A Proposal for a New Kind of Spontaneous Collapse Model

Abstract

Spontaneous collapse models are modifications of standard quantum mechanics in which a physical mechanism is responsible for the collapse of the wavefunction, thus providing a way to solve the so-called “measurement problem”. The two most famous of these models are the Ghirardi–Rimini–Weber (GRW) model and the Continuous Spontaneous Localisation (CSL) models. Here, we propose a new kind of non-relativistic spontaneous collapse model based on the idea of collapse points situated at fixed spacetime coordinates. This model shares properties of both GRW and CSL models, while starting from different assumptions. We show that it can lead to a dynamics quite similar to that of the GRW model while also naturally solving the problem of indistinguishable particles. On the other hand, we can also obtain the same master equation of the CSL models. Then, we show how our proposed model solves the measurement problem in a manner conceptually similar to the GRW model. Finally, we show how the proposed model can also accommodate for Newtonian gravity by treating the collapses as gravitational sources.

Appendices

Appendix 1 Probability Distribution of Flashes

In this Appendix, we give more details needed to prove the validity of Eq. (1).

Let us suppose that, at time $t=t_0$, the state of system S is given by $\big \vert {\psi _0} \big \rangle$. Up to the first collapse point the wavefunction evolves according to $U_1 = U_{x_1}U(t_1-t_0)$ where $U(t_1-t_0)$ is the evolution operator given by the Hamiltonian of system S and $U_{x_1}$ is the instanteous interaction with the collapse point located in $x_1$ at time $t_1$. The probability of getting a flash (or not) in $x_1$ and the subsequent wavefunction are given by

$$\begin{aligned} P(\xi _1) = \left\| \big \langle \xi _1\big \vert {U_1}\big \vert \psi _0,0_1\big \rangle \right\| ^2, \quad \big \vert {\psi ^{(\xi _1)}} \big \rangle = \frac{\big \langle \xi _1\big \vert {U_1}\big \vert \psi _0,0_1\big \rangle }{\sqrt{P(\xi _1)}}, \end{aligned}$$

(A1)

where $\xi$ stands for either 0 or 1 or, respectively, no flash and flash. Then, we have another spontaneous measurement with the second collapse point located in $x_2$. In this case, we have

$$\begin{aligned}{} P(\xi _2|\xi _1) &= \left\| \big \langle \xi _2\big \vert {U_2}\big \vert \psi ^{(\xi _1)},0_2\big \rangle \right\| ^2 = \left\| \big \langle \xi _2\big |U_2\frac{\big \langle \xi _1\big \vert {U_1}\big \vert \psi _0,0_1\big \rangle }{\sqrt{P(\xi _1)}}\big \vert {0_2} \big \rangle \right\| ^2 \nonumber \\ &= \frac{1}{P(\xi _1)} \left\| \big \langle \xi _1,\xi _2\big |U_2 U_1\big \vert {\psi _0,0_1,0_2} \big \rangle \right\| ^2, \end{aligned}$$

(A2)

where $U_2 = U_{x_2}U(t_2-t_1)$ and which implies that

$$\begin{aligned} P(\xi _1,\xi _2) = P(\xi _2 | \xi _1) P(\xi _1) = \left\| \big \langle \xi _1,\xi _2\big |U_2 U_1\big \vert {\psi _0,0_1,0_2} \big \rangle \right\| ^2. \end{aligned}$$

(A3)

The state is then given by

$$\begin{aligned} \big \vert {\psi ^{(\xi _1,\xi _2)}} \big \rangle = \frac{\big \langle \xi _2\big | U_2 \big \vert {\psi ^{(\xi _1)},0_2} \big \rangle }{\sqrt{P(\xi _2|\xi _1)}} = \frac{\big \langle \xi _1,\xi _2\big |U_2U_1\big \vert {\psi _0,0_1,0_2} \big \rangle }{\sqrt{P(\xi _1,\xi _2)}}. \end{aligned}$$

(A4)

By induction, it follows that

$$\begin{aligned} P(\xi _1,\dots ,\xi _n)= & {} \left\| \big \langle \xi _1,\dots ,\xi _n\big |U_n \dots U_1\big \vert {\psi _0,0_1,\dots ,0_n} \big \rangle \right\| ^2, \nonumber \\ \big \vert {\psi ^{(\xi _1,\dots ,\xi _n)}} \big \rangle= & {} \frac{\big \langle \xi _1,\dots ,\xi _n\big |U_n \dots U_1\big \vert {\psi _0,0_1,\dots ,0_n} \big \rangle }{\sqrt{P(\xi _1,\dots ,\xi _n)}}, \end{aligned}$$

(A5)

which is Eq. (1) of the main text.

Appendix 2 Derivation of the Stochastic Schrödinger Equation

The aim of this Appendix is to derive the stochastic Schrödinger equation given in the main text [Eq. (7)].

We consider the system wavefunction to be confined in a volume V and study its evolution for a time $\delta t$ which is very short with respect to the time scale of H, so that we can ignore it. First, we consider the joint dynamics of wavefunction and collapse points up to first order in $\gamma$. We assume that in the spacetime volume $c\delta t V$ there are, on average, N collapse points, where $N\gg 1$. In fact, the average number of collapse points in a given spacetime region is not a natural number but since it is much higher than one, we can round it down without harm, so we treat N as a natural number. Starting from the unitaries in Eq. (1) we get:

$$\begin{aligned}{} & {} U_N \dots U_1 \simeq \prod _{j=1}^N \left[ 1 -i \frac{\sqrt{\gamma }}{\hbar }{\hat{L}}\left( x_j\right) \sigma _x^{(j)} - \frac{\gamma }{2 \hbar ^2} {\hat{L}}^2 \left( x_j\right) \right] \\{} & {} \quad \simeq 1 -i \frac{\sqrt{\gamma }}{\hbar }\sum _{j=1}^N {\hat{L}}\left( x_j\right) \sigma _x^{(x_j)} -\frac{\gamma }{2 \hbar ^2} \sum _{j=1}^N {\hat{L}}^2 (x_j)\nonumber \\{} & {} \quad - \frac{\gamma }{\hbar ^2} \sum _{k>j}{\hat{L}}(x_k){\hat{L}}(x_j)\sigma _x^{(x_k)}\sigma _x^{(x_j)}. \end{aligned}$$

(B6)

Then, the probability of getting no flashes is

$$\begin{aligned}{} & {} P(0) \equiv P(0_1,\dots ,0_n) = \left\| \big \langle 0_1,\dots ,0_n\big |U_n \dots U_1\big \vert {\psi _0,0_1,\dots ,0_n} \big \rangle \right\| ^2 \\{} & {} \quad \simeq 1 - \frac{\gamma }{\hbar ^2} \sum _{j=1}^N \langle \psi _0\vert {{\hat{L}}^2 (x_j)}\vert \psi _0\rangle , \end{aligned}$$

(B7)

while the probability of getting a flash at point $x_j$ is given by

$$\begin{aligned} P(x_j) \equiv P(0_1,\dots ,1_j,\dots ,0_N) \simeq \frac{\gamma }{\hbar ^2}\langle \psi _0\vert {{\hat{L}}^2 (x_j)}\vert \psi _0\rangle . \end{aligned}$$

(B8)

We can see that at this order of perturbation $P(0) + \sum _j P(x_j) = 1$, which means that, at this order, the probability of getting two flashes must be neglected. The wavefunction in the two cases evolves as follows:

$$\begin{aligned}&\big \vert {\psi ^{(0)}} \big \rangle = \left[ 1 + \frac{\gamma }{2\hbar ^2}\mathop {\sum }\limits _{j=1}^N \left( \langle \psi _0\vert {{\hat{L}}^2 (x_j)}\vert \psi _0\rangle -{\hat{L}}^2 (x_j)\right) \right] \big \vert {\psi _0} \big \rangle , \nonumber \\&\quad \big \vert {\psi ^{(x_j)}} \big \rangle = \frac{-i {\hat{L}}(x_j)}{\sqrt{\langle \psi _0\vert {{\hat{L}}^2 (x_j)}\vert \psi _0\rangle }}\big \vert {\psi _0} \big \rangle . \end{aligned}$$

(B9)

Now, we consider the fact that the collapse points are uniformly distributed in the spacetime volume $c \delta t V$. We define $\mu = N/(c \delta t V)$ as the collapse point density in spacetime and we compute the probability of flash or no flash as

$$\begin{aligned}&P(0) \simeq 1 - \frac{\gamma N}{\hbar ^2 V} \int _V \textrm{d}^{3}{x} \langle \psi _0\vert {{\hat{L}}^2 (x)}\vert \psi _0\rangle \simeq 1 - \delta t\frac{c\gamma \mu }{\hbar ^2 } \int \textrm{d}^{3}{x} \langle \psi _0\vert {{\hat{L}}^2 (x)}\vert \psi _0\rangle , \nonumber \\&\quad P(1) \simeq \frac{\gamma N}{\hbar ^2 V} \int _V \textrm{d}^{3}{x} \langle \psi _0\vert {{\hat{L}}^2 (x)}\vert \psi _0\rangle \simeq \delta t\frac{c \gamma \mu }{\hbar ^2} \int \textrm{d}^{3}{x} \langle \psi _0\vert {{\hat{L}}^2 (x)}\vert \psi _0\rangle . \end{aligned}$$

(B10)

The probability of getting a flash around point x is

$$\begin{aligned} P(x\in \textrm{d}^{3}{x})\textrm{d}^{3}{x} \simeq \delta t\frac{c \gamma \mu }{\hbar ^2} \textrm{d}^{3}{x} \langle \psi _0\vert {{\hat{L}}^2 (x)}\vert \psi _0\rangle \end{aligned}$$

(B11)

Then, when there is no flash, or there is a flash at point x, the state changes according to

$$\begin{aligned}{} & {} \big \vert {\psi ^{(0)}} \big \rangle = \left[ 1 - \delta t\frac{c \gamma \mu }{2\hbar ^2}\int \textrm{d}^{3}{x} \left( {\hat{L}}^2 (x)-\langle \psi _0\vert {{\hat{L}}^2 (x)}\vert \psi _0\rangle \right) \right] \big \vert {\psi _0} \big \rangle , \nonumber \\{} & {} \quad \big \vert {\psi ^{(x)}} \big \rangle = \frac{-i {\hat{L}}(x)}{\sqrt{\langle \psi _0\vert {{\hat{L}}^2 (x)}\vert \psi _0\rangle }}\big \vert {\psi _0} \big \rangle . \end{aligned}$$

(B12)

Following Refs. [24, 48] the above leads to the stochastic equation

$$ \textrm{d}{\big \vert {\psi (t)} \big \rangle } = \int \textrm{d}^{3}{x} \left[ \textrm{d}{t}\frac{c \gamma \mu }{2\hbar ^2} \left( \big \langle {{\hat{L}}^2 (x)}\big \rangle -{\hat{L}}^2(x)\right) - \frac{\textrm{d}{N_x}}{\textrm{d}^{3}{x}}\left( 1 + \frac{i {\hat{L}}(x)}{\sqrt{\big \langle {{\hat{L}}^2 (x)}\big \rangle }}\right) \right] \big \vert {\psi (t)} \big \rangle ,$$

(B13)

where we have the Poisson process $\textrm{d}{N_x}$ such that ${\mathbb {E}}\left[ \textrm{d}{N_x}\right] =P(x\in \textrm{d}^{3}{x})\textrm{d}^{3}{x}$ (with $\delta t \rightarrow \textrm{d}{t}$). The Poisson process has the property $\textrm{d}{N_x}^2 = \textrm{d}^{3}{N_x}$ because its result can only be zero or one. Moreover, one has that $\textrm{d}{N_x}\,\textrm{d}{t} = 0$ and $\textrm{d}{N_x}\,\textrm{d}{N_y} = \delta (x-y)\,\textrm{d}^{3}{y}\,\textrm{d}{N_x}$. Equation (B13) can be seen as the spatially continuous version of Eq. (3.156) of Ref. [49], with the identification $a \rightarrow -i L$. Restoring the evolution due to the Hamiltonian H of the system we get Eq. (7) of the main text.

Appendix 3 Collapse Operators for Indistinguishable Particles

In this section we show how to re-obtain the collapse operator defined in Ref. [17] by starting from the smeared mass operator of Eq. (9). We recall that we consider the coarse-grain operation of the collapse points and the expansions at first order in $\gamma$ already effectuated. We start with a bosonic field and then explain why the same calculation holds for a fermionic field. The sum of the smeared number operators preceded by the factors $m_i/{m_R}$ gives the smeared mass operator and its application to wavefunctions with a precise number of particles for each kind of particle gives the same collapse operator defined in Ref. [17]. Indeed, if we consider a simplified toy model with just one kind of particles, we do not need the smeared mass operator and we can just use the smeared number operator.

We consider a single type of indistinguishable particles. We want the probability of a flash around point x to be proportional to the amount of particles so that [cf. Equation (5)] we require ${\hat{L}}^2 (x) \propto {\mathcal {N}}(x)$, where ${\mathcal {N}}(x)$ is the smeared number operator defined as follows:

$$\begin{aligned} {\mathcal {N}}(x) \equiv \int _{-\infty }^{+\infty }\textrm{d}{y}\,g(y-x) a^\dagger (y) a (y), \end{aligned}$$

(C14)

where a(x) and $a^\dagger (x)$ are, respectively, annihilation and creation operators at the spatial point x. The action of such operator on a state $\big \vert {\psi _n} \big \rangle$ with a fixed number n of particles is the following:

$$\begin{aligned} {\mathcal {N}}(x) \big \vert {\psi _n} \big \rangle&= \left[ \int _{-\infty }^{+\infty }\textrm{d}{y}\,\textrm{d}^{n}{z} \psi _n (z_1,\dots ,z_n) g(y-x) a^\dagger (y) a(y) a^\dagger (z_1) \dots a^\dagger (z_n)\right] \big \vert {0} \big \rangle ,\nonumber \\&= \left[ \int _{-\infty }^{+\infty }\textrm{d}{y}\, \textrm{d}^{n}{z} \psi _n (z_1,\dots ,z_n) g(y-x) a^\dagger (y) \left( \mathop {\sum }\limits _i \delta (y - z_i) \mathop {\prod }\limits _{j\ne i} a^\dagger (z_j)\right) \right] \big \vert {0} \big \rangle ,\nonumber \\&= \mathop {\sum }\limits _i \left[ \int _{-\infty }^{+\infty }\textrm{d}^{n}{z} \psi _n (z_1,\dots ,z_n) g(z_i-x) a^\dagger (z_1) \dots a^\dagger (z_n)\right] \big \vert {0} \big \rangle ,\nonumber \\&= \left[ \mathop {\sum }\limits _i g({\hat{z}}_i - x)\right] \big \vert {\psi _n} \big \rangle , \end{aligned}$$

(C15)

which implies that $\sqrt{{\mathcal {N}}(x)}$ acts exactly as the collapse operator defined in Refs. [17, 23], provided we choose the proper function g(x).

In the case of a fermionic field the result is the same. It is sufficient to notice that any minus sign obtained by moving the annihilation operator a(y) to the right is then countered by having to move the creation operator $a^\dagger (y)$ to the same position. In other words, one can consider that $a^\dagger (y) a(y) a^\dagger (z) = \delta (y-z)a^\dagger (y) + a^\dagger (z) a^\dagger (y) a(y)$.

Appendix 4 Derivation of the Master Equation

In this Appendix we show two different ways to derive Eq. (10) of the main text.

Starting from Eq. (B13) and exploiting the properties of the Poisson process $\textrm{d}{N}$, we can derive the stochastic equation for the density matrix

$$\begin{aligned} \begin{aligned}&\textrm{d}{\sigma } = {\big \vert \textrm{d}{\psi }\big \rangle }{\big \langle \textrm{d}{\psi }\big \vert } + {\big \vert \psi \big \rangle }{\big \langle \psi \big \vert } + {\big \vert \textrm{d}{\psi }\big \rangle }{\big \langle \textrm{d}{\psi }\big \vert } \int \textrm{d}^{3}{x} \left[ \textrm{d}{t} {\mathcal {Q}}(x) + \frac{\textrm{d}{N_x}}{\textrm{d}^{3}{x}}{\mathcal {L}}(x)\right] \sigma , \end{aligned} \end{aligned}$$

(D16)

where

$$\begin{aligned} {\mathcal {Q}}(x) \sigma = \frac{c \gamma \mu }{2\hbar ^2} \left( 2\big \langle {{\hat{L}}^2 (x)}\big \rangle \sigma - \big \{{{\hat{L}}^2(x)}\big \}{\sigma }\right) , \qquad {\mathcal {L}}(x) \sigma = \frac{{\hat{L}}\sigma {\hat{L}}}{\big \langle {{\hat{L}}^2}\big \rangle } - \sigma . \end{aligned}$$

(D17)

Then, the master equation for $\rho _t$ can be obtained by making the ensemble average of the stochatic density matrix. We make the substitution $\textrm{d}{N_x}/\textrm{d}^{3}{x} \rightarrow \textrm{d}{t} (c \gamma \mu /\hbar ^2) \big \langle {{\hat{L}}^2 (x)}\big \rangle$ and we get

$$\begin{aligned} \textrm{d}{\rho _t} = \textrm{d}{t}\frac{c\mu \gamma }{\hbar ^2} \int \textrm{d}^{3}{x} \left[ {\hat{L}}(x) \rho _t {\hat{L}}(x) - \frac{1}{2}\big \{{{\hat{L}}^2 (x)}\big \}{\rho _t}\right] . \end{aligned}$$

(D18)

Restoring the evolution due to the Hamiltonian H of the system we get Eq. (10) of the main text.

We can also derive Eq. (10) directly from Eq. (1) as follows. Denoting by $\big \vert {\psi _t} \big \rangle$ the state of the system at hand at time t, the interaction between a collapse point located in $x_c$ at time t and the system leads to

$$\begin{aligned} e^{-i\sqrt{\gamma } H_{x_c}/\hbar }\big \vert {\psi _t} \big \rangle \big \vert {0} \big \rangle \simeq \big \vert {\psi _t} \big \rangle \big \vert {0} \big \rangle - i \frac{\sqrt{\gamma }}{\hbar } {\hat{L}}(x_c)\big \vert {\psi _t} \big \rangle \big \vert {1} \big \rangle - \frac{\gamma }{2 \hbar ^2}{\hat{L}}^2 (x_c)\big \vert {\psi _t} \big \rangle \big \vert {0} \big \rangle , \end{aligned}$$

(D19)

where we assumed, again, that we can expand up to first order in $\gamma$. Then, the probability of getting a flash is given by

$$\begin{aligned} P (1) = \big \langle \psi _t,0\big |e^{+i\sqrt{\gamma } H_{x_c}/\hbar } {\big \vert 1\big \rangle }{\big \langle 1\big \vert }^{-i\sqrt{\gamma } H_{x_c}/\hbar } \big \vert {\psi _t,0} \big \rangle \simeq \frac{\gamma }{\hbar ^2} \langle \psi _t\vert {{\hat{L}}^2 (x_c)}\vert \psi _t\rangle , \end{aligned}$$

(D20)

and, indeed, the probability of getting no flash is just $1-P(1)$. The wavefunction evolves as follows:

$$\begin{aligned}{} & {} \big \vert {\psi _t} \big \rangle \xrightarrow {\text {Flash}} \big \vert {\psi _t^{(1)}} \big \rangle \simeq \frac{-i \sqrt{\gamma } {\hat{L}}(x_c) \big \vert {\psi _t} \big \rangle }{\hbar \sqrt{P (1)}} \nonumber \\{} & {} \quad \big \vert {\psi _t} \big \rangle \xrightarrow {\text {No Flash}} \big \vert {\psi _t^{(0)}} \big \rangle \simeq \frac{\left[ 1-(\gamma /2\hbar ^2){\hat{L}}^2 (x_c)\right] \big \vert {\psi _t} \big \rangle }{\sqrt{1-P (1)}}. \end{aligned}$$

(D21)

At the level of density matrix, this implies

$$\begin{aligned}{} & {} \rho _t \rightarrow P (1) {\big \vert \psi _t^{(1)}\big \rangle }{\big \langle \psi _t^{(1)}\big \vert } (1-P(1)){\big \vert \psi _t^{(0)}\big \rangle }{\big \langle \psi _t^{(0)}\big \vert }\nonumber \\{} & {} \quad \simeq \rho _t + \frac{\gamma }{\hbar ^2} \left[ {\hat{L}}(x_c) \rho _t {\hat{L}}(x_c) - \frac{1}{2}\big \{{{\hat{L}}^2 (x_c)}\big \}{\rho _t}\right] . \end{aligned}$$

(D22)

Let us assume now, as before, that the wavefunction is non-zero in a finite volume V. Then, in a time internal $\delta t$ we assume there are, on average, N collapse events (flash or no flash) in the spacetime volume $c \delta t V$, with $N \gg 1$. Keeping again only terms up to first order in $\gamma$, we obtain (neglecting the Hamiltonian evolution of the particle)

$$\begin{aligned} \rho _{t + \delta t} \simeq \rho _t + \frac{\gamma }{\hbar ^2} \sum _{j=1}^N \left[ {\hat{L}}(x_j) \rho _t {\hat{L}}(x_j) - \frac{1}{2}\big \{{{\hat{L}}^2 (x_j)}\big \}{\rho _t}\right] . \end{aligned}$$

(D23)

Since the collapse points are uniformly distributed in space-time, it follows that we can simplify the above expression by averaging over the distribution of the collapse points in $c\delta t V$. To each collapse we associate the uniform distribution and thus we obtain

$$\begin{aligned} \frac{\rho _{t + \delta t} -\rho _t }{\delta t} \simeq \frac{c\mu \gamma }{\hbar ^2} \int \textrm{d}^{3}{x} \left[ {\hat{L}}(x) \rho _t {\hat{L}}(x) - \frac{1}{2}\{{{\hat{L}}^2 (x)}\}{\rho _t}\right] , \end{aligned}$$

(D24)

where $\mu = N/(c\delta t V)$ is the collapse density in spacetime. By adding the Hamiltonian evolution and making the substitution $\delta t \rightarrow \textrm{d}{t}$, we get again Eq. (10) of the main text.

Appendix 5 Newtonian Gravity

1.1 Energy Impact of a Flash on a Single Particle

In this appendix, we compute the energy change of a single particle due to a flash. To do this, we choose coordinates such that the flash is situated in the origin. Then, the unitary operator $U_G$ assumes the form

$$\begin{aligned} U_G = \exp {\left\{ i r_m \int \textrm{d}^{3}{y}\frac{f_G(\vert {y}\vert )}{\vert {{\hat{x}}-y}\vert }\right\} }, \qquad r_m \equiv \frac{G m_R m}{\lambda _{\textrm{GRW}} \hbar }, \end{aligned}$$

(E25)

where we exploited the fact that, for a single particle, ${\hat{M}}(x) = m \delta ({\hat{x}}- x)$. In spherical coordinates, this unitary operator changes a wavefunction $\psi (r,\theta ,\phi )$ as follows

$$\begin{aligned}{} & {} U_G \rightarrow \exp \Bigg \{i r_m \int _0^{\infty } \textrm{d}{r_y} \int _0^{\pi } \textrm{d}{\theta _y} \int _{0}^{2\pi } \textrm{d}{\phi _y} \\{} & {} \quad \times \frac{r_y^2 \sin (\theta _y) f_G (r_y)}{\sqrt{r^2 + r_y^2 -2 r r_y \left[ \sin (\theta )\sin (\theta _y)\cos (\phi - \phi _y)+\cos (\theta )\cos (\theta _y)\right] }}\Bigg \}. \end{aligned}$$

(E26)

However, despite the appearances, the above function is only a function of r and does not depend on $\theta$ or $\phi$. So, we can write the unitary operator in spherical coordinate representation as $U_G (r) = \exp [i r_m F(r)]$. Setting $\theta =0$, F(r) assumes the following form (along with its derivative)

$$\begin{aligned} F(r)= & {} 2 \pi \int _0^{\infty } \textrm{d}{r_y}\, \frac{r_y f_G (r_y)}{r}\left( r+r_y -\vert {r-r_y}\vert \right) ,\nonumber \\= & {} 4 \pi \left[ \frac{1}{r}\int _0^r \textrm{d}{r_y}\, r_y^2 f_G (r_y) + \int _r^{\infty } \textrm{d}{r_y} \, r_y f_G (r_y)\right] ,\nonumber \\ F'(r)= & {} - \frac{4 \pi }{r^2} \int _0^r \textrm{d}{r_y}\, r_y^2 f_G (r_y). \end{aligned}$$

(E27)

The kinetic energy after a gravitational pull can then be calculated as follows

$$\begin{aligned}{} & {} \bigg \langle {\frac{{\hat{p}}^2}{2m}}\bigg \rangle = -\frac{\hbar ^2}{2m} \int _0^{\infty } \textrm{d}{r} \int _0^{\pi } \textrm{d}{\theta } \int _{0}^{2\pi } \textrm{d}{\phi } r^2 \sin (\theta ) \\{} & {} \quad \times \left[ \psi ^* (r,\theta ,\phi ) U_G^*(r) \nabla ^2 U_G (r)\psi (r,\theta ,\phi )\right] . \end{aligned}$$

(E28)

One can then see that

$$\begin{aligned} \nabla ^2 U_G \psi= & {} \frac{2 \psi }{r}\partial _r U_G + \psi \partial ^2_r U_G + 2 \left( \partial _r \psi \right) \left( \partial _r U_G\right) + U_G \nabla ^2 \psi ,\nonumber \\= & {} U_G \left[ - r_m^2 \left( F'(r)\right) ^2 \psi + i r_m \left( F''(r) \psi + \frac{2}{r} F'(r) \psi + 2 F'(r) \partial _r \psi \right) \right] \nonumber \\{} & {} \quad + U_G \nabla ^2 \psi ,\nonumber \\= & {} U_G \left[ - r_m^2 \left( F'(r)\right) ^2 \psi + i r_m \frac{1}{r^2}\partial _r \left( r^2 F'(r) \psi \right) + i r_m F'(r) \partial _r \psi \right] \nonumber \\{} & {} \quad + U_G \nabla ^2 \psi , \end{aligned}$$

(E29)

which leads to

$$\begin{aligned}{} & {} \bigg \langle {\frac{{\hat{p}}^2}{2m}}\bigg \rangle = -\frac{\hbar ^2}{2m} \int _0^{\infty } \textrm{d}{r} \int _0^{\pi } \textrm{d}{\theta } \int _{0}^{2\pi } \textrm{d}{\phi } \sin (\theta ) \\{} & {} \quad \times \left[ r^2 \psi ^*\nabla ^2\psi - r_m^2 r^2 \left( F'(r)\right) ^2 \vert {\psi }\vert ^2 + i r_m \psi ^* \partial _r \left( r^2 F'(r) \psi \right) \right. \nonumber \\{} & {} \quad \left. + i r_m r^2 F'(r) \psi ^*\partial _r \psi \right] . \end{aligned}$$

(E30)

The first term in the integral is just the energy prior to the gravitational pull while the other terms contain the energy due to it. We can integrate by parts the third term and see that it is minus the complex conjugate of the fourth. Therefore, we get

$$\begin{aligned}{} & {} \bigg \langle {\frac{{\hat{p}}^2}{2m}}\bigg \rangle = -\frac{\hbar ^2}{2m} \int _0^{\infty } \textrm{d}{r} \int _0^{\pi } \textrm{d}{\theta } \int _{0}^{2\pi } \textrm{d}{\phi } \nonumber \\ & \quad \times \Bigg \{ r^2 \sin (\theta ) \left[ \psi ^*\nabla ^2\psi - r_m^2 \left( F'(r)\right) ^2 \vert {\psi }\vert ^2 - 2 r_m F'(r) \mathrm{Im\big \{\psi ^* \partial _r \psi \big \}}\right] \Bigg \}. \end{aligned}$$

(E31)

One immediately see that if $F(r)=1/r$, the second term in the above integral may lead to a divergence. However, with the smearing of the collapse point as the gravitational source, this does not happen. For example, one could take $f_G (r) = (\pi r_G)^{-3/2} \exp [-r^2/r_G^2]$ to avoid this kind of divergences.

1.2 Single Particle Decoherence Dynamics

Let us analyse the dynamics of a single particle of mass m governed by Eq. (18). Since we are dealing with a single particle, the mass operator ${\hat{M}}(x)$ becomes ${\hat{M}}(x) \rightarrow m \delta (x-{\hat{x}})$ so that the unitary $U_G (x)$ and the collapse operator ${\hat{L}}(x)$ now read

$$\begin{aligned}{} & {} U_G (s) = \exp \left[ i r_m\int \textrm{d}^{3}z\frac{f_G(\vert {z-s}\vert )}{{\vert {{\hat{x}}-z}\vert }}\right] , \quad r_m \equiv \frac{G {m_R} m}{\lambda _{\textrm{GRW}}\hbar }, \nonumber \\{} & {} \quad {\hat{L}}(s) = \sqrt{\frac{m}{{m_R}}}f_C (\vert {{\hat{x}}-s}\vert ), \end{aligned}$$

(E32)

where $r_m$ is a new length scale associated to the gravitational effect of a flash and associated to the mass m of the particle, $f_C (\vert {x}\vert )$ is a real function, usually taken to be a Gaussian, characterised by the collapse radius $r_C$ [2, 13], while $f_G (\vert {x}\vert )$ is also a real function but characterised by a radius $r_G$. $r_G$ would be a new parameter but one could choose to set it equal to $r_C$. We leave it as a separate parameter for the sake of generality. For standard quantum particles it holds that $r_m \ll r_C$ [13].

In coordinate representation, neglecting the standard Hamiltonian of the particle, we get the single particle master equation

$$\begin{aligned} \frac{\textrm{d}}{\textrm{d}t} \big \langle x\big \vert {\rho _t}\big \vert y\big \rangle = \lambda _{\textrm{GRW}} \frac{m}{{m_R}}\Gamma (x,y)\big \langle x\big \vert {\rho _t}\big \vert y\big \rangle , \end{aligned}$$

(E33)

where

$$\begin{aligned}{} & {} \Gamma (x,y) = \int \textrm{d}^{3}{s} \Bigg \{ - \frac{f_C^2(\vert {x-s}\vert )}{2}- \frac{f_C^2(\vert {y-s}\vert )}{2} \nonumber \\{} & {} \quad +\exp \left[ i r_m \int \textrm{d}^{3}{z} \left( \frac{f_G(\vert {z-s}\vert )}{\vert {x-z}\vert }-\frac{f_G(\vert {z-s}\vert )}{\vert {y-z}\vert }\right) \right] f_C(\vert {x-s}\vert )f_C(\vert {y-s}\vert )\Bigg \}. \end{aligned}$$

(E34)

Notice how, by taking the function $f_C (x)$ to be the standard GRW choice, i.e., $f_C (x) = (\pi r_C^2)^{-3/4}\exp [-x^2/(2 r_C^2)]$, the equation for $\Gamma (x,y)-1$ becomes very similar to the one obtained in Ref. [13], Eqs. (10) and (11)¹².

Independently of the choice of $f_C (\vert {x}\vert )$, $\Gamma (x,x)=0$ and $\Gamma (x,y)=\Gamma ^*(y,x)$. We now prove that $\Gamma (x,y) \in {\mathbb {R}}$ and that $\Gamma (x,y)$ is actually a function of the single variable $d=\vert {x-y}\vert /2$. First, we observe that the last two terms in the integral are real so we only need to prove that the first term gives rise to a real term. Secondly, we define $F(\vert {x-s}\vert ) = \int \textrm{d}^{3}{z} \ f_G(\vert {z-s}\vert )/\vert {x-z}\vert$, where we can see that the result of the integral only depends on $\vert {x-s}\vert$ because we can choose, for each x and s, a coordinate system where $x=0$ and s has just one nonzero coordinate equal to $\vert {x-s}\vert$. After noticing this, we make the change of variable $s'= s - (x+y)/2$ which is a translation of the origin. Then, we rotate the coordinate system so that for a given couple of points (x, y) the point $d = (x-y)/2$ is along the z-axis with its z-coordinate being positive (or zero). It follows that, in cylindrical coordinates and defining $l(z,r,d) \equiv \sqrt{(z+d)^2+r^2}$, the above integral reads

$$\begin{aligned}{} & {} \Gamma (x,y) = 2 \pi \int _0^{\infty } \textrm{d}{r}\, r \Bigg \{ - \int _{-\infty }^{+\infty }\textrm{d}{z}\frac{f_C^2\left( l(z,r,-d)\right) }{2} - \int _{-\infty }^{+\infty }\textrm{d}{z}\frac{f_C^2\left( l(z,r,d)\right) }{2} \nonumber \\{} & {} \quad +\int _{0}^{+\infty } \textrm{d}{z}\exp {\{i r_m \left[ F \left( l(z,r,d)\right) -F \left( l(z,r,-d)\right) \right] \}}f_C\left( l(z,r,d)\right) f_C\left( l(z,r,-d)\right) \nonumber \\{} & {} \quad +\int _{0}^{+\infty } \textrm{d}{z}\exp {\{i r_m \left[ F \left( l(z,r,-d)\right) -F \left( l(z,r,d)\right) \right] \}}f_C\left( l(z,r,-d)\right) f_C\left( l(z,r,d)\right) \Bigg \}, \end{aligned}$$

(E35)

where we first splitted in two the integral in the z variable and then, in the second line, we made the substitution $z \rightarrow -z$ and re-ordered the integration extremes. We see that second and third line are the complex conjugate of each other. It follows then that $\Gamma (x,y)$ is real. Finally, with a slight abuse of notation, we can write

$$\begin{aligned}{} & {} \Gamma (d) = 2 \pi \int _{0}^{\infty } \textrm{d}{r}\, r \int _{-\infty }^{+\infty }\textrm{d}{z} \Bigg \{-\frac{f_C^2\left( l(z,r,-d)\right) }{2}-\frac{f_C^2\left( l(z,r,-d)\right) }{2} \nonumber \\{} & {} \quad +\cos \left( r_m \left[ F \left( l(z,r,-d)\right) -F \left( l(z,r,-d)\right) \right] \right) f_C\left( l(z,r,-d)\right) f_C\left( l(z,r,-d)\right) \Bigg \}. \end{aligned}$$

(E36)

Thus, as in Ref. [13], the only effect of self gravity, when neglecting the particle bare Hamiltonian, is to increase the rate of dephasing in position basis. Moreover, the dephasing rate of the density matrix $\rho (x,y)$ only depends on the distance $d = \vert {x-y}\vert /2$.

1.3 Single Particle Short Distance Decoherence Behaviour

We now turn to study the modifications to the behaviour at short distances due to the presence of the gravitational term. This calculation purpose is just to show the emergence of a dependence of the decoherence on the inverse of a power of $\lambda _{\textrm{GRW}}$, thus hinting at the possibility that a full model of such Newtonian gravity can pose testable, in principle, experimental bound from below for $\lambda _{\textrm{GRW}}$. To make this proof of principle calculation, we follow again Ref. [13], i.e., we choose $F(\vert {x-s}\vert ) = 1/\vert {x-s}\vert$ and we use the GRW collapse operator $f_C (x) = (\pi r_C^2)^{-3/4}\exp [-x^2/(2 r_C^2)]$. We, as expected, obtain the same qualitative result.

The first step is to write Eq. (E36) in spherical coordinates:

$$\begin{aligned}{} & {} \Gamma (d) + 1 = 2 \pi e^{-d^2/r_C^2}\int _{0}^{\infty } \textrm{d}{r}\, r^2 \int _0^{\pi } \textrm{d}{\theta } \nonumber \\{} & {} \quad \times \sin (\theta ) e^{-r^2/r_C^2}\cos \left( \frac{r_m }{\sqrt{r^2 + 2 d r \cos (\theta ) + d^2}}-\frac{r_m }{\sqrt{r^2 - 2 d r \cos (\theta ) + d^2}}\right) . \end{aligned}$$

(E37)

The functions inside the integral of Eq. (E37) are all bounded so that we can choose a small length $\epsilon$ such that the integration for $r \in [0,\epsilon ]$ gives an almost vanishing contribution. We then expand, at first order, the argument of the cosine for $d \ll \epsilon$ and get

$$\begin{aligned}{} & {} \cos \left( \frac{r_m }{\sqrt{r^2 + 2 d r \cos (\theta ) + d^2}}-\frac{r_m }{\sqrt{r^2 - 2 d r \cos (\theta ) + d^2}}\right) \nonumber \\{} & {} \quad \simeq \cos \left( \frac{2 r_m \cos (\theta ) d}{r^2}\right) , \end{aligned}$$

(E38)

which can be integrated as follows

$$\begin{aligned} \int _0^\pi \textrm{d}{\theta } \sin (\theta )\cos \left( \frac{2 r_m \cos (\theta ) d}{r^2}\right) &= -\left[ \frac{r^2}{2 r_m d} \sin \left( \frac{2 r_m d \cos (\theta )}{r^2}\right) \right] ^\pi _0 \nonumber \\ & = \frac{r^2}{r_m d} \sin \left( \frac{2 r_m d}{r^2}\right) . \end{aligned}$$

(E39)

Substituting this back leads to an integral that can be solved analytically (with $\epsilon =0$) but whose solution is quite cumbersome. Expanding this solution to second order in d gives

$$\begin{aligned} \Gamma (d) \simeq - \frac{32}{15}\frac{r_m^{3/2} }{r_C^3}d^{3/2} - \frac{1}{r_C^2}\left( 1-\frac{8}{3}\frac{r_m^2 }{r_C^2}\right) d^2. \end{aligned}$$

(E40)

The first leading order is much more important than the second when $d \ll r_m^3 /r_C^2, r_C^2/r_m$. The first leading order agrees with the result of Ref. [13]¹³.