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Does Gaussian function give bound states for a particle?

  1. Jun 2, 2014 #1
    Hello everyone. I was yesterday asked in an interview to draw a gaussian curve. I drew. And then they asked in what region would this give rise to bound states? I am really confused how to conclude if a function gives bound state or not. Please help.
    Thanks.
     
  2. jcsd
  3. Jun 2, 2014 #2

    fluidistic

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    I think your question is vague because we don't know what the Gaussian function represents. Is it a potential in which a particle is put into? Is it the wave function of a particle?
     
  4. Jun 2, 2014 #3
    Thanks for the reply. He didnt mention that. What if it is potential?
     
  5. Jun 2, 2014 #4
    And what if it is function?
     
  6. Jun 2, 2014 #5

    atyy

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  7. Jun 2, 2014 #6

    fluidistic

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    If it's the wave function you can't conclude because you need to know the potential. As atyy mentioned, you could have a Gaussian wave function if the particle is inside a quadratic potential and it would be in a bound state. But you could also have a Gaussian wave function for a free particle at some time t, and in such case there is no bound state.
    If it's the potential, there are no bound states possible. Unless the Gaussian is inverted (put a negative sign in front of the Gaussian function).
    In 1 dimension, the bound states, if they exist, are between the infimum value of the potential and the lesser of the assymptotes (assuming they exist) for when x tends to positive infinity and negative infinity.
    A condition that ensures the existence of at least one bound state is that ##\int V(x) dx <0## (assuming 2 other conditions on the potential that I forgot).
     
  8. Jun 2, 2014 #7
    Thanks for the reply fluidistic.
    But I did not get this point. Can you elaborate it a bit please?
    "In 1 dimension, the bound states, if they exist, are between the infimum value of the potential and the lesser of the assymptotes (assuming they exist) for when x tends to positive infinity and negative infinity."...
    Thanks
     
  9. Jun 2, 2014 #8

    fluidistic

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    Yes, I should have said "the eigenvalues (energies) corresponding to the bound states" instead of bound states.

    If you plot V(x) in function of x and if you assume that [itex]\lim _{x \to \pm \infty} V(x) =V_{\pm}[/itex] exist (##V_+## and ##V_-## could be worth + and - infinity), then the energy(ies) of the bound states can exist only in the interval ## (V_0, V_{min}]## where ##V_0## is the infimum of V(x) and ##V_{min}## is the minimum between ##V_+## and ##V_-##.

    Example 1: The quadratic potential (harmonic oscillator). In that case ##V_+=V_-=+\infty##. ##V_0=0## and ##V_{min}=+\infty##. So the energies corresponding to the bound states, if they exist, are in the interval ##(0, \infty )##.

    Example 2: The non-inverted Gaussian potential. In that case ##V_0=0##, ##V_\pm = 0=V_{min}## so there are no energy possible. Hence no possible bound state.

    Example 3:The inverted Gaussian potential. In that case ##V_0=-C## where C>0, ##V_{min}=0##. So the energies corresponding to bound states, if they exist, are between ##(-C,0]##. To ensure that they exist one would have to check whether the potential satisfies 2 criteria that I don't remember, plus that the condition ##\int _{-\infty} ^\infty V(x)dx <0##. This will always be satisfied with such an inverted Gaussian so there is at least 1 bound state with an energy between (-C,0].
     
  10. Jun 6, 2014 #9
    Thanks fluidistic. Crystal clear :)
     
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