and he forgot to mention in the assignment prompt that the problem was only for bound states of ψβ.
     
  9. Mar 9, 2016 #8
    [itex]\psi_{b}[/itex] is of the exact same form of [itex]\psi_{a}[/itex]
     
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