How Distorting the Trajectories of Quantum Particles Shapes the Statistical Properties of their Ensemble

Abstract

We present universal statistical relations between the center of mass of a quantum ensemble and its sub-ensembles when the Bohmian trajectories of the particles are distorted. These relations break the additivity property of the expectation into subadditivity of the distorted expectation, which obtains an evident difference between the various ways to infer quantum expectations. The results do not depend on the form of the Hamiltonian, and do not have a classical parallel. While an experimentalist can arrange the distortion of the trajectories to produce the proposed relations, such distortion may naturally appear in standard experimental setups.

Appendices

Appendix 1

From the definition of expectation of a continuous random variable X, we have $E\left( X\right) =\int _{-\infty }^{\infty }xdF_{X}\left( x\right) .$ Taking the transformation $q=F\left( x\right) ,$ we have $dq=f_{X}\left( x\right) dx,$ implying the desired formula for expectation $E\left( X\right) =\int _{0}^{1}x_{q}dq,$ where $x_{q}$ is the $q-th$ quantile of X,  $x_{q}=F_{X}^{-1}\left( q\right) .$

Appendix 2

Let us prove the subadditivity property of the distorted mean CoM. Our first step is to rewrite the function D,  in the integral form $D\left( q\right) =\int _{0}^{q}d\mu \left( \alpha \right) .$ By substituting this into (4), we get

$$\begin{aligned} \left\langle x\left( t\right) \right\rangle _{D}&=\int _{0}^{1}F_{X} ^{-1}\left( q\right) \int _{0}^{q}d\mu \left( \alpha \right) dq\\&=\int _{0}^{1}F_{X}^{-1}\left( q\right) \int _{0}^{1}d\mu \left( \alpha \right) \cdot H\left( q-\alpha \right) dq\nonumber \end{aligned}$$

(15)

where $H\left( q-\alpha \right)$ is the Heaviside step function

$$\begin{aligned} \left\langle x\left( t\right) \right\rangle _{D}&=\int _{0}^{1}\left( \int _{0}^{1}F_{X}^{-1}\left( q\right) H\left( q-\alpha \right) dq\right) d\mu \left( \alpha \right) \nonumber \\&=\int _{0}^{1}\left( 1-\alpha \right) d\mu \left( \alpha \right) \cdot \rho _{\alpha }\left( X\right) , \end{aligned}$$

(16)

where

$$\begin{aligned} \rho _{\alpha }\left( X\right) =\frac{1}{1-\alpha }\int _{\alpha }^{1}F_{X}^{-1}\left( q\right) dq, \end{aligned}$$

(17)

Let us note that (16) is the weighted-sum of all $\rho _{\alpha }\left( X\right)$ in the region $\alpha \in \left( 0,1\right) ,$ which can be proved via the relations

$$\begin{aligned} 1&=\int _{0}^{1}D\left( q\right) dq=\int _{0}^{1}dq\int _{0}^{q}d\mu \left( \alpha \right) =\int _{0}^{1}dq\int _{0}^{1}d\mu \left( \alpha \right) H\left( q-\alpha \right) \\&=\int _{0}^{1}dq\int _{0}^{1}d\mu \left( \alpha \right) H\left( q-\alpha \right) =\int _{0}^{1}d\mu \left( \alpha \right) \int _{\alpha }^{1}dq=\int _{0}^{1}d\mu \left( \alpha \right) \cdot \left( 1-\alpha \right) \nonumber \end{aligned}$$

(18)

as $\int _{0}^{1}d\mu \left( \alpha \right) \left( 1-\alpha \right) =1,$ $d\mu \left( \alpha \right) \left( 1-\alpha \right)$ serve as weights. Thus, if we prove that $\rho _{\alpha }$ is subadditive, it immediately implies that $\left\langle x\left( t\right) \right\rangle _{D}$ is subadditive. Now, for proving subadditivity, we consider the bivariate pdf $\rho _{X_{1},X_{2} }\left( x_{1},x_{2}\right)$ and, without loss of generality, we consider the sum $Z=X_{1}+X_{2}.$ We note that we can always consider convex sum with coefficients, with taking $X_{j}=m_{j}\widetilde{X}_{j}.$

From the definition of $\rho _{\alpha },$ we have

$$\begin{aligned}&\left( 1-\alpha \right) \cdot \left( \rho _{\alpha }\left( X_{1}\right) +\rho _{\alpha }\left( X_{2}\right) -\rho _{\alpha }\left( Z\right) \right) \\&=E\left( X_{1}\cdot H\left( F_{X_{1}}\left( X_{1}\right) \ge \alpha \right) +X_{2}\cdot H\left( F_{X_{2}}\left( X_{2}\right) \ge \alpha \right) -Z\cdot H\left( F_{Z}\left( Z\right) \ge \alpha \right) \right) \nonumber \end{aligned}$$

(19)

and after some algebraic calculations, we get

$$\begin{aligned}&\left( 1-\alpha \right) \cdot \left( \rho _{\alpha }\left( X_{1}\right) +\rho _{\alpha }\left( X_{2}\right) -\rho _{\alpha }\left( Z\right) \right) \\&=E\left( X_{1}\left( H\left( X_{1}\ge F_{X_{1}}^{-1}\left( \alpha \right) \right) -H\left( Z\ge F_{Z}^{-1}\left( \alpha \right) \right) \right) \right) \nonumber \\&+E\left( X_{2}\left( H\left( X_{2}\ge F_{X_{2}}^{-1}\left( \alpha \right) \right) -H\left( Z\ge F_{Z}^{-1}\left( \alpha \right) \right) \right) \right) .\nonumber \end{aligned}$$

(20)

Let us now define

$$\begin{aligned} K_{X_{1}}=\left( X_{1}-F_{X_{1}}^{-1}\left( \alpha \right) \right) \cdot \left( H\left( X_{1}\ge F_{X_{1}}^{-1}\left( \alpha \right) \right) -H\left( Z\ge F_{Z}^{-1}\left( \alpha \right) \right) \right) . \end{aligned}$$

(21)

In case $X>F_{X_{1}}^{-1}\left( \alpha \right)$ we have $H\left( X_{1}\ge F_{X_{1}}^{-1}\left( \alpha \right) \right) =1$ so, $K_{X_{1}}\ge 0$, and if $X_{1}<F_{X_{1}}^{-1}\left( \alpha \right)$ we have $H\left( X_{1}\ge F_{X_{1}}^{-1}\left( \alpha \right) \right) =0.$ For the equality $X_{1}=F_{X_{1}}^{-1}\left( q\right) ,$ we get $K_{X_{1}}=0.$ Thus, in general, $K_{X_{1}}\ge 0.$ This, of course, also holds for $X_{2}.$ We thus conclude that

$$\begin{aligned}&\left( 1-\alpha \right) \cdot \left( \rho _{\alpha }\left( X_{1}\right) +\rho _{\alpha }\left( X_{2}\right) -\rho _{\alpha }\left( Z\right) \right) \\&=E\left( X\left( H\left( X\ge F_{X}^{-1}\left( \alpha \right) \right) -H\left( Z\ge F_{Z}^{-1}\left( \alpha \right) \right) \right) \right) \nonumber \\&+E\left( Y\left( H\left( Y\ge F_{Y}^{-1}\left( \alpha \right) \right) -H\left( Z\ge F_{Z}^{-1}\left( \alpha \right) \right) \right) \right) \nonumber \\&\ge F_{X}^{-1}\left( \alpha \right) E\left( H\left( X\ge F_{X}^{-1}\left( \alpha \right) \right) -H\left( Z\ge F_{Z}^{-1}\left( \alpha \right) \right) \right) \nonumber \\&+F_{Y}^{-1}\left( q\right) E\left( H\left( Y\ge F_{Y}^{-1}\left( \alpha \right) \right) -H\left( Z\ge F_{Z}^{-1}\left( \alpha \right) \right) \right) \nonumber \\&=F_{X}^{-1}\left( \alpha \right) \cdot \left( \alpha -\alpha \right) +F_{Y}^{-1}\left( \alpha \right) \cdot \left( \alpha -\alpha \right) =0.\nonumber \end{aligned}$$

(22)

This shows the subadditivity of $\rho _{\alpha },$

$$\begin{aligned} \rho _{\alpha }\left( X_{1}+X_{2}\right) \le \rho _{\alpha }\left( X_{1}\right) +\rho _{\alpha }\left( X_{2}\right) . \end{aligned}$$

(23)

Now, for the case of N particles, the multivariate pdf is denoted by $\rho \left( x_{1},x_{2},...,x_{N};t\right) ,$ followed by the wavefunction of the coupled N particles, $\psi \left( x_{1},x_{2},...,x_{N};t\right) .$ The relations (8) can be immediately deduced by partitioning the CoM $\overline{x}_{{\text {col}}}:=\frac{1}{M}\sum _{j=1}^{N}m_{j}x_{j}$ to sum of two random variables, e.g., taking $\overline{x}_{{\text {col}} }:=\frac{1}{M}m_{1}X_{1}+Y,$ where $Y=\frac{1}{M}\sum _{j=2}^{N}m_{j}X_{j},$ we get the subadditivity relations $\rho _{\alpha }\left( X_{1}+Y\right) \le \rho _{\alpha }\left( X_{1}\right) +\rho _{\alpha }\left( Y\right) .$ We again, recall that such a property is well-studied in the literature of risk theory [7, 8].

Appendix 3

Following the Gaussian pdf (11), any weighted-sum $\sum _{j=1} ^{N}w_{j}X_{j}$ is a Gaussian random variable, with mean $\sum _{j=1}^{N} w_{j}\left\langle x_{j}\left( t\right) \right\rangle$ and variance $\sum _{j=1}^{N}w_{j}^{2}\sigma _{j}\left( t\right) ^{2}.$ For any Gaussian random variable with mean $\mu$ and variance $\sigma ^{2},$ the distorted expectation (10) can be computed explicitely. Namely,

$$\begin{aligned} \left\langle X\right\rangle _{D}&=E\left( X|X\ge x_{q_{0}}\right) =\frac{\int _{x_{q_{0}}}^{\infty }xf_{X}\left( x\right) dx}{1-q_{0}}\end{aligned}$$

(24)

$$\begin{aligned}&=\frac{\int _{x_{q_{0}}}^{\infty }x\frac{1}{\sqrt{2\pi }\sigma }\exp \left( -\frac{1}{2}\left( \frac{x-\mu }{\sigma }\right) ^{2}\right) dx}{1-q_{0}}, \end{aligned}$$

(25)

where $x_{q_{0}}=F_{X}^{-1}\left( q_{0}\right) .$ Taking the transformation $Z=\left( X-\mu \right) /\sigma ,$ we finally obtain

$$\begin{aligned} \left\langle X\right\rangle _{D}=\mu +\frac{\sigma }{\sqrt{2\pi }}\frac{\int _{x_{q_{0}}}^{\infty }z\exp \left( -\frac{1}{2}z^{2}\right) dz}{1-q_{0} }=\mu +\frac{\sigma }{1-q_{0}}\frac{1}{\sqrt{2\pi }}e^{-z_{q_{0}}^{2}/2}. \end{aligned}$$

(26)