7
$\begingroup$

My question relates to something that I´ve seen in many books and appears in all its glory here: Ryder, pg 198

My question is about eq. 6.74. Which I repeat below:

$$i \int {\cal D}\phi \frac{\delta \hat{Z} [\phi] }{\delta \phi} \exp \left(i\int J(x) \phi(x) dx\right)$$ $$= i\; \exp \left(i\int J(x) \phi(x) dx\right) \hat{Z}[\phi] \Bigg|_{\phi\rightarrow\infty}$$ $$+ \int {\cal D}\phi J(x) \hat{Z}[\phi] \exp \left(i\int J(x) \phi(x) dx\right).\tag{6.74}$$

$\phi$ is a scalar field, $J$ is a source, $x = x_{\mu}$ in 4D Minkowski space and $$\hat{Z}[\phi] = \frac{e^{iS}}{\int {\cal D}\phi\; e^{iS}}.$$

The author is clearly doing a integral by parts and the first term on the right hand side is a kind of surface term for the path integral. He then considers this term to be zero and the second one gives us:

$$i \int {\cal D}\phi \frac{\delta \hat{Z} [\phi] }{\delta \phi} \exp \left(i\int J(x) \phi(x) dx\right) = J(x) Z[J]$$

The trick thing here is that integral limits for $\int{\cal D}\phi$ are not very obvious (at least not to me). You are in fact summing up for all field configurations. So, there are actually two problems in my mind:

  1. For what configuration of $\phi$ is the surface term calculated? (the author says it is $\phi \rightarrow \infty$)

  2. Assuming the author is right about taking huge $\phi$: why is this term zero?

This applies to path integrals in general: can we do the usual trick of throwing out surface terms safely?

$\endgroup$
0
3
$\begingroup$

One mustn't confuse field space with physical space. The field space is some sort of manifold without boundaries (for a nonlinear sigma model), or $R^n$ for usual field theories, in either case, integration by parts works in Euclidean space, or if you add a little imaginary part to the propagators so that the action is decaying at large values of $\phi$.

The integration by parts in field space is simple--- there are no boundaries in field space, except at infinite field values, and the Euclidean or slightly off-Minkowsky action decays at infinity.

There is no relation to the integration by parts in physical space involved for instantons or other topological things.

$\endgroup$
2
  • $\begingroup$ So the point here is that the "surface terms" vanish when I integrate with ${\cal L} \rightarrow (1+i \epsilon){\cal L}$, right? $\endgroup$ – Forever_a_Newcomer Jul 30 '12 at 23:09
  • $\begingroup$ Yes, but not exactly, only because what you wrote down is not the infinitesimal continuation to Euclidean space. You need to do an infinitesimal Wick rotation, which also rotates the kinetic part of the action according to the time derivatives. To be precise, the slightly-Euclidean action is the thing you get when you eliminate the p's in $\int p \dot{q} - (1+i\epsilon)H $, for p the field momentum and q the field (or just this thing, with no elimination, for a first order action). $\endgroup$ – Ron Maimon Jul 30 '12 at 23:59
1
$\begingroup$

Simple way to get the identity

IMHO the matter has nothing to do with the behavior of $\phi$ at infinity. It has to do with a simple property of the path integral, namely, that the path integral of a functional derivative is zero (I shall explain it in a second).

To explain how it is employed, recall that functional derivatives obey the product rule. Therefore we have $$\dfrac{\delta}{\delta \phi(z)}\bigg[\hat{Z}[\phi]\exp\left(i \int d^dx \ J(x)\phi(x)\right)\bigg]=\dfrac{\delta\hat{Z}[\phi]}{\delta \phi(z)}\exp\left(i\int d^dx \ J(x)\phi(x)\right)+\\+\hat{Z}[\phi] iJ(z)\exp\left(i\int d^dx \ J(x)\phi(x)\right)\tag{1}.$$

Multiplying by $i$ and functionally integrating against $\mathfrak{D}\phi$ and employing the property I have claimed the LHS functionally integrates to zero. Therefore we obtain the identity $$i \int \mathfrak{D}\phi \dfrac{\delta\hat{Z}[\phi]}{\delta \phi(z)}\exp\left(i\int d^dx J(x)\phi(x)\right)=\int \mathfrak{D}\phi \hat{Z}[\phi]J(z)\exp\left(i \int d^d x \ J(x)\phi(x)\right)\tag{2}.$$

This is the equation in the OP with the first term already discarded since it corresponds to the functional integral of the functional derivative which I have just said that vanishes. Moreover we have been more careful to denote the free variable as $z$ instead of $x$ to avoid confusion with the integration variable in the exponent.

At first sight this would seem to be a result of assuming some sort of Stokes theorem holds here allowing the conversion of functional integrals of functional derivatives into boundary terms in function space together with the assumption that such boundary terms vanish. This is what I think the notation $\phi\to \infty$ means here: a term in the "boundary of function space".

Particularly I don't buy this argument (and honestly have no idea how to describe the boundary of function space in useful terms). Therefore I have a much simpler one to offer to explain this identity.

Functional integral of a functional derivative is zero

There is a simple argument for this if we assume translation invariance of the path integral measure $\mathfrak{D}\phi$ (which Physicists do assume all the time anyway). In the end of the day the claim is that if $F[\phi]$ is an arbitrary functional of $\phi$ which is functionally differentiable then $$\int \mathfrak{D}\phi \dfrac{\delta F[\phi]}{\delta \phi(z)}=0.\tag{1}$$

To understand this consider one infintiesimal variation $\delta \phi$ of the field configuration $\phi$ and evaluate the change in $F$:

$$\delta F[\phi]:=F[\phi+\delta\phi]-F[\phi]\tag{2}.$$

Now functionally integrate (2). Assuming linearity of the functional integral you'll have two functional integrals. In the first we change varaibles defining $\phi+\delta \phi = \phi'$. Assuming translation invariance of $\mathfrak{D}\phi$ we have $\mathfrak{D}\phi'=\mathfrak{D}\phi$ and therefore we have $$\int \mathfrak{D}\phi \delta F[\phi]=\int \mathfrak{D}\phi F[\phi+\delta \phi]-\int\mathfrak{D}\phi F[\phi]=\int \mathfrak{D}\phi' F[\phi']-\int \mathfrak{D}\phi F[\phi] = 0.\tag{3}$$

Therefore the functional integral of $\delta F$ is always zero. Now expand $\delta F$ in a functional Taylor series: $$\int \mathfrak{D}\phi \delta F[\phi] = \int \mathfrak{D}\phi \left[\int d^dz \dfrac{\delta F[\phi]}{\delta \phi(z)}\delta \phi(z)+\dfrac{1}{2!}\int d^dz_1 d^dz_2\dfrac{\delta^2 F[\phi]}{\delta \phi(z_1)\delta \phi(z_2)}\delta \phi(z_1)\delta \phi(z_2)+\cdots\right]\tag{4}$$

this thing must be zero. Therefore we equate it to zero term by term. The first term gives $$\int \mathfrak{D}\phi\int d^dz \dfrac{\delta F[\phi]}{\delta \phi(z)}\delta\phi(z)=0.\tag{5}$$

Finally assuming that we can interchange the path integral and the spacetime integral yields the equation $$\int d^dz \left(\int \mathfrak{D}\phi \dfrac{\delta F[\phi]}{\delta \phi(z)}\right)\delta \phi(z) =0. \tag{6}$$

Since the variation $\delta \phi(z)$ is arbitrary, equation (6) tells us that in the distributional sense equation (1) must hold.

This derivation of (1) is formal as many manipulations in Physics with path integrals but notice that the central assumption is translation invariance of $\mathfrak{D}\phi$ which is a standard assumption in Physics anyway. In the end, no need to talk about any boundary of function space nor any functional Stokes' theorem (which IMHO would also be formal and even more formal than this approach).

$\endgroup$
0
$\begingroup$

He is imagining the possible values of $\phi$ as a Euclidean space $\mathbb{R}^n$ with a "boundary sphere" $S^{n-1}$ at infinity. This is a reasonable thing to think about since we usually consider a UV cut-off and consider the limit as this cut-off goes to infinity. (This is just the ordinary way of thinking about integrals on noncompact spaces.)

This boundary term is zero when we assume that the action decays quickly enough as $\phi$ grows large. This assumption is necessary for his argument to work.

I hope this helps. I have always been a bit weary about these integration by parts arguments myself. I guess physics texts like these just always assume Stokes' law works.

These sorts of arguments can be subtle when the target of $\phi$ is not a contractible space like $\mathbb{R}^n$. For example, if $\phi$ is a map from $\mathbb{R}^4$ to $SU(2)$, which is topologically a 3-sphere, there are topologically nontrivial boundary configurations for $\phi$ labelled by $\pi_3(SU(2))=\mathbb{Z}$ . This is the case when one has Yang-Mills instantons.

One also must be careful of course when the spacetime has boundary. For example, when computing a propagator, we have a spacetime that looks like a tube: $M \times \mathbb{R}$, and we use the boundary condition that on one end of the tube we have one state and on the other end we have another state. Then the path integral computes the propagator between the states.

$\endgroup$
4
  • $\begingroup$ there is no UV cutoff on the field size, this is incorrect. Integration by parts in field space always works, even for nonlinear sigma models. The instanton issues are different, they involve integration by parts in space, not in field space. $\endgroup$ – Ron Maimon Jul 27 '12 at 23:21
  • $\begingroup$ You are right about the instantons being irrelevant. I confused myself. $\endgroup$ – Ryan Thorngren Jul 27 '12 at 23:43
  • 1
    $\begingroup$ However, if the boundary of the target space is not contractible there can be topologically distinct limits $\phi\rightarrow\infty$. $\endgroup$ – Ryan Thorngren Jul 27 '12 at 23:55
  • $\begingroup$ Oh, that's an interesting idea--- like a nonlinear sigma model with lots of weird ends! I can't imagine how you can have anything significant going on at the ends with a decaying action, you would need a topological action, or consider a 2d sigma model which can wander away to infinity. One can consider a manifold with boundary in sigma model or orbifold, there might be a nice example there. $\endgroup$ – Ron Maimon Jul 28 '12 at 4:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.