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Finding the Potential

  1. Dec 29, 2006 #1
    This one is from Liboff(p6.8)

    Given the wavefunction:
    [tex]\psi(x, t) = A exp[i(ax - bt)][/tex]
    What is the Potential field V(x) in which the particle is moving?
    If the momentum of the particle is measured, what value is found(in terms of a & b)?
    If the energy is measured, what value is found?

    My Work:

    [tex]\psi(x, t) = A exp[i(ax - bt)][/tex]
    I took the partial derivatives wrt to t and x:
    [tex]\frac{\partial \psi}{\partial t} = -(ib)\psi[/tex]

    [tex]\frac{\partial^2 \psi}{\partial x^2} = -a^2\psi[/tex]

    Time dependent Schrodinger's equation is:
    [tex]i\hbar \frac{\partial \psi}{\partial t} = -\frac{\hbar^2}{2m}\frac{\partial^2 \psi}{\partial x^2} + V(x)\psi[/tex]

    Substituting the above values in this equation:

    [tex]\hbar b \psi = \frac{\hbar^2 a^2}{2m}\psi + V(x)\psi[/tex]

    Dividing throughout by [itex]\psi[/itex] and rearranging, I get the potential field as:
    [tex]V(x) = \hbar\left(b - \frac{\hbar a^2}{2m}\right)[/tex]

    Am I going right? Before I can proceed furthur...
     
    Last edited: Dec 29, 2006
  2. jcsd
  3. Dec 29, 2006 #2
    :surprised

    The calculation seems correct, though.
     
  4. Dec 29, 2006 #3
    Sorry, I realised my mistake but I got disconnected before I could correct it. I've corrected it now. :biggrin:
     
  5. Dec 29, 2006 #4
    Now I have to get an expression for the momentum.
    [tex]p = \hbar k[/tex]
    [tex]V(x) = {1\over 2}kx^2[/tex]

    Equating this to the other expression of V(x) would mean an x2 term in the momentum expression. I am confused over this. :frown:
    [tex]k = {{2\hbar}\over x^2} \left(b - \frac{\hbar a^2}{2m}\right)[/tex]
     
  6. Dec 29, 2006 #5

    Gokul43201

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    Didn't you just find out that V(x) is a constant potential? Look at your answer for V(x) above. Why are you assuming that V(x) should now be a harmonic potential?

    Given a wavefunction, how do you calculate the values of observables?
     
  7. Dec 30, 2006 #6
    Use the operator relation?
    [tex]\hat p = -i\hbar\frac{\partial}{\partial x}[/tex]

    [tex]\hat H = -\frac{\hbar^2}{2m} \frac{\partial^2 }{\partial x^2}[/tex]
     
  8. Dec 30, 2006 #7
    Where does the above harmonic V(x) come from ?
    Your original way of working is correct. I didn't check all the algebra though.

    marlon
     
  9. Dec 30, 2006 #8
    Sorry! I made an erroneous assumption. :blushing: I think using the operator relation is the correct technique. Am I right?
     
  10. Dec 30, 2006 #9
    yes

    marlon
     
  11. Dec 30, 2006 #10

    Gokul43201

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    This is correct.

    This isn't right. The RHS is simply the kinetic energy, not the total energy.
     
  12. Dec 30, 2006 #11
    For the first part of working out the potential, you have to employ the TDSE, this is true.

    However, you should recognise that this is the wavefunction for a free particle, in the form:

    [psi] = Aexp[i(kx - wt)], where A = normalisation factor, p=[hbar]k, E = [hbar]w.

    Hence in this example, p = [hbar]a, E = [hbar]b.

    You may use the operator relations p[hat] = -i[hbar]d/dx, and H[hat] = i[hbar]d/dt to show this.

    Notice also how V = [hbar]w - ([hbar]k)2/2m = [hbar]w - p2/2m
    So V = E - KE
    i.e. Potential Energy = Total Energy - Kinetic Energy

    Which agrees with Total Energy = Kinetic Energy + Potential Energy
     
    Last edited: Dec 30, 2006
  13. Dec 31, 2006 #12
    Thanks for the responses Gokul, Marlon and Worzo! Things are making better sense now. :smile:
     
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