Functional Derivative in the Linear Sigma Model

Question

In the linear sigma model, the Lagrangian is given by

$$ \mathcal{L} = \frac{1}{2}\sum_{i=1}^{N} \left(\partial_\mu\phi^i\right)\left(\partial^\mu\phi^i\right) +\frac{1}{2}\mu^2\sum_{i=1}^{N}\left(\phi^i\right)^2-\frac{\lambda}{4}\left(\sum_{i=1}^{N}\left(\phi^i\right)^2\right)^2 \tag{11.65} $$ (for example see Peskin & Schroeder page 349).

When perturbatively computing the effective action for this Lagrangian the derivative $ \frac{\delta^2\mathcal{L}}{\delta\phi^k(x)\delta\phi^l(x)} $ needs to be computed. (for instance, Eq. (11.67) in P&S):

$$ \frac{\delta^2\mathcal{L}}{\delta\phi^k(x)\delta\phi^l(x)} ~=~ -\partial^2\delta^{kl} +\mu^2\delta^{kl}-\lambda\left[\phi^i\phi^i\delta^{kl}+2\phi^k\phi^l\right].\tag{11.67}$$

My question is, how is one supposed to handle the derivative term?

This seems to be completely implicit in the presentation of P&S, but from what I could gather it should go like so:

1) Because we are computing the effective action, $\mathcal{L}$ is actually under an integral and we can replace $\left(\partial_\mu\phi^i\right)\left(\partial_\mu\phi^i\right)$ with $-\left(\partial^\mu\partial_\mu\phi^i\right)\phi^i=-\left(\partial^2\phi^i\right)\phi^i$ using Stokes' theorem.

2) Then when performing the first derivative I get $\frac{\delta}{\delta\phi^l}\left[-\left(\partial^2\phi^i\right)\phi^i\right]=-\partial^2\phi^l$.

3) It is the second derivative I get stuck at, for as far as I can see, $\frac{\delta}{\delta\phi^k}\left[-\partial^2\phi^l\right]=0$, for there is only dependence on the 2nd derivative of $\phi^l$ and not $\phi^l$ itself. If, as is usual in field theory, the field and its derivatives are treated as independent dynamical variables, then the second derivative should also be an independent dynamical variable. How is it explained then, that the result of this computation should be $\frac{\delta}{\delta\phi^k}\left[-\partial^2\phi^l\right]=-\delta^{kl}\partial^2$?

Answer 1

Comment to the question (v2): P&S is using the notation of a 'same-spacetime' functional derivative. To illustrate this notation, let us, for simplicity, stay within the first variations and leave it to the reader to generalize to higher-order variations.

I) First of all, functional/variational derivatives should not be confused with partial derivatives. In practice, from an operational point of view (if we are not worried about mathematical details about existence and boundary terms), all we need to know is the following rules:

1. The formula $$\tag{A} \frac{\delta \phi^{\beta}(y)}{\delta\phi^{\alpha}(x)} ~=~\delta^{\beta}_{\alpha}~\delta^n(x-y), $$ where $n$ is the spacetime dimension.

2. Appropriate generalizations of elementary rules in calculus, such as, e.g., the chain rule, integration by parts, commutativity of derivatives, and the Dirac delta distribution.

For instance, by these rules 1 & 2, we have that

$$ \frac{\delta}{\delta\phi^{\beta}(y)} \frac{\partial}{\partial x^{\mu_1}}\ldots \frac{\partial}{\partial x^{\mu_r}}\phi^{\alpha}(x) ~=~ \frac{\partial}{\partial x^{\mu_1}}\ldots \frac{\partial}{\partial x^{\mu_r}} \frac{\delta}{\delta\phi^{\beta}(y)}\phi^{\alpha}(x)$$ $$\tag{B}~\stackrel{(A)}{=}~\delta_{\beta}^{\alpha}~\frac{\partial}{\partial x^{\mu_1}}\ldots \frac{\partial}{\partial x^{\mu_r}}\delta^n(x-y). $$

Similarly, by rules 1 & 2, we can deduce that the action

$$\tag{C}S~=~\int \mathrm{d}^nx ~{\cal L}(x) , \qquad {\cal L}(x)\equiv {\cal L}(\phi(x), \partial \phi(x), \ldots, x),$$

has the Euler-Lagrange expression as its functional derivative

$$ \tag{D}\frac{\delta S}{\delta\phi^{\alpha} (x)}~=~ \frac{\partial{\cal L}(x) }{\partial\phi^{\alpha} (x)} - \frac{\mathrm{d}}{\mathrm{d}x^{\mu}} \left(\frac{\partial{\cal L}(x) }{\partial\partial_{\mu}\phi^{\alpha} (x)} \right)+\ldots.$$

The ellipsis $\ldots$ in eqs. $(\text{C})$ and $(\text{D})$ denotes possible contributions from higher-order spacetime derivatives.

II) From formula $(\text{A})$, it becomes clear that it does not make sense to consider the functional derivative $\dfrac{\delta {\cal L}(x)}{\delta\phi^{\alpha} (x)}$ wrt. the same spacetime argument $x$, because that would lead to infinities, cf. $\delta^n(0)=\infty$. Nevertheless, it is tempting to introduce the notation of a 'same-spacetime' functional derivative

$$\tag{E}\frac{\delta {\cal L}(x)}{\delta\phi^{\alpha} (x)}~:=~ \frac{\partial{\cal L}(x) }{\partial\phi^{\alpha} (x)} - \frac{\mathrm{d}}{\mathrm{d}x^{\mu}} \left(\frac{\partial{\cal L}(x) }{\partial\partial_{\mu}\phi^{\alpha} (x)} \right)+\ldots.$$

We stress that eq. $(\text{E})$ is only a notational definition. It becomes meaningless if we try to interpret the lhs. of eq. $(\text{E})$ using the above rules 1 & 2.

III) Similarly, P&S talk about second-order 'same-spacetime' functional derivative

$$\tag{F}\frac{\delta^2 {\cal L}(x)}{\delta\phi^{\alpha} (x)\delta\phi^{\beta}(x)}.$$

We recommend first working out the ordinary second-order functional derivative

$$\tag{G}\frac{\delta^2 S}{\delta\phi^{\alpha} (x)\delta\phi^{\beta}(y)}$$

using rules 1 & 2. Then, it should be fairly straightforward to translate $(\text{G})$ into the 'same-spacetime' functional derivative language $(\text{F})$, if needed. [In particular, eq. $(\text{G})$ contains a $\delta^n(x-y)$ while eq. $(\text{F})$ does not.]

IV) Finally, we should mention that in field theory, one often suppresses the spacetime indices $x,y,\ldots$ by using DeWitt's condensed notation.

Answer 2

For reference sake, here is the computation of (G) mentioned above in @Qmechanic's answer:

The Lagrangian is given by: $$ \mathcal{L}\left[\phi\right]\left(z\right)=\frac{1}{2}\sum_{i=1}^{N}\left(\partial_{\mu}\phi^{i}\right)\left(\partial^{\mu}\phi^{i}\right)+\frac{1}{2}\mu^{2}\sum_{i=1}^{N}\left(\phi^{i}\right)^{2}-\frac{\lambda}{4}\left(\sum_{i=1}^{N}\left(\phi^{i}\right)^{2}\right)^{2} $$

and so the action is: $$ S\left[\phi\right]=\int d^{4}z\left\{ \frac{1}{2}\sum_{i=1}^{N}\left(\partial_{\mu z}\phi^{i}\left(z\right)\right)\left(\partial^{\mu}\,_{z}\phi^{i}\left(z\right)\right)+\frac{1}{2}\mu^{2}\sum_{i=1}^{N}\left(\phi^{i}\left(z\right)\right)^{2}-\frac{\lambda}{4}\left[\sum_{i=1}^{N}\left(\phi^{i}\left(z\right)\right)^{2}\right]^{2}\right\} $$

Thus the computation of equation (11.67) in P&S is, explicitly (not following the same-spacetime shorthand-notation mentioned above): \begin{eqnarray} \frac{\delta^{2}S\left[\phi\right]}{\delta\phi^{a}\left(x\right)\delta\phi^{b}\left(y\right)}&=&\frac{\delta^{2}}{\delta\phi^{a}\left(x\right)\delta\phi^{b}\left(y\right)}\int d^{4}z\left\{ \frac{1}{2}\sum_{i=1}^{N}\left(\partial_{\mu z}\phi^{i}\left(z\right)\right)\left(\partial^{\mu}\,_{z}\phi^{i}\left(z\right)\right)+\frac{1}{2}\mu^{2}\sum_{i=1}^{N}\left(\phi^{i}\left(z\right)\right)^{2}-\frac{\lambda}{4}\left[\sum_{i=1}^{N}\left(\phi^{i}\left(z\right)\right)^{2}\right]^{2}\right\} \\&\stackrel{\star}{=}&\frac{\delta}{\delta\phi^{a}\left(x\right)}\int d^{4}z\left\{ \sum_{i=1}^{N}\left(\left(\partial^{\mu}\,_{z}\phi^{i}\left(z\right)\right)\left(\partial_{\mu}\,_{z}\frac{\delta}{\delta\phi^{b}\left(y\right)}\phi^{i}\left(z\right)\right)\right)+\mu^{2}\sum_{i=1}^{N}\phi^{i}\left(z\right)\frac{\delta}{\delta\phi^{b}\left(y\right)}\phi^{i}\left(z\right)-\lambda\left[\sum_{i=1}^{N}\left(\phi^{i}\left(z\right)\right)^{2}\right]\left[\sum_{j=1}^{N}\phi^{j}\left(z\right)\frac{\delta}{\delta\phi^{b}\left(y\right)}\phi^{j}\left(z\right)\right]\right\} \\&=&\frac{\delta}{\delta\phi^{a}\left(x\right)}\int d^{4}z\left\{ \sum_{i=1}^{N}\left(\left(\partial^{\mu}\,_{z}\phi^{i}\left(z\right)\right)\left(\partial_{\mu}\,_{z}\delta^{ib}\delta\left(z-y\right)\right)\right)+\mu^{2}\sum_{i=1}^{N}\phi^{i}\left(z\right)\delta^{ib}\delta\left(z-y\right)-\lambda\left[\sum_{i=1}^{N}\left(\phi^{i}\left(z\right)\right)^{2}\right]\left[\sum_{j=1}^{N}\phi^{j}\left(z\right)\delta^{jb}\delta\left(z-y\right)\right]\right\} \\&=&\frac{\delta}{\delta\phi^{a}\left(x\right)}\int d^{4}z\left\{ \left(\partial^{\mu}\,_{z}\phi^{b}\left(z\right)\right)\left(\partial_{\mu}\,_{z}\delta\left(z-y\right)\right)+\mu^{2}\phi^{b}\left(z\right)\delta\left(z-y\right)-\lambda\left[\sum_{i=1}^{N}\left(\phi^{i}\left(z\right)\right)^{2}\right]\left[\phi^{b}\left(z\right)\delta\left(z-y\right)\right]\right\} \\&=&\int d^{4}z\left\{ \left(\partial^{\mu}\,_{z}\delta^{ab}\delta\left(x-z\right)\right)\left(\partial_{\mu}\,_{z}\delta\left(z-y\right)\right)+\mu^{2}\delta^{ab}\delta\left(x-z\right)\delta\left(z-y\right)-2\lambda\phi^{a}\left(z\right)\delta\left(x-z\right)\phi^{b}\left(z\right)\delta\left(z-y\right)-\lambda\left[\sum_{i=1}^{N}\left(\phi^{i}\left(z\right)\right)^{2}\right]\left[\delta^{ab}\delta\left(x-z\right)\delta\left(z-y\right)\right]\right\} \\&=&\int d^{4}z\left\{ -\delta^{ab}\left(\partial_{\mu}\,_{z}\partial^{\mu}\,_{z}\delta\left(x-z\right)\right)\delta\left(z-y\right)+\mu^{2}\delta^{ab}\delta\left(x-z\right)\delta\left(z-y\right)-\lambda\left\{ \left[\sum_{i=1}^{N}\left(\phi^{i}\left(z\right)\right)^{2}\right]\delta^{ab}+2\phi^{a}\left(z\right)\phi^{b}\left(z\right)\right\} \delta\left(x-z\right)\delta\left(z-y\right)\right\} \\&=&-\delta^{ab}\left(\partial_{\mu}\,_{y}\partial^{\mu}\,_{y}\delta\left(x-y\right)\right)+\mu^{2}\delta^{ab}\delta\left(x-y\right)-\lambda\left\{ \left[\sum_{i=1}^{N}\left(\phi^{i}\left(y\right)\right)^{2}\right]\delta^{ab}+2\phi^{a}\left(y\right)\phi^{b}\left(y\right)\right\} \delta\left(x-y\right) \end{eqnarray}

To prove $\star$, use the definition of the functional derivative as a Taylor expansion (where this whole enchalada started with in equation (11.58) in P&S): $$ \mathcal{L}[\phi(x)+\epsilon\,\eta(x)] = \sum_{n=0}^{\infty}\frac{1}{n!}\epsilon^n\left.\left(\frac{d}{d\epsilon} \mathcal{L}[\phi(x)+\epsilon\,\eta(x)]\right)\right|_{\epsilon=0} $$