Weinberg at page 300 of The Quantum Theory of Fields - Volume I says:

$L$ itself should be a space integral of an ordinary scalar function of $\Psi(x)$ and $\partial \Psi(x)/\partial x^\mu \,$, known as the Lagrangian density $\mathscr{L}$:

$$ L[\Psi(t), \dot{\Psi}(t)]= \int d^3x \, \mathscr{L}\bigr(\Psi({\bf x},t), \nabla \Psi({\bf x},t), \dot{\Psi}({\bf x},t)\bigl) $$

So he says that $\mathscr{L} \, $ is a function. But Gelfand and Formin at page one of their book Calculus of variations say:

By a functional we mean a correspondence which assigns a definite (real) number to each function (or curve) belonging to some class.

So from that I'd say it is a functional. The notes of quantum field theory of my professor stay on this side, explicitly calling the lagrangian density a functional.

I'm very confused at the moment. I always used this latter way of defining functionals (the Gelfand way) so Weinberg saying that $\mathscr{L}$ is a function confuses me.

Can someone makes some clarity about this?


4 Answers 4


The Lagrangian density is a function.

Consider the following examples: $$ A[f]=\int_0^1\mathrm dx\ f(x) $$ and $$ B(f(x))=f(x) $$

It is clear that $A$ is a functional, because for example $$ A[\sin]=1-\cos1=0.45\in\mathbb R $$ is a number, while $B$ is a function, because $$ B(\sin)=\sin $$ is not a number, but a function.

In your notation, $L$ is a functional, because given a certain field configuration, you get a number. But $\mathscr L$ is a function, because given a certain field configuration, you get another function, not a number.

In some cases, such as QED in the Coulomb gauge, you may want to include non-local terms in the Lagrangian density, which makes it into a function of some of its arguments, and a functional of the others. This is an exception to the rule above.

  • 1
    $\begingroup$ To make it precise - in field theory Lagrangian density and Lagrangian are function and functional, but not just of the field, but field $\Phi(x)$ and field time derivative $\partial\Phi(x)/\partial t$. $\endgroup$
    – Fedxa
    Feb 18, 2017 at 23:25
  • $\begingroup$ I'd like to know the reason behind the downvote. The fact that there are other, possibly better answers, doesn't mean that this one automatically becomes wrong and deserving of a downvote. If you like the other answers better, upvote them. Downvoting this one because there are better existing answers makes little sense (unless there is something wrong in my answer; but in that case I'd appreciate a comment explaining why...) $\endgroup$ Feb 22, 2017 at 14:15
  • 2
    $\begingroup$ Wait, in the integral for the Lagrangian L above, the time was never integrated over (just space). That means L (even after integrating a particular choice of field function) is going to be a function of time, not a number. So isn't L a function, not a functional? $\endgroup$ Aug 26, 2017 at 0:23
  • 1
    $\begingroup$ It seems like only the action should count as a functional to me, based on the requirement for a number output. $\endgroup$ Aug 26, 2017 at 0:33
  • $\begingroup$ Indeed. $L$ is neither a functional nor a function. $L[\phi](\cdot)$ is a function of time, $L[\cdot](t)$ is a functional, and $L[\phi](t)$ is a number. Explicitly, I'm defining $L$ to be $$L[\phi](t) \equiv \int\mathrm{d}t'\,\mathrm{d}^3x\, \delta(t'-t)\mathcal{L}\left(\phi(\mathbf{x},t'),\,\phi(\dot\phi(\mathbf{x},t')\right).$$ Making a strict decision whether or not to call $f(x)$ a function is often not done. Above I'm assuming a kind of "programmers' convention" where $f$ is a function, $f(x)$ is a value. $\endgroup$ Jan 16, 2019 at 15:39
  1. Within the framework of differential geometry, the construction goes as follows. Let there be given a 4-dimensional orientable spacetime manifold $M$. A field is a section $$\phi~\in~\Gamma(E) \tag{1}$$ in a bundle $(E,\pi,M)$.

  2. The Lagrangian$^1$ 4-form $\mathbb{L}$ is a bundle map $$ \begin{array}{rcl} J^1(M,E) &\stackrel{\mathbb{L}}{\longrightarrow}& \bigwedge^4T^{\ast}M \cr \searrow && \swarrow \cr &M&\end{array}\tag{2}$$ from the first$^2$ jet bundle $J^1(M,E)$ to the canonical bundle over $M$.

  3. In a local coordinates chart $U\subseteq M$, the Lagrangian 4-form is $$\left. \mathbb{L}\right|_U~=~{\cal L} ~\mathrm{d}^4x, \qquad \mathrm{d}^4x ~:=~\mathrm{d}x^0\wedge\mathrm{d}x^1\wedge\mathrm{d}x^2\wedge\mathrm{d}x^3, \tag{3}$$ where ${\cal L}$ is the Lagrangian density. In other words, the Lagrangian density ${\cal L}$ transforms$^3$ as a density under general coordinate transformations of spacetime $M$.

  4. The action functional $S:\Gamma(E)\to \mathbb{R}$ is defined as $$S[\phi]~:=~\int_{M} (j^1\phi)^{\ast} \mathbb{L}.\tag{4}$$ See also e.g. this and this Phys.SE posts.

  5. Let us for simplicity assume that $M$ has a globally defined coordinate system $U\subseteq \mathbb{R}^4$, and only use this from now on.

  6. Moreover, let us for simplicity assume that $\phi$ is a real scalar field. Then the total space is $E= M\times\mathbb{R}$, and the bundle $(E,\pi,M)$ is a trivial line bundle.

  7. Then the first jetbundle $$J^1(M,E)~\cong~ T^{\ast}M \times \mathbb{R}\tag{5}$$ has $4+1+4=9$ coordinates $$(x^0,x^1,x^2,x^3;~ u;~ u_0, u_1, u_2, u_3).\tag{6}$$ In particular, the Lagrangian density $${\cal L}: U \times \mathbb{R}^5~\longrightarrow ~\mathbb{R} \tag{7}$$ is a (density-valued) function, cf. OP's title question.

  8. Be aware that physicists often use the same notation for the Lagrangian density ${\cal L}$ and the pull-back $(j^1\phi)^{\ast} {\cal L}$, i.e. they often write $${\cal L}(x^0,x^1,x^2,x^3;~ u;~ u_0, u_1, u_2, u_3) \tag{8} $$ as $${\cal L}\left(x^0,x^1,x^2,x^3;~ \phi(x);~ \partial_0\phi(x), \partial_1\phi(x), \partial_2\phi(x), \partial_3\phi(x)\right). \tag{9} $$


$^1$ The Lagrangian 4-form $\mathbb{L}$ and the Lagrangian density ${\cal L}$ should not be confused with the Lagrangian $L$. More generally, the Lagrangian $L$ is a function (and equal to the Lagrangian density ${\cal L}$) in point mechanics; while the Lagrangian $L$ is a functional in field theory. See also this Phys.SE post.

$^2$ There is a straightforward generalization to higher-order Lagrangian systems in an (possibly non-orientable) spacetime $M$ of arbitrary dimension.

$^3$ Here we are being a bit cavalier about whether the Jacobian transformation law for a density should include an absolute value or not. This is not an issue if the spacetime $M$ is orientable.

  • $\begingroup$ Thank you, I'll read it all today but I'll have to work a bit to understand it completely. Your answers are always on point and very formal when requested, you are able to reduce incomprehension and hand waving for the sake of clarity. I wonder if it's something you developed on your own or if there are some reference for these topics treated in the correct and formal way you use to present here. I'm not a mathematician but sometimes a lot of confusion arises due to lack of some kind of "mathematical imprint" in the texbooks and lessons or "us" physicists, and I get lost. $\endgroup$ Feb 20, 2017 at 12:33

In general, a functional is a map $V \to F$ for a vector space $V$ over a field $F$. Such functionals live in the dual space $V^*$ if they are linear.

$V$ can be infinite-dimensional and if it is some space of functions, then a functional is a map of a function to some scalar. For example,

$$\mathcal{E}[f] = \int_0^1 f(x)^2 \, dx$$

is an example of a functional mapping $f(x) \mapsto \int_0^1 f(x)^2 \, dx.$ This is not the same as function composition, $f \circ g = f(g(x))$ which simply produces another function. Now, the Lagrangian density of the form, $\mathcal L (\phi, \partial_\mu \phi)$ takes some function $\phi$ and its derivatives, and produces a function. We could replace the notation with dummy variables and then use composition to write this operation equivalently.

On the other hand, the Lagrangian rather than Lagrangian density,

$$L = \int d^3 x \, \mathcal L (\phi, \partial_\mu \phi)$$

is a functional but providing you think of $t$ as being held fixed.

  • $\begingroup$ I think your last line is the key line - $t$ needs to held fixed. Otherwise, according to the definition in the first line, we can't call $L$ a functional. $\endgroup$
    – rainman
    Sep 26, 2021 at 5:22
  • $\begingroup$ @rainman I think you could define the underlying field (field in the maths sense not QFT sense) as $\mathbb C (t)$ if one wants to be pedantic to make it a functional. $\endgroup$
    – JamalS
    Sep 26, 2021 at 18:34

They're both correct. A function, in its most general sense, maps a domain to a range. In this sense, a functional is also a function; it just so happens that its domain happens to be a set of functions rather than numbers.


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.