Quantum Field Theory – Correct Way to Expand the Generating Functional

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Consider the following self-interacting real scalar field theory
$$
\mathcal{L} = \frac{1}{2}(\partial_\mu\phi)(\partial^\mu\phi) – \frac{1}{2}m^2\phi^2 – \frac{\lambda}{4!}\phi^4
$$
with $m^2 > 0$ and $\lambda > 0$. It is well-known that the generating functional of the full (i.e., interacting) theory can be written as
$$
Z[J] = \exp\Biggl\{-i\frac{\lambda}{4!} \int dx\ \frac{\delta^4}{\delta J(x)^4}\Biggr\}Z_0[J]
$$
where
$$
Z_0[J] = \mathcal{N} \exp\Biggl\{-\frac{1}{2} \int dz \int dw\ J(z) \Delta_F(z – w) J(w)\Biggr\}
$$
is the generating functional of the corresponding free theory. Here
$$
\Delta_F(z – w) = \int \frac{dp}{(2\pi)^d}\ \frac{i}{p^2 – m^2} e^{ip(z – w)}
$$
is Feynman's propagator of the scalar field $\phi$ and $\mathcal{N}$ is a normalization constant.

We can expand -in powers of $\lambda$– the exponential operator
$$
\exp\Biggl\{-i\frac{\lambda}{4!} \int dx\ \frac{\delta^4}{\delta J(x)^4}\Biggr\} = \sum^\infty_{\ell=0} \frac{(-i)^\ell \lambda^\ell}{(4!)^\ell\ell!} \int dx_1 \cdots \int dx_\ell\ \frac{\delta^{4\ell}}{\delta J(x_1)^4 \cdots \delta J(x_\ell)^4}.
$$
On the other hand, my problem is that:

I am not sure how to make a series expansion of the generating functional $Z_0[J]$ of the free theory.

I see two possibilities:

  1. According to equation (1.49) from this document (here, the author is working with the $\phi^3$-real scalar theory and the generating functional is called $W[J]$ instead of $Z[J]$), I should be able to make the expansion
    \begin{align}
    Z_0[J] &= \mathcal{N} \exp\Biggl\{-\frac{1}{2} \int dz \int dw\ J(z) \Delta_F(z – w) J(w)\Biggr\} \\
    &= \mathcal{N} \sum^\infty_{k=0} \frac{1}{k!} \Biggl(-\frac{1}{2} \int dz \int dw\ J(z) \Delta_F(z – w) J(w)\Biggr)^k \\
    &= \mathcal{N} \sum^\infty_{k=0} \frac{(-1)^k}{2^kk!} \int dz_1 \int dw_1 \cdots \int dz_k \int dw_k\ J(z_1)J(w_1) \cdots J(z_k)J(w_k) \Delta_F(z_1 – w_1) \cdots \Delta_F(z_k – w_k).
    \end{align}
    If this is correct, then we would have the functional expansion
    \begin{align}
    Z[J] &= \mathcal{N} \sum^\infty_{\ell=0} \sum^\infty_{k=0} \frac{(-1)^k(-i)^\ell \lambda^\ell}{(4!)^\ell\ell!2^kk!} \int dx_1 \cdots \int dx_\ell\ \int dz_1 \int dw_1 \cdots \int dz_k \int dw_k \\
    &\quad \times \Delta_F(z_1 – w_1) \cdots \Delta_F(z_k – w_k) \frac{\delta^{4\ell}}{\delta J(x_1)^4 \cdots \delta J(x_\ell)^4} J(z_1)J(w_1) \cdots J(z_k)J(w_k).
    \end{align}
    Consequently, some lowest-order terms are
    \begin{align}
    \frac{Z[J]}{\mathcal{N}} &= \lambda^0\Biggl\{1 – \frac{1}{2} \int dz_1 \int dw_1\ \Delta_F(z_1 – w_1)J(z_1)J(w_1) + \cdots\Biggr\} \\
    &\quad + \lambda\Biggl\{- \frac{i}{8} \int dx_1\ \Delta_F(x_1 – x_1)\Delta_F(x_1 – x_1) + \cdots\Biggr\} + \mathcal{O}(\lambda^2).
    \end{align}
  2. According to equation (92) from this document, the correct series expansion of $Z_0[J]$ is given by a Volterra series
    \begin{align}
    Z_0[J] &= \mathcal{N} \exp\Biggl\{-\frac{1}{2} \int dz \int dw\ J(z) \Delta_F(z – w) J(w)\Biggr\} \\
    &= \sum^\infty_{k=0} \frac{1}{k!} \int dz_1 \cdots \int dz_k\ J(z_1) \cdots \delta J(z_k) \frac{\delta^kZ_0[J]}{\delta J(z_1) \cdots \delta J(z_k)}\Biggr|_{J=0}.\end{align}
    If this is correct, then we would have the functional expansion
    \begin{align}
    Z[J] &= \sum^\infty_{\ell=0} \sum^\infty_{k=0} \frac{(-i)^\ell \lambda^\ell}{(4!)^\ell\ell!k!} \int dx_1 \cdots \int dx_\ell\ \int dz_1 \cdots \int dz_k \\
    &\quad \times \Biggl[\frac{\delta^{4\ell}}{\delta J(x_1)^4 \cdots \delta J(x_\ell)^4} J(z_1) \cdots J(z_k)\Biggr] \frac{\delta^kZ_0[J]}{\delta J(z_1) \cdots \delta J(z_k)}\Biggr|_{J=0}.
    \end{align}
    Consequently, some lowest-order terms are
    \begin{align}
    \frac{Z[J]}{\mathcal{N}} &= \lambda^0\Biggl\{1 – \frac{1}{2} \int dz_1 \int dz_2\ J(z_1)J(z_1)\Delta_F(z_1 – z_2) \\
    &\quad – \frac{1}{4!}\int dz_1 \int dz_2 \int dz_3 \int dz_4\ J(z_1)J(z_2)J(z_3)J(z_4) \\
    &\quad \times \Bigl[\Delta_F(z_1 – z_2)\Delta_F(z_3 – z_4) + \Delta_F(z_1 – z_3)\Delta_F(z_2 – z_4) + \Delta_F(z_1 – z_4)\Delta_F(z_2 – z_3)\Bigr] + \cdots\Biggr\} \\
    &\quad + \lambda\Biggl\{\cdots\Biggr\} + \mathcal{O}(\lambda^2).
    \end{align}
    Thus, as far as I can see both expansion are not yielding the same result. I would like to know which one is the correct one and why is the other wrong; or maybe both are equivalent but I can't see it.

EDIT: if both expansion for the generating functional $Z_0[J]$ are equal, i.e.
$$
\sum^\infty_{k=0} \frac{1}{k!} \int dz_1 \cdots \int dz_k\ J(z_1) \cdots \delta J(z_k) \frac{\delta^kZ_0[J]}{\delta J(z_1) \cdots \delta J(z_k)}\Biggr|_{J=0} = \mathcal{N} \sum^\infty_{k=0} \frac{(-1)^k}{2^kk!} \int dz_1 \int dw_1 \cdots \int dz_k \int dw_k\ J(z_1)J(w_1) \cdots J(z_k)J(w_k) \Delta_F(z_1 – w_1) \cdots \Delta_F(z_k – w_k)
$$
then, could we set the following identification?
$$
\frac{\delta^kZ_0[J]}{\delta J(z_1) \cdots \delta J(z_k)}\Biggr|_{J=0} = \mathcal{N} \frac{(-1)^k}{2^k} \int dw_1 \cdots \int dw_k\ J(w_1) \cdots J(w_k) \Delta_F(z_1 – w_1) \cdots \Delta_F(z_k – w_k)
$$

Best Answer

The two series for $Z_0[J]$ are equivalent. This generating functional is defined by : \begin{align} Z_0[J] &= \mathcal{N} \exp\Biggl\{-\frac{1}{2} \int dz \int dw\ J(z) \Delta_F(z - w) J(w)\Biggr\} \\ &= \mathcal{N} \sum^\infty_{k=0} \frac{1}{k!} \Biggl(-\frac{1}{2} \int dz \int dw\ J(z) \Delta_F(z - w) J(w)\Biggr)^k \\ \end{align} Then, the second series expansion is Taylor's formula : \begin{align} Z_0[J]&= \sum^\infty_{k=0} \frac{1}{k!} \int dz_1 \cdots \int dz_k\ J(z_1) \cdots J(z_k) \frac{\delta^kZ_0[J]}{\delta J(z_1) \cdots \delta J(z_k)}\Biggr|_{J=0}\end{align}

This formula actually holds for any functional $F[J]$ (which admits a formal series expansion) :

\begin{align} F[J]&= \sum^\infty_{k=0} \frac{1}{k!} \int dz_1 \cdots \int dz_k\ J(z_1) \cdots J(z_k) \frac{\delta^kF[J]}{\delta J(z_1) \cdots \delta J(z_k)}\Biggr|_{J=0}\end{align}

To prove Taylor's formula, we assume that we can expand $F$ as : $$ F[J] = \sum_{k=0}^\infty \int dz_1\ldots\int dz_kF_k(z_1,\ldots,z_k)J(z_1)\ldots J(z_k)$$

where the $F_k$ are integral kernels. The calculations should work even if the $F_k$ are higher order distributions, but this is not needed here.

Then, we compute its $k$th functional derivative at $J = 0$. This vanishes except on the term which contains exactly $k$ insertions of $J$. Therefore : $$\left.\frac{\delta^k F[J]}{\delta J(z_1)\ldots \delta J(z_k)}\right|_{J=0} = \sum_{\sigma \in \mathfrak S_k}F_k(z_{\sigma(1)},\ldots z_{\sigma(k)})$$ where the sum runs over all permutation of $\{1,\ldots,k\}$. When we integrate over $z_1,\ldots,z_k$, we can relabel the variables : \begin{align} \int dz_1 \cdots \int dz_k\ J(z_1) \cdots J(z_k) \frac{\delta^kF[J]}{\delta J(z_1) \cdots \delta J(z_k)}\Biggr|_{J=0} &=\sum_{\sigma \in \mathfrak S_k}\int dz_1 \cdots \int dz_k\ J(z_1) \cdots J(z_k)F_k(z_{\sigma(1)},\ldots z_{\sigma(k)}) \\ &= k! \int dz_1\ldots\int dz_kF_k(z_1,\ldots,z_k)J(z_1)\ldots J(z_k) \end{align} Dividing by $k!$ and summing over $k$, we see that Taylor's formula holds.

NB : the same calculations would have worked directly on $Z_0[J]$ but the precise expression for the kernels $F_k(z_1,\ldots,z_k)$ is hard to write down formally (Wick's theorem). Since it's precise form is not needed, it is easier to do the proof in a more general setting.

Edit : The expression for the functional derivatives of the free generating functional are given, for $n$ an even integer, by : \begin{gather} \left.\frac{\delta^{n} Z_0[J]}{\delta J(z_1) \ldots \delta J(z_{n})}\right|_{J=0} &= \left(-\frac 12\right)^n\sum_{\sigma \in\mathfrak S_{n}} \Delta_F(z_{\sigma(1)}-z_{\sigma(2)})\ldots \Delta_F(z_{\sigma(n-1)} -z_{\sigma(n)}) \end{gather} For $n$ odd, the functional derivative vanishes.

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