Is the Lagrangian of a quantum field really a functional?

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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?

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asked Apr 10, 2012
retagged Apr 19, 2014

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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.

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answered Apr 10, 2012 by (1,230 points)
Isn't it even simpler than that? At fixed time $t$ the Lagrangian is an ordinary functional on the space of classical fields $(\psi(t),\dot\psi(t))$.

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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.

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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.

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answered Apr 10, 2012 by (195 points)
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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

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

References:

1. S. Weinberg, The Quantum Theory of Fields, Vol 1.
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answered Apr 11, 2012 by (3,120 points)

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