• Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.


PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback


(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,054 questions , 2,207 unanswered
5,345 answers , 22,721 comments
1,470 users with positive rep
818 active unimported users
More ...

  Is the Lagrangian of a quantum field really a functional?

+ 5 like - 0 dislike

Weinberg says, pg 299, The quantum theory of fields v1, that "The Lagrangian is, in general, a functional $L[\Psi(t),\dot{\Psi}(t)$], of a set of generic fields $\Psi[x,t]$ and their time derivatives...". My hang-up concerns the use of the word functional.

For example, the free field Lagrangian density for a quantum scalar field is: $$-\frac{1}{2} \partial_{\mu}\Phi\partial^{\mu}\Phi-\frac{m^2}{2} \Phi^2$$ Since the terms appearing here contain operators, then the Lagrangian must take values in the set of operators. Yet, I thought that the definition of a "functional", was that it took vectors in a Hilbert space to real numbers, rather than operators. Furthermore, by the above definition, what vectors does the Lagrangian act on?

Should the above statement be interpreted: "The 'expectation value of the' Lagrangian is, in general, a functional...", where the expectation value is taken over some field state? I.e., the expectation value of the Lagrangian takes vectors from the Fock space of states to real numbers?

This post has been migrated from (A51.SE)
asked Apr 10, 2012 in Theoretical Physics by Jason (25 points) [ no revision ]
retagged Apr 19, 2014 by dimension10

3 Answers

+ 4 like - 0 dislike

A "functional" may also be a map of a function to another function. Hence in this case, $L:\mathcal{F}\times\mathcal{F}\rightarrow\mathcal{F};(\Psi(t),\dot\Psi(t))\mapsto L[\Psi(t),\dot\Psi(t)]$, $L$ maps functions in a function space $\mathcal{F}$ into $\mathcal{F}$ (this assumes that $\Psi(t)$ is well enough controlled, and that $L$ is not too exotic, for the map to be into the same function space).

Weinberg works with the path-integral formalism, in which time-ordering is used liberally. When time-ordering is used, products of operators under the time-ordering commute, so that we can work with operators under the time-ordering symbol as if they are functions. IMO, Weinberg is immersed enough in this way of thinking that he doesn't keep the distinction between function spaces and spaces of (noncommuting) operators as clear as he might. Strictly speaking, the introduction of time-ordering takes us out of the mathematical context of $C^\star$-algebras and Hilbert spaces into the context of $C^\star$-algebras, Hilbert spaces and time-ordering (and, perhaps, also anti-time-ordering). Algebraists have to some extent struggled to include time-ordering in an attractive mathematical system.

This post has been migrated from (A51.SE)
answered Apr 10, 2012 by Peter Morgan (1,230 points) [ no revision ]
Isn't it even simpler than that? At fixed time $t$ the Lagrangian is an ordinary functional on the space of classical fields $(\psi(t),\dot\psi(t))$.

This post has been migrated from (A51.SE)
I'd prefer to say that the Lagrangian maps a function of time $t$ to a different function of time $t$ than to say that at a fixed time $t$ the Lagrangian maps two real numbers to a different real number. But if you have a reason to do it the second way, that's OK.

This post has been migrated from (A51.SE)
+ 2 like - 0 dislike

In classical field theory, fields are simply sections of some fibre bundle $E$ over a spacetime $M$. To any such bundle, one can associate jet-bundles $J^kE$, sections of those contain derivatives of the fields, they form an inverse system, so you can define a bundle $J^\infty E$. A Lagrange density is then simply a map

$$ L \colon J^\infty(E) \to \Omega^n(M).$$

Given a section $\phi \colon M \to E$, you can compute the action:

$$ S[\phi] = \int_M L(j_\infty \phi),$$ where $j_\infty(\phi) = (\phi, \partial_i \phi, ...)$ is the prolongation of $\phi$.

There are other viewpoints of course, another is that the Lagrange density is an element in the variational bicomplex.

This post has been migrated from (A51.SE)
answered Apr 10, 2012 by orbifold (195 points) [ no revision ]
+ 1 like - 0 dislike

The term functional is used in at least two different meanings.

  1. One meaning is in the mathematical topic of functional analysis, where one in particular studies linear functionals. This meaning is not relevant for the discussion at page 299 in Ref. 1.

  2. Another meaning is in the topics of calculus of variations and (classical) field theory. This is the sense that is relevant here.

Since we are only discussing the classical action $S$ and not the full path integral, let us for simplicity forget about quantum aspects, such as, e.g., $\hbar$, Hilbert spaces, expectation values, etc.

Let us for simplicity assume that there is only one field $q$ (which we for semantical reasons will call a position field), and that it lives in $n$ spatial dimensions and one temporal dimension. The field $q$ is then a function $q:\mathbb{R}^{n+1}\to \mathbb{R}$. There is also a velocity field $v:\mathbb{R}^{n+1}\to \mathbb{R}$. The Lagrangian is a local functional

$$L[q(\cdot,t),v(\cdot,t)]~=~\int d^nx~{\cal L}\left(q(x,t),\partial q(x,t),\partial^2q(x,t), \ldots,\partial^Nq(x,t);\right. $$ $$\left. v(x,t),\partial v(x,t),\partial^2 v(x,t), \ldots,\partial^N v(x,t);x,t\right). $$

The Lagrangian density ${\cal L}$ is a function of these variables. Here $N\in\mathbb{N}$ is some finite order. Moreover, $\partial$ denote partial derivatives wrt. spatial variables $x$, not the temporal variable $t$.

Time $t$ plays the role of a passive spectator parameter, i.e, we may consider a specific Cauchy surface, where time $t$ has some fixed value, and where it makes sense to specify $q(\cdot,t)$ and $v(\cdot,t)$ independently. (If we consider more than one time instant, then the $q$ and $v$ profiles are not independent. See also e.g. this Physics.SE question.)

Weinberg is using the word functional because of the spatial dimensions. (In particular, if $n=0$, then Weinberg would have called the Lagrangian $L(q(t),v(t))$ a function of the instantaneous position $q(t)$ and the instantaneous velocity $v(t)$.)

It is important to treat $q(\cdot,t)$ (which Weinberg calls $\Psi(\cdot,t)$) and $v(\cdot,t)$ (which Weinberg calls $\dot{\Psi}(\cdot,t)$) for fixed time $t$ as two independent functions in order to make sense of the definition of the conjugate/canonical momentum $p(\cdot,t)$ (which Weinberg calls $\Pi(\cdot,t)$). The definition involves a functional/variational derivative wrt. to the velocity field, cf. eq. (7.2.1) in Ref. 1,

$$\tag{7.2.1} p(x,t)~:=~\frac{\delta L[q(\cdot,t),v(\cdot,t)]}{\delta v(x,t)}.$$

Let us finally integrate over time $t$. The action $S$ (which Weinberg calls $I$) is

$$\tag{7.2.3} S[q]~:=~\int dt~ \left. L[q(\cdot,t),v(\cdot,t)]\right|_{v=\dot{q}}$$

The corresponding Euler-Lagrange equation becomes

$$\tag{7.2.2} \left.\frac{d}{dt} \left(\frac{\delta L[q(\cdot,t),v(\cdot,t)]}{\delta v(x,t)}\right|_{v=\dot{q}}\right)~=~\left. \frac{\delta L[q(\cdot,t),v(\cdot,t)]}{\delta q(x,t)}\right|_{v=\dot{q}}$$


  1. S. Weinberg, The Quantum Theory of Fields, Vol 1.
This post has been migrated from (A51.SE)
answered Apr 11, 2012 by Qmechanic (3,120 points) [ no revision ]

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification

user contributions licensed under cc by-sa 3.0 with attribution required

Your rights