4
$\begingroup$

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? $$ \frac{\delta^kZ_0[J]}{\delta J(z_1) \cdots \delta J(z_k)}\Biggr|_{J=0} = \mathcal{N} \frac{(-1)^k}{2^k} \int dw_1 \cdots \int dw_k\ J(w_1) \cdots J(w_k) \Delta_F(z_1 - w_1) \cdots \Delta_F(z_k - w_k) $$

$\endgroup$

1 Answer 1

2
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ I understood your steps but unfortunately the only thing I see is a derivation of the Volterra series for $Z_0[J]$, i.e., how the kernels $F_k$ are related to de functional derivatives at $J=0$. I still can't see how the two series expansions of $Z_0[J]$ are equal. If this is true (which I'm also inclined to think so), then shouldn't we able to make the identification that I made in the edit of the question? $\endgroup$
    – SNC92
    Commented Jun 30, 2022 at 20:04
  • $\begingroup$ The equation in your edit cannot be true, since it contains some $J$ insertions in the RHS, which are set to zero. I edited in the correct formula. $\endgroup$ Commented Jun 30, 2022 at 21:37
  • 1
    $\begingroup$ Yes, absolutely. I thought I was comparing coefficients of series, but now I see I'm comparing two integrands which, of course, are not necessarily equal, i.e., $\int dx\ f(x) = \int dx\ g(x)\ \nRightarrow\ f(x) = g(x)$. Also, from your edit (Wick theorem) I was able to see that, after a change of variables, the Volterra series of $Z_0$ will reduce to its Taylor series. Thanks for your help! $\endgroup$
    – SNC92
    Commented Jul 1, 2022 at 2:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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