## 2022-04-04

### FOLLOW-UP: How to Tell Whether a Functional is Extremized

This post is a follow-up to an earlier post (link here) about how to tell whether a stationary point of a functional is a maximum, minimum, or saddle point. In particular, as I thought about it more, I realized that using the analogy to discrete vectors could help when formulating a more general expression for the second derivative of the nonrelativistic classical action for a single degree of freedom (i.e. the corresponding Hessian operator). Additionally, I thought of a few other examples of actions whose Hessian operators are positive-definite. Finally, I've thought more about how to express these equations for systems with multiple degrees of freedom (DOFs) as well as for fields and about how these ideas connect to the path integral formulation of quantum mechanics. Follow the jump to see more

## Second Derivative of the Nonrelativistic Classical Action for a Single Degree of Freedom

Going from from $S[x; t] = \int_{0}^{t} L(t', x, \dot{x})~\mathrm{d}t'$ to its discrete equivalent $S = \sum_{i} L(i, x_{i}, v_{i})$ means making use of the maps $$t \to i, x(t) \to x_{i}, \dot{x}(t) \to v_{i}, \delta(t - t') \to \delta_{ij}$$ applied consistently. Before proceeding, I should note that $$v_{i} = \sum_{j} D_{ij} x_{j}$$ means that $$v$$ implicitly depends on $$x$$, which is why partial derivatives with respect to $$x_{i}$$ will involve terms like $$\frac{\partial v_{k}}{\partial x_{i}}$$ even though these are supposedly partial derivatives. Unfortunately, I can't think of any better notation that would clarify this ambiguity.

With this, the first derivative is $\frac{\partial S}{\partial x_{j}} = \sum_{k} \left(\frac{\partial L}{\partial x_{k}}\frac{\partial x_{k}}{\partial x_{j}} + \frac{\partial L}{\partial v_{k}}\frac{\partial v_{k}}{\partial x_{j}}\right)$ which is evaluated as the Euler-Lagrange term $\frac{\partial S}{\partial x_{j}} = \frac{\partial L}{\partial x_{j}} - \sum_{k} D_{jk} \frac{\partial L}{\partial v_{k}}$ using the facts that $$\frac{\partial x_{k}}{\partial x_{j}} = \delta_{jk}$$ and $$\frac{\partial v_{k}}{\partial x_{j}} = D_{kj} = -D_{jk}$$. This means the second derivative is $\frac{\partial^{2} S}{\partial x_{i} \partial x_{j}} = \frac{\partial^{2} L}{\partial x_{j}^{2}} \delta_{ij} + \frac{\partial^{2} L}{\partial v_{j} \partial x_{j}} D_{ji} + \frac{\partial^{2} L}{\partial x_{i} \partial v_{i}} D_{ij} - \sum_{k} D_{ik} \frac{\partial^{2} L}{\partial v_{k}^{2}} D_{kj}$ where terms have been rearranged and the antisymmetry property $$D_{ij} = -D_{ji}$$ for the derivative operator has been used.

Reversing the aforementioned mapping means also using $$j \to t'$$, $$D_{ij} \to \frac{\mathrm{d}}{\mathrm{d}t} \delta(t - t')$$, and $$\sum_{j} D_{ij} u_{j} \to \frac{\mathrm{d}u}{\mathrm{d}t}$$ for any $$u_{i} \to u(t)$$. This means the Hessian operator in general is $\frac{\delta^{2} S}{\delta x(t) \delta x(t')} = \frac{\partial^{2} L}{\partial x^{2}}\bigg|_{x(t)} \delta(t - t') + \frac{\partial^{2} L}{\partial x \partial \dot{x}}\bigg|_{x(t)} \frac{\mathrm{d}}{\mathrm{d}t} \delta(t - t')$ $+ \frac{\partial^{2} L}{\partial x \partial \dot{x}}\bigg|_{x(t')} \frac{\mathrm{d}}{\mathrm{d}t'} \delta(t' - t) - \frac{\mathrm{d}}{\mathrm{d}t} \left(\frac{\partial^{2} L}{\partial \dot{x}^{2}}\bigg|_{x(t)} \frac{\mathrm{d}}{\mathrm{d}t} \delta(t - t')\right)$ which is a general Sturm-Liouville operator in the time domain. In the process of going from discrete vectors to continuous functions, I may have ignored some boundary terms & conditions associated with continuity, but I am otherwise reasonably confident that this expression is correct.

### Charged Particle in an Electromagnetic Field

As a specific example, I consider a charged particle in something that looks like an electromagnetic field, while keeping things simple by sticking to a single DOF. Such a particle has the Lagrangian $L = \frac{m}{2} \dot{x}^{2} - V(x) + \frac{q}{c} \dot{x}A(t, x) - q\phi(t, x)$ (which suggests the relations $$E(t, x) = -\frac{\partial \phi}{\partial x} - \frac{1}{c} \frac{\partial A}{\partial t}$$ and $$B(t, x) = \frac{\partial A}{\partial x}$$, but these fields don't necessarily obey Maxwell's equations per se, so the analogy has limitations). In such a case, the first derivative of the action is $\frac{\delta S}{\delta x(t)} = -\frac{\partial V}{\partial x(t)} - q\frac{\partial \phi}{\partial x(t)} - m\ddot{x}(t) - \frac{q}{c} \frac{\partial A}{\partial t}$ though the terms that go like $$\frac{q}{c} \dot{x} \frac{\partial A}{\partial x}$$ cancel because of the simplifications in this action. It is possible to compute the second derivative of the action similarly, but it looks messy, and because of the aforementioned simplifications, there isn't much of an obvious connection to actual electromagnetic theory.

## Other Examples of Actions with True Minima

The following contains just two examples of actions that are truly minimized at the physical trajectory. It is not meant to be an exhaustive list. The commonalities among these are that the potential $$V(x)$$ has a negative second derivative $$\frac{\partial^{2} V}{\partial x^{2}} < 0$$ for all $$x$$, so it can have at most one stationary point which must be a global maximum (and if it doesn't, then it asymptotically approaches a global maximum).

The first example of an action with a true minimum at the physical trajectory is $S[x; t] = \int_{0}^{t} \left(\frac{m}{2} \dot{x}^{2} + \frac{\kappa}{2} x^{2}\right)~\mathrm{d}t'$ which corresponds to an "inverted harmonic oscillator" $$V(x) = \frac{k}{2} x^{2}$$ with $$k = -\kappa < 0$$. The second derivative of the action is therefore given by $\frac{\delta^{2} S}{\delta x(t) \delta x(t')} = \left(\kappa - m\frac{\mathrm{d}^{2}}{\mathrm{d}t^{2}}\right)\delta(t - t')$ independent of the trajectory itself (which corresponds to exponentially growing & decaying solutions $$x = A\exp\left(t\sqrt{\frac{\kappa}{m}}\right) + B\exp\left(-t\sqrt{\frac{\kappa}{m}}\right)$$ with respect to time). This has eigenvalues $$\kappa + m\omega^{2}$$ as $\frac{\delta^{2} S}{\delta x(t) \delta x(t')} = \int_{-\infty}^{\infty} (\kappa + m\omega^{2})\exp(-\mathrm{i}\omega(t - t'))~\frac{\mathrm{d}\omega}{2\pi}$ is the Fourier expansion. The eigenvalues are strictly positive for all $$\omega$$, so the action is truly minimized at the physical trajectories.

The second example of an action with a true minimum at the physical trajectory is $S[x; t] = \int_{0}^{t} \left(\frac{m}{2} \dot{x}^{2} + V_{0} \exp(x/x_{0})\right)~\mathrm{d}t'$ which corresponds to a negative exponential potential $$V(x) = -V_{0} \exp(x/x_{0})$$ for $$V_{0} > 0$$. With the initial conditions $$x(0) = 0$$ and $$\dot{x}(0) = 0$$, the energy is conserved at $$V_{0}$$, so defining $$t_{0} \equiv \frac{1}{x_{0}} \sqrt{\frac{V_{0}}{m}}$$ and the variables $$y \equiv x/x_{0}$$, $$\tau \equiv t/t_{0}$$, and $$\dot{y} \equiv \frac{\mathrm{d}y}{\mathrm{d}\tau}$$ means that $$\frac{1}{2} \dot{y}^{2} = \exp(y) - 1$$ must always hold. This leads to the solution $$y(\tau) = \ln\left(1 + \tan^{2} \left(\frac{\tau}{\sqrt{2}}\right)\right)$$, so the Hessian operator $$\frac{\delta^{2} S}{\delta x(t) \delta x(t')} = \left(\frac{V_{0}}{x_{0}^{2}} \left(1 + \tan^{2}\left(\frac{t}{t_{0}\sqrt{2}}\right)\right) - m\frac{\mathrm{d}^{2}}{\mathrm{d}t^{2}}\right)\delta(t - t')$$ is positive-definite at the physical trajectory (although its spectrum is a bit complicated) as it corresponds to the nonrelativistic quantum mechanical problem of a single particle in the potential $$V(x) \sim 1 + \tan^{2}(x)$$, which is a convex & positive potential leading to a discrete & positive set of eigenvalues.

## Stationary Action with Multiple Degrees of Freedom

For a system with $$N$$ DOFs $$q_{i}$$ labeled by the index $$i$$, the action is written as $S[q_{1}, \ldots, q_{N}; t] = \int_{0}^{t} L(t', q_{1}, \ldots, q_{N}, \dot{q}_{1}, \ldots, \dot{q}_{N})~\mathrm{d}t'$ and its stationary points can be found as follows. Using the same discretization procedures as before with respect to $$t$$ (but changing notation slightly so that $$t$$ itself will be treated as a discrete index as needed as $$i$$ has been reserved for the $$N$$ DOFs) and defining $$r_{i} = \dot{q}_{i}$$ in continuous time or $$r_{i,t} = \sum_{t'} D_{t,t'} q_{i,t'}$$ in discrete time, then $$S = \sum_{t'} L_{t'}(q_{1}, \ldots, q_{N}, r_{1}, \ldots, r_{N})$$ must hold. This means that $$\frac{\partial S}{\partial q_{i,t}} = \sum_{t'} \left(\frac{\partial L_{t'}}{\partial q_{i,t'}} \frac{\partial q_{i,t'}}{\partial q_{i,t}} + \frac{\partial L_{t'}}{\partial r_{i,t'}} \frac{\partial r_{i,t'}}{\partial q_{i,t}}\right)$$ must hold. As $$\frac{\partial q_{i,t'}}{\partial q_{i,t}} = \delta_{t',t}$$ and $$\frac{\partial r_{i,t'}}{\partial q_{i,t}} = D_{t',t}$$, then $$\frac{\partial S}{\partial q_{i,t}} = \frac{\partial L_{t}}{\partial q_{i,t}} - \sum_{t'} D_{t,t'} \frac{\partial L_{t'}}{\partial r_{i,t'}}$$ must hold. In continuous time, this becomes $\frac{\delta S}{\delta q_{i}(t)} = \frac{\partial L}{\partial q_{i}(t)} - \frac{\mathrm{d}}{\mathrm{d}t} \frac{\partial L}{\partial \dot{q}_{i}(t)}$ and this must vanish for each $$i$$ to yield the equations of motion of the system.

The second derivative can be computed as follows. In discrete time, the first derivative can be rewritten with different indices as $$\frac{\partial S}{\partial q_{j,t'}} = \frac{\partial L_{t'}}{\partial q_{j,t'}} - \sum_{t''} D_{t',t''} \frac{\partial L_{t''}}{\partial r_{j,t''}}$$. This leads to $$\frac{\partial^{2} S}{\partial q_{i,t} \partial q_{j,t'}} = \frac{\partial ^{2} L_{t'}}{\partial q_{i,t'} \partial q_{j,t'}} \delta_{t,t'} + \frac{\partial^{2} L_{t'}}{\partial r_{i,t'} \partial q_{j,t'}} \frac{\partial r_{i,t'}}{\partial q_{i,t}}$$ $$- \sum_{t''} D_{t',t''} \left(\frac{\partial ^{2} L_{t''}}{\partial q_{i,t''} \partial r_{j,t''}} \frac{\partial q_{i,t''}}{\partial q_{i,t}} + \frac{\partial ^{2} L_{t''}}{\partial r_{i,t''} \partial r_{j,t''}} \frac{\partial r_{i,t''}}{\partial q_{i,t}}\right)$$. These can be evaluated as $$\frac{\partial r_{i,t'}}{\partial q_{i,t}} = D_{t',t}$$, $$\frac{\partial q_{i,t''}}{\partial q_{i,t}} = \delta_{t'',t}$$, and $$\frac{\partial r_{i,t''}}{\partial q_{i,t}} = D_{t'',t}$$, yielding $$\frac{\partial^{2} S}{\partial q_{i,t} \partial q_{j,t'}} = \frac{\partial ^{2} L_{t'}}{\partial q_{i,t'} \partial q_{j,t'}} \delta_{t,t'} + \frac{\partial^{2} L_{t'}}{\partial r_{i,t'} \partial q_{j,t'}} D_{t',t}$$ $$+ \frac{\partial^{2} L_{t}}{\partial q_{i,t} \partial r_{j,t}} D_{t,t'} - \sum_{t''} D_{t,t''} \frac{\partial ^{2} L_{t''}}{\partial r_{i,t''} \partial r_{j,t''}} D_{t'',t'}$$ in discrete time. In continuous time, the expression then is $\frac{\delta^{2} S}{\delta q_{i}(t) \delta q_{j}(t')} = \frac{\partial^{2} L}{\partial q_{i}(t) \partial q_{j}(t)} \delta(t - t') + \frac{\partial^{2} L}{\partial \dot{q}_{i}(t') \partial q_{j}(t')} \frac{\mathrm{d}}{\mathrm{d}t'} \delta(t' - t)$ $+ \frac{\partial^{2} L}{\partial q_{i}(t) \partial \dot{q}_{j}(t)} \frac{\mathrm{d}}{\mathrm{d}t} \delta(t - t') - \frac{\mathrm{d}}{\mathrm{d}t} \left(\frac{\partial^{2} L}{\partial \dot{q}_{i}(t) \partial \dot{q}_{j}(t)} \frac{\mathrm{d}}{\mathrm{d}t} \delta(t - t')\right)$ for the Hessian operator. This operator is expressed in terms of the composite indices $$((t, i), (t', j))$$ and is symmetric with respect to interchange of the composite indices. For the action to be truly minimized, this operator must be positive-definite.

## Lagrangian and Hamiltonian Formulations with Fields

Generalizing formulas for discrete DOFs to fields that depend on time & space as $$(t, x)$$ (henceforth using only $$d = 1$$ spatial dimension unless otherwise specified) can be done from two different perspectives. The first, which I will call the relativistic view, starts from formulas for a single DOF with action $S[q; t] = \int_{0}^{t} L(t', q, \dot{q})~\mathrm{d}t'$ and "upgrades" the dynamical variables to depend on both time & space, such that the action $S[q] = \int \mathcal{L}\left(t, x, q, \frac{\partial q}{\partial t}, \frac{\partial q}{\partial x}\right)~\mathrm{d}x~\mathrm{d}t$ is a functional of the field $$q(t, x)$$ (via the Lagrangian density $$\mathcal{L}$$); this view puts time & space on an equal footing, and the net effect is to simply introduce more partial derivatives into relevant formulas. The second, which I will call the many body view, starts from formulas for $$N$$ DOFs and takes the limit $$N \to \infty$$, associating each DOF with a point in space $$x$$.

In my opinion, a significant advantage of the many-body view is that it allows for arbitrary spatially nonlocal many-body interactions. For instance, in an action that depends on a large but finite number of DOFs, there could be a term in the Lagrangian for each DOF index $$i$$ that goes like $$\exp(q_{i}\tan(q_{i + 100} - q_{i - 33})/q_{i + 74})$$. As another example, an action that depends on infinite DOFs could be written as $$S = \int \left(\frac{\rho}{2} \left(\frac{\partial q}{\partial t}\right)^{2} - \frac{\tau}{2} \left(\frac{\partial q}{\partial x}\right)^{2} - \alpha\int \frac{q(t, x)q(t, x')}{|x - x'|^{\beta}}~\mathrm{d}x'\right)~\mathrm{d}x~\mathrm{d}t$$. In both of these cases, the dependence on arbitrarily separated indices or spatial coordinates means that the action can't be expressed only as $$S[q] = \int \mathcal{L}\left(t, x, q, \frac{\partial q}{\partial t}, \frac{\partial q}{\partial x}\right)~\mathrm{d}x~\mathrm{d}t$$, so the latter expression can be seen as a special case of the more general situation, as the latter expression implies that spatial interactions, like temporal interactions, are semilocal (depending only on the field and neighboring points via first derivatives).

This in turn shows a significant disadvantage of the many-body view, namely that it is not possible to write the action generically as the integral of a Lagrangian while allowing for the possibility of arbitrary spatially nonlocal many-body interactions (though it is always possible to imagine a specific form of the action that looks like some sort of integral, such as the one above involving $$|x - x'|^{-\beta}$$); the best that can be done is to say $$S[q] = \int L\left(t, q, \frac{\partial q}{\partial t}\right)~\mathrm{d}t$$ and that any dependence on the spatial derivatives $$\frac{\partial q}{\partial x}$$ is implicit (as would be the case for discrete DOFs), but this requires much more care to be taken when evaluating functional derivatives (as well as partial derivatives). Furthermore, most physical systems involving interacting classical fields (so as to sidestep any questions about quantum nonlocality) fundamentally involve only local interactions, and any nonlocality is a consequence of tracing over some degrees of freedom. For example, interactions via the Coulomb potential can be reverse-engineered to yield purely semilocal equations of motion involving things like the Poisson equation for the electric potential as a new field to be considered in the action. Thus, it can often be justified to use an action encoding only semilocal interactions, and this is what will be done henceforth unless otherwise specified. That said, this doesn't necessarily mean that the relativistic view is "better"; both views should henceforth be kept in mind too.

### Stationary Action with Fields

If the form $$S[q] = \int L\left(t, q, \frac{\partial q}{\partial t}\right)~\mathrm{d}t$$ is used, with any dependence on the spatial dependence of $$q$$ being implicit, then following the steps for multiple degrees of freedom yields $\frac{\delta S}{\delta q(t, x)} = \frac{\delta L}{\delta q(t, x)} - \frac{\partial}{\partial t} \frac{\delta L}{\delta \frac{\partial q(t, x)}{\partial t}}$ as the first derivative of the action and $\frac{\delta^{2} S}{\delta q(t, x) \delta q(t', x')} = \frac{\delta^{2} L}{\delta q(t, x) \delta q(t, x')} \delta(t - t') + \frac{\delta^{2} L}{\delta \frac{\partial q(t', x)}{\partial t'} \delta q(t', x')} \frac{\partial}{\partial t'} \delta(t' - t)$ $+ \frac{\delta^{2} L}{\delta q(t, x) \delta \frac{\partial q(t, x')}{\partial t}} \frac{\partial}{\partial t} \delta(t - t') - \frac{\partial}{\partial t} \left(\frac{\delta^{2} L}{\delta \frac{\partial q(t, x)}{\partial t} \delta \frac{\partial q(t, x')}{\partial t}} \frac{\partial}{\partial t} \delta(t - t')\right)$ as the second derivative. In the case of $$N$$ DOFs labeled by $$i$$, the operator was defined with respect to the composite index $$(t, i)$$, so now in the continuous case with DOFs labeled by spatial index $$x$$ per the many-body view, the operator is defined with respect to the composite continuous index $$(t, x)$$.

Regarding the above derivation, there are two important points to note. First, in both the first and second derivatives, although $$L$$ may have an explicit dependence on $$t$$, each instance of $$\frac{\partial}{\partial t}$$ should act on the full expression, including every instance of $$q$$, and not just on the explicit time dependence (which may be absent); this is a shortcoming of the partial derivative notation, but using the notation of a total derivative $$\frac{\mathrm{d}}{\mathrm{d}t}$$ may be misleading for other reasons, and I can't find better alternatives. Second, the derivatives of $$L$$ are functional derivatives for the aforementioned reason that there may be implicit dependence on $$q$$ through spatial derivatives $$\frac{\partial}{\partial x}$$, so those need to be accounted for too beyond just the partial derivatives $$\frac{\partial}{\partial q(t, x)}$$.

Now, if the form $$S[q] = \int \mathcal{L}\left(t, x, q, \frac{\partial q}{\partial t}, \frac{\partial q}{\partial x}\right)~\mathrm{d}x~\mathrm{d}t$$ is used, then the first derivative can be found as follows. First, the fields $$r(t, x) = \frac{\partial q(t, x)}{\partial t}$$ and $$s(t, x) = \frac{\partial q(t, x)}{\partial x}$$ will be defined for notational convenience. Next, discretizing $$t$$ and $$x$$ yields $$S[q] = \sum_{t', x'} \mathcal{L}_{t', x'}(q_{t', x'}, r_{t', x'}, s_{t', x'})$$ where $$r_{t, x} = \sum_{t', x'} D_{t, t'} \delta_{x, x'} q_{t', x'}$$ and $$s_{t, x} = \sum_{t', x'} \delta_{t, t'} D_{x, x'} q_{t', x'}$$. This means that $$\frac{\delta S}{\delta q_{t, x}} = \sum_{t', x'} \left(\frac{\partial \mathcal{L}_{t', x'}}{\partial q_{t', x'}} \frac{\partial q_{t', x'}}{\partial q_{t, x}} + \frac{\partial \mathcal{L}_{t', x'}}{\partial r_{t', x'}} \frac{\partial r_{t', x'}}{\partial q_{t, x}} + \frac{\partial \mathcal{L}_{t', x'}}{\partial s_{t', x'}} \frac{\partial s_{t', x'}}{\partial q_{t, x}}\right)$$. Using the aforementioned relations (along with the trivial relation $$q_{t, x} = \sum_{t', x'} \delta_{t, t'} \delta_{x, x'} q_{t', x'}$$) and the properties $$D_{t, t'} = -D_{t', t}$$ & $$D_{x, x'} = -D_{x', x}$$ gives $$\frac{\delta S}{\delta q_{t, x}} = \frac{\partial \mathcal{L}_{t, x}}{\partial q_{t, x}} - \sum_{t'} D_{t, t'} \frac{\partial \mathcal{L}_{t', x}}{\partial r_{t', x}} - \sum_{x'} D_{x, x'} \frac{\partial \mathcal{L}_{t, x'}}{\partial s_{t, x'}}$$ as the first derivative in the discrete case. This means the first derivative of the action is $\frac{\delta S}{\delta q(t, x)} = \frac{\partial \mathcal{L}}{\partial q(t, x)} - \frac{\partial}{\partial t} \frac{\partial \mathcal{L}}{\partial \left(\frac{\partial q(t, x)}{\partial t}\right)} - \frac{\partial}{\partial x} \frac{\partial \mathcal{L}}{\partial \left(\frac{\partial q(t, x)}{\partial x}\right)}$ in the continuous case, and its vanishing gives the equations of motion for the field $$q(t, x)$$.

The second derivative can be computed by starting from the discrete expression $$\frac{\delta S}{\delta q_{t', x'}} = \frac{\partial \mathcal{L}_{t', x'}}{\partial q_{t', x'}} - \sum_{t''} D_{t', t''} \frac{\partial \mathcal{L}_{t'', x'}}{\partial r_{t'', x'}} - \sum_{x''} D_{x', x''} \frac{\partial \mathcal{L}_{t', x''}}{\partial s_{t', x''}}$$ after relabeling indices. Applying the functional derivative $$\frac{\delta}{\delta q_{t, x}}$$ should be done to each side of the equation term by term. The first term is $$\frac{\delta}{\delta q_{t, x}} \frac{\partial \mathcal{L}_{t', x'}}{\partial q_{t', x'}} = \frac{\partial^{2} \mathcal{L}_{t', x'}}{\partial q_{t', x'}^{2}} \frac{\partial q_{t', x'}}{\partial q_{t, x}} + \frac{\partial^{2} \mathcal{L}_{t', x'}}{\partial r_{t', x'} \partial q_{t', x'}} \frac{\partial r_{t', x'}}{\partial q_{t, x}} + \frac{\partial^{2} \mathcal{L}_{t', x'}}{\partial s_{t', x'} \partial q_{t', x'}} \frac{\partial s_{t', x'}}{\partial q_{t, x}}$$, which yields $$\frac{\partial^{2} \mathcal{L}_{t', x'}}{\partial q_{t', x'}^{2}} \delta_{t', t} \delta_{x', x} + \frac{\partial^{2} \mathcal{L}_{t', x'}}{\partial r_{t', x'} \partial q_{t', x'}} D_{t', t} \delta_{x', x} + \frac{\partial^{2} \mathcal{L}_{t', x'}}{\partial s_{t', x'} \partial q_{t', x'}} \delta_{t', t} D_{x', x}$$. The second term is $$\frac{\delta}{\delta q_{t, x}} \frac{\partial \mathcal{L}_{t'', x'}}{\partial r_{t'', x'}} = \frac{\partial^{2} \mathcal{L}_{t'', x'}}{\partial q_{t'', x'} \partial r_{t'', x'}} \frac{\partial q_{t'', x'}}{\partial q_{t, x}} + \frac{\partial^{2} \mathcal{L}_{t'', x'}}{\partial r_{t'', x'}^{2}} \frac{\partial r_{t'', x'}}{\partial q_{t, x}} + \frac{\partial^{2} \mathcal{L}_{t'', x'}}{\partial s_{t'', x'} \partial r_{t'', x'}} \frac{\partial s_{t'', x'}}{\partial q_{t, x}}$$, which yields $$\frac{\partial^{2} \mathcal{L}_{t'', x'}}{\partial q_{t'', x'} \partial r_{t'', x'}} \delta_{t'', t} \delta_{x', x} + \frac{\partial^{2} \mathcal{L}_{t'', x'}}{\partial r_{t'', x'}^{2}} D_{t'', t} \delta_{x', x} + \frac{\partial^{2} \mathcal{L}_{t'', x'}}{\partial s_{t'', x'} \partial r_{t'', x'}} \delta_{t'', t} D_{x', x}$$. The third term is $$\frac{\delta}{\delta q_{t, x}} \frac{\partial \mathcal{L}_{t', x''}}{\partial s_{t', x''}} = \frac{\partial^{2} \mathcal{L}_{t', x''}}{\partial q_{t', x''} \partial s_{t', x''}} \frac{\partial q_{t', x''}}{\partial q_{t, x}} + \frac{\partial^{2} \mathcal{L}_{t', x''}}{\partial r_{t', x''} \partial s_{t', x''}} \frac{\partial r_{t', x''}}{\partial q_{t, x}} + \frac{\partial^{2} \mathcal{L}_{t', x''}}{\partial s_{t', x''}^{2}} \frac{\partial s_{t', x''}}{\partial q_{t, x}}$$, which yields $$\frac{\partial^{2} \mathcal{L}_{t', x''}}{\partial q_{t', x''} \partial s_{t', x''}} \delta_{t', t} \delta_{x'', x} + \frac{\partial^{2} \mathcal{L}_{t', x''}}{\partial r_{t', x''} \partial s_{t', x''}} D_{t', t} \delta_{x'', x} + \frac{\partial^{2} \mathcal{L}_{t', x''}}{\partial s_{t', x''}^{2}} \delta_{t', t} D_{x'', x}$$. Putting these together yields $$\frac{\delta^{2} S}{\delta q_{t, x} \delta q_{t', x'}} = \frac{\partial^{2} \mathcal{L}_{t, x}}{\partial q_{t, x}^{2}} \delta_{t, t'} \delta_{x, x'} + \frac{\partial^{2} \mathcal{L}_{t', x}}{\partial r_{t', x} \partial q_{t', x}} D_{t', t} \delta_{x, x'} + \frac{\partial^{2} \mathcal{L}_{t, x'}}{\partial s_{t, x'} \partial q_{t, x'}} \delta_{t, t'} D_{x', x}$$ $$+ \frac{\partial^{2} \mathcal{L}_{t, x}}{\partial q_{t, x} \partial r_{t, x}} D_{t, t'} \delta_{x, x'} + \frac{\partial^{2} \mathcal{L}_{t, x'}}{\partial s_{t, x'} \partial r_{t, x'}} D_{t, t'} D_{x', x} - \sum_{t''} D_{t, t''} \frac{\partial^{2} \mathcal{L}_{t'', x}}{\partial r_{t'', x}^{2}} D_{t'', t'} \delta_{x, x'}$$ $$+ \frac{\partial^{2} \mathcal{L}_{t, x}}{\partial q_{t, x} \partial s_{t, x}} \delta_{t, t'} D_{x, x'} + \frac{\partial^{2} \mathcal{L}_{t', x}}{\partial r_{t', x} \partial s_{t', x}} D_{t', t} D_{x, x'} - \sum_{x''} D_{x, x''} \frac{\partial^{2} \mathcal{L}_{t, x''}}{\partial s_{t, x''}^{2}} D_{x'', x'} \delta_{t, t'}$$ in the discrete treatment. In the continuous treatment, the Hessian operator is $\frac{\delta^{2} S}{\delta q(t, x) \delta q(t', x')} = \frac{\partial^{2} \mathcal{L}}{\partial q(t, x)^{2}} \delta(t - t') \delta(x - x')$ $+ \frac{\partial^{2} \mathcal{L}}{\partial \left(\frac{\partial q(t', x)}{\partial t'}\right) \partial q(t', x)} \delta(x - x') \frac{\partial}{\partial t'} \delta(t' - t)$ $+ \frac{\partial^{2} \mathcal{L}}{\partial \left(\frac{\partial q(t, x')}{\partial x'}\right) \partial q(t, x')} \delta(t - t') \frac{\partial}{\partial x'} \delta(x' - x)$ $+ \frac{\partial^{2} \mathcal{L}}{\partial q(t, x) \partial \left(\frac{\partial q(t, x)}{\partial t}\right)} \delta(x - x') \frac{\partial}{\partial t} \delta(t - t')$ $- \frac{\partial}{\partial t} \left(\frac{\partial^{2} \mathcal{L}}{\partial \left(\frac{\partial q(t, x)}{\partial t}\right)^{2}} \frac{\partial}{\partial t} \delta(t - t')\right) \delta(x - x')$ $+ \frac{\partial^{2} \mathcal{L}}{\partial \left(\frac{\partial q(t, x')}{\partial x'}\right) \partial \left(\frac{\partial q(t, x')}{\partial t}\right)} \frac{\partial}{\partial t} \delta(t - t') \frac{\partial}{\partial x'} \delta(x' - x)$ $+ \frac{\partial^{2} \mathcal{L}}{\partial q(t, x) \partial \left(\frac{\partial q(t, x)}{\partial x}\right)} \delta(t - t') \frac{\partial}{\partial x} \delta(x - x')$ $+ \frac{\partial^{2} \mathcal{L}}{\partial \left(\frac{\partial q(t', x)}{\partial t'}\right) \partial \left(\frac{\partial q(t', x)}{\partial x}\right)} \frac{\partial}{\partial t'} \delta(t' - t) \frac{\partial}{\partial x} \delta(x - x')$ $- \frac{\partial}{\partial x} \left(\frac{\partial^{2} \mathcal{L}}{\partial \left(\frac{\partial q(t, x)}{\partial x}\right)^{2}} \frac{\partial}{\partial x} \delta(x - x')\right) \delta(t - t')$ which, like its discrete counterpart, is symmetric under interchange of the composite indices $$(t, x) \leftrightarrow (t', x')$$. This operator must be positive-definite for the fields solving the physical equations of motion for the action to be minimized for that field configuration.

As an example, consider the Lagrangian density $\mathcal{L}\left(t, x, q, \frac{\partial q}{\partial t}, \frac{\partial q}{\partial x}\right) = \frac{\rho}{2} \left(\frac{\partial q}{\partial t}\right)^{2} - \frac{\tau}{2} \left(\frac{\partial q}{\partial x}\right)^{2} - f(x, t)q - \frac{\kappa}{2} q^{2}$ which corresponds to a field with a positive mass and a positive propagation speed that is linearly coupled to an external forcing function. The first derivative of the action can be read off as $\frac{\delta S}{\delta q(t, x)} = -f(x, t) -\kappa q - \rho \frac{\partial^{2} q}{\partial t^{2}} + \tau \frac{\partial^{2} q}{\partial x^{2}}$ and its vanishing corresponds to the equations of motion (i.e. the forced massive Klein-Gordon equation, though with a different speed and mass). The second derivative of the action can be read off as $\frac{\delta^{2} S}{\delta q(t, x) \delta q(t', x')} = \left(-\kappa - \rho \frac{\partial^{2}}{\partial t^{2}} + \tau \frac{\partial^{2}}{\partial x^{2}}\right) \delta(t - t')\delta(x - x')$ and this operator is diagonalized in the basis of traveling waves $$\exp(\mathrm{i}(kx - \omega t))$$ such that the eigenvalues are $$-\kappa + \rho \omega^{2} - \tau k^{2}$$; the eigenvalues change sign depending on the values of $$\omega$$ and $$k$$, both of which can be arbitrarily large or small in magnitude and both of which can take either sign, so the operator is not positive-definite, meaning that the equations of motion correspond to a saddle point in the action.

### Hamiltonian Formulation for Fields

This part doesn't have to do with the question of whether the action is minimized at a stationary point corresponding to the equations of motion, but it does illustrate the difference between the relativistic versus many-body views of classical fields. To do so, consider the case of a single DOF. The Hamiltonian can be expressed in terms of the Lagrangian as $H = p\dot{q} - L$ where $p = \frac{\partial L}{\partial \dot{q}}$ is the canonically conjugate momentum. The Hamiltonian is related to the Lagrangian by a Legendre transformation, so inverting that means that the action $S[q, p; t] = \int_{0}^{t} \left(p\dot{q} - H(t', q, p)\right)~\mathrm{d}t'$ can be written as a functional of both the coordinate $$q$$ and conjugate momentum $$p$$. The first derivatives therefore are $\frac{\delta S}{\delta q(t)} = -\dot{p} - \frac{\partial H}{\partial q}$ (having used integration by parts or, equivalently, the antisymmetry of the time derivative operator) and $\frac{\delta S}{\delta p(t)} = \dot{q} - \frac{\partial H}{\partial p}$ which, when vanishing, reproduce Hamilton's equations of motion for the canonical phase space DOFs $$(q, p)$$.

The many-body view comes from the generalization of formulas for $$N$$ degrees of freedom labeled $$i$$. The action $S[q_{1}, \ldots, q_{N}, p_{1}, \ldots, p_{N}; t] = \int_{0}^{t} \left(\sum_{i = 1}^{N} p_{i}\dot{q}_{i} - H(t', q_{1}, \ldots, q_{N}, p_{1}, \ldots, p_{N})\right)~\mathrm{d}t'$ has first derivatives $\frac{\delta S}{\delta q_{i}(t)} = -\dot{p}_{i} - \frac{\partial H}{\partial q_{i}}$ and $\frac{\delta S}{\delta p_{i}(t)} = \dot{q}_{i} - \frac{\partial H}{\partial p_{i}}$ which, when vanishing, reproduce Hamilton's equations of motion for each canonical phase space DOF in terms of itself & all of the others. If the labels $$i$$ are replaced by spatial coordinates $$x$$ but the time parameter $$t$$ is still taken to be special as per the many-body view, then the field $$q(t, x)$$ and its conjugate momentum $$p(x, t)$$ enter the action $S[q, p; t] = \int_{0}^{t} \left(\int p(t, x)\frac{\partial q(t, x)}{\partial t}~\mathrm{d}x - H[q, p; t']\right)~\mathrm{d}t'$ where the Hamiltonian is itself a functional containing implicit spatial dependence due to the possibility of arbitrary spatially nonlocal many-body interactions. The first derivatives of the action $\frac{\delta S}{\delta q(t, x)} = -\frac{\partial p(t, x)}{\partial t} - \frac{\delta H}{\delta q(t, x)}$ and $\frac{\delta S}{\delta p(t, x)} = \frac{\partial q(t, x)}{\partial t} - \frac{\delta H}{\delta p(t, x)}$ are, when vanishing, Hamilton's equations of motion for the field & its conjugate momentum, in which the derivatives of the Hamiltonian are functional derivatives as the Hamiltonian is itself a functional.

Things become a little simpler to understand with a Hamiltonian density encoding spatially semilocal interactions. In particular, it is possible at each $$(t, x)$$ to perform a Legendre transformation from the Lagrangian density to the Hamiltonian density according to $\mathcal{H}\left(t, x, q, p, \frac{\partial q}{\partial x}\right) = p(t, x)\frac{\partial q(t, x)}{\partial t} - \mathcal{L}\left(t, x, q, \frac{\partial q}{\partial t}, \frac{\partial q}{\partial x}\right)$ where it is clear that $$t$$ takes special precedence over $$x$$ as the dependence on $$\frac{\partial q}{\partial x}$$ is explicit even after the Legendre transformation from $$\frac{\partial q}{\partial t}$$ to $$p$$. Inverting this Legendre transformation and defining $H = \int \mathcal{H}~\mathrm{d}x$ means that the action is $S = \int_{0}^{t} \int \left(p(t', x)\frac{\partial q(t', x)}{\partial t'} - \mathcal{H}\left(t', x, q, p, \frac{\partial q}{\partial x}\right)\right)~\mathrm{d}x~\mathrm{d}t'$ and the first derivatives are $\frac{\delta S}{\delta q(t, x)} = -\frac{\partial p(t, x)}{\partial t} - \frac{\partial \mathcal{H}}{\partial q(t, x)}$ and $\frac{\delta S}{\delta p(t, x)} = \frac{\partial q(t, x)}{\partial t} - \frac{\partial \mathcal{H}}{\partial p(t, x)}$ which, when vanishing, yield Hamilton's equations of motion for the field and its conjugate momentum.

The aforementioned discussion of Hamilton's equations for classical fields privileges $$t$$ as per the many-body view. If instead the relativistic view is used, using index notation $$x^{\mu}$$ & $$\frac{\partial}{\partial x^{\mu}}$$ (where $$x^{(t)} = t$$ & $$x^{(x)} = x$$, the subscripts are put in parentheses to distinguish from exponents, and Einstein summation notation will not be used), the canonical stress energy tensor $T^{\mu}_{\; \nu}(t, x, q, p^{(t)}, p^{(x)}) = p^{\mu}(t, x) \frac{\partial q(t, x)}{\partial x^{\nu}} - \delta^{\mu}_{\; \nu} \mathcal{L}\left(t, x, q, \frac{\partial q}{\partial t}, \frac{\partial q}{\partial x}\right)$ is the Legendre transformation of the Lagrangian density treating $$(t, x)$$ on the same footing, although it is not symmetric under interchange of the coordinate indices $$(\mu, \nu)$$. Inverting the Legendre transformation by changing the index $$\nu$$ to $$\mu$$ and taking the sum over that index yields $\mathcal{L} = \frac{1}{d + 1} \sum_{\mu} \left(p^{\mu} \frac{\partial q}{\partial x^{\mu}} - T^{\mu}_{\; \mu}\right)$ where $$d$$ is the number of spatial dimensions (with $$d = 1$$ consistently through this post), though the fact that the Lagrangian density is being multiplied by a uniform prefactor means that prefactor will not affect the equations of motion. The first derivatives of the action are $\frac{\delta S}{\delta q(t, x)} = \frac{1}{d + 1} \sum_{\mu} \left(-\frac{\partial p^{\mu}(t, x)}{\partial x^{\mu}} - \frac{\partial T^{\mu}_{\; \mu}}{\partial q(t, x)}\right)$ for $$q$$, $\frac{\delta S}{\delta p^{(t)}(t, x)} = \frac{1}{d + 1} \left(\frac{\partial q(t, x)}{\partial t} - \sum_{\mu} \frac{\partial T^{\mu}_{\; \mu}}{\partial p^{(t)}(t, x)}\right)$ for $$p^{(t)}$$, and $\frac{\delta S}{\delta p^{(x)}(t, x)} = \frac{1}{d + 1} \left(\frac{\partial q(t, x)}{\partial x} - \sum_{\mu} \frac{\partial T^{\mu}_{\; \mu}}{\partial p^{(x)}(t, x)}\right)$ for $$p^{(x)}$$. Each of these first derivatives vanishing gives the equation of motion for the corresponding canonical phase space field.

## Connection to the Path Integral Formulation of Quantum Mechanics

All of this rigmarole in deriving & explaining the second derivative of the action can be justified in part by connecting it to the path integral formulation of quantum mechanics. In particular, for a single nonrelativistic DOF evolving under a Lagrangian or Hamiltonian that does not explicitly depend on $$t$$, the probability amplitude for a state localized at position $$x_{1}$$ at time $$t_{1}$$ to evolve to a state localized at position $$x_{2}$$ at time $$t_{2}$$ is given by $\langle x_{2}|\exp(-\mathrm{i}(t_{2} - t_{1})\hat{H}/\hbar)|x_{1} \rangle = \int \exp(\mathrm{i} S[x]/\hbar)~\mathcal{D}x$ where on the left-hand side, $$\hat{H}$$ is the quantum Hamiltonian, while on the right-hand side, $$S[x]$$ is the classical action and the integral is a functional integral over all trajectories which satisfy $$x(t_{1}) = x_{1}$$ and $$x(t_{2}) = x_{2}$$. If $$X(t)$$ is the physical classical trajectory, then expanding the action gives $S[x] = S[X] + \int_{t_{1}}^{t_{2}} \int_{t_{1}}^{t_{2}} (x(t) - X(t)) \frac{\delta^{2} S[y]}{\delta y(t) \delta y(t')}\bigg|_{y = X} (x(t') - X(t'))~\mathrm{d}t'~\mathrm{d}t + \ldots$ as the first derivative vanishes by construction at the stationary point. The fact that the argument of the exponential in the functional integral has an overall multiplicative factor of $$\mathrm{i}$$ is the reason that the nature of the stationary point doesn't really matter: whether it is a saddle point, minimum, or maximum in the action, contributions around that trajectory will constructively interfere, and even if the action looks like a hyperbolic paraboloid with many trajectories of much higher action & many other trajectories of much lower action, the contributions from all of those other trajectories together will destructively interfere. (I expect that the situation will look similar in field theory, but I won't comment on that explicitly because of the subtleties of the treatment of time in relativistic quantum field theory compared to nonrelativistic quantum field theory.) This contrasts with the case from classical statistical field theory where, for example, the partition function may be written as $Z = \int \exp(-\beta H[m])~\mathcal{D}m$ where the Hamiltonian $$H$$ is a real-valued functional of the real-valued field $$m(x)$$ (assuming one spatial dimension with coordinate $$x$$ in this case); in such a case, using Laplace's method to approximate the functional integral as a sum over Gaussian integrals each around stationary points of the Hamiltonian requires that those stationary points be minima.

### Connections Between Functional Integral Constraints and Lagrange Multipliers

Further examination of the aforementioned functional integral involving $$\exp(\mathrm{i}S[x]/\hbar)$$ can yield some useful insights connecting functional integral constraints and Lagrange multipliers. In particular, implicit in the statement that the functional integral is taken over all trajectories which satisfy $$x(t_{1}) = x_{1}$$ and $$x(t_{2}) = x_{2}$$ is also the fact that the action shouldn't be evaluated for $$t < t_{1}$$ or $$t > t_{2}$$ and that all contributions to the integral which would have resulted from the inclusion of such paths violating the aforementioned constraints should be ignored. (For the rest of this section, $$x_{1}$$ will be renamed $$a$$ and $$x_{2}$$ will be renamed $$b$$ to avoid confusion in the discrete case where $$x(t)$$ is replaced by subscript indexing as in $$x_{t}$$.)

Enforcement of these constraints may be initially easier to understand in the discretized case. If $$S = \sum_{t = t_{1}}^{t_{2}} L_{t}(x_{t}, v_{t})$$ where $$v_{t} = \sum_{t' = -\infty}^{\infty} D_{t, t'} x_{t'}$$, then the functional integral becomes a multidimensional integral $$\int \exp(\mathrm{i}S/\hbar) \prod_{t} \mathrm{d}x_{t}$$. The constraints $$x(t_{1}) = a$$, $$x(t_{2}) = b$$, and $$x_{t} = 0$$ for $$t \notin [t_{1}, t_{2}]$$ can be written in the discrete case by simply using the definition of $$S$$ and plugging in the constraints at $$t_{1}$$ and $$t_{2}$$. That said, even if $$S$$ only includes contributions from $$t \in [t_{1}, t_{2}]$$, the multidimensional integral includes integration over $$\mathrm{d}x_{t}$$ for $$t$$ outside of that interval, so the limits of the product need to be restricted to $$t$$ in that interval. In the discrete case, this is easily done with the replacement of the infinitesimal volume $$\prod_{t = -\infty}^{\infty} \mathrm{d}x_{t} \to \prod_{t = t_{1} + 1}^{t_{2} - 1} \mathrm{d}x_{t}$$ (to exclude the endpoints of the interval, as those correspond to the fixed boundary values of the trajectory).

Difficulties arise when going to the continuous case though, as there is no easy way to make such explicit replacements. Instead, it is instructive to reverse-engineer explicit constraints in the discrete case that can be modified appropriately for the continuous case. First, restricting trajectories such that $$x(t_{1}) = a$$ and $$x(t_{2}) = b$$ can be done by multiplying the integrand $$\exp(\mathrm{i}S[x]/\hbar)$$ by $$\delta(x_{t_{1}} - a) \delta(x_{t_{2}} - b)$$; each Dirac delta function of the trajectory at each of these two specific time indices enforces equality with the specified boundary values. Second, restricting trajectories such that $$x_{t} = 0$$ for $$t \notin [t_{1}, t_{2}]$$ can be done by multiplying the new integrand $$\exp(\mathrm{i}S/\hbar) \delta(x_{t_{1}} - a) \delta(x_{t_{2}} - b)$$ by the product of Dirac delta functions $$\prod_{t = -\infty}^{\infty} \delta\left(\left(1 - \Theta\left(t - t_{1}\right)\Theta\left(t_{2} - t\right)\right)x_{t} \right)$$ where $$\Theta$$ is the Heaviside step function (defined here such that $$\Theta(0) = 1$$). Third, the summation limits can be replaced in the definition of the action with the expression $$S[x] = \sum_{t = -\infty}^{\infty} \Theta\left(t - t_{1}\right)\Theta\left(t_{2} - t\right)L_{t}(x_{t}, v_{t})$$, which ensures that the action vanishes for $$t \notin [t_{1}, t_{2}]$$ and that the integral over $$x_{t}$$ for each $$t$$ outside of the desired interval yields 1. With those changes, the multidimensional integral can be written as $$\int \exp(\mathrm{i}S/\hbar) \delta(x_{t_{1}} - a) \delta(x_{t_{2}} - b) \times$$ $$\left(\prod_{t = -\infty}^{\infty} \delta\left(\left(1 - \Theta\left(t - t_{1}\right)\Theta\left(t_{2} - t\right)\right)x_{t} \right)\right) \prod_{t = -\infty}^{\infty} \mathrm{d}x_{t}$$ in the discrete case.

In the continuous case, the following modifications are made. The action is now expressed as $S[x] = \int_{-\infty}^{\infty} \Theta(t - t_{1})\Theta(t_{2} - t)L(t, x, \dot{x})~\mathrm{d}t$ after expanding the limits of integration. In the functional integral, the factor $$\delta(x_{t_{1}} - a) \delta(x_{t_{2}} - b)$$ becomes $$\delta(x(t_{1}) - a) \delta(x(t_{2}) - b)$$; note that each Dirac delta function is a typical Dirac delta function in that each depends on a single variable, either $$x(t_{1})$$ or $$x(t_{2})$$, but each can also be considered a functional of the function $$x(t)$$ by mapping the function $$x$$ to a real number given by a Dirac delta function that depends on the function evaluated for specific values of $$t$$. Finally, it will be helpful to define the entity $g[x; t] = \left(1 - \Theta\left(t - t_{1}\right)\Theta\left(t_{2} - t\right)\right)x(t)$ so that the expression $$\prod_{t = -\infty}^{\infty} \delta(g[x; t])$$ in the discrete case becomes $$\delta[g[x; t]]$$ in the continuous case; this is a Dirac delta functional, as discussed in a previous post [link here], because evaluating $$g$$ for a given function $$x$$ yields a function of $$t$$ which then goes into a Dirac delta function that is multiplied for each value of $$t$$. Putting all of these together gives $\langle x_{2}|\exp(-\mathrm{i}(t_{2} - t_{1})\hat{H}/\hbar)|x_{1} \rangle$ $= \int \exp(\mathrm{i} S[x]/\hbar) \delta(x(t_{1}) - a)\delta(x(t_{2}) - b)\delta[g[x; t]] \mathcal{D}x$ as the functional integral form of the probability amplitude, where now all constraints are explicit so the functional integral is over the whole space of functions $$x(t)$$.

Connections to Lagrange multipliers become evident by expressing each Dirac delta function or functional in terms of its respective complex- or functional-valued Fourier expansion. The constraints on the trajectory at the boundary values can be expressed as $\delta(x(t_{1}) - a) = \int_{-\infty}^{\infty} \exp(\mathrm{i}k_{1} \times (x(t_{1}) - a))~\frac{\mathrm{d}k_{1}}{2\pi}$ with a similar expression for $$\delta(x(t_{2}) - b)$$ (using $$k_{2}$$ as the corresponding variable in the Fourier integral). The integrand can be further expressed as $\exp(\mathrm{i}k_{1} \times (x(t_{1}) - a)) = \exp\left(\mathrm{i}k_{1} \int_{-\infty}^{\infty} (x(t) - a)\delta(t - t_{1})~\mathrm{d}t\right)$ (and likewise for the expression involving $$x(t_{2})$$ and $$k_{2}$$). The Dirac delta functional whose argument is $$g[x; t]$$ can be expressed as $\delta[g[x; t]] \sim \exp\left(\mathrm{i} \int_{-\infty}^{\infty} k(t)\left(1 - \Theta(t - t_{1})\Theta(t_{2} - t)\right)x(t)~\mathrm{d}t \right)~\mathcal{D}k$ where $$k(t)$$ is the functional Fourier variable that constrains $$x$$ at each $$t$$ to vanish for $$t \notin [t_{1}, t_{2}]$$; the use of $$\sim$$ instead of $$=$$ is because the denominator would have a factor $$(2\pi)^{N}$$ leading to the integral vanishing in the limit $$N \to \infty$$. Thus, the functional integral can be expressed as $\langle x_{2}|\exp(-\mathrm{i}(t_{2} - t_{1})\hat{H}/\hbar)|x_{1} \rangle \sim$ $\int \exp(\mathrm{i}F[x, k; t_{1}, t_{2}, a, b, k_{1}, k_{2}])~\mathcal{D}x~\mathcal{D}k~\frac{\mathrm{d}k_{1}~\mathrm{d}k_{2}}{(2\pi)^{2}}$ where the integrals over $$k_{1}$$ and $$k_{2}$$ each go over the entire real line and the functional integrals over $$x(t)$$ and $$k(t)$$ each go over the whole space of real-valued functions. This is written in terms of the functional $F[x, k; t_{1}, t_{2}, a, b, k_{1}, k_{2}] = \int_{-\infty}^{\infty} f(t, x(t), \dot{x}(t), k(t), t_{1}, t_{2}, a, b, k_{1}, k_{2})~\mathrm{d}t$ using the notation $f(t, x(t), \dot{x}(t), k(t), t_{1}, t_{2}, a, b, k_{1}, k_{2}) =$ $\frac{1}{\hbar} \Theta(t - t_{1})\Theta(t_{2} - t)L(t, x(t), \dot{x}(t)) + k_{1} \times (x(t) - a)\delta(t - t_{1}) +$ $k_{2} \times (x(t) - b)\delta(t - t_{2}) + k(t)(1 - \Theta(t - t_{1})\Theta(t_{2} - t))x(t)$ for brevity. Thus, it can be seen that the Lagrangian has additive constraints enforced by Lagrange multipliers: the Lagrange multiplier $$k_{1}$$ enforces the constraint that $$x$$ must equal $$a$$ at the instant $$t_{1}$$, the Lagrange multiplier $$k_{2}$$ enforces the constraint that $$x$$ must equal $$b$$ at the instant $$t_{2}$$, and the Lagrange multiplier function $$k(t)$$ enforces the constraint that $$x$$ at every instant $$t$$ must vanish for $$t \notin [t_{1}, t_{2}]$$. These additive constraints to the Lagrangian exactly turn into multiplicative functional constraints in the functional integral. Moreover, there is no particular reason that the Lagrange multipliers $$k_{1}$$ and $$k_{2}$$ must be separate from the Lagrange multiplier function $$k(t)$$, as they could be defined such that $$k_{1} = k(t_{1})$$ and $$k_{2} = k(t_{2})$$. With this in mind, the functional integral can be expressed as $\langle x_{2}|\exp(-\mathrm{i}(t_{2} - t_{1})\hat{H}/\hbar)|x_{1} \rangle \sim \int \exp(\mathrm{i}F[x, k; t_{1}, t_{2}, a, b])~\mathcal{D}x~\mathcal{D}k$ after dispensing with the previous integrals over $$k_{1}$$ and $$k_{2}$$. This is written in terms of the functional $F[x, k; t_{1}, t_{2}, a, b] = \int_{-\infty}^{\infty} f(t, x(t), \dot{x}(t), k(t), t_{1}, t_{2}, a, b)~\mathrm{d}t$ using the notation $f(t, x(t), \dot{x}(t), k(t), t_{1}, t_{2}, a, b) = \frac{1}{\hbar} \Theta(t - t_{1})\Theta(t_{2} - t)L(t, x(t), \dot{x}(t)) +$ $k(t) \left((x(t) - a)\delta(t - t_{1}) + (x(t) - b)\delta(t - t_{2}) + (1 - \Theta(t - t_{1})\Theta(t_{2} - t))x(t)\right)$ for brevity. Writing $$f$$ in this way makes clear that $$k(t)$$ can act as the single Lagrange multiplier function enforcing the boundary values as well as vanishing of $$x(t)$$ for $$t \notin [t_{1}, t_{2}]$$. Additionally, writing $$f$$ in this way shows the symmetry between $$x(t)$$ and $$k(t)$$, as the Dirac delta functions retain the values of $$k(t)$$ when evaluated at $$t_{1}$$ and $$t_{2}$$, while the Heaviside step functions retain the values of $$k(t)$$ only for $$t \in [t_{1}, t_{2}]$$. Thus, using the Fourier expansions of the functional integral constraints makes the connection to Lagrange multipliers clear, albeit at the expense that the quantity $$\langle x_{2}|\exp(-\mathrm{i}(t_{2} - t_{1})\hat{H}/\hbar)|x_{1} \rangle$$ is proportional but not exactly equal to the functional integral (though the need to perform the functional integral on a computer, which will be present in most cases, will require a discrete representation that makes concerns about proportionality moot).