Why do we consider Lagrangian densities in QFT?

My question is: Why do we consider Lagrangian densities in QFT (as opposed to Lagrangians as in classical mechanics)?

Is it simply because of the following?

We wish the theories to be Lorentz invariant and as in Special Relativity space and time are no longer independent of one another we must consider the physics at each spacetime point $x^{\mu}$, implying that the theory should be locally described by a density.

Additionally, to obey Poincare invariance the Lagrangian density should not be explicitly dependent on $x^{\mu}$, and for it to be Lorentz invariant it must be dependent on spatial derivatives $\nabla\phi$ of the fields as well as temporal derivatives $\partial_{t}\phi$ (as time derivatives in one reference frame will correspond to a combination of time and spatial derivatives in another frame).

(Physically does the fact that $\mathscr{L}$ depends on $\nabla\phi$ as well as $\partial_{t}\phi$ because the field $\phi$ is defined at each spacetime point and so a fluctuation in $\phi$ at a point $x^{\mu}$ will produce temporal and spatial derivatives at that point which will affect fields in its immediate neighbourhood (i.e. infinitesimally close to it)?)

• The difference here is not one between classical and quantum theory, but one between "discrete" and field theory. There's also classical field theory (Lagrangian density for electrodynamics). – ACuriousMind Jul 7 '15 at 13:59
• @ACuriousMind So the Lagrangian is necessarily a density because it depends on fields (which have a value at each spacetime point) and so necessarily has a value at each spacetime point. We then obtain the classical Lagrangian by integrating over a spatial volume, hence the term Lagrangian density?! Why does one need to make the Lagrangian density a function of spatial derivatives as well though? – Will Jul 7 '15 at 14:02
• ...Is the point that locality requires that the dynamics of the field (described by the Lagrangian) should only depend on the states of fields in an infinitesimally small neighbourhood of the spacetime point at which the Lagrangian is being evaluated at, and therefore should depend on the field and its temporal and spatial derivatives at that point (so that we can account for the change in the field as we move an infinitesimally amount from that point)? – Will Jul 7 '15 at 14:14
• It's not forbidden to include the spatial derivatives, and so we include them. It then turns out Lagrangians dependent on spatial derivatives describe physical systems. There's nothing more going on. – ACuriousMind Jul 7 '15 at 14:28
• In addition to the justification that "everything that is not forbidden is compulsory", there's also the fact that a Lagrangian that depended only on time derivatives couldn't possibly lead to a Lorentz-invariant description of nature. – Michael Seifert Jul 7 '15 at 14:50

The difference is that in classical mechanics positions are exactly the fields you are looking at, whereas in general field theories they are the variables the actual fields depend on.

In classical mechanics the solution of the dynamics is given by the knowledge of the position and the velocity $(q(t),\dot{q}(t))$ at any time $t$. Time plays the role of the path parameter that describes the solution on the fibre bundle of the configuration space and therefore it is the only free parameter you are allowed to integrate on; hence the action must be of the form $$S[q,\dot{q}]=\int_{\gamma}dt\,L(q(t),\dot{q}(t);t).$$ In field theories you essentially play the same game: fields are maps $x^{\mu}\to\phi(x^{\mu})$ depending on both positions and time (whereas previously the only field was the position itself, depending on time, notice the difference). You may want to mirror the above action as $$S[\phi,\dot{\phi}] = \int_{\mathcal{D_t}}dt\,L(\alpha(t),\dot{\alpha}(t);t)$$ having in mind that the new dummy variables $\alpha(t),\dot{\alpha}(t)$ result in turn from the integration of the fields on the position variables as $$S[\phi,\partial\phi] = \int_{\mathcal{D_t}}dt\, \left(\int_{\mathcal{D'}}d^3x\,\mathscr{L}(\phi(x,t),\partial\phi(x,t);x,t) \right)$$ collecting the above you may simply define the action as $$S[\phi,\partial\phi] = \int_{\mathcal{D}}d^4x\,\mathscr{L}(\phi(x,t),\partial\phi(x,t);x,t).$$ We call Lagrangian (density) whatever function appears in the integral. Density just means that you have something you want to integrate over the variables that you have, no more than that. The standard Lagrangian in classical mechanics is a density as well with respect to the time variable, only we do not call it such (although it is). Then, as you pointed out, for the action to be Lorentz invariant you can drop the explicit dependencies on $(x,t)$, so that the Lagrangian (or the density, as you want to call it) ends up being invariant under translations in the Poincaré group. Further requirements can be added to include any other invariance we want.

• Why is the Lagrangian also dependent on spatial derivatives of the fields though? Is this required for Lorentz invariance (as time derivatives in one frame will correspond to a combination of time and spatial derivatives in another frame)? Also, why is the Lagrangian only considered local if it is a function of a single spacetime point? Why can't it be a function of two spacetime points that are causally connected? (Is it because the Lagrangian is evaluated at a single point in time and thus the only case in which it can be local is if it is evaluated at a single spatial point) – Will Jul 7 '15 at 13:04
• Because it's just the most general way it can be. Fields depend on $(x,t)$ and the most general derivatives are derivatives with respect to any of those variables: spatial is no more special than time. However, what do you mean by the Lagrangian is local? The Lagrangian is a function of the fields, which in turn depend on point by point as functions themselves. – gented Jul 7 '15 at 14:17
• Is the reason why locality requires the value of the Lagrangian at a given spacetime point to be dependent on the value of the field (and its derivatives) at that point because if it were not, then the description of the dynamics of the field (provided by the Lagrangian) at a given spacetime point would require knowledge of behaviour of the field at other points in spacetime which the condition that the dynamics are only dependent on the behaviour of the field at a given point (and its behaviour in the infinitesimal neighbourhood of that point)? – Will Jul 7 '15 at 14:29
• There is absolutely no locality involved in the definition of the action principle. It is just the integral of a function of the fields in the most general possible way. – gented Jul 7 '15 at 14:31
• Ok, it's just that I've read that we require a physical theory to obey locality and this corresponds to requiring that the value of the Lagrangian density (describing the theory) at each point is dependent only on the value of the field (and its derivatives) at that point?! – Will Jul 7 '15 at 14:34

The use of the Lagrangian density is a convenience, and it is not directly related to causality or relativity, and neither strictly to quantum theories. What I mean is that it is possible to formulate non-relativistic (quantum or classical) field theories using exactly the same language.

The difference between "mechanics" and "field theory" is that, instead of using particles as the fundamental objects of the theory, we use fields i.e. functions $\phi:X\to \mathbb{C}$. So, while in mechanics the phase space is a finite dimensional structure (e.g. a manifold) in field theory it is an infinite dimensional one (e.g. the space of measurable functions).

The physical motivation is that there are systems that are impossible to describe with a finite number of degrees of freedom (e.g. the elctromagnetic field). Special relativity adds an important motivation, on the quantum side: particles can be created and destroyed, therefore your relativistic quantum space must contain all the possible configurations with an arbitrary number $n$ of particles; this is conveniently described mathematically considering the one-particle QM space as the classical phase space, and build the so called Fock-Cook (second) quantization upon it. Again, an infinite dimensional phase-space ($L^2(\Omega)$, for some suitable $\Omega$) has to be considered.

Once you are in an infinite-dimensional phase space setting, the introduction of the Lagrangian density is quite natural, and it is a matter of convenience. Let $\mathbb{C}^X$ be the set of functions from $X$ to $\mathbb{C}$, and your infinite dimensional phase space (or position-velocity space for lagrangian formulation). Mimicking the finite dimensional situation, you want to build up a function(al) of the variables of the system $\phi\in \mathbb{C}^X$ that full encodes the dynamical informations. We call it Lagrangian function $L:\mathbb{C}^X\to \mathbb{R}$ (usually taken to be real-valued). For the moment it is not important to worry about the distinction between variables (fields $\phi$) and their derivatives ($\dot{\phi}$); think of them as "independent variables" inside $\mathbb{C}^X$ (as position and velocity are in finite dimensions).

Now, we have an object $L(\cdot)$ that encodes the dynamical informations, and has to be evaluated on functions $\phi\in \mathbb{C}^X$. What happens in most situations in practice, is that the Lagrangian consists of two formal operations, one that provides a pointwise information for each $x\in X$ (depending on $\phi$), and another that puts all those informations together providing the complete evaluation $L(\phi)$. The first is the Lagrangian density $\mathscr{L}:\mathbb{C}^X\to \mathbb{R}^X$ (because usually it is real-valued), the second is the "integration" or we may call it in general the global evaluation $\mathbb{E}_{gl}: \mathbb{R}^X\to \mathbb{R}$. This splitting results in $L=\mathbb{E}_{gl}\circ \mathscr{L}$, and it is convenient if you want to separate the "functional pointwise manipulations", done in the lagrangian density, from the global evaluation of these manipulations w.r.t. the totality of points.

However, apart from convenience, this is just the standard setting of infinite dimensional phase spaces; no relativistic or quantum considerations involved a priori.

• Excellent answer. +1 Though it does seem like someone could ask if the "What happens in most situations in practice" happens because of locality. – Timaeus Sep 16 '15 at 14:37
• @Timaeus That would be an interesting question indeed. Anyways, the Lagrangian density can be defined also in non-relativistic field theories (even if, admittedly, it is not strictly necessary to introduce fields in that case). I think e.g. of the lagrangian of Hartree equation, that second quantized gives the dynamics of many non-relativistic bosons. And in that case also the Lagrangian can be written as the same two formal operations. – yuggib Sep 16 '15 at 15:51
• I thought that was what this question was asking. Otherwise you just have an $L$ defined on fields with no obvious break into $L=\mathbb{E}_{gl}\circ \mathscr{L}.$ – Timaeus Sep 16 '15 at 15:54
• @Timaeus Well, since you do the same subdivision in non-relativistic theories (where it is not necessary to preserve local causal properties), I don't think the splitting is necessarily because of those causality requirements ("locality", as I think it was intended in the question). Anyways, I agree that it is natural to think of the value of a quantity at a precise point (in a local way), and then to obtain the global information by an additional operation. – yuggib Sep 16 '15 at 16:02
• It's hard to tell what sense of locality the OP is concerned about. Check out physics.stackexchange.com/q/204016 where locality is defined as the exact split you are talking about, but with the density depending only on the fields and its derivatives at the point. – Timaeus Sep 16 '15 at 16:06