An exploration of a special case in the relationship between Fisher information and quantum potential in the causal interpretation of quantum mechanics

Abstract

Fisher information is a cornerstone of both statistical inference and physical theory, leading to debate about whether its latter role is active or passive. Motivated by connections between Fisher information, entropy, and the quantum potential in the de Broglie–Bohm causal interpretation of quantum mechanics, the purpose of this article is to derive the position probability density when there a is a close and ubiquitous bonding of Fisher information and quantum potential. This is done by exploring a case in which a particle moves in a straight line and the integrands in Fisher information and expected quantum potential are proportional. It is found that in this case the probability density given by the Schrödinger wave equation has a Laplace distribution and that quantum potential is a negative constant at all points and times. It is noted that the rate of change of the entropy of the particle is bounded above by a limit that is proportional to the square roots of both Fisher information and the absolute value of quantum potential. Unlike Fisher information, quantum potential is a measure of a real physical potential, and it is proposed that it is quantum potential that puts an upper bound on the rate of change of the particle’s entropy and that, being negative in this case, may also act to contain the particle on its straight-line path. It is suggested that Fisher information does not have an active role in the physics, at least in this case, and only provides information about entropy.

Appendices

Appendix

Solution of the differential Eq. (4.1a) and derivation of the pdf (4.2c)

The differential equation that must be satisfied to derive the amplitude of the wavefunction and hence the pdf when the integrand of the Fisher information is proportional to that of the quantum potential at each point x at each time t, expressed in shorthand notation where the variable is x, is $( R^{\prime } )^{2} = {{RR}}^{\prime \prime }$, with support − ∞  ≤ x ≤ ∞ to allow the change of variable in Eq. (2.2).

The solution that needs to be demonstrated as such is¹

$$ R = \lambda e^{{ - \lambda \left| {x - \theta } \right|}} = \lambda e^{{ - \lambda \left( {x - \theta } \right)}}\,\,\,{\text{if}}\,\,\,x \ge \theta\,\,\,{\text{and}}\,\,\,\lambda e^{{ - \lambda \left( {\theta - x} \right)}} \quad {\text{if }}x < \theta $$

This solution yields.

$$ \begin{gathered} R^{\prime } = - \lambda^{2} e^{{ - \lambda \left( {x - \theta } \right)}}\,{\text{and}}\,\lambda^{2} e^{{ - \lambda \left( {\theta - x} \right)}}\,\,\text{respectively, and} \hfill \\ R^{\prime \prime } = \lambda^{3} e^{{ - \lambda \left( {x - \theta } \right)}}\,{\text{and}}\;\lambda^{3} e^{{ - \lambda \left( {\theta - x} \right)}} \hfill \\ ( R^{\prime } )^{2} = \lambda^{4} e^{{ - 2\lambda \left( {x - \theta } \right)}}\,{\text{and}}\,\lambda^{4} e^{{ - 2\lambda \left( {\theta - x} \right)}}\,{\text{and finally}} \hfill \\ ( R^{\prime } )^{2} = R^{\prime } R^{\prime \prime }\,{\text{in both cases}} \hfill \\ \end{gathered} $$

(QED)

From the solution, we get $R^{2} = \lambda^{2} {\text{e}}^{{ - 2\lambda \left( {x - \theta } \right)}}$, $x \ge \theta\,\,{\text{joined to}}$ $\lambda^{2} {\text{e}}^{{ - 2\lambda \left( {\theta - x} \right)}}$, $x < \theta$.

and the normalising constant is then

$$ \mathop \int \limits_{\theta }^{\infty } \lambda^{2} e^{{ - 2\lambda \left( {x - \theta } \right)}} dx + \mathop \int \limits_{ - \infty }^{\theta } \lambda^{2} e^{{ - 2\lambda \left( {\theta - x} \right)}} dx = \left\{ {\left[ - \frac{\lambda }{2}e^{{ - 2\lambda \left( {x - \vartheta } \right)}} \right]_{\vartheta }^{\infty } = \left[ {0 - \left( { - \frac{\lambda }{2}} \right)} \right] = \frac{\lambda }{2}} \right\} + \left\{ {\left[ \frac{\lambda }{2}e^{{ - 2\lambda \left( {\vartheta - x} \right)}} \right]_{ - \infty }^{\vartheta } = \left[ { \frac{\lambda }{2} - 0} \right] = \frac{\lambda }{2}} \right\} = \lambda $$

Normalising $R^{2}$ and multiplying by $\frac{2}{2}$ to put the pdf in the form of the Laplace distribution give:

$$ p(x|\theta ,\lambda ) = \frac{1}{2}2\lambda e^{{ - 2\lambda \left( {x - \theta } \right)}} ,x \ge \theta\,\,\,{\text{joined}}\,{\text{to}}\,\,\,\frac{1}{2}2\lambda e^{{ - 2\lambda \left( {\theta - x} \right)}} , x < \theta $$

(QED)

From Eq. (2.4c), it can be seen by inspection that the Fisher information in each component of this pdf is $4\lambda^{2}$, making a total of $8\lambda^{2}$. Finally, from Eq. (1.1) and the above differential equation representing (4.1a), the quantum potential $Q = - \frac{{\hbar^{2} }}{2m}\frac{{R^{\prime \prime } }}{R} = - \frac{{\hbar^{2} }}{2m}\frac{{\lambda^{3} }}{\lambda } = - \frac{{\hbar^{2} }}{2m}\lambda^{2}$ for each of the two components, so in total $Q = - \frac{{\hbar^{2} }}{2m}2\lambda^{2}$ and quantum potential is constant at all points and times. As a result, the expected quantum potential E[Q] also equals $- \frac{{\hbar^{2} }}{2m}2\lambda^{2}$ and $I(t) = - \frac{8m}{{\hbar^{2} }}E\left[ Q \right]$.