Global conserved quantities for point particle coupled to a Schrodinger field

Question

We have a box (represented by a potential $V$) with a classical particle in it. If the box has a finite inertia and it's floating in space, then it shakes as the particle bumps on the walls. The total energy and momentum are both conserved (if collisions are elastic, which is the case as we are using a potential to represent the box's walls).

Now, let's imagine the same system but "quantum" (the particle in the box is now represented by a classical Schrodinger field $\psi$). We all know the usual Schrodinger equation for $\psi$ in an external potential $V$, but what if the center of the potential well acts-reacts to the motion of $\psi$?

I imagine that we have to deal with some problem of this type ($\hbar=1$):

$$ i\partial_t \psi(x,t) = H(t)\psi(x,t) =-\frac{1}{2 m }\partial_x^2 \psi(x,t) + V(x-x_0(t)) \psi(x,t) $$

where $x$ is the coordinate in the "laboratory" reference frame, while $x_0(t)$ is the moving center of the box. For example, our box may be represented by a potential $V(y)$ centered at $y=0$, so that putting $y=x-x_0(t)$ accounts for the dynamics of the floating box.

Now, we need an additional equation for the floating box. I guess that we should conserve the total momentum ($m_0$ is the box's mass):

$$ \frac{d}{dt} \bigg[\, m_0 \, \dot{x}_0(t) + \int dx \, \psi(x,t)^\dagger (-i\partial_x) \psi(x,t) \bigg] = 0 $$

or

$$ \dot{x}_0(t) = \frac{p_{tot}}{ m_0 } + \frac{i}{ m_0 } \int dx \, \psi(x,t)^\dagger \partial_x \psi(x,t) $$

for a constant and assigned total momentum of the system $p_{tot}$. This may be fine, but this equation is not in the form $m \ddot{x}_0=F[x_0,\psi]$. The force $F$ should be related to $V$, but how?

Now, all this stuff is quite heuristic: I have not derived it from first principles. It is not even clear to me how to show that $p_{tot}$ should be conserved in the first place, even though I have heuristically used it to define the equation of motion of the box.

Is there a total energy that is conserved? Something like:

$$ E_{tot} = \dfrac{1}{2} m_0 \, \dot{x}_0(t)^2 + \int dx \, \psi(x,t)^\dagger H(t)\psi(x,t) $$

again, this is built on intuition, and it's not clear to me how to show that the dynamics (if the equations that I wrote are correct) conserves it. I think that everything should be more clear (including the presence of constants of motion) if it is possible to write down an action principle (see the update below!) for the dynamical degrees of freedom $\psi$ (the classical Schrodinger field) and $x_0$ (the classical point particle that represents the box).

Regarding the physical interest for this kind of system: The classical field $\psi$ represents a BEC of ultracold atoms (i.e. $\psi =\langle \hat\psi\rangle$ ) interacting with the potential trap $V$, but the trap is a floating one, so its center $x_0$ reacts dynamically as the BEC scatters off the trap's walls.

Update (see the comments): if instead of a classical particle at $x_0$ (i.e. the box) we have two quantum particles, then the problem is similar to the two-body problem in QM. This is not what I am interested in. My question is more: "how to define the (classical) dynamics of a point particle interacting with a classical scalar (complex) field?". This is typically done in classical electrodynamics, where $A^\mu$ evolves together with the charged point particles, see e.g. this answer. However, I have never seen a classical point particle coupled to a scalar field (note that the Newtonian gravitational potential $\phi$ for a configuration of point masses is not an example of "point particles"-"classical scalar field" interaction because $\phi$ is not dynamical).

Edit (variational principle): I figured out that we can obtain the dynamics by considering a total Lagrangian that consists of three parts, \begin{equation} L= L_\psi + L_0 + L_{0\psi} \, , \end{equation} where $L_\psi$ is the usual Lagrangian for the Schrodinger field \begin{equation} L_{\psi} = \int \big\{ Im [ \psi \partial_t \psi^* ] -\frac{1}{2m}|\partial_x \psi |^2 \big\} \, d x \, . \end{equation} Then, we have that $ L_0 $ for the classical particle that represents the box's center of mass is simply \begin{equation} L_0 = \frac{ m_0 }{2} |\dot{ x}_0(t)|^2 \end{equation} Finally, the coupling between the box and the Schrodinger field is \begin{equation} L_{0\psi} = - \int dx \, V( x - x_0 ) |\psi(x,t)|^2 \, . \end{equation} The action $I$ for the system is \begin{equation} I = \int d t \, L(x_0, \dot{x}_0, \psi,\partial_t\psi,\partial_x \psi) \end{equation} and the Euler-Lagrange equations read \begin{align} & \frac{\delta I}{\delta x_0 } = \frac{\partial L}{\partial x_0} - \frac{d}{dt} \frac{\partial L}{\partial \dot{x}_0} =0 \\ & \frac{\delta I}{\delta \psi^* } = \frac{\partial L}{\partial \psi^*} - \partial_t \frac{\partial L}{\partial (\partial_t\psi^*)} - \partial_x \frac{\partial L}{\partial (\partial_x \psi^*)} =0 \end{align} Explicit calculation gives, respectively, \begin{align} & i \, \partial_t \psi(x,t)= V(x-x_0(t)) \psi(x,t) -\frac{1}{2m} \partial_x^2 \, \psi(x,t) \\ & m_0 \, \ddot{x}_0 = - \int d x \,|\psi(x,t)|^2\, \partial_x V(x-x_0(t)) \end{align} Given this scheme, how to show that there is a total energy and a total momentum that are conserved? Probably for the energy $E_{tot}$ we should compute the Hamiltonian, and for the total momentum $p_{tot}$?

Finding the Hamiltonian: starting from the Lagrangian it is not obvious which is the associated Hamiltonian (this could be a strategy to find out the conserved energy). This is because the Lagrangian is first order in $\partial_t \psi$. See this paper and this question.

NOTE: I found good insights in this article.

Answer 1

Let us consider two quantum-mechanical particles, with coordinates $x$ and $X$ and momenta $p$ and $P$, interacting cia a central potential. Their joint Hamiltonian can be written as $$ H=\frac{p^2}{2m} + \frac{P^2}{2M} + V(|x-X|) $$ one can write the two-particle Schrödinger equation $$ \partial_t \Psi(x, X) = H\Psi(x,X), $$ and transform the variables to the center-of-mass and the relative coordinates, using the usual transformations: $$ y_c = \frac{mx + MX}{m+M},\\ y_r = x-X. $$ This procedure is usually done in quantum mechanics textbooks when solving the hydrigen atom, except that here we are dealing with one dimension and the potential is rectangular (such a potential is used sometimes to described nucleaons, e.g., deuterium).

If we want one particle to behave as classical, we could couple it to a bath of the oscillators and enforce its decoherence using one of the many existing methods (e.g., using quantum Langevin equations or master equations or a Keldysh Green's function approach, Caldeira-Legget, etc) - this is going to be by far the most difficult part of it.

Answer 2

To check the conservation of energy and momentum from the two given equations:

\begin{align} & i \, \partial_t \psi(x,t)= V(x-x_0(t)) \psi(x,t) -\frac{1}{2m} \partial_x^2 \, \psi(x,t) \tag{1} \\ & m_0 \, \ddot{x}_0 = + \int d x \,|\psi(x,t)|^2\, \partial_x V(x-x_0(t)) = -\int d x \,|\psi(x,t)|^2\, \partial_{x_0} V(x-x_0(t)) \tag{2} \end{align}

Starting from Eq.(2) the semi-classical equation of motion: \begin{align} m_0 & \frac{d}{dt}\dot{x}_0 = -\frac{d}{dx_0}\int d x |\psi(x,t)|^2\, V(x-x_0(t)) =-\frac{dt}{dx_0}\frac{d}{d t}\int d x |\psi(x,t)|^2\, V(x-x_0(t)) \\ m_0 & \dot x_0 \frac{d}{dt}\dot{x}_0 = - \frac{d}{d t} \langle V(x-x_0(t))\rangle \\ \frac{d}{dt}& \left\{ \frac{1}{2}m_0 \dot{x_0}^2 + \langle V(x-x_0(t))\rangle \right\} = 0 \tag{3} \end{align} where the braket denotes an quantum expectation value.

Now, lets examine the $\langle V(x-x_0(t))\rangle$ from the Eq.(1) \begin{align} i \int dx \psi^* \frac{\partial}{\partial t} \psi & = \int dx \psi^* V(x-x_0(t)) \psi -\frac{1}{2m} \int dx\psi^* \frac{\partial^2 \psi}{\partial x^2} \\ \langle E \rangle & = \langle V(x-x_0(t))\rangle +\langle \frac{p^2}{2m} \rangle \\ \langle V(x-x_0(t))\rangle & =\langle E \rangle -\langle \frac{p^2}{2m} \rangle \tag{4} \end{align}

Substitute Eq.(4) in Eq.(3): \begin{align} \frac{d}{dt} & \left\{ \frac{1}{2}m_0 \dot{x_0}^2 + \langle E \rangle - \langle \frac{p^2}{2m} \rangle \right\} = 0 \\ \langle E \rangle & + \frac{1}{2}m_0 \dot{x_0}^2 - \langle \frac{p^2}{2m} \rangle = C \tag{5} \end{align} Eq.(5) shows how the energy changes between the two parties.

To check the conservation of momentum:

The Eq.(2) give the time derivative of the momentum of the box: \begin{align} \frac{d}{dt}m_0\dot{x}_0 & = \int d x \,|\psi(x,t)|^2 \frac{\partial}{\partial x} V(x-x_0(t))\\ \frac{d P_0}{dt} & = \langle \frac{\partial}{\partial x} V(x-x_0(t)) \rangle \tag{6} \end{align}

Then, we change the momentum change from $\frac{dp}{dt} = i \left[H, P\right]$. The commutator:

\begin{align} \left[H, P\right] & = \left[\frac{P^2}{2m} + V(x-x_0), P\right]=\left[V(x-x_0), P\right]\\ & = V(x-x_0) \frac{1}{i}\frac{\partial}{\partial x} - \frac{1}{i}\frac{\partial}{\partial x}V(x-x_0) =-\frac{1}{i}\frac{\partial V(x-x_0)}{\partial x} \end{align}

Thus, the $\frac{dp}{dt}$: $$ \tag{7} \frac{dp}{dt} = i \left[H, P\right] = -\frac{\partial V(x-x_0)}{\partial x} $$

Compare Eq.(6) with Eq.(7) (adding the expectation average), we have \begin{align} & \frac{d \langle p \rangle }{dt} = - \frac{d P_0}{dt}.\\ & \frac{d}{dt} \left\{\langle p \rangle + P_0 \right\} = 0. \tag{8} \end{align} Eq.(8) shows the conservation of momentum.