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Dimensional analysis of the SED

  1. Nov 27, 2012 #1
    I have an exercise at the moment where I am supposed to put the Schrödingerin dimensionless form (the exact exercise is attached). I must admit that this idea of dimensional analysis is quite new to me. I don't understand how you can write the SED in the dimensionless form described. Therefore I could use some hints from one of you :) They introduce this new x' = x/x0. To put the SED in the given form are we then supposed to substitute x = x'x0 in the SED? If so I don't see how the h^2/2m disappears.

    Attached Files:

  2. jcsd
  3. Nov 27, 2012 #2
    The second derivative also changes:
    \frac{\partial^2}{\partial x^2} \stackrel{?}{=} k \, \frac{\partial^2}{\partial x'^2}
    How is k related to x0?
  4. Nov 28, 2012 #3
    I guess the derivative somehow would throw a factor of 1/x02 in but I am overall unsure about what is done in this exercise.
    Do we switch variables from: x-> x'/x0? In that case everything that is named x should just be changed to x' and I can't see what sense that would make. On the other hand we could subsitute x= x'x0 but I don't see that going anywhere. Can you in detail explain what the idea is?
  5. Nov 28, 2012 #4
    I'm not here to do your homework. You are right that you get a factor of [itex]1/x^2_0[/itex] in front of the second derivative. Now, if you multiply by
    \frac{m x^2_0}{\hbar^2}
    the coefficient in front of the second derivative becomes [itex]-1/2[/itex], as required in the problem. You can read off what [itex]\tilde{V}_0[/itex] and [itex]\tilde{E}[/itex].
  6. Nov 29, 2012 #5
    you are not here for my homework? No, I guess not but you are here to help me, and I think you can trust me on the fact that I have really tried to think this over but can't make sense of it:
    So I ask again, in a more elaborative way, what is that is done:
    Do we switch variables from x-> x'? In that way I don't see how the factor of 1/x02 comes in since you would basically just replace every x by x'? You could I suppose plug in x = x'x0 but that doesn't seem to make sense either. Can't you see the problem? A student pointed out the problem today too and the teacher agreed in a way and said something I didn't really get.
    If you really don't want to help me fine, but please don't reply to my posts in future times :)
    Last edited: Nov 29, 2012
  7. Nov 29, 2012 #6
    Please read through what has been already written carefully, instead of rambling on things that do not make sense.
  8. Nov 29, 2012 #7
    I have! But your comments merely state that the derivative changes which throws in a factor of 1/x02. So in principle I can solve the problem yes. But I would like to understand what it is you do to obtain this. Why this aggressive attitude? I asked a question - if you don't want to help me then simply don't respond. And if things I say doesn't make sense maybe that should hint you at the fact that I don't understand the general procedure in this exercise.
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