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Solving the radial Schroedinger equation with a linear potentail

  1. Aug 11, 2013 #1
    1. The problem statement, all variables and given/known data

    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.
     
  2. jcsd
  3. Aug 11, 2013 #2

    TSny

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    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: Aug 12, 2013
  4. Aug 12, 2013 #3
    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.
     
  5. Aug 12, 2013 #4

    TSny

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    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 [itex]\widetilde{u}[/itex](r) such that u(r) = λ[itex]\widetilde{u}[/itex](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 [itex]\widetilde{u}[/itex](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 [itex]\widetilde{E}[/itex], 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.
     
  6. Aug 12, 2013 #5
    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)?
     
  7. Aug 13, 2013 #6

    TSny

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    [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: Aug 13, 2013
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