Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Bound state

  1. Dec 25, 2008 #1

    KFC

    User Avatar

    Assume the potential in question is

    [tex]
    V = \left\{
    \begin{matrix}
    \infty, \qquad x<0 \\
    -V_0, \qquad 0\leq x \leq a \\
    0, \qquad x>a
    \end{matrix}
    \right.
    [/tex]

    where [tex]V_0[/tex] is positive.

    if we need to find the bound state, we consider the energy is less than the potential. But the potential withn [0, a] is negative, is that mean the energy will be more negative (i.e. [tex]|E| > V_0[/tex]) ?

    Generally, the Schrodinger equation will be written of the following form

    [tex]
    \frac{d^2\psi}{dx^2} + k^2\psi = 0
    [/tex]

    where

    [tex]k = \sqrt{\frac{2m}{\hbar^2}(E-V)}[/tex]. The general solution is of the form

    [tex]\psi = A\exp(ikx) + B\exp(-ikx)[/tex]

    For x<0 or x>a region, the wavefunction must decay because the potential is larger than E, [tex]k=\sqrt{\frac{2m}{\hbar^2}(E-V)} = i \sqrt{\frac{2m}{\hbar^2}(V-E)} = i\kappa[/tex], the solutions in those region become

    [tex]\psi = A\exp(\kappa x) + B\exp(-\kappa x)[/tex]

    But within [0, a], if we want to find bound state, energy must be negative and [tex]|E|>V_0[/tex], so

    [tex]k = \sqrt{\frac{2m}{\hbar^2}(-|E|-(-V_0))} = \sqrt{\frac{2m}{\hbar^2}(V_0-|E|)} [/tex]

    but this also lead to k be imaginary number, that is, the solution within [0, a] is decaying again??? But I think the solution in that region should be oscillating. Where am I get wrong?
     
  2. jcsd
  3. Dec 25, 2008 #2
    first of all your "general solution" isn't a general solution. it's the solution in the well and you produce the correct solution for x>a by luck.

    energy should be less than zero but greater than -v_0.

    for x<0 the wavefunction doesn't exist. write out the entire time independent shrodinger eqn like this:

    [tex]\frac{-h^2}{2m}\frac{d^2 \psi(x)}{dx^2}+V(x) \psi (x) = E \psi(x)[/tex]

    for each region and don't skip steps and you should end up with

    [tex]k=\pm\sqrt{\frac{2m}{\hbar^2}(-E+V_0)}=\pm\sqrt{-\frac{2m}{\hbar^2}(V_0 - E)}[/tex]

    inside the halfwell which produces standing waves in the well. note E is negative by assumption.
     
  4. Dec 25, 2008 #3

    KFC

    User Avatar

    Thanks for reply.

    I don't understand why not. In that solution form, k is varied from region to region. In rightmost region, k is imaginary which decaying solution. So why can I say in all regions they have same form of solution but with different k?

    This is my doubt. I wonder what's the requirement of bound state? Is E<V or |E|<|V| ?
    If the potential V is positive, it will trap the particle has energy less than the potential, it is quite trivial. But how to understanding the situation when the potential is negative?

     
  5. Dec 25, 2008 #4
    but you have it wrong. in right most region k is real and therefore the solution decays, and in the well k is imaginary and hence oscillates. have you taken a differential equations class?
    the requirement for a bound state is |E|<|V|. you can understand the situation when the potential is negative by realizing that theres no such thing as absolute potentials, only relative potentials and therefore the whole situation is the same if we shift the potential up everywhere by [itex]V_0[/itex]. then for [itex]0<x<a ~~ V(x)=0[/itex], for [itex]x>a ~~ V(x) = V_0 [/itex] and the for [itex]x<0 ~~ V(x) = \infty [/itex]
     
  6. Dec 25, 2008 #5

    KFC

    User Avatar

    Thanks for explanation :)

     
  7. Dec 25, 2008 #6
    no prob. just out of curiosity what is your native language?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Bound state
  1. Bound States (Replies: 5)

  2. Bound states (Replies: 7)

Loading...