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Finding minimum potential

  1. Sep 6, 2009 #1
    1. The problem statement, all variables and given/known data

    A particle of mass m moves in 3-d in the potential well

    [tex] V(r)=-V_0 [/tex] at [tex] r<r_0[/tex]

    where [tex] V0 [/tex] and [tex]r_0[/tex] are positive constants. If there exists a state in which the particle is bound to the potential well, the wave function for the bound state with the lowest energy is spherically symmetric and the radial wave satisfies equations

    [tex] -h-bar^2/2m*(d^2/dr^2)*u(r)+V(r)*u(r)=Eu(r)[/tex]

    [tex] u=\varphi*r[/tex]

    Find the minimum value of the depth [tex] V_0
    [/tex] for which there exists a bound state. (recall that the radial function satisfies the condition u(0)=0 , because [tex]\varphi[/tex](r)= u(r)/r has to be regular at the origin

    2. Relevant equations


    [tex] -h-bar^2/2m*(d^2/dr^2)*u(r)+V(r)*u(r)=E*u(r)[/tex]



    [tex] r^2=(x^2+y^2+z^2) [/tex]


    3. The attempt at a solution

    I don't know what they mean when they state ' [tex]\varphi[/tex](r)= u(r)/r has to be regular at the origin'; I don't know why they want you to find a minimum value for V_0 since it is already given in the problem

    Should I apply seperation of variables where [tex] u=R(r)*THETA(\vartheta)*\Phi(\phi)[/tex]
    and transform should i differentiate r^2 with respect to x?
     
    Last edited: Sep 6, 2009
  2. jcsd
  3. Sep 6, 2009 #2

    Redbelly98

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    They mean that u(r)/r approaches a finite value (not ∞) as r→0.

    They are not just asking for a minimum value of Vo. They ask for the minimum value for which a bound state exists.

    [strike]Note that the particle's total energy is E. What is the relation between E and Vo in the case of a bound state? [/strike]

    (EDIT) What must be true about the total energy E in the case of a bound state?

    Since r is the only variable in your differential equation, you do not need separation of variables. That has already been done to arrive at the equation:
    [/URL]
     
    Last edited by a moderator: May 4, 2017
  4. Sep 6, 2009 #3
    [tex] u -> 0[/tex] as [tex] r -> 0 [/tex] but [tex][\phi [/tex] -> ∞

    Total Bound energy must be less than zero, and [tex] V_0 > E [/tex]
    .
     
  5. Sep 6, 2009 #4

    Redbelly98

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    Yes, u will approach 0 at r=0. Or we can just say that u(0)=0.

    Phi does not approach infinity. But since you just need to solve the differential equation for u, we needn't worry about phi here.

    Correct that E<0. But also E>-Vo. In other words:

    -Vo < E < 0​
     
  6. Sep 6, 2009 #5
    I don't understand why phi isn't approaching infinity; but should I plug in u=r*phi(r) into the Schrodinger equation?
     
  7. Sep 6, 2009 #6
    [itex]0 \leq \phi < 2 \pi[/itex] as [itex]\phi[/itex] is just the azimuthal angle.

    since [itex]u=r \phi[/itex], the reason [itex] u \rightarrow 0[/itex] is because [itex]r \rightarrow 0[/itex] regardless of what [itex]\phi[/itex] does.
     
  8. Sep 6, 2009 #7

    Redbelly98

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    The wavefunction phi must be continuous and have a continuous 1st derivative everywhere. Therefore it can't approach infinity at r=0 or anywhere else.

    No, just solve the equation for u. Making the substitution you suggest would overly complicate things.

    Basically, this tells us that the 2nd derivative of u(r) is a constant times u(r). The solution depends on whether that constant is negative or positive, but it is pretty straightforward to solve this.
     
  9. Sep 6, 2009 #8
    you mean u(r) is a constant times r right? so I would take the first and second derivatives of u(r) to find a solution for u(r)?
     
  10. Sep 6, 2009 #9

    Redbelly98

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    No, I meant what I said. Since V and E are constants,

    d2u / du2 = constant × u

    where the "constant" here involves Vo, E, and hbar2/2m.

    Yes.
    .

    By the way, I'll just mention where this all is going. Solving the differential equation for u(r), you'll eventually end up with an equation for the total energy E, involving the other constants in the problem (Vo, ro, etc.) as well.

    The key will be to figure out what values of Vo allow a solution where E agrees with the condition for a bound state,

    -Vo < E < 0​
     
  11. Sep 6, 2009 #10
    I am confused. I though I would be taking the second derivative of u with respect to r, being that u=phi(r)*r?
     
  12. Sep 6, 2009 #11

    Redbelly98

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    You have a differential equation in u, the same equation you posted under "Relavent equations" in your post #1. Simply solve it to find u!

    If you want to know what phi is, you can always use

    phi(r) = u(r) / r

    after you have found u(r).


    p.s. I'm logging off for a while, good luck :smile:
     
  13. Sep 6, 2009 #12
    [tex]-h-bar^2/2m*u_rr[/tex] [tex] +[/tex] [tex] -E-V_0 [/tex]=0 is a homogeneous equation and I can now find a solution for u .
     
  14. Sep 8, 2009 #13
    bump! would my final solution be a homogeneous solution and would it my homogeneous solution look like this:

    -h-bar^2/2m*u_rr=(E+V)*u and subsequently find a solution for u.
     
  15. Sep 8, 2009 #14

    Redbelly98

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    That should be (E - V)*u on the right hand side. And yes, you need to solve that differential equation to find u.
     
  16. Sep 9, 2009 #15
    would my bounds range from 0 to [tex] V_0[/tex]
     
  17. Sep 10, 2009 #16

    Redbelly98

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    If you mean the bounds on E, then no. See the very end of post #9 for the bounds on E.
     
  18. Sep 10, 2009 #17

    Redbelly98

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    By the way, the solution u(r) will be different for the two regions r<r0 and r>r0, since the value of (E-V) is different for those two regions. You'll have to solve the differential equation for each region.
     
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