Question

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Given a geodesic equation in a quantum mechanical context. How does make a relation to the field equation? (Or how does one guess the relation between curvature and mass)?

Motivation

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Proceeding from an equivalence principle directly to the geodesic equation is known. In the same spirit we use the Heisenberg picture to define velocity $\hat v$:

$$ \hat v = \frac{dU^\dagger x U}{dt} = U^\dagger\frac{[H,x]}{- i \hbar}U$$

where $U^\dagger$ is the unitary operator and $H$ is the Hamiltonian.

Now we can again differentiate to get acceleration $\hat a$:

$$ \hat a = \hat U^\dagger\frac{[[\hat H, \hat x], \hat x]}{-i \hbar} \hat U = \hat U^\dagger\frac{(\hat H^2 \hat x + \hat x \hat H^2 - 2\hat H \hat x \hat H)}{\hbar^2} \hat U $$

We can simplify the calculation by splitting the Hamiltonian into potential $ \hat V $ and kinetic energy $ \hat T $: $\hat H = \hat T + \hat V$

By noticing (one can also calculate this) that the acceleration of an object in a constant potential is $0$:

$$ \hat 0 = \hat T^2 x + x \hat T^2 - 2 \hat T \hat x \hat T $$

We also know $ [\hat V, \hat x] = 0 $ as potential is a function of position. Thus, we can simplify acceleration as:

$$ \hat a \hbar^2 = \hat V \hat T \hat x + \hat x \hat T \hat V - \hat V \hat x \hat T - \hat T \hat x \hat V $$

Note this acceleration operator also commutes with position. Also acceleration only makes sense in Quantum Mechanics and therefore we are in the regime $v\ll c$:

$$ [\hat a , \hat x ]=0$$

To make this physical law in uniformity with general covariance. We replace the $\nabla^2 $ with the Laplace–Beltrami operator:

$$ \hat T = \frac{1}{2m} \Delta_{LB}\ = \frac{1}{2m} \dfrac{1}{\sqrt{\left\lvert g\right\rvert}}\,\partial_i\,\Big(\sqrt{\left\lvert g\right\rvert} \,g^{ij}\,\partial_j \Big) $$

And redefine $r$ to be a three vector in any co-ordinate system (for example in Cartesian coordinates $\hat r = (x,y,z)$ and $\hat a = (a_x , a_y, a_z$):

$$ \hat a \hbar^2 = \hat V \frac{1}{2m} \dfrac{1}{\sqrt{\left\lvert g\right\rvert}}\,\partial_i\,\Big(\sqrt{\left\lvert g\right\rvert} \,g^{ij}\,\partial_j \Big) \hat r + \hat r \frac{1}{2m} \dfrac{1}{\sqrt{\left\lvert g\right\rvert}}\,\partial_i\,\Big(\sqrt{\left\lvert g\right\rvert} \,g^{ij}\,\partial_j \Big)\hat V $$

$$ - \hat V \hat r \frac{1}{2m} \dfrac{1}{\sqrt{\left\lvert g\right\rvert}}\,\partial_i\,\Big(\sqrt{\left\lvert g\right\rvert} \,g^{ij}\,\partial_j \Big) - \frac{1}{2m} \dfrac{1}{\sqrt{\left\lvert g\right\rvert}}\,\partial_i\,\Big(\sqrt{\left\lvert g\right\rvert} \,g^{ij}\,\partial_j \Big) \hat r \hat V $$

We will now proceed with the final step of this derivation. Suppose that no particles are accelerating in the neighbourhood of a point-event with respect to a freely falling coordinate system:

$$ 0 = \hat V \frac{1}{2m} \dfrac{1}{\sqrt{\left\lvert g\right\rvert}}\,\partial_i\,\Big(\sqrt{\left\lvert g\right\rvert} \,g^{ij}\,\partial_j \Big) \hat r + \hat r \frac{1}{2m} \dfrac{1}{\sqrt{\left\lvert g\right\rvert}}\,\partial_i\,\Big(\sqrt{\left\lvert g\right\rvert} \,g^{ij}\,\partial_j \Big)\hat V $$

$$ - \hat V \hat r \frac{1}{2m} \dfrac{1}{\sqrt{\left\lvert g\right\rvert}}\,\partial_i\,\Big(\sqrt{\left\lvert g\right\rvert} \,g^{ij}\,\partial_j \Big) - \frac{1}{2m} \dfrac{1}{\sqrt{\left\lvert g\right\rvert}}\,\partial_i\,\Big(\sqrt{\left\lvert g\right\rvert} \,g^{ij}\,\partial_j \Big) \hat r \hat V $$

Hence, given a metric in the limit $c \to \infty$ will follow the geodesic equation above. One would have to solve for the potential $V$ and substitute that in $H\psi = (T + V)\psi$