Solving the radial Schroedinger equation with a linear potentail

  • Thread starter Thread starter forceface
  • Start date Start date
  • Tags Tags
    Linear Radial
forceface
Messages
9
Reaction score
0

Homework Statement



The problem is number three (page 3) on the following link.http://panda.unm.edu/pandaweb/graduate/prelims/QM_S12.pdf
I was going to type it out but it got too messy.

2. The attempt at a solution
What I want to know is the relevant unitless variable substitution that the problem suggests. I tried x(r)=u(r)/r^2 thinking that u(r) is in units of r^2. This just led to a very messy diff equ. So If anyone has any other ideas on what I could use, or if this is the right substitution and i merely made a mistake.
 
Physics news on Phys.org
Hello forceface. Welcome to PF!

Transforming to dimensionless variables just means letting ##r = a\tilde{r}## where ##a## is a constant (to be determined) that has the dimensions of length. So, ##\tilde{r}## is a dimensionless variable.

Likewise, let ##E = b\tilde{E}## where ##b## is a constant (to be determined) that has the dimensions of energy. Thus, ##\tilde{E}## is a dimensionless quantity.

See if you can rewrite the Schrodinger equation in terms of ##\tilde{r}## and ##\tilde{E}## and ##a## and ##b##. Then choose the constants ##a## and ##b## so that the Schrodinger equation becomes dimensionless of the form $$(-\frac{d^2}{d\tilde{r}^2} + \tilde{r})u = \tilde{E}u$$
 
Last edited:
This clears it all up, thank you. I was getting caught on the units of u(r) because I was thinking that it was not dimensionless but infact it is. So the idea behind this problem to convert everything else in this equation into dimensionless variables and then from that a relation can be obtained between the dimensionless E and the E in the original equation. The term relating the two E's is related to the allowed energies. But this doesn't mean we have solved for the allowed energies, maybe just an order of magnitude or something.
 
The wavefunction u(r) is not dimensionless since |u(r)|2 represents a probability per unit length. However, the dimensions of u(r) are not important in this question because u(r) appears on both sides of the Schrodinger equation so that its dimensions automatically cancel out. In other words, you could introduce a dimensionless \widetilde{u}(r) such that u(r) = λ\widetilde{u}(r) where λ is some constant with dimensions (length)-1/2, but λ would just cancel out when you rewrote the equation in terms of \widetilde{u}(r).

As you say, going over to the dimensionless form of the Schrodinger equation still doesn't tell you what the energy levels are. But, it does tell you that if you solved the dimensionless equation for the dimensionless energy levels \widetilde{E}, then the energy levels for the original problem would be ##E = (\frac{\hbar^2 k^2}{2m})^{1/3}\widetilde{E}##. That provides some very useful information. For example, if you replace the particle with a different particle with 8 times as much mass, then all of the energy levels would be reduced by 1/2. So you can see how the energy levels scale with the various parameters of the system.
 
Well if u(r) is defined at u(r)=r*R(r), where R(r) is the radial part of the wave function, what are the units of R(r)?
 
[EDITED] If R(r) is the radial part of the wavefunction: ψ = R(r)Y(θ,∅), then R(r) would have dimensions of length-3/2. So, u would have dimensions of length -1/2.

But I don't think the dimensions of u(r) makes any difference regarding this problem.
 
Last edited:
##|\Psi|^2=\frac{1}{\sqrt{\pi b^2}}\exp(\frac{-(x-x_0)^2}{b^2}).## ##\braket{x}=\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dx\,x\exp(-\frac{(x-x_0)^2}{b^2}).## ##y=x-x_0 \quad x=y+x_0 \quad dy=dx.## The boundaries remain infinite, I believe. ##\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dy(y+x_0)\exp(\frac{-y^2}{b^2}).## ##\frac{2}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,y\exp(\frac{-y^2}{b^2})+\frac{2x_0}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,\exp(-\frac{y^2}{b^2}).## I then resolved the two...
Hello everyone, I’m considering a point charge q that oscillates harmonically about the origin along the z-axis, e.g. $$z_{q}(t)= A\sin(wt)$$ In a strongly simplified / quasi-instantaneous approximation I ignore retardation and take the electric field at the position ##r=(x,y,z)## simply to be the “Coulomb field at the charge’s instantaneous position”: $$E(r,t)=\frac{q}{4\pi\varepsilon_{0}}\frac{r-r_{q}(t)}{||r-r_{q}(t)||^{3}}$$ with $$r_{q}(t)=(0,0,z_{q}(t))$$ (I’m aware this isn’t...
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...
Back
Top