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Please help with Quantum Mechanics Dirac Delta Problems!

  1. Sep 16, 2010 #1
    These problems are from Introductory Quantum Mechanics (Liboff, 4th Ed.)
    Note: I'm using "D" as the dirac delta function.

    3.9 (a) Show that D( sqrt(x) ) = 0

    This has me stumped.
    It is my understanding that the Dirac function is 0, everywhere, except at x=0.
    So, how can I show this to be true, when at x=0 (sqrt(0) = 0), the dirac function is 1?

    3.9 (b) Evaluate D( sqrt(x^2 - a^2) )

    My first hunch is that this function is:

    1 , at ABS(a) = ABS(x) (ABS - abosulte value)
    0 , everywhere else.

    However, since I don't get part (a), I'm guessing that my evaluation of (b) probably isn't correct.

    Any pointers would be GREATLY appreciated.
  2. jcsd
  3. Sep 16, 2010 #2


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    Actually, at x=0, the delta function is infinite, not 1. It's the integral that is equal to 1:
    [tex]\int_{-\infty}^{+\infty}\delta(x - x_0)\mathrm{d}x = 1[/tex]
    where [itex]x_0[/itex] is some number, or more generally,
    [tex]\int_{x_0-\epsilon}^{x_0+\epsilon}\delta(x - x_0)\mathrm{d}x = 1[/tex]
    where [itex]\epsilon[/itex] is a positive real constant.

    Here's the thing, though: the delta function is not actually a function, not in the usual sense of a rule that maps numbers to other numbers. It's really a distribution, which means that it only has a meaning when it's used as a factor in an integral. So the real definition of the Dirac delta distribution is something like
    [tex]\int_{x_0-\epsilon}^{x_0+\epsilon}\delta(x - x_0)f(x)\mathrm{d}x = f(x_0)[/tex]
    As a consequence, you can't argue that [itex]\delta(\sqrt{x})[/itex] is nonzero at [itex]x = 0[/itex] simply because [itex]\sqrt{0} = 0[/itex]. You have to actually put it into an integral:
    [tex]\int_{x_0-\epsilon}^{x_0+\epsilon}\delta(\sqrt{x})\mathrm{d}x = ?[/itex]
    To solve this, you can use u-substitution, setting [itex]u = \sqrt{x}[/itex]. I trust that you can take it from here.

    There is a chain rule for the delta distribution which is usually more convenient than actually doing that integral (especially for part b):
    [tex]\delta(g(x)) = \sum_{x_i:g(x_i)=0}\frac{\delta(x - x_i)}{g'(x_i)}[/tex]
    Basically for each root [itex]x_i[/itex] of [itex]g(x)[/itex], you compute a term [itex]\delta(x - x_i)/g'(x_i)[/itex] and add them up. I'm fairly sure this can be derived from the integral definition above.
  4. Sep 17, 2010 #3
    Thanks for the help.
    I’m still working on the finer points of using LaTex…I hope this comes out correctly.
    Liboff suggests using the form
    (this should be integral{f(x)D(x)dx} = f(0) - I'm not sure why it's not showing this way).
    \int_{-\infty}^{+\infty}f(x)\delta(x)\mathrm{d}xf(x) = f(0)

    I used substitution here:
    u = \sqrt{x}

    du = \frac{1}{2}x\x^{\frac{-1}{2}}dx
    (this keeps showing as a y...even though I'm using x)

    And ended up with
    2\int f(u^{2})\delta(u)udu
    (limits at infinity)

    I did a lot of substituting back and forth with u, x, du, and dx (not sure if this is “legal”).
    I went back and worked some of the early problems in this chapter (proofs of Dirac identities/properties). It seems like once you get to a form (as shown just above), you can say that this is equal to 0, as the integral of the delta function (over infinity) is equal to f(0), or
    [f(u^{2})u]\ |_{u=0} = 0
    Does this seems correct?
  5. Sep 17, 2010 #4


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    When you preview a post, the forum handles the LaTeX images incorrectly, so it might not show what you actually typed. You should be able to see the actual images by refreshing the page.

    Anyway, I'm not sure I quite follow what you're doing. Did you mean to write
    [tex]\int f(x)\delta(\sqrt{x})\mathrm{d}x[/tex]
    for the first integral? If so, I think you've got the idea.
  6. Sep 17, 2010 #5
    Yes, that's what I meant for the first integral...I got frustrated and kept adding f(x) everywhere just to see if the preview would change. I must've forgot to erase one.

    Thanks for the heads-up about the previewer; at least now I know it's not something that I'm doing.

    I think that I'm going to use the same approach for the (b) question in the OP.
    The first problem I ran into was that using u for x^2-a^2 still leaves a Dirac function with u^(1/2). Is it permissable to use yet another substution, such as z = u^(1/2) (as long as everything is changed accordingly)?

    Forgive me for all the calculus questions...it's been a while.
  7. Sep 19, 2010 #6


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    Sure, it's fine to substitute multiple times. Substitution just turns one integral into another integral; there's nothing special about the new integral that would prevent substitution from working on it.

    By the way, sorry about the delayed response; I was out of town over the weekend.
  8. Sep 20, 2010 #7
    Thanks for all the help with this one.

    Thanks for the chain rule. The dirac chain rule you listed is actually listed in the textbook; it's only listed as a problem for the reader to prove; it's not explicitly listed as a Dirac Delta property to know, love, and apply.

    Anyway, using the chain rule for (a) clearly and quickly proves that this is = 0.
    I won't post the work for (b), as I'm pretty sure this is zero as well. Using g(x) = sqrt(x^2 - a^2) yields g'(x) that gives 0, for the two roots of g(x).

    I think I'm beginning to understand how this textbook was written. Only the basics/fundamentals are provided, leaving all "commonly known" properties, identities, relationships, etc. for the reader to prove and work through on their own (given as problems).
  9. Sep 20, 2010 #8
  10. Sep 20, 2010 #9


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    Yes, that's already been covered (it's exactly the formula I gave earlier in the thread).
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