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 $(i, j, k, l)$.


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?


1 Answer 1


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.

  • $\begingroup$ 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, 2017 at 1:14
  • 1
    $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Jul 30, 2017 at 5:24

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.