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Particle in xy plane

  1. May 19, 2015 #1

    Maylis

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    1. The problem statement, all variables and given/known data
    upload_2015-5-19_22-23-32.png upload_2015-5-19_22-23-51.png
    upload_2015-5-19_22-24-17.png
    upload_2015-5-19_22-24-34.png

    2. Relevant equations


    3. The attempt at a solution
    a) I know that I am allowed to separate variables, such that ##\psi (r, \theta) = R(r) \Theta(\theta)##, but the part with the exponential is yet to be seen. So I put it into the Hamiltonian,
    $$H = - \frac {\hbar^{2}}{2 \mu} \Big ( \frac {1}{R} \frac {\partial^{2} R}{\partial r^{2}} + \frac {1}{R} \frac {1}{r} \frac {\partial R}{\partial r} + \frac {1}{r^{2}} \frac {1}{\Theta} \frac {\partial^{2} \Theta}{\partial \theta^{2}} \Big ) + V $$

    Since V is a function of r, it is not a constant, so I can't just call every term with derivatives as constants, can I? This is what I would normally be able to do, particularly if V = 0. So how do I proceed from here?
     
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  3. May 19, 2015 #2

    mfb

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    You can take another angular derivative, then all the non-trivial R terms will go away.
    The problem statement is misleading, there are solutions that cannot be written in that way, e.g. superpositions of states with different m.
     
  4. May 19, 2015 #3

    Maylis

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    What is the other angular derivative you are talking about? I just see the one in my equation.
     
  5. May 19, 2015 #4

    mfb

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    "Another" as in "one more". You can calculate ##\frac{dH}{d\theta}##.
     
  6. May 19, 2015 #5

    Maylis

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    Wouldn't that give some sort of triple derivative on the right side of my equation for the third term in the parenthesis?
     
  7. May 19, 2015 #6

    mfb

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    Sure.
     
  8. May 19, 2015 #7

    vela

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    It might help you to write out the complete equation.
    $$-\frac{\hbar^2}{2\mu}\left(\frac{R''}{R} + \frac{R'}{rR} + \frac{\Theta'}{r^2\Theta}\right) + V = E.$$ Multiply by ##r^2## and you can separate the terms into those depending on ##r## and those that depend only on ##\theta##.
     
  9. May 20, 2015 #8

    Maylis

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    Okay, so I can separate the equation

    $$ -(H-V)r^{2} \frac {2 \mu}{\hbar^{2}} - \frac {r^{2}}{R} R'' - \frac {r}{R} R' = \frac {\Theta''}{\Theta} $$
    But how can I solve this?
     
  10. May 21, 2015 #9

    mfb

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    The left side is independent of theta, which means the whole equation has to have the same value for all values of theta. You can set the left side equal to some constant (it will depend on r but not on theta). This is the same as setting the derivative (with respect to theta) to zero.
     
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