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Functional differentiation

  1. Apr 25, 2014 #1

    Matterwave

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    Hi guys, I'm not sure where to put this question, so I'll just put it here. If a mod knows of a better place, just point me to it, thanks.

    I'm looking at the functional differentiation equation:

    $$\left.\frac{dF[f+\tau h]}{d\tau}\right|_{\tau=0}\equiv \int\frac{\delta F[f]}{\delta f(x)}h(x)dx$$

    I understand how this equation works (relatively well) for functionals of the form:

    $$F[f]=\int \mathscr{L}(f,f',f'',...)dx$$

    But there is another kind of functional that we often see in quantum field theory, which looks more like:

    $$F[f]=\mathcal{N}\int \left([e^{i\int \mathcal{L}(\phi(x),\phi'(x),...)f(x)dx}\right)\mathscr{D}\phi$$

    Here, we are integrating over functions ##\phi## rather than over the real line. Let's assume for the purposes of the discussion that this latter functional, taken as a whole (with the normalization constant), is well defined.

    Superficially at least, this second functional seems to be a different beast than the first, since we are integrating over functions rather than a real variable. However, this functional is a functional in that it takes functions and maps them to real numbers. In this second case of functionals, I'm not sure how to apply the rule for functional differentiation. Does it still work?

    Looking at the simplest functional of the second type, my analysis is as follows. Let's consider:

    $$F[f]=\mathcal{N}\int \left(e^{i\int \phi(x)f(x)dx}\right)\mathscr{D}\phi$$

    Now, let's construct ##F[f+\tau h]##:

    $$F[f+\tau h]=\mathcal{N}\int \left(e^{i\int[f(x)+\tau h(x)]\phi(x)dx}\right)\mathscr{D}\phi$$

    Taking the derivative ##d/d\tau## and the limit ##\tau\rightarrow 0##:

    $$\left.\frac{dF[f+\tau h]}{d\tau}\right|_{\tau=0}=\mathcal{N}\int\left(e^{i\int \phi(x)f(x)dx}i\int h(t)\phi(t)dt\right)\mathscr{D}\phi$$

    The problem I am left with here is that we are not left with an expression of the form in the first equation. We have a extra ##\mathcal{N}\int\mathscr{D}\phi## outside of everything. Taking the assumption that we can switch orders of integrals over the dummy variable t and the functions ##\phi## (which I am very dubious about), and looking at the first expression above, we might guess:

    $$\frac{\delta F[f]}{\delta f(x)}=i\mathcal{N}\int\left(\phi(x)e^{i\int \phi(t)f(t)dt}\right)\mathscr{D}\phi$$

    Even if we took this expression at face value, something seems to be very amiss since usually ##\delta F/\delta f## is a function, whereas here we just have a number (due to the integration over all ##\phi##)...

    Some help would be appreciated @_@
     
    Last edited: Apr 25, 2014
  2. jcsd
  3. Apr 27, 2014 #2
    At the level of theoretical physics the answer seems to be that yes the order can be changed, and I'm sure that your suspicion can be dealt with. I would advice to have a closer look at some simpler example. For example, suppose we considered functions

    [tex]
    f:\{1,2,\ldots, N\}^K\to\mathbb{R}
    [/tex]

    and then a functional

    [tex]
    F(f) = \sum_{a\in\{1,2,\ldots, N\}^K}f(a)
    [/tex]

    Then consider the case

    [tex]
    f(a) = \sum_{k=1}^K a_k^2
    [/tex]

    Now we can ask if the change of order

    [tex]
    F(f) = \sum_{a\in\{1,2,\ldots, N\}^K} \sum_{k=1}^K a_k^2\quad
    \overset{???}{=}\quad \sum_{k=1}^K \sum_{a\in\{1,2,\ldots, N\}^K} a_k^2\quad\quad (1)
    [/tex]

    is right. The answer is perhaps not obvious, because the domains of summation look different from what one usually sees, but if you study this, you can see that it is right. For the proof first assume that we have chosen some identification

    [tex]
    \{1,2,\ldots,N\}^K \sim \{1,2,\ldots, N^K\}
    [/tex]

    The total number of different mappings [itex]a:\{1,2,\ldots, K\}\to \{1,2,\ldots, N\}[/itex] is [itex]N^K[/itex] so we can put them in some order and have them identified with numbers. Then we can define a mapping

    [tex]
    g:\{1,2,\ldots, N^K\}\times\{1,2,\ldots, K\}\to\mathbb{R},\quad g(i,k) = a_k^2\quad\textrm{where}\; a\sim i
    [/tex]

    Now the equality marked with question marks in (1) is the same as

    [tex]
    \sum_{i=1}^{N^K} \sum_{k=1}^K g(i,k) = \sum_{k=1}^K \sum_{i=1}^{N^K} g(i,k)
    [/tex]

    which we know to be true.

    The intuition can be helped with a simple example [itex]N=2[/itex] and [itex]K=2[/itex]. We can also identify [itex]\{1,2\}\sim \{\alpha,\beta\}[/itex] so the calculations look more general.

    Let's look at the left side of (1). In outer sum the function [itex]a[/itex] will go through four values, which are

    [tex]
    a=(\alpha,\alpha),(\alpha,\beta),(\beta,\alpha),(\beta,\beta)
    [/tex]

    with each fixed [itex]a[/itex], the inner sum is obvious, so the quantity on left is by definition

    [tex]
    (\alpha^2+\alpha^2) + (\alpha^2+\beta^2) + (\beta^2+\alpha^2) + (\beta^2+\beta^2)\quad\quad (2)
    [/tex]

    Let's look at the right side of (1). If we fix [itex]n=1[/itex], in the inner sum the [itex]a_n^2[/itex] (now [itex]a_1^2[/itex]) will go through values

    [tex]
    a_1^2 = \alpha^2,\alpha^2,\beta^2,\beta^2
    [/tex]

    in the same order as above. If we fix [itex]n=2[/itex], in the inner sum the [itex]a_n^2[/itex] will go through values

    [tex]
    a_2^2 = \alpha^2,\beta^2,\alpha^2,\beta^2
    [/tex]

    So the quantity on the right side of (1) is by definition

    [tex]
    (\alpha^2+\alpha^2+\beta^2+\beta^2) + (\alpha^2+\beta^2+\alpha^2+\beta^2)\quad\quad (3)
    [/tex]

    Clearly the (2) and (3) are the same.

    This formula looks right, but you have only made a mistake when interpreting it. Notice that the "number" you are talking about depends on [itex]x[/itex]. For example, if I define a quantity

    [tex]
    Q(k) = \sum_{a\in\Omega} a_k^2
    [/tex]

    where [itex]\Omega\subset\{1,2,\ldots,N\}^K[/itex], the [itex]Q(k)[/itex] will depend on [itex]k[/itex] even though there is a sum over [itex]a[/itex]. So similarly your [itex]\frac{\delta F}{\delta f(x)}[/itex] depends on [itex]x[/itex] even though there is an integral over [itex]\phi[/itex].
     
    Last edited: Apr 27, 2014
  4. Apr 28, 2014 #3

    Matterwave

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    Hmmm, I'm not sure I follow your first argument. What do you mean by the first line:

    $$f:\{1,2,\ldots, N\}^K\to\mathbb{R}$$

    This is a function of N variables? Or something different? What's the superscript K? I am not familiar with this notation, sorry.

    I got your second point, that seems to make sense. Thanks.
     
  5. Apr 29, 2014 #4
    I just realized that my example could have been confusing because the [itex]f[/itex] in my example is not in the same role as your [itex]f[/itex]. I was only focusing on replacing the integrals with sums.

    Anyway, your question: The set [itex]\{1,2,\ldots, N\}^K[/itex] can be interpreted in two ways, but both are equivalent. The first is the ordinary cartesian product:

    [tex]
    \{1,2,\ldots, N\}\times \{1,2,\ldots, N\}\times\cdots\times \{1,2,\ldots, N\}
    [/tex]

    K times. If [itex]a[/itex] is its element, then [itex]a=(a_1,a_2,\ldots, a_K)[/itex] such that [itex]a_k\in\{1,2,\ldots, N\}[/itex] for all k.

    The second is that [itex]X^Y[/itex] always means the set of mappings [itex]Y\to X[/itex]. The number K can be identified with the set [itex]\{0,1,\ldots, K-1\}[/itex] or [itex]\{1,2,\ldots, K\}[/itex], and then the set [itex]\{1,2,\ldots, N\}^K[/itex] is the set of mappings [itex]\{1,2,\ldots, K\}\to\{1,2,\ldots, N\}[/itex], which are the same thing as the previously mentioned vectors [itex]a[/itex].

    Clarification on the similarities:

    The [itex]\phi:\mathbb{R}^D\to\mathbb{R}[/itex] in the original example is like [itex]a:\{1,2,\ldots, K\}\to \{1,2,\ldots, N\}[/itex] in my example.

    The integral [itex]\int\phi (x) f(x)dx[/itex] is like the sum [itex]f(a) = \sum_{k=1}^K a_k^2[/itex]. Notice that in the integral [itex]x[/itex] goes through its domain [itex]\mathbb{R}^D[/itex], and in the sum the [itex]k[/itex] goes through its domain similarly.

    The integral [itex]\int e^{i\int \phi(x) f(x)dx}\mathcal{D}\phi[/itex] is similar to the sum [itex]\sum_{a\in \{1,2,\ldots, N\}^K} f(a)[/itex]. So here the mapping [itex]\phi[/itex] goes through its possible values (different value meaning different mapping), and similarly the [itex]a[/itex], which can be seen as a mapping, goes through its possible values.
     
  6. Apr 29, 2014 #5
    I'm pretty sure this is supposed to be a functional integral, since we are integrating over all functions (whatever that's supposed to mean) and maps a number (assuming that it doesn't diverge, which I think for most cases it does) to a functional.
    So perhaps this could be interpreted as a function of a functional?

    I don't mean to jack the thread, but how are functional integrals actually defined?

    Edit: Didn't see this
    Right but how is that sum at the end supposed to converge?
     
    Last edited: Apr 29, 2014
  7. May 2, 2014 #6

    Matterwave

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    Ok, I'm sufficiently convinced...@_@
     
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