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I am currently reading through Weinberg's Lectures on Quantum Mechanics. Chapter 11 deals with field theory:

Correspondingly, the Lagrangian $L(t)$ is a functional of $\psi_n(\vec{x}, t)$ and $\dot{\psi}_n(\vec{x}, t)$, depending on the form of all of the functions $\psi_n(x, t)$ and $\dot{\psi}_n(x, t)$ for all $\vec{x}$, but at a fixed time $t$.

I understand the statement above in the following way: the Lagrangian, as a functional, takes the fields $\psi_n$ and $\dot{\psi}_n$ and a specific value of $t$ as input and produces some output that is not dependent of $\vec{x}$.

Is this "interpretation" correct and if not, what else does is mean? At least it would make sense to me, especially when the Lagrangian is defined as

$L(t) = \int \, d^3x \cal{L}\big(\psi(\vec{x},t), \vec{\nabla}\psi(\vec{x},t),\dot{\psi}(\vec{x},t)\big)$.

Qmechanic
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maxxam
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    That is correct. It takes the functions $\psi_n(\vec{x})$ and ${\dot \psi}_n(\vec{x})$ which are the values of $\psi_n$ and ${\dot \psi}_n$ at a particular value of $t$ as inputs. The output is a number independent of $\vec{x}$. – Prahar Jul 13 '21 at 08:21

1 Answers1

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  1. Perhaps it would be more pedagogical to use the notation $q$ and $v$ as the notation for the fields $\psi$ and $\dot{\psi}$, respectively, because they are independent fields: $\mathbb{R}^{n+1}\to\mathbb{R}$ in the Lagrangian functional $L[q(\cdot,t),v(\cdot,t);t]$.

    On the other hand, in the action $S[q]$ the 2 fields are actually dependent. For the explanation in the simpler situation of point mechanics, see e.g. this Phys.SE post.

  2. OP interpretation is correct. The Lagrangian is a map $$ F(\mathbb{R}^n,\mathbb{R}) \times F(\mathbb{R}^n,\mathbb{R}) \times \mathbb{R}~~\stackrel{L}{\longrightarrow}~~\mathbb{R}, $$ where $F(\mathbb{R}^n,\mathbb{R})$ denotes an appropriate class of functions: $\mathbb{R}^n\to\mathbb{R}$. In particular, spatial derivatives $\vec{\nabla}\psi(\vec{x},t)$ are also allowed in a first-order Lagrangian density ${\cal L}$.

  3. For a higher-order Lagrangian density ${\cal L}$, see e.g. my related Phys.SE answer here.

Qmechanic
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