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finding minimum potential |
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| Sep6-09, 12:29 AM | #1 |
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finding minimum potential
finding minimum potential
1. The problem statement, all variables and given/known data A particle of mass m moves in 3-d in the potential well LaTeX Code: V(r)=-V_0 at LaTeX Code: r<r_0 where LaTeX Code: V0 and LaTeX Code: r_0 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 LaTeX Code: -h-bar^2/2m*(d^2/dr^2)*u(r)+V(r)*u(r)=Eu(r) LaTeX Code: u=\\varphi*r Find the minimum value of the depth LaTeX Code: V_0<BR> for which there exists a bound state. (recall that the radial function satisfies the condition u(0)=0 , because LaTeX Code: \\varphi (r)= u(r)/r has to be regular at the origin 2. Relevant equations LaTeX Code: -h-bar^2/2m*(d^2/dr^2)*u(r)+V(r)*u(r)=E*u(r) LaTeX Code: r^2=(x^2+y^2+z^2) 3. The attempt at a solution I don't know what they mean when they state ' LaTeX Code: \\varphi (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 LaTeX Code: u=R(r)*THETA(\\vartheta)*\\Phi(\\phi) and transform should i differentiate r^2 with respect to x? Report Post Edit/Delete Message |
| Sep6-09, 01:52 AM | #2 |
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First, let me clean up the tex:
Potential is [tex] V(r) = -V_0 [/tex] for [tex] r < r_0 [/tex] [tex] u(r) [/tex] satisfies the equation [tex] -\hbar^2/2m \frac{d^2 u}{d r^2} + V(r) u(r) = E u(r)[/tex] where [tex] u(r) \equiv \varphi(r) r[/tex] What they mean by regular at the origin is "smooth at the origin". For example the linear function [tex] f(r) = r [/tex] is not smooth at the origin when viewed as a function of [tex] x, y, [/tex] and [tex] z [/tex]. To see this write: [tex] f(r) = r = (x^2 + y^2 + z^2)^{1/2} [/tex]. Let's see what this function looks like along the [tex]x[/tex] axis where [tex]y=z=0[/tex]. Along this axis [tex]f(x) = (x^2)^{1/2} = |x|[/tex]. This function has a cusp at [tex]x=0[/tex] and so is not smooth. On the other hand [tex]f(r)=r^2[/tex] is smooth. But [tex] f(r) = r^3 [/tex] is not smooth. One thus concludes that smooth functions of "r" must have a Taylor series at the origin that proceeds in even powers of "r". Now as far as the solution of the equation is concerned. It is a linear ordinary differential equation so you should be able to solve it yourself. You are looking for bound states, i.e., solutions for which the wavefunction vanishes (please check this; I am not a physicist) for [tex] r > r_0 [/tex]. Good luck |
| Sep6-09, 02:27 AM | #3 |
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ignore this post. i posted a more clearer post.
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