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I How to compute second-order variation of an action?

  1. May 9, 2017 #1
    Starting with the action for a free scalar field $$S[\phi]=\frac{1}{2}\int\;d^{4}x\left(\partial_{\mu}\phi(x)\partial^{\mu}\phi(x)-m^{2}\phi^{2}(x)\right)=\int\;d^{4}x\mathcal{L}$$ Naively, if I expand this to second-order, I get $$S[\phi+\delta\phi]=S[\phi]+\int\;d^{4}x\frac{\delta S[\phi(x)]}{\delta\phi(x)}\delta\phi(x)+\int\;d^{4}x d^{4}y\frac{\delta^{2} S[\phi(x)]}{\delta\phi(x)\delta\phi(y)}\delta\phi(x)\delta\phi(y)$$ Now, assuming that ##\phi(x)## satisfies the equations of motion (EOM), then the first-order term vanishes, however, I'm unsure how to calculate the second-order variation. So far, my attempt is $$\delta^{2}S=\int\;d^{4}x d^{4}y\frac{\delta^{2} S[\phi(x)]}{\delta\phi(x)\delta\phi(y)}\delta\phi(x)\delta\phi(y)=\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\\=\int\;d^{4}x d^{4}y\left(\frac{\partial^{2}\mathcal{L}}{\partial\phi(x)\partial\phi(y)}\delta\phi(x)\delta\phi(y)+2\frac{\partial^{2}\mathcal{L}}{\partial\phi(x)\partial(\partial_{\mu}\phi(y)}\delta\phi(x)\delta(\partial_{\mu}\phi(y))+\frac{\partial^{2}\mathcal{L}}{\partial(\partial_{\mu}\phi(x))\partial(\partial_{\mu}\phi(y))}\delta(\partial_{\mu}\phi(x))\delta(\partial_{\mu}\phi(y))\right)$$ However, I am unsure how to progress (integration by parts doesn't seem to work as nicely in this case), as naively it seems as though the only term that would survive is ##\frac{\partial^{2}\mathcal{L}}{\partial\phi(x)\partial\phi(y)}##, but I've seen references stating that ##\frac{\delta^{2} S[\phi(x)]}{\delta\phi(x)\delta\phi(y)}## is of the form ##\frac{\delta^{2} S[\phi(x)]}{\delta\phi(x)\delta\phi(y)}=\Box +m^{2}##.

    Any help would be much appreciated.
     
  2. jcsd
  3. May 14, 2017 #2
    Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.
     
  4. May 17, 2017 #3
    First I'll rewrite S by integrating by part:

    $$ S = \frac12 \int d^4 x \partial_\mu \phi \partial^\mu \phi - m^2 \phi^2 = Boundary term - \int d^4 x \phi ( \partial^2 + m^2) \phi $$

    Whenever you have to think of functional derivatives I find it easiest to do it in discrete space:

    If

    $$S[\phi] = \sum_{x y} \phi_x A_{xy} \phi_y$$

    then the second order variation

    $$ {\delta S \over \delta \phi_x \delta \phi_y} \equiv {\partial S \over \partial \phi_x \partial \phi_y} = A_{xy}$$

    In the continuum, your operator $A_{xy}$ is just $\partial^2 + m^2$ as derived above. That just means:
    $$ {\delta S \over \delta \phi_x \delta \phi_y} = \partial^2 + m^2$$
     
  5. May 17, 2017 #4
    Thanks for the response. Having read up on this a bit more since I first wrote this post, shouldn't there be a ##\delta##-function floating around that the ##\Box## operator acts on?

    This is what I've come up with since for the more general case of a Lagrangian for a scalar field with up to first-order derivatives in the field. Starting from $$\delta\mathcal{L}=\frac{\partial\mathcal{L}}{\partial\phi(x)}\delta\phi(x)+
    \frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi(x))}\partial_{\mu}(\delta\phi(x))+\mathcal{O}(\delta\phi^{2})$$ we have that to first-order: $$\frac{\delta S}{\delta\phi(y)}=\int\,d^{4}z\left[
    \frac{\partial\mathcal{L}}{\partial\phi(z)}\delta^{(4)}(y-z)+
    \frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi(z))}\partial_{\mu}(\delta^{(4)}(y-z))
    \right]\\=\int\,d^{4}z\left[
    \frac{\partial\mathcal{L}}{\partial\phi(z)}-\partial_{\mu}\left(
    \frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi(z))}\right)
    \right]\delta^{(4)}(y-z)\\
    =\frac{\partial\mathcal{L}}{\partial\phi(y)}-\partial_{\mu}\left(
    \frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi(y))}\right)\qquad\qquad\qquad\qquad
    $$ where the boundary term vanishes since ##\delta^{(4)}(y-z)## is zero on the boundary of the spacetime volume we're integrating over (because it has compact support within the the boundary of the spacetime volume), and we have used that $$\frac{\delta\phi(x)}{\delta\phi(y)}=\delta^{(4)}(x-y)$$
    Given this we can then take the second order variation by noting that
    $$
    \delta\left[\frac{\partial\mathcal{L}}{\partial\phi(y)}-\partial_{\mu}\left(
    \frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi(y))}\right)\right]=\frac{\partial^{2}\mathcal{L}}{\partial\phi(y)^{2}}\delta\phi(y)+\frac{\partial^{2}\mathcal{L}}{\partial\phi(y)\partial(\partial_{\mu}\phi(y))}\partial_{\mu}(\delta\phi(y))-
    \partial_{\mu}\left(
    \frac{\partial^{2}\mathcal{L}}{\partial\phi(y)\partial(\partial_{\mu}\phi(y))}\delta\phi(y)\right)\\-
    \partial_{\mu}\left(
    \frac{\partial^{2}\mathcal{L}}{\partial(\partial_{\mu}\phi(y))\partial(\partial_{\nu}\phi(y))}\partial_{\nu}(\delta\phi(y))\right)$$ such that
    $$\frac{\delta^{2}S}{\delta\phi(x)\delta\phi(y)}=
    \frac{\partial^{2}\mathcal{L}}{\partial\phi(y)^{2}}\delta^{(4)}(x-y)+\frac{\partial^{2}\mathcal{L}}{\partial\phi(y)\partial(\partial_{\mu}\phi(y))}\partial_{\mu}(\delta^{(4)}(x-y))-
    \partial_{\mu}\left(
    \frac{\partial^{2}\mathcal{L}}{\partial\phi(y)\partial(\partial_{\mu}\phi(y))}\delta^{(4)}(x-y)\right)\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad
    \\-
    \partial_{\mu}\left(
    \frac{\partial^{2}\mathcal{L}}{\partial(\partial_{\mu}\phi(y))\partial(\partial_{\nu}\phi(y))}\partial_{\nu}(\delta^{(4)}(x-y))\right)
    \\=\left[\frac{\partial^{2}\mathcal{L}}{\partial\phi(y)^{2}}-
    \partial_{\mu}\left(
    \frac{\partial^{2}\mathcal{L}}{\partial\phi(y)\partial(\partial_{\mu}\phi(y))}\right)
    \right]\delta^{(4)}(x-y)-\partial_{\nu}\left(
    \frac{\partial^{2}\mathcal{L}}{\partial(\partial_{\nu}\phi(y))\partial(\partial_{\mu}\phi(y))}\right)\partial_{\mu}(\delta^{(4)}(x-y))-\frac{\partial^{2}\mathcal{L}}{\partial(\partial_{\mu}\phi(y))\partial(\partial_{\nu}\phi(y))}\partial_{\mu}\partial_{\nu}(\delta^{(4)}(x-y))
    $$
    Thus, in the case where ##\mathcal{L}=\frac{1}{2}\partial_{\mu}\phi\partial^{\mu}\phi-\frac{1}{2}\phi^{2}##, we see that $$\frac{\delta^{2}S}{\delta\phi(x)\delta\phi(y)}=-m^{2}\delta^{(4)}(x-y)-\Box\left(\delta^{(4)}(x-y)\right)=-\left(\Box +m^{2}\right)\delta^{(4)}(x-y)$$ Would this analysis be correct at all? I'm struggling to find any notes that go into detail about it all. I'm trying to understand it in an effort to get to grips with effective actions and the background field method for finding the one-loop effective action.
     
    Last edited: May 17, 2017
  6. May 18, 2017 #5
    Another problem I have, is that the above result changes if I use a slightly different method. Indeed, $$\delta^{2}S=\int\,d^{4}x\Big[\frac{\partial^{2}\mathcal{L}}{\partial\phi(x)^{2}}\delta\phi(x)\delta\phi(x)+2\frac{\partial^{2}\mathcal{L}}{\partial\phi(x)\partial(\partial_{\mu}\phi(x))}\partial_{\mu}(\delta\phi(x))\delta\phi(x)\qquad+\frac{\partial^{2}\mathcal{L}}{\partial(\partial_{\mu}\phi(x))\partial(\partial_{\nu}\phi(x))}\partial_{\mu}(\delta\phi(x))\partial_{\nu}(\delta\phi(x))\Big]\\ =\int\,d^{4}x\Big[\frac{\partial^{2}\mathcal{L}}{\partial\phi(x)^{2}}\delta\phi(x)\delta\phi(x)+2\frac{\partial^{2}\mathcal{L}}{\partial\phi(x)\partial(\partial_{\mu}\phi(x))}\partial_{\mu}(\delta\phi(x))\delta\phi(x)\qquad-\partial_{\nu}\left(\frac{\partial^{2}\mathcal{L}}{\partial(\partial_{\mu}\phi(x))\partial(\partial_{\nu}\phi(x))}\partial_{\mu}(\delta\phi(x))\right)\delta\phi(x)\Big]$$ where I have integrated by parts and used that ##\delta\phi=0## on the boundary. Then one can employ the trick ##\delta\phi(x)=\int\,d^{4}y\delta^{(4)}(x-y)\delta\phi(y)## to recast this as $$\delta^{2}S=\int\,d^{4}xd^{4}y\,\delta^{(4)}(x-y)\Big[\frac{\partial^{2}\mathcal{L}}{\partial\phi(x)^{2}}\delta\phi(x)\delta\phi(y)+2\frac{\partial^{2}\mathcal{L}}{\partial\phi(x)\partial(\partial^{(x)}_{\mu}\phi(x))}\partial^{(x)}_{\mu}(\delta\phi(x))\delta\phi(y)\qquad\qquad\qquad\qquad\qquad\qquad-\partial^{(x)}_{\nu}\left(\frac{\partial^{2}\mathcal{L}}{\partial(\partial^{(x)}_{\mu}\phi(x))\partial(\partial^{(x)}_{\nu}\phi(x))}\partial^{(x)}_{\mu}(\delta\phi(x))\right)\delta\phi(y)\Big]$$ which upon integrating by parts once in the second term and twice in the third term, noting that ##\partial^{(x)}_{\mu}(\delta\phi(y))=0## (where the superscript ##x## denotes that the derivative acts on the variable ##x## only), and neglecting boundary terms (which will vanish due to the variation in the field vanishing on the boundary), we have: $$\delta^{2}S=\int\,d^{4}xd^{4}y\Big[\frac{\partial^{2}\mathcal{L}}{\partial\phi(x)^{2}}\delta^{(4)}(x-y)-2\partial^{(x)}_{\mu}\left(\frac{\partial^{2}\mathcal{L}}{\partial\phi(x)\partial(\partial_{\mu}\phi(x))}\delta^{(4)}(x-y)\right)\qquad\qquad\qquad\qquad\qquad\qquad-\partial^{(x)}_{\mu}\left(\frac{\partial^{2}\mathcal{L}}{\partial(\partial^{(x)}_{\mu}\phi(x))\partial(\partial^{(x)}_{\nu}\phi(x))}\partial^{(x)}_{\nu}(\delta^{(4)}(x-y))\right)\Big]\delta\phi(x)\delta\phi(y)$$ Thus, comparing this with the definition of the second-order variation of the action $$\delta^{2}S=\int\,d^{4}xd^{4}y\frac{\delta^{2}S}{\delta\phi(x)\delta\phi(y)}\delta\phi(x)\delta\phi(y)$$ we see that $$\frac{\delta^{2}S}{\delta\phi(x)\delta\phi(y)}=\frac{\partial^{2}\mathcal{L}}{\partial\phi(x)^{2}}\delta^{(4)}(x-y)-2\partial^{(x)}_{\mu}\left(\frac{\partial^{2}\mathcal{L}}{\partial\phi(x)\partial(\partial_{\mu}\phi(x))}\delta^{(4)}(x-y)\right)\qquad\qquad\qquad\qquad\qquad\qquad-\partial^{(x)}_{\mu}\left(\frac{\partial^{2}\mathcal{L}}{\partial(\partial^{(x)}_{\mu}\phi(x))\partial(\partial^{(x)}_{\nu}\phi(x))}\partial^{(x)}_{\nu}(\delta^{(4)}(x-y))\right)\\ =\Bigg[\frac{\partial^{2}\mathcal{L}}{\partial\phi(x)^{2}}-2\partial^{(x)}_{\mu}\left(\frac{\partial^{2}\mathcal{L}}{\partial\phi(x)\partial(\partial_{\mu}\phi(x))}\right)\Bigg]\delta^{(4)}(x-y)\qquad\qquad\qquad\qquad\qquad-\Bigg[2\frac{\partial^{2}\mathcal{L}}{\partial\phi(x)\partial(\partial_{\mu}\phi(x))}+\partial^{(x)}_{\nu}\left(\frac{\partial^{2}\mathcal{L}}{\partial(\partial^{(x)}_{\mu}\phi(x))\partial(\partial^{(x)}_{\nu}\phi(x))}\right)\Bigg]\partial^{(x)}_{\mu}(\delta^{(4)}(x-y))-\frac{\partial^{2}\mathcal{L}}{\partial(\partial^{(x)}_{\mu}\phi(x))\partial(\partial^{(x)}_{\nu}\phi(x))}\partial^{(x)}_{\mu}\partial^{(x)}_{\nu}(\delta^{(4)}(x-y))$$ Which clearly isn't equal to the expression I arrived at in the post above. What am I missing? Clearly the final result shouldn't depend on how one calculates it (unless the calculation is wrong of course).
     
    Last edited: May 18, 2017
  7. May 18, 2017 #6
    Hi Frank, sorry, but it's hard for me to follow your math. Thinking about this is much simpler when you treat the case $\phi(x)$ becomes a finite dimension vector (sth you can put in matlab and code). Do you at least understand the finite dimensional version of your equation I wrote?

    If $S(\phi_i)$ was a scalar function of many variables $S(\phi_1, \phi_2...)$ (basically, you just discretized space x to a finite number of points), then all you're trying to calculate is:

    $${\delta S(\phi) \over \delta \phi(x) \phi(y)} \equiv {\partial^2 \over \partial \phi_i \partial \phi_j } S$$

    The quantity on the right is a 2 dimensional matrix called the Hessian (do you understand why?). It is a linear operator.
    Similarly, the action is a number generated from a vector $\phi_1, \phi_2...$ as follows:
    $$S = \sum_{ij} \phi_i A_{ij} \phi_j$$

    Just looking at the way I wrote the action, you immediately see the Hessian is $A_{ij}$. In other words, the strategy to calculate what you want is to rewrite the action as
    $$S = \sum_{space} \phi \times (some operator) \times \phi$$.

    This is what I did by integration by part.

    When you generalize to a continuum number of indexes, ${\delta S(\phi) \over \delta \phi(x) \phi(y)}$ is also an operator (it's just a very big matrix, in this case infinite dimensional). That matrix is the
    operator $$ \Box + m^2$$.

    I'm being pedantic here, but it's important you understand which things are functions (aka vectors), which are operators (aka matrices.)
     
  8. May 19, 2017 #7

    Orodruin

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    The immediate continuous generalisation of
    would be an integral of the form
    $$
    S = \int \int \phi(x) A(x,y) \phi(y) dx\, dy,
    $$
    where ##A## is some operator depending on ##x## and ##y##. In this sense, yes, there should be a ##\delta(x-y)## such that
    $$
    S = \int \int \phi(x) (\Box + m^2) \delta(x-y) \phi(y) dx\,dy = \int\int \delta(x-y) \phi(y) (\Box + m^2) \phi(x) dx\, dy
    = \int \phi(x) (\Box + m^2) \phi(x) dx.
    $$
     
  9. May 19, 2017 #8
    It's because it contains all the information about the second-order partial derivatives of ##\phi_{i}##, right?

    I understand that ##A_{ij}## is (the matrix representation of) an operator, and similarly in the continuum case, that ##\Box +m^{2}## is also an operator acting on the space of functions ##\phi(x)##. Shouldn't their be a ##\delta##-function there in the continuum case though, i.e. $$\frac{\delta^{2}S}{\delta\phi(x)\delta\phi(y)}=(\Box+m^{2})\delta^{(4)}(x-y)$$ This comes from noting that the second-order functional derivative of ##S## is defined in terms of its second-order variation, i.e. $$\delta^{2}S=\int\,d^{4}xd^{4}y\,\delta\phi(x)\frac{\delta^{2}S}{\delta\phi(x)\delta\phi(y)}\delta\phi(y)$$ hence why my calculations in the previous two posts were aiming to get in this form so I could imply the form of ##\frac{\delta^{2}S}{\delta\phi(x)\delta\phi(y)}## in the slightly more general case. I was kind of hoping to use the definition of the functional derivative to derive the second order functional derivative of an action with a slightly more general Lagrangian.

    Is any of what I wrote in my previous two posts (#4 and #5) correct at all? I've found it confusing since none of the books I've read give an explicit derivation and there seems to be an inconsistency - some include the ##\delta##-function and some don't, but it seems that since the second-order functional derivative contains the derivatives of the field at two different points, ##x## and ##y## , that there should be a ##\delta##-function there such that, $$\frac{\delta^{2}S}{\delta\phi(x)\delta\phi(y)}=(\Box+m^{2})\delta^{(4)}(x-y)$$ Can ##\delta^{(4)}(x-y)## be treated as the continuum case of the components of the identity matrix, such that the quantity ##(\Box+m^{2})\delta^{(4)}(x-y)## is the continuum version of the components of a diagonal matrix ##\Box +m^{2}##? And then in general ##\frac{\delta^{2}S}{\delta\phi(x)\delta\phi(y)}## are the continuum version of the components of a matrix?
     
    Last edited: May 19, 2017
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