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Continuity of Function and Derivative at boundary.

  1. Oct 18, 2009 #1
    I am reading both Griffiths and Gasiorowicz and I can't get either of them to tell me why the continuity of the derivative of the natural log of the amplitude

    [tex]\frac{d(ln(u(x)))}{dx}=\frac{1}{u(x)}\frac{du(x)}{dx}[/tex]

    or put a different way

    [tex]\frac{1}{u(a^{-})}\frac{du(a^{-})}{dx}= \frac{1}{u(a^{+})}\frac{du(a^{+})}{dx}[/tex]

    at a boundary a implies that

    [tex]u(a^{-})=u(a^{+})[/tex]

    and

    [tex]\frac{du(a^{-})}{dx}=\frac{du(a^{+})}{dx}[/tex]


    Why is this true?



    It looks to me that there still is some ambiguity (as in the top equation is neccessary but not sufficient) because

    [tex]\frac{1}{u(a^{-})}\frac{du(a^{-})}{dx}= \frac{1}{u(a^{+})}\frac{du(a^{+})}{dx}[/tex]

    can be reconfigured to

    [tex]\frac{u(a^{+})}{u(a^{-})}=\frac{\frac{du(a^{+})}{dx}}{\frac{du(a^{-})}{dx}}[/tex]

    Which just means that the ratios have to be equal. But say for instance that [tex]u(a^{+})[/tex] is twice as big as [tex]u(a^{-})[/tex] so that the ratio is 2. Then all that means is the ratio of the slopes at a has to be 2. So, I don't see where they are limited to be identical.
     
    Last edited: Oct 18, 2009
  2. jcsd
  3. Oct 19, 2009 #2

    Avodyne

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    It isn't true.

    What is true is that continuity of both u and du/dx can be derived from the Schrodinger equation. And then, since the overall normalization is not fixed by the Schrodinger equation, it's equivalent to just consider continuity of the log derivative.
     
  4. Oct 19, 2009 #3
    Thank you! I appreciate the reply. I am glad to hear that I wasn't missing something in the math. I understand that Schrodinger's equation forces the continuity of Psi and it's derivative. But there still is a gap in my understanding, I do not know what is meant by "since the overall normalization is not fixed by the Schrodinger equation" and then how the log derivative follows from that. Can you expand on that just a little more?

    Is the "normalization" that you are referring to the normalization of |Psi|^2 over all space?
     
    Last edited: Oct 19, 2009
  5. Oct 20, 2009 #4

    Avodyne

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    Oops, what I said isn't quite true either.

    You need both conditions to get both the energy eigenvalue and the eigenfunction. But if we are interested in only the eigenvalue (or we want to get that first), then taking the ratio removes the unknown info about the eigenfunction.

    And yes, by normalization I mean the integral of psi^2.
     
  6. Oct 20, 2009 #5
    So the problem for me lies in this: It is our task to match two different functions and their derivatives at the boundary because Schrodinger's equation tells us it has to be that way. If we then just find two solutions in different regions that meet at a boundary then they must be continuous as well as their derivatives because the Schrodinger equation tells us it must be that way. But the book says that having the very first equation (composite equation) be true necessarily means the other two equations are true and I just don't see how that happens.

    In other words, we need to find two solutions that fit into Schrodinger's constraints at the boundary. These solutions to start with have arbitrary constants all over the place so they can take on any finite value at the boundary. Then we apparently use the combined equation to simplify the finding out of these constants. Instead of doing two constraints we apply just one that apparently contains both. I don't understand how that combined equation can contain both for the reasons given in the first post.

    Can you explain exactly how the normalization combined with this composite equation guarantees that we will find the constants that give us identical boundaries?



    *Another view.*

    Suppose we do the normalization integral, then we will get a sum of products/etc of constants that have to equal 1. The let's say we pick these constants so that they purposely solve the normalization equation, but also so that they do not satisfy the boundary rules that the Schrodinger equation insists on. Is it possible to do this? If not why are we guaranteed that we can't do this?
     
    Last edited: Oct 20, 2009
  7. Oct 20, 2009 #6

    Avodyne

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    You don't see how that happens because the book is wrong.

    Oops, I did it again! This is only correct in special situations, where, before matching, you know the wave function in each region up to overall normalization. This happens if the potential is even, and you therefore know that the energy eigenfunctions are even or odd.

    But, more generally, you don't know this, and then you do indeed have to separately match the wave function and its derivative at each boundary.
     
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