Probability densities of free particles
In the proposed local-realistic model for quantum mechanics, when the rules of motion for single particle instances are applied to an ensemble of similarly prepared particles, give rise to probability densities. In a previous post, we have discussed probability distribution for position in the case of free particles without quantum forces, that is, particles that are emitted from a single source.
Here we shall discuss how this probability density function changes in the presence of quantum forces. The derivation is rather straightforward. We shall use the already discussed large-time approximation and replace the position x with its average value as the argument of the probability density,
=\frac{1}{2}&space;\left&space;|&space;\frac{dv_0}{d\left&space;\langle&space;x&space;\right&space;\rangle}&space;\right&space;| "\rho(\left \langle x \right \rangle; t)=\frac{1}{2} \left | \frac{dv_0}{d\left \langle x \right \rangle} \right |")
with ⟨x⟩ = Vt, where we take the "trained" (i.e., steady-state) value for V. We also know from the elementary rules of motion that the momentum propensity is the sum of the source momentum and of the momenta of all particle bosons captured,
} "V=v_0-\sum_{\ell,\lambda}v_Q^{(\ell\lambda)}")
We have seen in this post that, after particle training, the PBM are
}=\frac{\sqrt{P_0^{(\ell\lambda)}}}{\delta^{(\ell\lambda)}}\omega_{xt}^{(\ell\lambda)} "v_Q^{(\ell\lambda)}=\frac{\sqrt{P_0^{(\ell\lambda)}}}{\delta^{(\ell\lambda)}}\omega_{xt}^{(\ell\lambda)}")
}=\frac{\sin{\left(\pi\delta^{(\ell\lambda)}\frac{x-(x_0^{(\ell)}+x_0^{(\lambda)})/2)}{t}-\epsilon^{(\ell\lambda)}\right)}}{\pi} "\omega_{xt}^{(\ell\lambda)}=\frac{\sin{\left(\pi\delta^{(\ell\lambda)}\frac{x-(x_0^{(\ell)}+x_0^{(\lambda)})/2)}{t}-\epsilon^{(\ell\lambda)}\right)}}{\pi}")
}P_0^{(\lambda)}}}{\pi\delta^{(\ell\lambda)}}\sin{\left(\pi\delta^{(\ell\lambda)}\frac{x-(x_0^{(\ell)}+x_0^{(\lambda)})/2)}{t}-\epsilon^{(\ell\lambda)}\right)} "V=v_0-\sum_{\ell,\lambda}\frac{\sqrt{P_0^{(\ell)}P_0^{(\lambda)}}}{\pi\delta^{(\ell\lambda)}}\sin{\left(\pi\delta^{(\ell\lambda)}\frac{x-(x_0^{(\ell)}+x_0^{(\lambda)})/2)}{t}-\epsilon^{(\ell\lambda)}\right)}")
and, consequently, the position pdf reads
=\frac{1+\sum_{\ell,\lambda}\sqrt{P_0^{(\ell)}P_0^{(\lambda)}}\cos{\left(\pi\delta^{(\ell\lambda)}\frac{x-(x_0^{(\ell)}+x_0^{(\lambda)})/2)}{t}-\epsilon^{(\ell\lambda)}\right)}}{2t} "\rho(x;t)=\frac{1+\sum_{\ell,\lambda}\sqrt{P_0^{(\ell)}P_0^{(\lambda)}}\cos{\pi\left(\delta^{(\ell\lambda)}\frac{x-(x_0^{(\ell)}+x_0^{(\lambda)})/2)}{t}-\epsilon^{(\ell\lambda)}\right)}}{2t}")
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| Illustration by user Geek3, Wikimedia Commons, https://commons.wikimedia.org/wiki/File:Hydrogen_eigenstate_n5_l2_m1.png |
Here we shall discuss how this probability density function changes in the presence of quantum forces. The derivation is rather straightforward. We shall use the already discussed large-time approximation and replace the position x with its average value as the argument of the probability density,
with ⟨x⟩ = Vt, where we take the "trained" (i.e., steady-state) value for V. We also know from the elementary rules of motion that the momentum propensity is the sum of the source momentum and of the momenta of all particle bosons captured,
We have seen in this post that, after particle training, the PBM are
where P0 is the source probability function and δ is the path difference.
We have also seen in this post that, after lattice training, the LBM are
We have also seen in this post that, after lattice training, the LBM are
where we have introduced the phase difference as explained in this post and denoted the average position with x for simplicity.
By wrapping up all these equations, we obtain the sought relation between the average position and the source momentum,
and, consequently, the position pdf reads
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