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Turning points

  1. Nov 27, 2006 #1
    How can I find the turning points for the one dimensional Morse potential

    [tex]V(x) = D(e^{-2ax}-2e^{-ax})[/tex]

    ??
     
  2. jcsd
  3. Nov 27, 2006 #2

    HallsofIvy

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    It would be a good start to state the DEFINITION of "turning point"!
     
  4. Nov 27, 2006 #3
    That's probably my problem then, what is the definition of "turning point"?
     
  5. Nov 27, 2006 #4

    dextercioby

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    Can you graph the Morse potential ?

    Daniel.
     
  6. Nov 27, 2006 #5
    Yes I have graphed the potential.
     
  7. Nov 27, 2006 #6
    Should I invert the function and find the min and max for x(V) ?
     
  8. Nov 27, 2006 #7

    dextercioby

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    Turning points are related to the "classically forbidden regions", they are boundaries for these regions. It's easier to see if the problem is unidimensional and you can graph the potential.

    Daniel.
     
  9. Nov 27, 2006 #8
    Yes, but how do I determine the turning points?
     
  10. Nov 27, 2006 #9

    dextercioby

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    By solving an impossible equation. Joking, just solve

    [tex] V(x)=E [/tex]

    .The eqn is not impossible. It can be brought to an algebraic one, a quadratic one, even.

    Daniel.
     
  11. Nov 27, 2006 #10
    But isn't
    [tex]E = \frac{1}{2}m \dot{x}^2 + V(x)[/tex]?
     
  12. Nov 27, 2006 #11

    dextercioby

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    Nope and yes. I assumed you wish to find the classical turning points of the Morse potential and for that you need to solve the eqn i wrote. At these turning points the classical KE is zero.

    Daniel.
     
  13. Nov 27, 2006 #12
    Yes of course. But then I've only got V(x) = V(x) ??
     
  14. Nov 27, 2006 #13

    dextercioby

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    Nope, V(x) is equal to E, the total, given, energy. The E is the a generic notation for the the variable used to index the spectrum of the quantum Hamiltonian.

    Daniel.
     
  15. Nov 27, 2006 #14
    Ok, but I'm studying classical mechanics so I think I have to use another approach...
     
  16. Nov 27, 2006 #15

    dextercioby

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    Alright, then, find the solution from other source and compare to what i said and suggested.

    Daniel
     
  17. Nov 27, 2006 #16
    I'm not aware of another approach. You start off with a certain amount of energy, which is a constant in a conserved system. But the classical turning points are when [tex]\dot{x} = 0[\tex], and then a moment later the velocity changes signs (i.e. the particle goes from going to the right to going to the left), so then you do what Dexter suggests, and hopefully understand why you're doing it.
     
  18. Nov 27, 2006 #17
    I understand this now. I get the equation

    [tex]D(e^{-2ax}-2e^{-ax})-E=0[/tex]

    Any tricks on how to solve this?
     
  19. Nov 28, 2006 #18

    dextercioby

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    Yes, i already did tell you b4. Just substitute [itex] e^{-ax} =t [/itex] and then c what you get.

    Daniel.
     
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