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Potential question

  1. Oct 27, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the energies for the bound states of a potential well with the following potential:
    V(x)= +∞ for x<0
    -V0 for 0<x<a
    0 for x>a


    2. Relevant equations
    Time Independent Schrodinger Equation.

    3. The attempt at a solution
    I really do not know where to start; I've read through my text, looked through my notes, and am unsure of what to do. I think that you have to solve the SE; but I'm not sure how to, or if that is what you have to do. That seems to be the common thing through all of the physical systems we have discussed in class. Thanks in advance.
     
  2. jcsd
  3. Oct 27, 2013 #2

    CAF123

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    Gold Member

    What does it mean for the particle to be in a bound state? Write the TISE in the region 0 < x < a and x > a and then solve those equations to obtain the behavior of the wave function in each region. Use the boundary conditions to determine the constants, with which you can extract the allowed energies.
     
  4. Oct 27, 2013 #3
    Thank you CAF123; that makes sense. I must admit, I have not had a diff equations class yet, so I am very confused by them. However, I do understand what you mean by applying the TISE to each region of the well. Thanks again.
     
  5. Oct 27, 2013 #4
    With CAF123's help, i was able to come up with the following differential equations for each region of the potential well.
    REGION 1.)
    [itex]\frac{\partial^2ψ}{\partial x^2}[/itex]=[itex]\frac{-2m}{h^2}[/itex]Eψ

    REGION 2.)
    [itex]\frac{\partial^2ψ}{\partial x^2}[/itex]=[itex]\frac{-2m}{h^2}[/itex](E+V0

    REGION 3.)
    This region blows up to infinity; since the well was defined as having infinite potential (V(x)=∞) with x< 0, adding infinity to the SE blows it up. This just heads to infinity.

    I am unsure of what to do next; I know that this has to be solved, I just don't know how to do it. Any pointers would be great. Thanks.
    PS: h is not just planck's constant, but it is h-bar. Thanks again.
     
  6. Oct 27, 2013 #5

    CAF123

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    Gold Member

    This is correct, although you do not need partial derivatives here. ##\psi## is an eigenstate of the Hamiltonian and as such does not depend on time, only x, so a total derivative is fine here.

    The region x < 0 is a non-physical region, so we set ##\psi = 0## here.

    Perhaps check your notes or a textbook. It is quite common for the infinite potential to be worked and then the step potential. The solutions in the regions are oscillatory motion and decaying exponential.
     
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