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The Dirac delta in squere root of the absolute value

  1. Jan 25, 2012 #1
    Dear Forum Users,

    I have got more math question rather then the physics question. Does someone know if:

    [itex]\mid d(x)\mid^2[/itex]

    equals just [itex]d(x)[/itex], here [itex]d(x)[/itex] is just the Dirac delta ?

    best regards,
    nykon
     
  2. jcsd
  3. Jan 25, 2012 #2

    strangerep

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    It's undefined. In general, you cannot sensibly multiply distributions at the same point.
     
  4. Jan 25, 2012 #3

    Hurkyl

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    Minor clarification -- strangerep means distributions of the same variable. It's fine to multiply them if they are of independent variables, such as [itex]\delta(x)\delta(y)[/itex] or even [itex]\delta(x) \delta(x-y)[/itex]. (assuming x and y are independent)
     
  5. Jan 26, 2012 #4

    bhobba

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    That's true - although I have read somewhere while it is thought you can't give a meaning to such things in general it hasn't actually been proven yet - actually I seem to recall it was in Theory of Distributions: A Non-technical Introduction by Richards and Youn, a copy of which I have.

    Thanks
    Bill
     
    Last edited: Jan 27, 2012
  6. Jan 27, 2012 #5

    strangerep

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    That book does not have "look inside" enabled on Amazon, so I can't browse it easily.

    But... the way you've stated it sounds like a non-sequitur. In maths, one can only prove things based on the precise definitions and axioms one starts with.

    Could you look up the precise quote and context, pls?
     
  7. Jan 27, 2012 #6

    bhobba

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    Got the book and refreshed my memory about what is said. Its in chapter 7 and says, regarding multiplying any two distributions together, 'there is strong evidence such a general definition does not exist'. However they later express the view 'Finally we mention the thought, and hope, that non standard analysis (a theory that allows the use of infinitesimal quantities) may someday provide a mush simpler treatment of these questions. However that must lie outside the scope of this book.'

    In math you often think things are a lot more certain than statements like the above where it is freely admitted people, at this point in time, don't really know. This is not good but the reality of the situation. In QFT you sometimes see expressions containing terms like the Dirac Delta function squared but it is well known in math, at least to the best of my knowledge, no one has figured out how to properly define it. - I suppose thats the reason they say there is strong evidence it can't be done.

    In QFT the way I view it is to think of the Dirac delta function as the integral of e^2i pi x t over t from infinity to - infinity. However, in the spirit of a cutoff I think of it as not being over infinity but a very large number so for all practical purposes its behaves like a delta function but is a perfectly good function that you can take the square of etc.

    Thanks
    Bill
     
  8. Jan 27, 2012 #7

    jambaugh

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    nykon,
    To clarify some points. The Dirac delta is not truly a function (hence asking what [itex]\delta(x) ^2[/itex] means is not that simple).

    If you think of real valued functions as elements of an abstract vector space then you can define an inner (dot) product on (some) of them by:
    [tex] f\bullet g = \int_{-\infty}^{\infty}f(x)g(x)dx[/tex]
    (or we can choose other limits of integration.)
    Now when dealing with vector spaces you can define the dual space of linear mappings from vectors to numbers, called linear functionals. In enumerating the linear functionals on a space, if it is finite dimensional all of them can be expressed using the dot product with respect to other vectors: [itex] \phi(\mathbf{v})= \mathbf{u}\bullet \mathbf{v}[/itex] for some [itex] \mathbf{u}[/itex].

    This is a very convenient fact but it doesn't hold once you consider infinite dimensional vector spaces like typical function spaces. As an example evaluation of a function at 0 is itself a linear functional. You plug in a function f and get out a number f(0). You can't find another function [itex]g[/itex] such that: [itex] f(0) = g\bullet f[/itex]. So we invent a symbolic representation of such, call it [itex]\delta[/itex] which isn't a function but takes the place of a function in the notation. Its only defined when used with the inner product (inside that integral above) to mean the specific linear functional:
    [tex] f \mapsto f(0)=\delta\bullet f = \int_{-\infty}^\infty \delta(x)f(x)dx[/tex]

    This and similar notations (such as the derivatives of the delta or dually shifted delta) are called distributions or sometimes generalized functions.

    As to your question, firstly we can express the delta distribution as the limit of actual functions as you've likely seen in texts. As such it may be meaningful to say [itex]|\delta(x)|=\delta(x)[/itex] based on properties. However to give [itex]\delta(x)^2[/itex] meaning we would have to be able to evaluate:
    [tex]\int_{-\infty}^\infty \delta(x)\delta(x)dx = \delta(0)[/tex]
    Since the Dirac delta really isn't a function this value isn't meaningful and so neither is the notation of the square of the delta.

    In summary, since the space of functions is infinite dimensional there are more linear mappings from functions to numbers than can be expressed by integrating them times other functions. To express some of these we extend the idea of functions to distributions of which the Dirac delta is an example. In this context squaring distributions such as the Dirac delta is not meaningful. Note however we CAN shift, scale, add and take derivatives of distributions, e.g.
    [tex]\int_{\mathbb{R}} \delta'(x-a)f(x)dx = f'(a)[/tex]
    Regards,
    James
     
  9. Jan 27, 2012 #8

    DrDu

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    Nevertheless the square of the delta function (or something related) shows up regularly in physical problems.
    Consider for example a unitary transformation [itex]U(x)=\exp(i\pi\theta(x))[/itex] and the Hamiltonian [itex]H=p^2/2m[/itex]. What do you get for [itex] U^+HU[/itex]?
     
  10. Jan 27, 2012 #9
    In any case, whenever the square of the Dirac delta distribution shows up, it is important to investigate why. Are your assumtions correct, and everything well-defined? E.g. maybe you have used a step function where only continuous or C^1 functions are allowed, etc.
     
  11. Jan 27, 2012 #10

    bhobba

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    That's true - but the rigorous theory of the Delta function, distribution theory, to the best of my knowledge anyway, leaves it undefined. As I said, the way I handle it, in applied problems, is to think of it as an ordinary function that for all practical purposes behaves like the Delta function but really isn't. It sort of like considering dx etc as actual quantities so you can do informal manipulations.

    I don't know if it appears in the example above - but I have seen it appear in QFT problems eg some of the early chapters of Zee.

    Thanks
    Bill
     
    Last edited: Jan 27, 2012
  12. Jan 27, 2012 #11

    strangerep

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    ??? I must be missing something...
    [tex]
    U^+ H U ~=~ H ~+~ U^+[H,U]
    [/tex]
    so if we take CCRs to be [itex][x,p]=i[/itex], which implies [itex][p,f(x)] = -i f'(x)[/itex] for a large class of analytic functions f, then we have (redefining [itex]\pi\theta \rightarrow \theta[/itex] to reduce the mess)...

    [tex]
    2m [H,U] ~=~ p [p,U] + [p,U] p
    [/tex]
    [tex]
    [p,U] ~=~ \left[p, e^{i\theta(x)}\right]
    ~=~ e^{i\theta(x)} (-i) i\theta'(x)
    ~=~ U \theta'(x)
    [/tex]
    Hence
    [tex]
    U^* [H,U] ~=~ \frac{1}{2m} \Big(U^* p U \theta'(x) + \theta'(x) p \Big)
    ~=~ \frac{1}{2m} \Big(p \theta'(x) + U^*[p,U] + \theta'(x) p \Big)
    [/tex]
    [tex]
    ~~~~~~=~ \frac{1}{2m} \Big(p \theta'(x) + \theta'(x) + \theta'(x) p \Big)
    [/tex]
    So where is the Dirac delta squared? What have I overlooked?
     
  13. Jan 27, 2012 #12

    strangerep

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    Thanks.

    OK, so it's just speculative wishful thinking on their part. Nonstandard analysis has been around for a long time now, but so far no one (afaik) has applied to QM in a way that shows it to be superior in practice from conventional techniques.

    But we can say for sure that the desired rigorous definition does not exist at this time.

    Can you give a specific example, pls?

    Generalized functions are typically limits of sequences of ordinary functions. But to have a good theory, the essential properties must survive in the limit. Pointwise multiplication does not, and this can (eg) compromise Lorentz invariance.
     
  14. Jan 27, 2012 #13

    Hurkyl

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    It's a lot more like claiming that you can divide by zero because you think of 0 as being a really small positive number.

    It can work very well if what you're actually doing is similar to what you pretend is happening -- e.g. you can deduce [itex]\lim_{x \to 0} x/x = 1[/itex] by simply plugging in 0, and then dividing a "really small positive number" by itself. But most other times you'll be disasterously wrong.

    Incidentally, dx is a "quantity" -- a differential form.


    Non-standard analysis doesn't help you multiply distributions in exactly the same way it doesn't help you divide by zero. What it might do is help you avoid having to consider multiplying tempered distributions, in a similar fashion to how it helps you avoid having to consider division by zero. (i.e. because the calculation you're doing leaves you with a nonzero infinitesimal in the denominator)
     
    Last edited: Jan 27, 2012
  15. Jan 28, 2012 #14

    bhobba

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    I have seen it in Zee's textbook on QFT, and it somtimes arises in QFT but I couldnt locate it easily. However its is used with gay abandon in Chapter 7 of QFT Demystified - The Feynman Rules where you have Dirac Delta Functions multiplied by other delta functions with equations of the form integral something delta(p1 + p2 - q) delta(q - p3 + p4) dq then claim it can be done by inspection by applying the second delta function to the first so you have something (with p3 - p4 substituted for q) delta (p1 + p2 - p3 + p4) which you can do if you believe in the fiction it is an ordinary function - which of course it isn't.

    I know the rigorous definition of delta functions and other generalized functions. The formal manipulations you see with them are sometimes not justified - welcome to applied math.

    Thanks
    Bill
     
    Last edited: Jan 28, 2012
  16. Jan 28, 2012 #15

    bhobba

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    Actually most of the time its perfectly OK. Yes I know dx can be considered a differential form. I am thinking of the case where you think of dt as a small but actual number like in for example deriving the poison distribution. You write down the equation of what happens in the small time interval and ignore higher powers of dt. It works just fine. Rigorous - no way - but you see it all the time in applied literature. You can however make it rigorous by introducing terms of O(2) and taking a limit at the end. But often applied guys don't worry about it.

    Thanks
    Bill
     
  17. Jan 28, 2012 #16

    DrDu

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    I should have mentioned that theta is Heavisides step function whose derivative gives the delta function.
    Besides this, the last line above is not correct, specifically you seem to have lost a theta' in the second term, which is precisely the delta-square term.

    That doesn't seem to relevant as long as x ranges from -infinity to +infinity. But when you replace x by an angle phi, this becomes the Aharonov Bohm Hamiltonian in a special gauge. The strength of the delta function is then the phase shift of an electron upon encircling a magnetic flux line.

    The above example also shows how the delta squared has to be interpreted/avoided.
    The hamiltonian is defined only on functions with a discontinuity (what is certainly not in any of the function spaces on which distributions are defined normally), so that their derivativative will cancel the delta square.
     
    Last edited: Jan 28, 2012
  18. Jan 28, 2012 #17

    Hurkyl

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    I was talking about the "pretend 0 is a tiny positive number".

    Manipulating dx as an algebraic "quantity" is perfectly okay, because it is an algebraic quantity.

    The applied guys won't accept it either. If they don't look into a proof, then they are going to go for empirical verification. (they might go for empirical validation anyways even when given a proof)

    But honestly, the reason you don't see anything further done is because the aren't isn't to derive the Poisson distribution; it is to help one gain confidence the Poisson distribution is what it claims to be.




    Incidentally, your example is not a dx. I believe you can arrange it to be a [itex]\Delta x[/itex], though. (e.g. a tangent vector, if we use the differential geometric semantics of infinitesimal)

    And I would bet there is a perfectly rigorous argument that is a rather direct translation of "you expect twice as many hits in intervals twice as wide" -- i.e. a rigorous argument whose idea is more intuitive than the mumbled "and then ignore second-order effects" you suggest.
     
    Last edited: Jan 28, 2012
  19. Jan 28, 2012 #18

    strangerep

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    OK.

    Oops! So weird how one can be blind to things that are obvious when pointed out. (sigh)

    Hmm. I know A-B of course, but haven't seen it treated this way. So I should probably read up on that. Can you suggest a suitable reference?

    :biggrin:

    OK. That's all making sense now. One must choose the space of functions and its dual quite carefully.

    Cheers.
     
  20. Jan 28, 2012 #19

    bhobba

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    I must disagree with you there. I did a degree in applied math and, at the time, in my hubris, was very concerned with rigour - doing my epsilonics was my favourite subject and I continued it onto Hilbert spaces and stuff. Take my word for it - when questions like that are raised their eyes roll back, they tell you stories about they know books that, while totally rigorous, you would not read, of disrespect of applied guys that use the Dirac Delta function with gay abandon and people like Von Neumann that chided them for it, and other ditties. Read the introduction of Von Neumanns Mathematical Foundation's of QM to see what I mean.

    Here is another paper of interest in this discussion:
    http://arxiv.org/pdf/quant-ph/9907069v2.pdf

    The basic point of the applied guys is it works - so there must be a reason for it and the pure guys need to investigate it. Trouble is issues such as squaring a delta function still remain.

    These days however I am rather sanguine about the issue, resolving stuff like multiplying two delta functions together by taking them to be ordinary functions that for all practical purposes are delta functions.

    Thanks
    Bill
     
    Last edited: Jan 28, 2012
  21. Jan 29, 2012 #20

    DrDu

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    In an idealized description of the AB effect, the magnetic flux [itex]\Phi[/itex] is concentrated in an infinitely small region around r=0 (using polar coordinates). The line integral [itex]\oint A (r,\phi)r d\phi=\Phi[/itex] whence for r not equal 0 [itex] A(r,\phi)=\Phi \delta (\phi)/2\pi r [/itex] is a valid choice for the vector potential. The hamiltonian for a particle in a magnetic field is [itex]H=(p-eA)^2/2m[/itex]. Expressing this in polar coordinates and splitting off some squareroot of r factor from the wavefunction should yield the hamiltonian we were discussing for the phi dependence.
     
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