Question
---
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
---
Proceeding from an equivalence principle directly to the geodesic equation is known. In the same spirit we use the Heisenberg picture to define velocity ˆv:
ˆv=dU†xUdt=U†[H,x]−iℏU
where U† is the unitary operator and H is the Hamiltonian.
Now we can again differentiate to get acceleration ˆa:
ˆa=ˆU†[[ˆH,ˆx],ˆx]−iℏˆU=ˆU†(ˆH2ˆx+ˆxˆH2−2ˆHˆxˆH)ℏ2ˆU
We can simplify the calculation by splitting the Hamiltonian into potential ˆV and kinetic energy ˆT: ˆH=ˆT+ˆV
By noticing (one can also calculate this) that the acceleration of an object in a constant potential is 0:
ˆ0=ˆT2x+xˆT2−2ˆTˆxˆT
We also know [ˆV,ˆx]=0 as potential is a function of position. Thus, we can simplify acceleration as:
ˆaℏ2=ˆVˆTˆx+ˆxˆTˆV−ˆVˆxˆT−ˆTˆxˆV
Note this acceleration operator also commutes with position. Also acceleration only makes sense in Quantum Mechanics and therefore we are in the regime v≪c:
[ˆa,ˆx]=0
To make this physical law in uniformity with general covariance. We replace the
∇2 with the Laplace–Beltrami operator:
ˆT=12mΔLB =12m1√∣g∣∂i(√∣g∣gij∂j)
And redefine r to be a three vector in any co-ordinate system (for example in Cartesian coordinates ˆr=(x,y,z) and ˆa=(ax,ay,az):
ˆaℏ2=ˆV12m1√∣g∣∂i(√∣g∣gij∂j)ˆr+ˆr12m1√∣g∣∂i(√∣g∣gij∂j)ˆV
−ˆVˆr12m1√∣g∣∂i(√∣g∣gij∂j)−12m1√∣g∣∂i(√∣g∣gij∂j)ˆrˆV
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=ˆV12m1√∣g∣∂i(√∣g∣gij∂j)ˆr+ˆr12m1√∣g∣∂i(√∣g∣gij∂j)ˆV
−ˆVˆr12m1√∣g∣∂i(√∣g∣gij∂j)−12m1√∣g∣∂i(√∣g∣gij∂j)ˆrˆV
Hence, given a metric in the limit
c→∞ will follow the geodesic equation above. One would have to solve for the potential
V and substitute that in
Hψ=(T+V)ψ