1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Quantum potential problem -- Particle confined in 1 dimension

  1. Jan 22, 2016 #1
    1. The problem statement, all variables and given/known data
    A particle with mass m and electric charge e is confined to move in one dimension along the x -axis. It experiences the following potential: V(x) = infinity when x<0, V(x) = -e^2/4*pi*ε*x when x≥0
    For the region x ≥ 0 , by substituting in the Schrödinger equation, show that the wave function
    u(x) = C*x*exp(-αx) can be a satisfactory solution of the Schrodinger equation so long as the constant α is suitably chosen. Determine the unique expression for α in terms of m , e and other fundamental constants. Note that C is a normalisation constant.then Show that the energy of the particle represented by u (x) is given by
    E = -m*e^4/ 2(4*pi*ε*ħ)^2
    This has me really stumped, cant show that the energy is equal to the equation above.
    2. Relevant equations
    Schrodinger equation

    3. The attempt at a solution
    attempt:
    d^2(u(x))/dx^2 = ((E - V(x))2m/ħ^2)u(x) rearranging the Schrodinger equation
    and
    d^2(u(x))/dx^2 = (α^2 - 2α/x)C*x*exp(-αx)
    so
    ((E - V(x))2m/ħ^2) = (α^2 - 2α/x) equating coefficients
    this is where i get stuck. Any help apreciated
     
    Last edited: Jan 22, 2016
  2. jcsd
  3. Jan 22, 2016 #2

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    (***) You might want to check your differentiation.
     
  4. Jan 22, 2016 #3
    I have, i cant see the mistake; i even wolfram alpha'd it and it gave me the same awnser
     
  5. Jan 22, 2016 #4

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    There must be one term without a factor of x. ##f(x) = xg(x) \ \Rightarrow \ f''(x) = 2g'(x) + xg''(x)##
     
  6. Jan 22, 2016 #5
    Your right, sorry, i copied it down wrong. I have been working with the correct derivative though and i still cant do it.
    question now corrected
     
  7. Jan 22, 2016 #6

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Hint: when are two polynomials equal?
     
  8. Jan 22, 2016 #7

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You may also have mistyped the answer. I get ##E = \frac{-me^4}{2(4\pi \epsilon \hbar)^2}##
     
  9. Jan 22, 2016 #8
    Yeah that was the answer.
    thanks for the help
     
  10. Jan 22, 2016 #9

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Hello gabz, :welcome:

    Interesting exercise.
    In the future, Could you use brackets in appropriate places when posting ? For example, I assume your potential is $$
    V(x) = -{e^2\over 4 \pi \epsilon_0 \; x} \ \rm ?
    $$
     
  11. Jan 22, 2016 #10
    yeah, sorry about that. Made quite a few mistakes, ill be more careful next time
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Quantum potential problem -- Particle confined in 1 dimension
  1. Problem on dimensions (Replies: 0)

Loading...