Free particles
1. Newtonian mechanics
Free particles in Newtonian mechanics are simply point particles with no force applied to them, so that $\vec{F} = 0$. In other words,
\begin{eqnarray} \ddot{\vec{x}}(t) = 0 \end{eqnarray}This is fairly easily solved for the velocity as
\begin{equation} \dot{\vec{x}}(t) = \vec{v}_0 \end{equation}For some constant of integration $\vec{v}_0$ (corresponding to the initial velocity : $\dot{\vec{x}}(0) = \vec{v}_0$), and for the position
\begin{equation} \vec{x}(t) = \vec{v}_0 t + \vec{x}_0 \end{equation}for some constant of integration $\vec{x}_0$, corresponding to the initial position ($\vec{x}(0) = \vec{x}_0$).
2. Lagrangian mechanics
Dynamics
The Lagrangian of a free particle is
\begin{equation} L = \frac{m}{2} \dot{\vec{x}} \cdot \dot{\vec{x}} \end{equation}with action
\begin{equation} S = \int_{t_i}^{t_f} dt\ \frac{m}{2} \dot{\vec{x}} \cdot \dot{\vec{x}} \end{equation}This action is defined for some function $\vec{x}$, on the space of $C^1$ functions mapping $\vec{x} : [t_a, t_b] \to \mathbb{R}^n$, with appropriate boundary/initial value conditions. The variation of the action within that function space gives
\begin{eqnarray} \delta S &=& \int_{t_i}^{t_f} dt \left[\frac{\partial L}{\partial \vec{x}} \delta \vec{x} + \frac{\partial L}{\partial \dot{\vec{x}}} \delta \dot{\vec{x}} \right] \end{eqnarray}which is, integrating by parts,
\begin{eqnarray} \delta S &=& \int_{t_i}^{t_f} dt \left[ (\frac{\partial L}{\partial \vec{x}} - \frac{d}{dt}\frac{\partial L}{\partial \dot{\vec{x}}}) \delta \vec{x} \right] + \left[ \frac{\partial L}{\partial \dot{\vec{x}}} \cdot \delta \vec{x} \right]^{t_2}_{t_1} \end{eqnarray}As the perturbation $\delta \vec{x}$ is always $0$ on the boundaries, this leads us, by the fundamental lemma of the calculus of variation, to
\begin{equation} \frac{\partial L}{\partial \vec{x}} - \frac{d}{dt}\frac{\partial L}{\partial \dot{\vec{x}}} = 0 \end{equation} \begin{equation} - \frac{d}{dt} (m \dot{\vec{x}}(t)) = 0 \end{equation} \begin{equation} \ddot{\vec{x}}(t) = 0 \end{equation}We can solve this in two ways. As an initial value problem, $\vec{x}(0) = \vec{x}_0$, $\dot{\vec{x}}(0) = \vec{v}_0$, the solution is the same as classically,
\begin{equation} \vec{x}(t) = \vec{v}_0 t + \vec{x}_0 \end{equation}As a boundary value problem, $\vec{x}(t_a) = \vec{x}_a$, $\vec{x}(t_b) = \vec{x}_b$, we get
\begin{equation} \vec{x}(t) = \frac{\vec{x}_b - \vec{x}_a}{t_b - t_a} t + \frac{1}{2} \left[ \vec{x}_a + \vec{x}_b - \frac{\vec{x}_b - \vec{x}_a}{t_b - t_a} (t_a + t_b)\right] \end{equation}Once we have the actual solution, we can also compute the actual extremal action if we wish to.
\begin{eqnarray} S &=& \int_{t_i}^{t_f} dt \frac{m}{2} \vec{v}_0 \cdot \vec{v}_0\\ &=& \frac{m}{2} \vec{v}_0 \cdot \vec{v}_0 [t_f - t_i]\\ \end{eqnarray}Symmetries
In the case of point particles, we can find the full set of symmetries of the action by hand. The set of all automorphisms of the $\pi : \mathbb{R}^n \to \mathbb{R}$ bundle is
\begin{equation} \Phi(t, \vec{x}) = (\alpha(t), \vec{f}(\vec{x}, t)) \end{equation}which acts on sections as
\begin{eqnarray} \Phi(t, \vec{x}(t)) &=& (\alpha(t), \vec{f}(\vec{x}(t), t))\\ &=& (s, \vec{f}(\vec{x}(\alpha^{-1}(s)), \alpha^{-1}(s))) \end{eqnarray}with the new variable $s = \alpha(t)$. Then the action becomes, with the help of some variable change,
\begin{eqnarray} S(\Phi(\vec{x})) &=& \int_{\alpha(t_a)}^{\alpha(t_b)} (\frac{d}{dt}\vec{f}(\vec{x}(\alpha^{-1}(s)), \alpha^{-1}(s)) \cdot (\frac{d}{dt}\vec{f}(\vec{x}(\alpha^{-1}(s)), \alpha^{-1}(s))) \frac{dt}{ds} ds\\ &=& \int_{\alpha(t_a)}^{\alpha(t_b)} (\frac{ds}{dt} \frac{d}{ds}\vec{f}(\vec{x}(\alpha^{-1}(s)), \alpha^{-1}(s)) \cdot (\frac{ds}{dt} \frac{d}{ds}\vec{f}(\vec{x}(\alpha^{-1}(s)), \alpha^{-1}(s))) \frac{dt}{ds} ds\\ &=& \int_{\alpha(t_a)}^{\alpha(t_b)} \dot{\alpha}(\alpha(s)) \left[ \dot{\vec{x}}(t) \frac{\partial}{\partial \vec{x}}\vec{f}(\vec{x}(\alpha^{-1}(s)), \alpha^{-1}(s)) + \frac{\partial}{\partial s}\vec{f}(\vec{x}(\alpha^{-1}(s)), \alpha^{-1}(s)) \right] \cdot (\frac{d}{ds}\vec{f}(\vec{x}(\alpha^{-1}(s)), \alpha^{-1}(s))) ds\\ \end{eqnarray}The action is invariant under the Galilean group, as well as the Bargmann group if we wish. This can be shown by considering the various transformations :
Time translations : For the transformation $t \to t' = t + T$, the action transforms as
\begin{eqnarray} S &=& \int_{t_i}^{t_f} dt'\ \frac{m}{2} \dot{\vec{x}}(t') \cdot \dot{\vec{x}}(t')\\ &=& \int_{t_i}^{t_f} dt\ \frac{m}{2} \dot{\vec{x}}(t + T) \cdot \dot{\vec{x}}(t + T)\\ &=& \int_{t_i}^{t_f} dt\ \frac{m}{2} \dot{\vec{x}}(t + T) \cdot \dot{\vec{x}}(t + T) \end{eqnarray}van Vleck determinant
The extremal action for the boundary conditions $\vec{x}(t_i) = \vec{x}_i$, $\vec{x}(t_i) = \vec{x}_i$
\begin{eqnarray} S(\vec{x}_i, t_i; \vec{x}_f, t_f) &=& \frac{m}{2} \frac{\| \vec{x_f} - \vec{x}_i \|^2}{t_f - t_i} \end{eqnarray}There is only a single extremal path for free particles, therefore there is a unique van Vleck determinant, which is
\begin{eqnarray} \Delta(\vec{x}_i, t_i; \vec{x}_f, t_f) &=& (-1)^n \det (\frac{\partial^2 S}{\partial q_i \partial q_f})\\ &=& (-1)^n \frac{m}{2} \frac{\det( 2(\vec{x}_f + \vec{x}_i - 1) )}{t_f - t_i} \end{eqnarray}3. Hamiltonian mechanics
The momentum associated with the Lagrangian is
\begin{eqnarray} \vec{p} &=& \frac{\partial L}{\partial \dot{\vec{x}}}\\ &=& m \dot{\vec{x}} \end{eqnarray}The map from momentum to velocity is invertible (since the Hessian of the Lagrangian with respect to the velocities is simply $\mathrm{Hess}(L) = m \delta$, which is of full rank), via the function
\begin{equation} \vec{v}(\vec{p}) = \frac{\vec{p}}{m} \end{equation}So that we can write the Hamiltonian out as the Legendre transform
\begin{eqnarray} H &=& \vec{v}(\vec{p}) \cdot \vec{p} - L\\ &=& \frac{\vec{p}\cdot\vec{p}}{m} - \frac{m}{2} \vec{v}(\vec{p}) \cdot \vec{v}(\vec{p}) \\ &=& \frac{\vec{p}\cdot\vec{p}}{2m} \end{eqnarray}The equations of motion are given by, as usual
\begin{eqnarray} \frac{\partial H}{\partial \vec{x}} &=& -\dot{\vec{p}}\\ &=& 0\\ \frac{\partial H}{\partial \vec{p}} &=& \dot{\vec{x}}\\ &=& \frac{\vec{p}}{m} \end{eqnarray}with the solution
\begin{eqnarray} \vec{p}(t) &=& \vec{p}_0\\ \vec{x}(t) &=& \frac{\vec{\vec{p}_0}}{m} t + \vec{x}_0 \end{eqnarray}For later purpose, it may be useful to compute our Poisson brackets. With our canonical coordinates, this is
\begin{eqnarray} \left\{ f, g \right\} &=& \frac{\partial f}{\partial \vec{x}} \cdot \frac{\partial g}{\partial \vec{p}} - \frac{\partial g}{\partial \vec{x}} \cdot \frac{\partial f}{\partial \vec{p}} \end{eqnarray}It then follows that
\begin{eqnarray} \left\{ x^i, p^j \right\} &=& \delta^{ij}\\ \left\{ x^i, x^j \right\} &=& 0\\ \left\{ p^i, p^j \right\} &=& 0 \end{eqnarray}4. Dirac parametrized system
As an example of a simple parametrized system, it is possible to turn the point particle into a parametrization invariant Lagrangian, by considering the degrees of freedom $(t, \vec{x})$, parametrized by $\tau$, in which case the action is
\begin{equation} S = \int_{t_i}^{t_f} d\tau\ \frac{m}{2} \frac{\dot{\vec{x}}(\tau) \cdot \dot{\vec{x}}(\tau)}{\dot{t}(\tau)} \end{equation}This action is invariant under the reparametrization $\tau \to \tau' = f(\tau)$, for some diffeomorphism $f$, in which case, we get via change of variables
\begin{eqnarray} S' &=& \int_{\tau'_1}^{\tau'_2} d\tau' \frac{m}{2} \frac{\dot{\vec{x}}(\tau') \cdot \dot{\vec{x}}(\tau')}{\dot{t}(\tau')}\\ &=& \int_{\tau_1}^{\tau_2} d\tau' f'(\tau) \frac{m}{2} \frac{\dot{\vec{x}}(f(\tau)) \cdot \dot{\vec{x}}(f(\tau))}{\dot{t}(f(\tau))}\\ \end{eqnarray}We can verify that our system is underdetermined by computing the equations of motion :
\begin{eqnarray} \frac{\partial L}{\partial \vec{x}} &=& 0\\ \frac{\partial L}{\partial t} &=& 0\\ \frac{\partial L}{\partial \dot{\vec{x}}} &=& m \frac{\dot{\vec{x}}(\tau)}{\dot{t}(\tau)}\\ \frac{\partial L}{\partial \dot{t}} &=& -\frac{m}{2} \frac{\dot{\vec{x}}(\tau) \cdot \dot{\vec{x}}(\tau)}{\dot{t}^2(\tau)} \end{eqnarray}So that the Euler-Lagrange equations become
\begin{eqnarray} \frac{d}{dt}\left(\frac{\dot{\vec{x}}(\tau)}{\dot{t}(\tau)}\right) &=& 0\\ \frac{d}{dt}\left(\frac{\dot{\vec{x}}(\tau) \cdot \dot{\vec{x}}(\tau)}{\dot{t}^2(\tau)} \right) &=& 0 \end{eqnarray} \begin{eqnarray} \frac{\dot{\vec{x}}(\tau)}{\dot{t}(\tau)} &=& \vec{C}_1\\ \frac{\dot{\vec{x}}(\tau) \cdot \dot{\vec{x}}(\tau)}{\dot{t}^2(\tau)} &=& C_2 \end{eqnarray}We'll note that the first equation reduces to the second by taking its norm :
\begin{eqnarray} \frac{\dot{\vec{x}}(\tau)}{\dot{t}(\tau)} \cdot \frac{\dot{\vec{x}}(\tau)}{\dot{t}(\tau)} &=& \vec{C}_1 \cdot \vec{C}_1\\ &=& C_2 \end{eqnarray}We therefore have that for $4$ degrees of freedom, we only have three independent equations.
\begin{equation} \dot{\vec{x}}(\tau) = \dot{t}(\tau) \vec{C}_1 \end{equation}Our solution is therefore
\begin{equation} \vec{x}(\tau) = \vec{x}_0 + \int d\tau\ \dot{t}(\tau) \vec{C}_1 \end{equation}The basic gauge we can select here is $t = \tau$, in which case we simply get our previous case back. Now let's see the Hamiltonian method applied here. Our momenta are
\begin{eqnarray} \vec{p} &=& m \frac{\dot{\vec{x}}(\tau)}{\dot{t}(\tau)}\\ p_t &=& -\frac{m}{2} \frac{\dot{\vec{x}}(\tau) \cdot \dot{\vec{x}}(\tau)}{\dot{t}^2(\tau)} \end{eqnarray}As our problem is under-determined, we cannot invert our momenta back to velocities. We are going to need to apply some constraints to our system. The important quantity to consider here is the rank of the matrix
\begin{equation} \frac{\partial^2 L}{\partial \dot{q}^a \partial \dot{q}^b} \end{equation}with $q = (t, \vec{x})$. This matrix turns out to be
\begin{equation} \frac{\partial^2 L}{\partial \dot{q}^a \partial \dot{q}^b} = \begin{pmatrix} m \frac{\dot{\vec{x}}(\tau) \cdot \dot{\vec{x}}(\tau)}{\dot{t}^3(\tau)} & -\frac{m}{2} \frac{\dot{\vec{x}}(\tau)}{\dot{t}^2(\tau)} \\ m \frac{1}{\dot{t}(\tau)} & \frac{m}{\dot{t}(\tau)} \delta_{ij} \end{pmatrix} \end{equation}...
The obvious constrain we get from the form of our momenta is
\begin{equation} \Phi = \frac{\vec{p} \cdot \vec{p}}{2m} + p_t = 0 \end{equation}...
The Hamiltonian vanishes
\begin{equation} H = \vec{p} \cdot \dot{\vec{x}} + p_t \dot{t} - L = 0 \end{equation}meaning that only the constraint term will remain in the extended Hamiltonian
\begin{equation} H_E = v \Phi \end{equation}The Dirac brackets of our theory are
Bundle methods
For more abstract ways in classical mechanics, we can define the configuration space of a point particles by the tangent bundle. Given Euclidian space $E = \mathbb{R}^n$, our tangent bundle is $TE \cong \mathbb{R}^n \times \mathbb{R}^n$. Then the particle is described by the map $\gamma : \mathbb{R} \to E$, and our action is a function $S : TE \to \mathbb{R}$,
Special relativity
The special relativistic version of free particles can be done in any number of formalisms, similarly to the non-relativistic case. The simplest one being simply the relativistic version of Newton's second law
\begin{equation} f^\mu = mc^2 \ddot{x}^\mu(\tau) \end{equation}which is here simply $\ddot{x}^\mu(\tau) = 0$. Similarly as before, this has the solution
\begin{equation} x^\mu(\tau) = v^\mu_0 \tau + x_0^\mu \end{equation}Our particle is a standard slower than light particle if our initial velocity $v$ is timelike : $g(v,v) < 0$, in which case the velocity itself will be timelike. Similarly, it will be a null particle if $g(v,v) = 0$, and a tachyonic particle if $g(v,v) > 0$.
The action associated with a timelike point particle is
\begin{equation} S = -m \int_{\tau_1}^{\tau_2} ds = -m \int_{\tau_1}^{\tau_2} d\tau \sqrt{\dot{x}^\mu(\tau) \dot{x}_\mu(\tau)} \end{equation}with the usual variation
\begin{eqnarray} \delta S &=& \int_{\tau_1}^{\tau_2} d\tau \left[ \frac{\partial L}{\partial x^\mu} \delta x^\mu + \frac{\partial L}{\partial \dot{x}^\mu} \delta \dot{x}^\mu \right]\\ &=& \int_{\tau_1}^{\tau_2} d\tau \frac{\partial L}{\partial \dot{x}^\mu} \delta \dot{x}^\mu \end{eqnarray} \begin{eqnarray} \frac{\partial L}{\partial \dot{x}^\mu} &=& -m \frac{\partial \sqrt{l}}{\partial l} \frac{\partial l}{\partial \dot{x}^\mu}\\ &=& -m \frac{1}{2 \sqrt{\dot{x}^\mu(\tau) \dot{x}_\mu(\tau)}} \dot{x}_\mu(\tau) \end{eqnarray}EoM :
\begin{eqnarray} -\frac{m}{2} \frac{d}{d\tau} \left[\frac{1}{\sqrt{\dot{x}^\mu(\tau) \dot{x}_\mu(\tau)}} \dot{x}_\mu(\tau) \right] &=& 0\\ &=& -\frac{m}{2} \left[ \frac{d}{d\tau} (\frac{1}{\sqrt{\dot{x}^\mu(\tau) \dot{x}_\mu(\tau)}}) \dot{x}_\mu(\tau) + \frac{1}{\sqrt{\dot{x}^\mu(\tau) \dot{x}_\mu(\tau)}} \ddot{x}_\mu(\tau) \right]\\ &=& -\frac{m}{2} \left[ - \frac{1}{(\dot{x}^\mu(\tau) \dot{x}_\mu(\tau))^{3/2}} (\ddot{x}^\mu(\tau) \dot{x}_\mu(\tau)) \dot{x}_\mu(\tau) + \frac{1}{\sqrt{\dot{x}^\mu(\tau) \dot{x}_\mu(\tau)}} \ddot{x}_\mu(\tau) \right] \end{eqnarray}Moving along a few terms, this gives us
\begin{eqnarray} \left[ - \frac{\ddot{x}^\mu(\tau) \dot{x}_\mu(\tau)}{(\dot{x}^\mu(\tau) \dot{x}_\mu(\tau))} \dot{x}_\mu(\tau) + \ddot{x}_\mu(\tau) \right] = 0 \end{eqnarray}This is not quite what we expected, compared to the usual form of a free particle. This is because we forgot to take into account the parametrization invariance. Let's consider for instance the curve $x^\mu(\tau) = (\tau, 0, 0, 0)$. This is obviously a solution to both equations. By parametrization invariance, then, so should it be under the new parametrization $\tau' = \tau^3 + \tau$, with the curve $x^\mu(\tau) = (\tau^3 + \tau, 0, 0, 0)$. But here, our original equations of motion become
\begin{equation} \ddot{x}^\mu(\tau) = (6 \tau, 0, 0, 0) \end{equation}On the other hand, we do have
\begin{eqnarray} - \frac{\ddot{x}^\mu(\tau) \dot{x}_\mu(\tau)}{(\dot{x}^\mu(\tau) \dot{x}_\mu(\tau))} \dot{x}_\mu(\tau) &=& - \frac{6 \tau \times 3 \tau^2}{3 \tau^2 * 3 \tau^2} \dot{x}_\mu(\tau)\\ &=& - 2 \tau^{-1} \dot{x}_\mu(\tau)\\ &=& (- 6 \tau, 0, 0, 0) \end{eqnarray}which does compensate this term. To obtain a gauge-fixed equation, we can simply fix $x^0 = $
Parametrization invariance
\begin{eqnarray} \frac{\partial^2 L}{\partial \dot{x}^\mu \dot{x}^\nu} &=& -\frac{m}{2} \frac{\partial}{\partial \dot{x}^\nu} \frac{1}{ \sqrt{\dot{x}^\\sigma(\tau) \dot{x}_\sigma(\tau)}} \dot{x}_\mu(\tau)\\ &=& \end{eqnarray}Polyakov action
Another action we can use for the relativistic point particle is the Polyakov action
\begin{equation} S[X, e] = \int_\Sigma (\frac{1}{e(\tau)} \dot{X}^\mu(\tau) \dot{X}_\mu\nu(\tau) - e(\tau) m^2) d\tau \end{equation} \begin{eqnarray} \frac{\partial L}{\partial X^\mu} &=& 0\\ \frac{\partial L}{\partial \dot{X}^\mu} &=& \frac{2}{e(\tau)}\dot{X}^\mu(\tau)\\ \frac{\partial L}{\partial e} &=& -\frac{1}{e^2(\tau)} \dot{X}^\mu(\tau) \dot{X}_\mu(\tau) - m^2 \\ \frac{\partial L}{\partial \dot{e}} &=& 0 \end{eqnarray}General relativity
There are a few actions we can use in general relativity, as well as the choice of whether or not we consider a test particle or if our particle reacts to our spacetime. Obviously it is well known that the motion of a point particle in a spacetime is a geodesic, obeying the geodesic equation
\begin{equation} \ddot{x}^\mu(\tau) + {\Gamma^\mu}_{\alpha\beta}(x(\tau)) \dot{x}^\alpha(\tau) \dot{x}^\beta(\tau) = 0 \end{equation}but we'll see how to derive these.
Nambu-Goto action
The Nambu-Goto action is given by the embedding $X$ of a curve $\gamma$ in our spacetime, such that we minimize the arc length of our curve.
\begin{eqnarray} S &=& m \int_{\tau_1}^{\tau_2} d\mu[X_* g]\\ &=& m \int_{\tau_1}^{\tau_2} d\tau \sqrt{g_{\mu\nu}(X) \dot{X}^\mu(\tau) \dot{X}^\nu(\tau)} \end{eqnarray}Polyakov action
Backreaction
In a spacetime with otherwise no other sources of matter (and disregarding other potential sources of curvature like gravitational waves), the resulting spacetime is the Schwarzschild metric. This is not trivial to prove rigorously, as the stress-energy tensor is here a distribution and has its singular support on a $1$-dimensional submanifold, meaning that the Geroch-Traschen condition doesn't apply : the metric cannot be expressed by a tensor distribution. We'll have to do with the Colombeau algebra method for this.
First let's consider a point particle's stress energy tensor. If we express it as a distribution, it will be
\begin{equation} T^{\mu\nu} = \mu \int d^\tau \dot{X}^\mu \dot{X}^\nu \delta^{(n)}(x - X(\tau)) \end{equation}Non-relativistic quantum mechanics (canonical quantization)
The canonical quantization of non-relativistic free particles is fairly straightforward, as we have the basic Poisson brackets
\begin{eqnarray} \left\{ x_i, p_j \right\} &=& \delta_{ij}\\ \left\{ x_i, x_j \right\} &=& 0 \\ \left\{ p_i, p_j \right\} &=& 0 \end{eqnarray}The canonical quantization then gives us
\begin{eqnarray} \left[ \hat{x}_i, \hat{p}_j \right] &=& i \hbar \delta_{ij} \hat{I}\\ \left[ \hat{x}_i, \hat{x}_j \right] &=& 0 \\ \left[ \hat{p}_i, \hat{p}_j \right] &=& 0 \end{eqnarray}By the Stone-von Neumann theorem, for finite dimensions, all sets of operators satisfying those conditions are unitarily equivalent, and in particular for the Hilbert space $\mathcal{H} = L^2(\mathbb{R}^n, d\mu)$, with $\psi(\vec{x}) \in L^2(\mathbb{R}^n, d\mu)$, we can use
\begin{eqnarray} \hat{\vec{x}} \psi(\vec{x}) &=& \vec{x} \psi(\vec{x})\\ \hat{\vec{p}}_j \psi(\vec{x}) &=& -i\hbar \vec{\nabla} \psi(\vec{x}) \end{eqnarray}Then we can convert our Hamiltonian according to this quantization :
\begin{eqnarray} \hat{H} &=& \frac{\hat{\vec{p}} \cdot \hat{\vec{p}}}{2m}\\ &=& -\frac{\hbar^2}{2m} \Delta \end{eqnarray}From Schrödinger's equation, we get therefore
\begin{equation} -\frac{\hbar^2}{2m} \Delta \psi(t, \vec{x}) = i \hbar \frac{\partial}{\partial t} \psi(t, \vec{x}) \end{equation}As we're in the space of square integrable functions over $\mathbb{R}^n$, we can use the Fourier transform to obtain
\begin{equation} -\frac{\hbar^2}{2m} \| \vec{k} \|^2\tilde{\psi}(t, \vec{k}) = i \hbar \frac{\partial}{\partial t} \tilde{\psi}(t, \vec{k}) \end{equation}with the solution
\begin{equation} \tilde{\psi}(t, \vec{k}) = e^{-i\frac{\hbar}{2m} \frac{\| \vec{k} \|^2}{}} \end{equation}Another simple enough way to get solutions for this equation is to consider the generalized eigenvectors associated with our operators. In our case, we'll take the space on which $\hat{p}$ is defined, which is the dense subset of $C^1$ functions of $L^2(\mathbb{R}^n, d\mu)$ where the derivative is itself an $L^2$ function, which is the $H^1(\mathbb{R}^n) = W^{1,2}(\mathbb{R}^n)$ Sobolev space.
\begin{equation} \psi(t, \vec{x}) = \int f(p) e^ipx \end{equation}For more specific cases, let's consider the closest states to a classical free particle, which is a coherent state. A coherent state is a state of minimal uncertainty, that is,
\begin{equation} \Delta x \cdot \Delta p = \frac{\hbar}{2} \end{equation}Such that $\Delta x \approx \Delta p$, up to some dimensional factor. This makes it a wavepacket with the least spread in position and momentum space.
Non-relativistic quantum mechanics (path integral quantization)
For the path integral quantization, we just consider the Wick-rotated Feynman integral
\begin{equation} W(\vec{x}_i, \vec{x}_f) = \int_{x_i}^{x_f} d\mu[x(t)] \end{equation}with $d\mu[x(t)]$ the Gaussian measure.
Non-relativistic quantum mechanics (deformation quantization)
To deform the classical phase space formulation of our system into a quantum system, let's consider the algebra defined by the Moyal product
\begin{equation} f \star g = fg + \sum_{i = 1}^\infty \hbar^n C_n(f,g) \end{equation}Non-relativistic quantum mechanics (BRST quantization)
By using our parametrized version of our free particle, we can use the BRST quantization to hopefully find the same quantum system as before.
Bohmian mechanics
While quite possibly not correct in some aspects, Bohmian mechanics is entirely fit to describe the motion of point particles. There are many variants of it, but let's consider the basic one.
The system is composed of two parts : the particle itself, $\vec{x}$, and the guiding wave $\psi$. The guiding wave obeys the Schrödinger equation,
\begin{equation} i\hbar \frac{\partial \psi}{\partial t} = -\frac{\hbar^2}{2m} \Delta \psi \end{equation}while the particle obeys the guidance equation,
\begin{equation} \dot{\vec{x}} = \frac{\hbar}{m} \Im(\frac{\vec{\nabla}\psi}{\psi}) \end{equation}Relativistic quantum mechanics
Relativistic quantum mechanics is famed for working out poorly, but this is mainly due to the non-rigorous way in which it is usually quantized. Fortunately, using constraint quantization it is possible to work it out.
\begin{equation} \{ X^\mu, P^\nu \} = \delta \end{equation}