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Variational Method and Bound States

  1. Dec 4, 2013 #1
    1. The problem statement, all variables and given/known data
    Consider a potential function [itex] V(x)[/itex] such that:
    $$
    \begin{cases}
    V(x)\leq 0\text{ for }x\in[-x_0,x_0] \\
    V(x)=0 \text{ for }x\not\in[-x_0,x_0]
    \end{cases}
    $$
    Show, using the variational method that:

    (a) In the 1-dimensional case [itex]\lambda^2V(x)[/itex] always possesses at least one bound state.
    (b) In the 3-dimensional spherically symmetric case, [itex]V(|\vec r|)[/itex], it possesses no bound states if [itex]\lambda^2[/itex] is made sufficiently small.



    2. The attempt at a solution
    The idea is to use the variational method, i.e.:
    $$
    \frac{\langle\psi|H|\psi\rangle}{\langle \psi|\psi\rangle}\geq E_{ground},
    $$
    to show that the average value of the energy is negative and hence the ground state energy is negative.

    In my first attempt I used a gaussian trial function:
    $$
    \psi(x)=\left(\frac{2a}{\pi}\right)^{1/4}e^{-ax^2}.
    $$
    However the problem is that it turns out that the average value of the kinetic energy is [itex]\hbar^2a/2m[/itex] and so, in order to determine whether the average energy is negative or not we need to know [itex]a[/itex] explicitly. However, this is not possible since we have no insight on the actual shape of the potential.

    It seems to me that this thing will turn out to be a problem for every trial function. How can I do that?

    Thank you
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 6, 2016 #2

    Orodruin

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    In the 1D case the kinetic energy is proportional to ##a##. The potential energy is negative and goes as ##\sqrt{a}## when ##a## is small. Hence, if ##a## is small enough, the energy in that state will be negative, implying the existence of at least one bound state.

    In the 3D case, the normalisation of the wave function gets additional factors of ##a## and the argument changes.
     
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