Question
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Prove or disprove the below toy model for non-relativistic quantum gravity (in absence of other forces) is agreeable (does it violate) with the equivalence principle and Newtonian gravity
\begin{equation}
\frac{-1}{\psi^2 (x,y,z)} (\nabla \psi) \cdot (\alpha \psi) + \frac{1}{\psi(x,y,z)} (\alpha \cdot \nabla) \psi = 4 \pi G \rho_m
\end{equation}
Where $\alpha = (\frac{-i}{m \hbar}([H,p_x],[H,p_y],[H,p_z]] )$, $H$ is the Hamiltonian, $\rho_m$ is the mass density, $p_x,p_y,p_z$ is the momentum in the $x,y,z$ directions, $\psi$ is the wave-function (of say an electron in a gravitational field).
Given a particular mass density one can find a corresponding potential energy and thereby find the energy-eigenvalues as done in normal quantum mechanics.
How the Toy model was constructed
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Let, us first begin with the equivalence principle's famous thought experiment. Someone is stuck within a lift, drops, an electron and thereby is unable to conclude if it fell due gravity or him being accelerated by say a rocket attached to the lift. Einstein's conclusion was that no matter what one did one would never be able to distinguish between the two situations. But what if the person in the lift starts to measure the the standard deviation of the electron's acceleration by shining light on it?? Answer: He will get some standard deviation but will but will be unable to differentiate as to why this phenomena is occurring. But, note if this were a purely classical result one would conclude that the standard deviation attribute to an experimental errors. However, what about quantum mechanics? What would it tell us about the standard deviation?
The Quantum Analysis
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We continue motivated by the above thought experiment. First we define the velocity operator in quantum mechanics using Heisenberg's equation of motion:
\begin{equation}
\dot x = v = \frac{[H,x]}{i \hbar} = [\frac{p^2}{2m},x] = \frac{p}{m}
\end{equation}
Now to define the acceleration operator. One defines the following:
\begin{equation}
\dot v = \alpha =\frac{[H,v]}{i \hbar} = \frac{-i}{\hbar}[V(x),\frac{p}{m}]
\end{equation}
Where $V(x)$ is the potential energy, $H$ is the Hamiltonian, $x$ is the position, $\alpha$ is the acceleration operator and $v$ is the velocity operator.
We note:
$$ [x, \alpha] = 0$$
While $$ [p, \alpha] \neq 0$$
Note using the above an uncertainty principle can be built but that will not be explicitly stated here.
Measuring the acceleration
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Let's continue with the experiment and work in $3$ dimensions with standard Cartesian co-ordinates $x,y,z$. The person inside the lift measures acceleration this measurement is independent of the fact of what is causing the acceleration (gravity or an external force).
Let us assume the acceleration measured a point $x$:
\begin{equation}
\alpha_x \psi(x,y,z) = a(x,y,z) \psi(x,y,z) \implies \frac{1}{\psi(x,y,z)} \alpha \psi(x,y,z) = a(x,y,z)
\end{equation}
where $a(x,y,z)$ is the acceleration eigenvalue of $\psi$ as a function of $x,y,z$
Let us measure the acceleration at $x + \epsilon$:
\begin{equation}
\alpha_x \psi(x+ \epsilon,y,z) = a(x+ \epsilon,y,z) \psi(x+ \epsilon,y,z) \implies \frac{1}{\psi(x+ \epsilon,y,z)} \alpha \psi(x+\epsilon,y,z) = a(x+ \epsilon,y,z)
\end{equation}
Subtracting the above $2$ equations and dividing by $\epsilon$:
\begin{equation}
\frac{\frac{1}{\psi(x+ \epsilon,y,z)} \alpha_x \psi(x+\epsilon,y,z) - \frac{1}{\psi(x,y,z)} \alpha_x \psi(x,y,z)}{\epsilon} = \frac{a(x+ \epsilon,y,z) - a(x,y,z)}{\epsilon}
\end{equation}
Where $\epsilon$ is a function of position and hence commutes with acceleration.
\begin{equation}
\frac{(\frac{1}{\psi(x,y,z)} - \frac{\epsilon}{\psi^2(x,y,z)}\frac{\partial \psi}{\partial x}) \alpha_x (\psi(x,y,z)+ \epsilon \frac{\partial \psi}{\partial x}) - \frac{1}{\psi(x,y,z)} \alpha_x \psi(x,y,z)}{\epsilon} = \frac{a(x+ \epsilon,y,z) - a(x,y,z)}{\epsilon}
\end{equation}
Taking the limit $\epsilon$ to $0$ :
\begin{equation}
\frac{-1}{\psi^2 (x,y,z)} \frac{\partial \psi}{\partial x} \alpha_x \psi + \frac{1}{\psi(x,y,z)}\alpha_x\frac{\partial \psi}{\partial x} = \lim_{\epsilon \to 0} \frac{a(x+ \epsilon,y,z) - a(x,y,z)}{\epsilon}
\end{equation}
Proceeding in a similar manner for the co-ordinates $y$ and $z$:
\begin{equation}
\frac{-1}{\psi^2 (x,y,z)} \frac{\partial \psi}{\partial y} \alpha_y \psi + \frac{1}{\psi(x,y,z)}\alpha_y\frac{\partial \psi}{\partial y} = \lim_{\epsilon \to 0} \frac{a(x,y+ \epsilon,z) - a(x,y,z)}{\epsilon}
\end{equation}
\begin{equation}
\frac{-1}{\psi^2 (x,y,z)} \frac{\partial \psi}{\partial z} \alpha_z \psi + \frac{1}{\psi(x,y,z)}\alpha_z\frac{\partial \psi}{\partial z} = \lim_{\epsilon \to 0} \frac{a(x,y,z+ \epsilon) - a(x,y,z)}{\epsilon}
\end{equation}
Adding all $3$ equations.
\begin{equation}
\frac{-1}{\psi^2 (x,y,z)} (\nabla \psi) \cdot (\alpha \psi) + \frac{1}{\psi(x,y,z)} (\alpha \cdot \nabla) \psi = \nabla \cdot a
\end{equation}
Where $\alpha$ is $(\alpha_x,\alpha_y,\alpha_z)$ and $a$ is the acceleration vector. Note $a$ must be $-g$ where $g$ is the gravitational field for an outside observer. We can now use Guass's law for gravity.
\begin{equation}
\frac{-1}{\psi^2 (x,y,z)} (\nabla \psi) \cdot (\alpha \psi) + \frac{1}{\psi(x,y,z)} (\alpha \cdot \nabla) \psi = - \nabla \cdot g = 4 \pi G \rho_m
\end{equation}
Writing it explicitly:
\begin{equation}
\frac{-1}{\psi^2 (x,y,z)} (\nabla \psi) \cdot (\alpha \psi) + \frac{1}{\psi(x,y,z)} (\alpha \cdot \nabla) \psi = - \nabla \cdot g = 4 \pi G \rho_m
\end{equation}
Where $\alpha = (\frac{-i}{m \hbar}([H,p_x],[H,p_y],[H,p_z]] ) $
Conclusion
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Writing the above explicitly:
\begin{equation}
\frac{-1}{\psi^2 (x,y,z)} (\nabla \psi) \cdot (\alpha \psi) + \frac{1}{\psi(x,y,z)} (\alpha \cdot \nabla) \psi = 4 \pi G \rho_m
\end{equation}
Where $\alpha = (\frac{-i}{m \hbar}([H,p_x],[H,p_y],[H,p_z]] )$, $H$ is the Hamiltonian, $\rho_m$ is the mass density, $p_x,p_y,p_z$ is the momentum in the $x,y,z$ directions.
Hence, in conclusion:
\begin{equation}
\frac{-1}{\psi^2 (x,y,z)} (\nabla \psi) \cdot (\alpha \psi) + \frac{1}{\psi(x,y,z)} (\alpha \cdot \nabla) \psi = 4 \pi G \rho_m
\end{equation}
\begin{equation}
H\psi_n = - \hbar^2 \frac{\nabla^2}{2m} \psi_n + V_{\text{eff}} \psi_n = E_n \psi_n \end{equation}
\begin{equation}
\langle \psi | \psi \rangle =1 \end{equation}
All $3$ equations must be solved simultaneously in order to determine the effective potential and the wavefunction. Hence, given any empirically determined mass density one can find an effective potential and it's corresponding wavefunction.
Q&A (4871)
