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A particle in a spherical symmetric potential

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

    A particle that moves in three dimensions is trapped in a deep spherically symmetric potential V(r):

    V(r)=0 at r<r_0

    infinity at r>= r_0

    where r_0 is a positive constant. The ground state wave function is spherically symmetric , so the radial wave function u(r) satistifies the 1-dimensional schrodinger energy eigenvalue equation (-[tex]\hbar[/tex]2/2m)*d^2/dr^2*u(r)+V(r)*u(r)=Eu(r) with the boundary condition u(0)=0 (eqn u(r)=r*phi(r)

    a) Explain why, in the potential well in the equation V(r)=0 at r<r_0

    infinity at r>= r_0, the wave function is forced to vanish at r=r_0

    b) Using the known boundary conditions on the radial wave function u(r) at r=0 and r=r_0 , find the ground state energy of the particle of this potential well

    2. Relevant equations

    (-[tex]\hbar[/tex]2/2m)*d^2/dr^2*u(r)+V(r)*u(r)=Eu(r)
    u(r)=r*phi(r)


    3. The attempt at a solution

    a)I don't exactly know what they mean by vanish in this context. Does V vanish when V=0? I know in a classically allowed region, when a particle just sits in a well and there is no external energy being transferred to the particle, then V=0.

    b) To begin this problem, I should probably plug u(r) into (-[tex]\hbar[/tex]2/2m)*d^2/dr^2*u(r)+V(r)*u(r)=Eu(r). Not sure how following this procedure will help me find the value for the ground state energy.
     
  2. jcsd
  3. Sep 1, 2009 #2
    For A), the question is why the wavefunction must vanish (the angular parts are irrelevant; only u(r) really need vanish) at r=r0 where the potential jumps to be unboundedly large. The wavefunction does not "seep into" the region of infinite potential. The reason this must be can be seen in Schroedinger's equation: What happens when we try to put in V=infinity? What would d^2/dr^2*u or E have to be to satisfy the equation, and why would this be physically ruled out? On the other hand, if we just demand u=0 starting at r0, then the equation does not even need to be solved in a non-trivial way...it is trivially solved already.
    For B), this kind of reasoning is similar in many 1-dimensional problems. First think of just the well region where V=0. Then in that region you'd get free particle plane waves as solutions to the equation. But of course the particle isn't free because the V=0 region is not all there is to this particle's world; it sees the V=infinity jump. But we know that the conditions on the function u(r) are: u(r=0)=0=u(r=r0). So that just restricts the types of wavefunctions u(r) can be: Not just any exp(ikr) type plane wave, but a subset of them...what type of waveforms? Remember that the waveform you write down must be chosen so that the boundary conditions u(r=0)=0=u(r=r0) could be satisfied. You will also have to write down conditions on k such that the conditions are satisfied, just like in other 1-dimensional problems.

    What is the ground state function, u(r), then? (Hint: note that we can write k in terms of known quantities). Putting u(r) in the Schr. eqn for the region r=[0,r0] will give you an eqn for E in terms of the mass, r0 and h-bar (Hint: What is V for this region?).
     
  4. Sep 2, 2009 #3
    If V is at infinity, then C= (2m/h-bar^2)*(V(x)-E)=(1/phi)*(d^2/dx^2) approaches infinity as well which means that wave function bends far away from the axis and the ground state function approaches zero.=> u=0, since u=r*phi_0


    wouldn't u=sin(k*pi) , k being an integer; so now I would be able to put u in my schrodinger equation to find V?
     
  5. Sep 2, 2009 #4
    er... bump! DId I apply the correct line of reasoning and appproach to the problem?
     
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