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In the book Quantum Field Theory, Ryder defines the vacuum-to-vacuum transition amplitude in the presence of the source $J(x)$ as $$Z[J] = \int \mathcal{D}\phi \, \exp\left\lbrace i\int \, d^4x \, \left[\mathcal{L}(\phi) + J(x) \phi(x) + \frac{i}{2} \epsilon\phi^2 \right]\right\rbrace \propto \langle0, \infty|0,-\infty\rangle.\tag{6.1}$$

This is Eq. (6.1) of the book.

In discrete form, $$Z[J] = \lim_{N \to \infty} \int \prod_{n=1}^{N^4} d\phi_n \, \exp\left\lbrace i\sum_{n=1}^{N^4} \delta^4 \left( \mathcal{L}_n + \phi_nJ_n + \frac{i}{2} \epsilon \phi_n^2 \right) \right\rbrace,\tag{6.2}$$ where $n$ stands for the four indices $(i, j, k, l)$.

Question

What does $d\phi_n$ mean?

I think that $$d\phi_n = \frac{\partial\phi_n}{\partial x_i}dx_i + \frac{\partial\phi_n}{\partial y_j}dy_j + \frac{\partial\phi_n}{\partial z_k}dz_k + \frac{\partial\phi_n}{\partial t_l}dt_l, $$ where $\phi(x_i, y_j, z_k, t_l) \equiv \phi_n$ and, the other symbols have their usual meanings. Is this correct?

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Your last equation (v3) is not what Ryder meant. Think about it this way:

  1. The scalar field $\phi: \{1, \ldots, N\}^4 \to \mathbb{R}$ is a map from a discretized spacetime region $\{1, \ldots, N\}^4$ to a target space $\mathbb{R}$.

  2. For fixed lattice index $n\in \{1, \ldots, N\}^4$, the integration variable $\phi_n\in \mathbb{R}$ is a real number that parametrizes the target space $\mathbb{R}$.

  3. Concretely, the symbol $\mathrm{d}\phi_n$ stands for the integration measure (i.e. the standard Lebesgue measure over $\mathbb{R}$) in the integral $$\int_{\mathbb{R}}\mathrm{d}\phi_n ~f(\phi_n) .$$ Here $f(\phi_n)$ is some integrand.

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  • $\begingroup$ The problem is for fixed lattice index $n \in \mathbb {Z}^4$, $\phi_n$ is just a constant real number. Then identifying $d\phi_n$ as the integration measure seems a little difficult for me. Can you please explain a bit more? $\endgroup$ – rainman Jul 30 '17 at 1:14
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    $\begingroup$ I updated the answer. $\endgroup$ – Qmechanic Jul 30 '17 at 5:24

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