- #1
Xyius
- 501
- 4
Hello,
This latest homework I have been doing has been very confusing to me and I have spent hours trying to complete it. Here is a problem that I really don't know where to start. If anyone could just point me in the right direction, or let me know if my ideas are correct or not, it would be so great!
Consider a particle in a central potential of the form..
V(r)=-U\frac{a}{r}e^{-\frac{r}{a}}
Where U>0 and a>0.
(i) Show that, if ##U >> \frac{\hbar^2}{ma^2}##, in the zeroth order approximation the eigenkets and eigenenergies of the low-lying levels have the same form as in the Coulomb potential. Determine the eigenenergies to zeroth order.
(ii) Compute all corrections to those eigenenergies in first order perturbation theory. Hint: Estimate the order of magnitude of second order corrections and compute all first order corrections that are greater.
The problem says the following equations might be useful.
<r^k>=\int_0^{∞}dr r^{2+k}[R_{nl}(r)]^2
<r>=\left( \frac{a_0}{2Z} \right)[3n^2-l(l+1)]
<r^2>=\left( \frac{a_0^2 n^2}{2Z^2} \right)[5n^2+1-3(l+1)]
<1/r>=\frac{Z}{n^2 a_0}
<1/r^2> = \frac{Z^2}{n^3a_0^2\left( l+\frac{1}{2} \right)}
For some reason, "zeroth order" is confusing me. We talked about calculating things to first and second order, and I am having some sort of "disconnect" as to what I am supposed to do. The only thing I can think of is the following.
Expand the exponential term in the central potential function.
V(r) \approx -U\frac{a}{r}\left[ 1-\frac{r}{a}+\frac{r^2}{2!a^2}+ \dots \right]
The zeroth order would be only the first term. Meaning the potential function would become..
V(r) \approx -U\frac{a}{r}
Which is of the same form as the Coulomb potential. I can then just make a substitution in the final results of the Coulomb potential and obtain the eigenenergies and eigenvectors. But I don't know where the ##U >> \frac{\hbar^2}{ma^2}## comes into play. Also, I don't understand the "Hint" in part (ii).
If anyone could steer me in the right direction it would be great!
This latest homework I have been doing has been very confusing to me and I have spent hours trying to complete it. Here is a problem that I really don't know where to start. If anyone could just point me in the right direction, or let me know if my ideas are correct or not, it would be so great!
Homework Statement
Consider a particle in a central potential of the form..
V(r)=-U\frac{a}{r}e^{-\frac{r}{a}}
Where U>0 and a>0.
(i) Show that, if ##U >> \frac{\hbar^2}{ma^2}##, in the zeroth order approximation the eigenkets and eigenenergies of the low-lying levels have the same form as in the Coulomb potential. Determine the eigenenergies to zeroth order.
(ii) Compute all corrections to those eigenenergies in first order perturbation theory. Hint: Estimate the order of magnitude of second order corrections and compute all first order corrections that are greater.
Homework Equations
The problem says the following equations might be useful.
<r^k>=\int_0^{∞}dr r^{2+k}[R_{nl}(r)]^2
<r>=\left( \frac{a_0}{2Z} \right)[3n^2-l(l+1)]
<r^2>=\left( \frac{a_0^2 n^2}{2Z^2} \right)[5n^2+1-3(l+1)]
<1/r>=\frac{Z}{n^2 a_0}
<1/r^2> = \frac{Z^2}{n^3a_0^2\left( l+\frac{1}{2} \right)}
The Attempt at a Solution
For some reason, "zeroth order" is confusing me. We talked about calculating things to first and second order, and I am having some sort of "disconnect" as to what I am supposed to do. The only thing I can think of is the following.
Expand the exponential term in the central potential function.
V(r) \approx -U\frac{a}{r}\left[ 1-\frac{r}{a}+\frac{r^2}{2!a^2}+ \dots \right]
The zeroth order would be only the first term. Meaning the potential function would become..
V(r) \approx -U\frac{a}{r}
Which is of the same form as the Coulomb potential. I can then just make a substitution in the final results of the Coulomb potential and obtain the eigenenergies and eigenvectors. But I don't know where the ##U >> \frac{\hbar^2}{ma^2}## comes into play. Also, I don't understand the "Hint" in part (ii).
If anyone could steer me in the right direction it would be great!