What is the minimum potential for a bound state in a 3D potential well?

  • Thread starter Thread starter noblegas
  • Start date Start date
  • Tags Tags
    Minimum Potential
noblegas
Messages
266
Reaction score
0
finding minimum potential
1. Homework Statement

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. Homework 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 separation 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
 
Last edited:
Physics news on Phys.org
First, let me clean up the tex:

Potential is V(r) = -V_0 for r &lt; r_0

u(r) satisfies the equation

-\hbar^2/2m \frac{d^2 u}{d r^2} + V(r) u(r) = E u(r) where u(r) \equiv \varphi(r) r

What they mean by regular at the origin is "smooth at the origin". For example the linear function f(r) = r is not smooth at the origin when viewed as a function of x, y, and z.

To see this write: f(r) = r = (x^2 + y^2 + z^2)^{1/2}. Let's see what this function looks like along the x axis where y=z=0. Along this axis f(x) = (x^2)^{1/2} = |x|. This function has a cusp at x=0 and so is not smooth. On the other hand f(r)=r^2 is smooth. But f(r) = r^3 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
r &gt; r_0.

Good luck
 
Last edited:
ignore this post. i posted a more clearer post.
 
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
The value of H equals ## 10^{3}## in natural units, According to : https://en.wikipedia.org/wiki/Natural_units, ## t \sim 10^{-21} sec = 10^{21} Hz ##, and since ## \text{GeV} \sim 10^{24} \text{Hz } ##, ## GeV \sim 10^{24} \times 10^{-21} = 10^3 ## in natural units. So is this conversion correct? Also in the above formula, can I convert H to that natural units , since it’s a constant, while keeping k in Hz ?
Back
Top