Relativistic quantum field theory of stochastic dynamics in the Hilbert space

Let us start our discussion from a simple physical system - one-dimensional harmonic oscillator whose action reads

$$I_{0}=\int^{t^{\prime}}_{t_{0}}dt\left[\frac{1}{2}m\left(\frac{dx}{dt}\right)^{2}-\frac{1}{2}m\omega^{2}x^{2}\right],$$

(1)

where $m$ is the mass, $x(t)$ is the position of particle at time $t$, and $\omega$ is the oscillating frequency. In classical mechanics, given the initial and final positions, e.g. $x(t_{0})=x_{0}$ and $x(t^{\prime})=x^{\prime}$, one can find the path of particle by minimizing the action.

Suppose there is a random force acting on the particle. We hope that the dynamics keeps Markovian, which means that the particle’s future state depends only upon its current state but is independent of its past state. This requirement puts strong constraint on the random force which must be white. A mathematically elegant way of describing a white-noise force is to add a random term into the action, which then becomes

$$I_{W}=I_{0}+\lambda\int^{t^{\prime}}_{t_{0}}dW_{t}\ x(t),$$

(2)

where $\lambda$ is the strength of random coupling, and $W_{t}$ is the Wiener process. The integral over Wiener process is the so-called Itô integral, defined as

$$\int^{t^{\prime}}_{t_{0}}dW_{t}\ x(t)=\lim_{\Delta t\to 0}\ \sum_{j=0}^{N-1}\Delta W_{t_{j}}x(t_{j}),$$

(3)

where $t_{0}<t_{1}<\cdots<t_{N-1}<t_{N}=t^{\prime}$ with $t_{j+1}-t_{j}=\Delta t$ is a partition of the interval $[t_{0},t^{\prime}]$, and $\Delta W_{t_{j}}=W_{t_{j}+\Delta t}-W_{t_{j}}$ is an infinitesimal increment. Note that the Wiener process is non-differentiable everywhere, hence, we cannot express $dW_{t}$ as $\displaystyle\frac{dW_{t}}{dt}dt$ because $\displaystyle\frac{dW_{t}}{dt}$ does not exist. As a consequence, we cannot see the action as an integral of some Lagrangian, because the latter is not well-defined.

A defining property of the Wiener process is that the increments $\Delta W_{t_{j}}$ are all independent Gaussian random variables with zero mean and $\Delta t$ variance. Therefore, when doing the calculations involving an action like Eq. (2), one can start from its time-discretization form in which $\left\{\Delta W_{t_{0}},\Delta W_{t_{1}},\cdots,\Delta W_{t_{N-1}}\right\}$ is a set of independent Gaussians, and only take the limit $\Delta t\to 0$ in the final results.

Next we minimize the action (2) to find the classical equation of motion. Bearing in mind that $I_{W}$ is a random number whose randomness comes from $W_{t}$, we define the minimization of $I_{W}$ in a pathwise way. For each path of $W_{t}$, or in the discretized version, for each value of the vector $\left(W_{t_{0}},W_{t_{1}},\cdots,W_{t_{N-1}}\right)$, we minimize $I_{W}$ to obtain the path $x(t)$ which then becomes a functional of $W_{t}$. Since $W_{t}$ is a random path, $x(t)$ is also a random path (or stochastic process). In probability theory, one usually says that the set of paths of $W_{t}$ forms a sample space, and the probability of a path $x(t)$ or other quantities can be defined by seeing them as a functional of $W_{t}$.

The minimization of $I_{W}$ results in a SDE

$$mdv=-m\omega^{2}xdt+\lambda dW_{t},$$

(4)

where $v=dx/dt$ is the velocity. Note that without the random force ($\lambda=0$), the minimization of action gives the Euler-Lagrangian equation. But in the presence of randomness, Eq. (4) is recognized as the Langevin equation. Next we focus on how to quantize this theory.

We start from the action (2) and go through the canonical quantization process. Notice that we keep viewing $W_{t}$ as a classical noise during the quantization, thereafter, our formalism is different from the quantum Langevin equation Breuer02 . First, we perform the Legendre transformation to obtain a Hamilton equation that is equivalent to the equation of motion (4). Since the action is in a quadratic form of velocity and the velocity is not coupled to $W_{t}$, we then define the canonical momentum as $p(t)={\delta I_{W}}/{\delta v(t)}=mv(t)$, where $\delta/\delta v$ is the functional derivative. $p(t)$ is understood as a functional of $W_{t}$ and then a random variable. In the same reason as Lagrangian, the Hamiltonian is not well-defined, instead, we can only define something like its time integral, namely the Hamiltonian integral reading

$$\begin{split}\mathcal{H}&=\int_{t_{0}}^{t^{\prime}}dt\ v(t)p(t)-I_{W}\\ &=\int_{t_{0}}^{t^{\prime}}dt\ \left[\frac{p(t)^{2}}{2m}+\frac{1}{2}m\omega^{2}x(t)^{2}\right]-\lambda\int_{t_{0}}^{t^{\prime}}dW_{t}\ x(t).\end{split}$$

(5)

Again, $\mathcal{H}$ being a random variable must be seen as a functional of $W_{t}$.

Now taking the variation of $\mathcal{H}$ with respect to $x(t)$ and $p(t)$, we obtain

$$\begin{split}\delta\mathcal{H}=&\int_{t_{0}}^{t^{\prime}}\left(dt\frac{p(t)}{m}\right)\ \delta p(t)\\ &+\int_{t_{0}}^{t^{\prime}}\left(dt\ m\omega^{2}x(t)-\lambda dW_{t}\right)\ \delta x(t).\end{split}$$

(6)

In Eq. (6), the bracketed terms describe the change of $\mathcal{H}$ caused by $\delta p(t)$ and $\delta x(t)$, respectively. They play the roles of $\frac{\displaystyle\delta\mathcal{H}}{\displaystyle\delta p(t)}dt$ and $\frac{\displaystyle\delta\mathcal{H}}{\displaystyle\delta x(t)}dt$. The Hamilton equation is traditionally built by letting $\delta\mathcal{H}/\delta p(t)$ and $\delta\mathcal{H}/\delta x(t)$ equal $dx(t)/dt$ and $-dp(t)/dt$, respectively. But in the presence of randomness, we have to avoid using $\delta\mathcal{H}/\delta p(t)$ or $\delta\mathcal{H}/\delta x(t)$ which are not well-defined. Instead, we let the bracketed terms equal $dx(t)$ and $-dp(t)$, respectively, and then obtain

$$\begin{split}&dx(t)=\frac{p(t)}{m}dt,\\ &-dp(t)=m\omega^{2}x(t)dt-\lambda dW_{t},\end{split}$$

(7)

where $dx(t)=x(t+dt)-x(t)$ and $dp(t)=p(t+dt)-p(t)$ are the differentials of canonical coordinate and momentum, respectively. It is easy to verify that Eq. (7) is equivalent to the Hamilton equation of a harmonic oscillator in the case $\lambda=0$. As $\lambda$ is finite, Eq. (7) is equivalent to the equation of motion (4). Therefore, Eq. (7) is the replacement of the Hamilton equation in the presence of random forces, just as the Langevin equation (4) is the replacement of the Euler-Lagrangian equation.

The quantization is to simply replace the canonical coordinate $x(t)$ and momentum $p(t)$ by the operators $\hat{x}(t)$ and $\hat{p}(t)$, respectively. We need these operators satisfy exactly the same equation:

$$\begin{split}&d\hat{x}(t)=\frac{\hat{p}(t)}{m}dt,\\ &-d\hat{p}(t)=m\omega^{2}\hat{x}(t)dt-\lambda dW_{t}.\end{split}$$

(8)

This can be done by defining the commutator $\left[\hat{x}(t),\hat{p}(t)\right]=i$. We choose the unit $\hbar=1$ throughout this paper. And the infinitesimal transformation from $\hat{x}(t)$ or $\hat{p}(t)$ to $\hat{x}(t+dt)$ or $\hat{p}(t+dt)$, respectively, now becomes a unitary transformation expressed as

$$\begin{split}&\hat{x}(t+dt)=e^{id\hat{\mathcal{H}}_{t}}\hat{x}(t)e^{-id\hat{\mathcal{H}}_{t}},\\ &\hat{p}(t+dt)=e^{id\hat{\mathcal{H}}_{t}}\hat{p}(t)e^{-id\hat{\mathcal{H}}_{t}},\end{split}$$

(9)

where $d\hat{\mathcal{H}}_{t}$ is the differential of Hamiltonian integral which equals the Hamiltonian multiplied by $dt$ as $\lambda=0$. By reexpressing the right-hand side of Eq. (5) in a time-discretization form, we easily find

$$d\hat{\mathcal{H}}_{t}=\left[\frac{\hat{p}(t)^{2}}{2m}+\frac{1}{2}m\omega^{2}\hat{x}(t)^{2}\right]dt-\lambda\hat{x}(t)dW_{t}.$$

(10)

Substituting Eq. (10) into Eq. (9) and using the commutator relation, we recover Eq. (8).

Therefore, the canonical quantization still works even in the presence of random forces. But it must be adapted. Due to the non-differentiability of Wiener process, some physical quantities become non-differentiable so that we cannot use their derivatives, instead, we use their differentials. This explains why we do not use the Lagrangian and Hamiltonian which are the derivatives of $I_{W}$ and $\mathcal{H}$, respectively, because $I_{W}$ and $\mathcal{H}$ are non-differentiable as $\lambda\neq 0$.

In quantum mechanics, Eq. (8) or the equivalent Eq. (9) are understood as the equations of operators in the Heisenberg picture. In this picture, the wave function does not change with time. The next step is to choose a reference time $t_{0}$ and change into the Schrödinger picture. We use $\hat{x}(t_{0})\equiv\hat{x}$ and $\hat{p}(t_{0})\equiv\hat{p}$ as the coordinate and momentum operators, respectively, in the Schrödinger picture. They are connected to the operators in the Heisenberg picture by a unitary evolution reading

$$\begin{split}&\hat{x}(t^{\prime})=\hat{U}^{\dagger}(t^{\prime},t_{0})\hat{x}(t_{0})\hat{U}(t^{\prime},t_{0}),\\ &\hat{p}(t^{\prime})=\hat{U}^{\dagger}(t^{\prime},t_{0})\hat{p}(t_{0})\hat{U}(t^{\prime},t_{0}),\end{split}$$

(11)

where, by iteratively using Eq. (9), we find

$$\hat{U}(t^{\prime},t_{0})=\lim_{\Delta t\to 0}e^{-i\Delta\hat{\mathcal{H}}_{t_{0}}}e^{-i\Delta\hat{\mathcal{H}}_{t_{1}}}\cdots e^{-i\Delta\hat{\mathcal{H}}_{t_{N-1}}}$$

(12)

with $t_{j}=t_{0}+j\Delta t$ and $t^{\prime}=t_{N}$. Note that $\Delta\hat{\mathcal{H}}_{t_{j}}$ is the change of Hamiltonian integral over the interval $\left[t_{j},t_{j}+\Delta t\right]$. In this paper, we use the differential symbol $d$ as we want to emphasize an infinitesimal change, but the difference symbol $\Delta$ for a finite change.

The wave function in the Schrödinger picture evolves unitarily according to

$$\ket{\psi(t^{\prime})}=\hat{U}(t^{\prime},t_{0})\ket{\psi(t_{0})},$$

(13)

where $\ket{\psi(t_{0})}$ is the wave function at the reference time which equals the wave function in the Heisenberg picture. As $\lambda\neq 0$, the infinitesimal unitary transformation $e^{-id\hat{\mathcal{H}}_{t}}$ depends on $dW_{t}$, being then time-dependent. We are dealing with an evolution similar to that governed by a time-dependent Hamiltonian. According to the definition (12), an infinitesimal evolution from $t$ to $t+dt$ is $\hat{U}(t+dt,t)=e^{-id\hat{\mathcal{H}}_{t}}$, where $d\hat{\mathcal{H}}_{t}$ in Eq. (10) is expressed in terms of $\hat{x}(t)$ and $\hat{p}(t)$. Using Eq. (11), we replace $\hat{x}(t)$ and $\hat{p}(t)$ by the corresponding operators in the Schrödinger picture, respectively, and then find $\hat{U}(t+dt,t)=\hat{U}^{\dagger}(t,t_{0})e^{-id\hat{\mathcal{H}}^{S}_{t}}\hat{U}(t,t_{0})$, where

$$d\hat{\mathcal{H}}^{S}_{t}=\left(\frac{\hat{p}^{2}}{2m}+\frac{1}{2}m\omega^{2}\hat{x}^{2}\right)dt-\lambda\hat{x}dW_{t}$$

(14)

is the differential of Hamiltonian integral (10) but with all the operators replaced by their time-independent versions in the Schrödinger picture. We call $d\hat{\mathcal{H}}^{S}_{t}$ the Hamiltonian differential in the Schrödinger picture. Using the unitarity of $\hat{U}$ and noticing $\hat{U}(t+dt,t_{0})=\hat{U}(t,t_{0})\hat{U}(t+dt,t)$, we obtain

$$\hat{U}(t+dt,t_{0})=e^{-id\hat{\mathcal{H}}^{S}_{t}}\hat{U}(t,t_{0}),$$

(15)

By iteratively using Eq. (15), we find

$$\hat{U}(t^{\prime},t_{0})=\lim_{\Delta t\to 0}e^{-i\Delta\hat{\mathcal{H}}^{S}_{t_{N-1}}}\cdots e^{-i\Delta\hat{\mathcal{H}}^{S}_{t_{1}}}e^{-i\Delta\hat{\mathcal{H}}^{S}_{t_{0}}}.$$

(16)

The evolution operator is now reexpressed in a form that we are familiar with. In this expression, the time dependence of $\Delta\hat{\mathcal{H}}^{S}_{t}$ comes only from $\Delta W_{t}$ but not from the operators.

From Eq. (15) and (13), we easily derive

$$\ket{\psi(t+dt)}=e^{-id\hat{\mathcal{H}}^{S}_{t}}\ket{\psi(t)}.$$

(17)

This equation tells us how the wave function evolves. The unitarity of $e^{-id\hat{\mathcal{H}}^{S}_{t}}$ guarantees the normalization of wave function in a pathwise way, that is $\braket{\psi(t)}{\psi(t)}=1$ for arbitrary $t$ and path of $W_{t}$. We calculate the differential of wave function. Notice that one must keep the second order terms in the Taylor series of $e^{-id\hat{\mathcal{H}}^{S}_{t}}$, because the exponent contains not only $dt$ but also $dW_{t}$ and $\left(dW_{t}\right)^{2}\sim dt$ is in the first order of $dt$ according to the Itô calculus. The result is

$$\begin{split}d\ket{\psi(t)}=&-idt\left(\frac{\hat{p}^{2}}{2m}+\frac{1}{2}m\omega^{2}\hat{x}^{2}\right)\ket{\psi(t)}\\ &+i\lambda dW_{t}\ \hat{x}\ket{\psi(t)}-\frac{\lambda^{2}}{2}dt\ \hat{x}^{2}\ket{\psi(t)}.\end{split}$$

(18)

In the case $\lambda=0$, Eq. (18) reduces to the Schrödinger equation. As $\lambda\neq 0$, Eq. (18) becomes a SDE which describes a random unitary evolution in the Hilbert space. Notice that $\ket{\psi(t)}$ is a functional of $W_{t}$, being then a random vector.

To see how the random term in Eq. (18) works, we temporarily neglect the Hamiltonian term (the first term in the right-hand side of the equal sign), and then the solution becomes $\psi(t,x)=e^{i\lambda x\left(W_{t}-W_{t_{0}}\right)}\psi(t_{0},x)$, where $\psi(t,x)=\braket{x}{\psi(t)}$ is the wave function in real space. This simple analysis indicates that the random force after quantization tends to contribute a random phase to the wave function.

It is worth comparing Eq. (18) with the collapse model - QMUPL. The difference is that the coupling between $\hat{x}$ and the Wiener process in Eq. (18) is purely imaginary but it is real in the QMUPL model, and there is a lack of the nonlinear term $\bra{\psi(t)}\hat{x}\ket{\psi(t)}dW_{t}$ in Eq. (18). Such a difference leads to a different fate of the wave packet, as will be further discussed in next.

Let us shortly discuss the master equation of density matrix. The density matrix is defined as $\hat{\rho}(t)=\text{E}\left(\ket{\psi(t)}\bra{\psi(t)}\right)$ where $\text{E}\left(\cdot\right)$ denotes the mean over all the paths of $W_{t}$. From Eq. (18), it is straightforward to derive

$$\begin{split}\frac{d\hat{\rho}(t)}{dt}=&-i\left[\frac{\hat{p}^{2}}{2m}+\frac{1}{2}m\omega^{2}\hat{x}^{2},\hat{\rho}(t)\right]\\ &+\lambda^{2}\left(\hat{x}\hat{\rho}(t)\hat{x}-\frac{1}{2}\hat{x}^{2}\hat{\rho}(t)-\frac{1}{2}\hat{\rho}(t)\hat{x}^{2}\right).\end{split}$$

(19)

This is a typical Lindblad equation with the first term in the right-hand side describing the unitary evolution governed by the Hamiltonian and the second term describing the decoherence. Indeed, Eq. (19) is the master equation for single-particle Brownian motion in the macroscopic and high-temperature limit Zurek03 . It is typically employed to describe an environment-induced decoherence when position is the instantaneous pointer observable. To see the decoherence, we temporarily neglect the first term, and then the solution of Eq. (19) becomes

$$\rho_{x,y}(t)=e^{-\frac{\lambda^{2}}{2}\left(x-y\right)^{2}t}\rho_{x,y}(0),$$

(20)

where $\rho_{x,y}(t)=\bra{x}\hat{\rho}(t)\ket{y}$ is the density matrix element in the coordinate space. It is clear that the diagonal elements with $x=y$ keep invariant but the off-diagonal elements decay exponentially with a rate proportional to the squared distance. The model (2), which is originally proposed for demonstrating our approach, can also be used to simulate a master equation which describes the rapid destruction of nonlocal superposition and emergent classicality of the eigenstates of position.

The density matrix gives only a statistical description of the quantum state. To see how a wave packet evolves, we need solve the SDE (18). A possible method is to assume a form of the wave function (such as a Gaussian function), obtain the SDE of undetermined parameters, and write down the solution in terms of Wiener process Bassi05 . This method is difficult to be generalized to QFT. Here we choose a different way - the path integral approach.

Let us calculate the propagator

$$\braket{t^{\prime}x^{\prime}}{t_{0}x_{0}}=\bra{x^{\prime}}\hat{U}(t^{\prime},t_{0})\ket{x_{0}},$$

(21)

where $t^{\prime}>t_{0}$ is an arbitrary ending time. Expressing $\hat{U}(t^{\prime},t_{0})$ in terms of infinitesimal unitary evolutions and utilizing the complete relations in the coordinate and momentum spaces, we obtain

$$\begin{split}\braket{t^{\prime}x^{\prime}}{t_{0}x_{0}}=&\lim_{\Delta t\to 0}\int\left(\prod_{j=1}^{N-1}dx_{j}\right)\left(\prod_{j=0}^{N-1}dp_{j}\right)\\ &\times\prod_{j=0}^{N-1}\left(\braket{x_{j+1}}{p_{j}}\bra{p_{j}}e^{-i\Delta\hat{\mathcal{H}}^{S}_{t_{j}}}\ket{x_{j}}\right),\end{split}$$

(22)

where $x^{\prime}=x_{N}$, $t^{\prime}=t_{N}$ and $\braket{x_{j+1}}{p_{j}}=e^{ip_{j}x_{j+1}}/\sqrt{2\pi}$. The Baker-Campbell-Hausdorff formula tells us

$$e^{-i\Delta\hat{\mathcal{H}}^{S}_{t_{j}}}=e^{-i\frac{\hat{p}^{2}}{2m}\Delta t}e^{-i\left(\frac{1}{2}m\omega^{2}\hat{x}^{2}\Delta t-\lambda\hat{x}\Delta W_{t}\right)}e^{\hat{X}},$$

(23)

where $\hat{X}$ is a series of operators with the lowest order terms proportional to $\Delta t\Delta W_{t}$ or $\Delta t^{2}$, therefore, $e^{\hat{X}}$ can be dropped in the limit $\Delta t\to 0$ according to the Itô calculus. Indeed, in the Itô calculus, except for the linear terms, only $\Delta W_{t}^{2}$ need to be kept in the second order terms, and all the third and higher order terms can be neglected. Integrating out the momentum $p_{0},\cdots,p_{N-1}$, we obtain

$$\begin{split}&\braket{t^{\prime}x^{\prime}}{t_{0}x_{0}}\\ =&\lim_{\Delta t\to 0}\left(\frac{2\pi i\Delta t}{m}\right)^{-N/2}\int\left(\prod_{j=1}^{N-1}dx_{j}\right)\\ &exp\bigg{\{}i\displaystyle\sum_{j=0}^{N-1}\bigg{[}\left(\frac{1}{2}m\left(\frac{x_{j+1}-x_{j}}{\Delta t}\right)^{2}-\frac{1}{2}m\omega^{2}x_{j}^{2}\right)\Delta t\\ &+\lambda x_{j}\Delta W_{t_{j}}\bigg{]}\bigg{\}}.\end{split}$$

(24)

The exponent in Eq. (24) is recognized as $iI_{W}$ in the limit $\Delta t\to 0$. We can then rewrite it by employing the path integral notation:

$$\braket{t^{\prime}x^{\prime}}{t_{0}x_{0}}=\int Dx(t)\ e^{iI_{W}}.$$

(25)

It is encouraging to see that the path integral approach is working and leads to the same Feynman integral formula even in the presence of white noise. In general, if the Wiener process is coupled to the coordinate but not to the velocity, then the definition of momentum keeps the same and the random term can be treated as an additional potential, thereafter, the Feynman’s formula keeps the same.

Next we work out the propagator for the special case $\omega=0$. In Eq. (24), we can see $iI_{W}$ as a quadratic function of the vector $x=(x_{1},\cdots,x_{N-1})^{T}$ and reexpress it as

$$iI_{W}=-\displaystyle\frac{1}{2}\frac{m}{i\Delta t}\ x^{T}Ax+iB^{T}x+C,$$

(26)

where

$$A=\left(\begin{array}[]{cccc}2&-1&0&\cdots\\ -1&2&-1&\cdots\\ 0&-1&2&\cdots\\ \cdots&\cdots&\cdots&\cdots\end{array}\right)$$

(27)

is a $(N-1)$-dimensional tridiagonal matrix,

$$B=\left(\begin{array}[]{c}\lambda\Delta W_{t_{1}}-\displaystyle\frac{m}{\Delta t}x_{0}\\ \lambda\Delta W_{t_{2}}\\ \cdots\\ \cdots\\ \lambda\Delta W_{t_{N-2}}\\ \lambda\Delta W_{t_{N-1}}-\displaystyle\frac{m}{\Delta t}x^{\prime}\end{array}\right)$$

(28)

is a $\left(N-1\right)$-dimensional vector, and $A$, $B$ and $C$ are all independent of $x_{j}$ for $j=1,\cdots N-1$. To calculate the multi-variant Gaussian integral in Eq. (24), we use the formula

$$\begin{split}&\int\left(\prod_{j=1}^{N-1}dx_{j}\right)e^{iI_{W}(x_{1},x_{2},\cdots,x_{N-1})}\\ &=\left(\frac{m}{i2\pi\Delta t}\right)^{-\frac{N-1}{2}}\left(\text{Det}A\right)^{-1/2}e^{iI_{W}(\bar{x}_{1},\bar{x}_{2},\cdots,\bar{x}_{N-1})},\end{split}$$

(29)

where $\text{Det}A$ is the determinant of $A$, and $\left(\bar{x}_{1},\cdots,\bar{x}_{N-1}\right)$ is the stationary point of $I_{W}$ defined by $\partial I_{W}/\partial x_{j}=0$ for each $j$. $\bar{x}(t_{j})=\bar{x}_{j}$ can be also seen as the classical path which minimizes $I_{W}$, and $\bar{I}_{W}=I_{W}(\bar{x})$ is the classical action.

Notice that the Wiener process is linearly coupled to the canonical coordinate in our model. This is a very useful property in the calculation. Because the coefficient matrix $A$ contains no $\Delta W_{t}$, its determinant or inverse are exactly the same as those in the absence of randomness, hence, the familar path integral formula can be directly applied. Usually, the path integral results in a simple expression only if the random term is linear.

By using mathematical induction, we easily prove $\text{Det}A=N$. And according to Eq. (26), the stationary point of $I_{W}$ is $\bar{x}=-\left({\Delta t}/{m}\right)A^{-1}B$. By some tidy calculation, we find that the matrix elements of the inverse of $A$ can be expressed as $A^{-1}_{j,j^{\prime}}=\frac{1}{N}\left[N\text{min}(j,j^{\prime})-jj^{\prime}\right]$ with $j,j^{\prime}=1,\cdots,N-1$ and $\text{min}(j,j^{\prime})$ denoting the smaller one between $j$ and $j^{\prime}$. The stationary point is found to be

$$\begin{split}\bar{x}_{j}=&\frac{(N-j)x_{0}+jx^{\prime}}{N}\\ &-\frac{\lambda\Delta t}{m}\sum_{j^{\prime}=1}^{N-1}\Delta W_{t_{j^{\prime}}}\left(\text{min}(j,j^{\prime})-\frac{jj^{\prime}}{N}\right).\end{split}$$

(30)

The first term is recognized as the position of the particle at time $t_{j}$ if it is moving at a constant velocity from the coordinates $(t_{0},x_{0})$ to $(t^{\prime},x^{\prime})$, as being expected since the classical path without random forces is a straight line in the spacetime. And the second term gives the contribution of the random force to the path. By using Eq. (29) and (30), we finally evaluate Eq. (24) to be

$$\braket{t^{\prime}x^{\prime}}{t_{0}x_{0}}=\sqrt{\frac{m}{2\pi i(t^{\prime}-t_{0})}}e^{i\bar{I}_{W}},$$

(31)

where $\bar{I}_{W}$ reads

$$\begin{split}\bar{I}_{W}=&\lim_{\Delta t\to 0}\bigg{\{}\frac{1}{2}m\frac{\left(x^{\prime}-x_{0}\right)^{2}}{t^{\prime}-t_{0}}+\sum_{j=0}^{N-1}\lambda\Delta W_{t_{j}}\frac{x_{0}(t^{\prime}-t_{j})+x^{\prime}(t_{j}-t_{0})}{t^{\prime}-t_{0}}\\ &-\frac{\lambda^{2}}{2m}\sum_{j,j^{\prime}=0}^{N-1}\Delta W_{t_{j}}\Delta W_{t_{j^{\prime}}}\frac{\left[t^{\prime}-\text{max}(t_{j},t_{j^{\prime}})\right]\left[\text{min}(t_{j},t_{j^{\prime}})-t_{0}\right]}{t^{\prime}-t_{0}}\bigg{\}}\\ =&\frac{1}{2}m\frac{\left(x^{\prime}-x_{0}\right)^{2}}{t^{\prime}-t_{0}}+\lambda\int^{t^{\prime}}_{t_{0}}dW_{t}\frac{x_{0}(t^{\prime}-t)+x^{\prime}(t-t_{0})}{t^{\prime}-t_{0}}-\frac{\lambda^{2}}{2m}\int^{t^{\prime}}_{t_{0}}dW_{t_{1}}\int^{t^{\prime}}_{t_{0}}dW_{t_{2}}\frac{\left[t^{\prime}-\text{max}(t_{1},t_{2})\right]\left[\text{min}(t_{1},t_{2})-t_{0}\right]}{t^{\prime}-t_{0}}.\end{split}$$

(32)

Using the independent-increment property of $W_{t}$, it is easy to see that $\bar{I}_{W}$ depends only upon the difference $t^{\prime}-t_{0}$ but is independent of the initial time. As $\lambda=0$, only the first term of $\bar{I}_{W}$ survives and our result repeats the well-known propagator of a free particle. As $\lambda\neq 0$, the propagator is different from the free-particle one by a random phase that is the combination of second and third terms of $\bar{I}_{W}$.

Using the propagator, we can study the evolution of a general wave packet. Suppose the quantum state is pure with a wave function ${\psi}(t_{0},x_{0})$ at the initial time $t_{0}$, the wave function at an arbitrary later time $t^{\prime}$ is then

$$\psi(t^{\prime},x^{\prime})=\int dx_{0}\braket{t^{\prime}x^{\prime}}{t_{0}x_{0}}\psi(t_{0},x_{0}).$$

(33)

In the study of the decoherence of a single particle, the initial state is usually assumed to be a Gaussian wave packet centered at $q_{0}$ with an averaged momentum $p_{0}$, reading $\psi(t_{0},x_{0})=\left(\pi\sigma_{0}^{2}\right)^{-1/4}exp\left[-\frac{\left(x_{0}-q_{0}\right)^{2}}{2\sigma_{0}^{2}}+ip_{0}x_{0}\right]$ with $\sigma_{0}$ denoting the initial packet width. Let us see how this wave packet evolves in course of time. The propagator being independent of the choice of $t_{0}$ makes the calculation easier (we can simply set $t_{0}=0$). By using Eq. (33), we obtain the expression of $\psi(t^{\prime},x^{\prime})$. We are only interested in the probability distribution of the particle’s position, which is

$$\left|\psi(t^{\prime},x^{\prime})\right|^{2}=\left[\pi\sigma^{2}(t^{\prime})\right]^{-1/2}e^{-\frac{\left(x^{\prime}-q(t^{\prime})\right)^{2}}{\sigma^{2}(t^{\prime})}}.$$

(34)

The packet width $\sigma(t^{\prime})=\sigma_{0}\sqrt{1+\frac{\left(t^{\prime}-t_{0}\right)^{2}}{m^{2}\sigma_{0}^{4}}}$ is increasing linearly with time as the evolution time is much larger than $m\sigma_{0}^{2}$. While the center of packet is at

$$q(t^{\prime})=q_{0}+\frac{p_{0}+p_{1}}{m}(t^{\prime}-t_{0}).$$

(35)

with $p_{1}=\lambda\displaystyle\int_{t_{0}}^{t^{\prime}}dW_{t}\left[1-(t-t_{0})/(t^{\prime}-t_{0})\right]$ being a functional of $W_{t}$. Eq. (35) tells us that the packet center is changing at a velocity ${(p_{0}+p_{1})}/{m}$ which contains a constant part $p_{0}/m$ and a random part $p_{1}/m$ with $p_{1}$ being just the accumulated change of momentum caused by the random force.

It is now clear that the packet is widening at a speed independent of the random force. While the Hamiltonian controls the widening of the wave packet, the effect of the random force is to randomize the position of the wave packet’s center.

The action (2) is the simplest action with Markovian property that after quantization can describe a random unitary evolution in Hilbert space. The quantization techniques developed in this section can be easily generalized to a field theory.

Finally, we discuss the symmetry of model (2). We focus on the time translational symmetry. In general, the path of $W_{t}$ is time-dependent, therefore, the model (2) does not have the time translational symmetry in a pathwise way. For a specific path of $W_{t}$, if we perform the coordinate transformations $\tilde{t}=t+\tau$ and $\tilde{x}=x$, then in the new coordinates $(\tilde{t},\tilde{x})$, the particle’s path becomes $\tilde{x}(\tilde{t})=\tilde{x}(t+\tau)=x(t)$ and the action becomes

$$\begin{split}\tilde{I}_{W}=&\int^{\tilde{t}^{\prime}}_{\tilde{t}_{0}}d\tilde{t}\left[\frac{1}{2}m\left(\frac{d\tilde{x}}{d\tilde{t}}\right)^{2}-\frac{1}{2}m\omega^{2}\tilde{x}^{2}\right]\\ &+\lambda\int^{\tilde{t}^{\prime}}_{\tilde{t}_{0}}dW_{\tilde{t}}\ \tilde{x}(\tilde{t})\end{split}$$

(36)

with $\tilde{t}_{0}=t_{0}+\tau$ and $\tilde{t}^{\prime}={t}^{\prime}+\tau$. The first term of $\tilde{I}_{W}$ is equal to that of $I_{W}$ in Eq. (2), indicating that the action keeps invariant under time translation as $\lambda=0$. But due to $dW_{t}\neq dW_{\tilde{t}}$, the second term of $\tilde{I}_{W}$ is different from that of $I_{W}$, therefore, the time translational symmetry is explicitly broken by the random force.

On the other hand, we should not forget that $dW_{t}$ and $dW_{\tilde{t}}$ are both random numbers with exactly the same distribution, because the Wiener process has stationary independent increments. Indeed, the vector of increments from $t_{0}$ to $t^{\prime}$, i.e. $\left(\Delta W_{t_{0}},\Delta W_{t_{1}},\cdots\Delta W_{t_{N-1}}\right)$ has exactly the same probability distribution as $\left(\Delta W_{\tilde{t}_{0}},\Delta W_{\tilde{t}_{1}},\cdots\Delta W_{\tilde{t}_{N-1}}\right)$ that is the increments from $\tilde{t}_{0}$ to $\tilde{t}^{\prime}$. In probability theory, we say that the two vectors are equal in distribution to each other. As emphasized above, $I_{W}$ is a functional of $\Delta W_{t}$ within the interval $(t_{0},t^{\prime})$, and $\tilde{I}_{W}$ is the same functional but of $\Delta W_{\tilde{t}}$ in the interval $(\tilde{t}_{0},\tilde{t}^{\prime})$. As a consequence, $I_{W}$ and $\tilde{I}_{W}$ are equal in distribution to each other, written as

$$I_{W}\stackrel{{\scriptstyle d}}{{=}}\tilde{I}_{W}.$$

(37)

The action keeps invariant in the meaning of probability distribution. We say that the action has a statistical symmetry.

The statistical symmetry of action has important consequences. First, since the Hamiltonian integral is the Legendre transformation of action, $\hat{\mathcal{H}}$ and then $d\hat{\mathcal{H}}$ are both statistically invariant under time translation. Then the Heisenberg equation (8) must be also statistically invariant, so is the unitary evolution operator $\hat{U}(t^{\prime},t_{0})$. In the Schrödinger picture, the evolution operator $e^{-id\mathcal{H}^{S}_{t}}$ and then the SDE of wave function (18) have the statistical symmetry. And because the density matrix is the expectation value of $\ket{\psi(t)}\bra{\psi(t)}$, its dynamical equation (19) must have an explicit time translational symmetry, as been easily seen. Therefore, a statistical symmetry of action indicates an explicit symmetry of master equation. This conclusion stands for the other symmetries.

By using the path integral method, we have expressed the propagator $\braket{t^{\prime}x^{\prime}}{t_{0}x_{0}}$ as the integral of the exponential of action. It is straightforward to see that the propagator has a statistical time translational symmetry. In other words, the probability distribution of $\braket{t^{\prime}x^{\prime}}{t_{0}x_{0}}$ depends only upon the time difference $t^{\prime}-t_{0}$, being independent of the initial time $t_{0}$. This can be seen in the expression (32).

In a model of random unitary evolution, the conventional symmetry in quantum mechanics is replaced by the statistical symmetry. Once if the action has some statistical symmetry, by using the aforementioned quantization technique, the resulting equation of motion or propagator will have the same symmetry. This theorem can help us to construct a stochastic quantum theory with more complicated symmetries, e.g. the Lorentz symmetry.

Let us start from the simplest Lorentz-invariant action

$$I_{0}=\int d^{4}x\left(-\frac{1}{2}\partial_{\mu}\phi(x)\partial^{\mu}\phi(x)-\frac{1}{2}m^{2}\phi^{2}(x)\right),$$

(38)

where $x=(x^{0},x^{1},x^{2},x^{3})$ is the 1+3-dimensional spacetime coordinates, $\phi(x)$ is a real scalar field and $\partial_{\mu}={\partial}/{\partial x^{\mu}}$. The sign of metric is chosen to be $\left(-,+,+,+\right)$. And we use the units $c=\hbar=1$. In QFT, $I_{0}$ is the action of a free boson of spin-zero and mass $m$.

According to the aforementioned action approach, we need to add to $I_{0}$ a random term without breaking the Lorentz symmetry. And it is reasonable to first try a linear coupling between $\phi$ and some random field. Following Eq. (2), we formally write down the new action as

$$I_{W}=I_{0}+\lambda\int dW(x)\ \phi(x),$$

(39)

where $dW(x)$ is the generalization of $dW_{t}$ in the 1+3-dimensional spacetime and $\lambda$ is the coupling constant. However, the mathematically precise definition of $dW(x)$ is not easy to see. In above, we see $dW_{t}$ as the differential of the Wiener process. One might naively think $dW(x)$ to be also the differential of some stochastic process $W(x)$. Unfortunately, it is difficult if not impossible to find such a $W(x)$. Because there are infinite paths connecting two different points (say $x_{1}$ and $x_{2}$) in a multi-dimensional spacetime. If we see $W(x)$ as a function of $x$, then $W(x_{2})-W(x_{1})$ must be an unambiguously defined random number, and can be obtained by accumulating the infinitesimal increments along arbitrary path connecting $x_{1}$ and $x_{2}$. But there is no way to guarantee that the sum of increments along two different paths are the same, because these increments are random numbers and they should be independent of each other (the Wiener process has independent increments). To further clarify the above idea, let us consider $W(x^{0}+\Delta x^{0},x^{1}+\Delta x^{1})-W(x)$. Here we omit the coordinates $x^{2}$ and $x^{3}$ by simply assuming them to be constants. There are two paths connecting $x$ and $(x^{0}+\Delta x^{0},x^{1}+\Delta x^{1})$ with the intermediate point being $(x^{0},x^{1}+\Delta x^{1})$ and $(x^{0}+\Delta x^{0},x^{1})$, respectively. It is obvious to see

$$\begin{split}&\left[W(x^{0}+\Delta x^{0},x^{1}+\Delta x^{1})-W(x^{0},x^{1}+\Delta x^{1})\right]\\ &+\left[W(x^{0},x^{1}+\Delta x^{1})-W(x)\right]\\ =&\left[W(x^{0}+\Delta x^{0},x^{1}+\Delta x^{1})-W(x^{0}+\Delta x^{0},x^{1})\right]\\ &+\left[W(x^{0}+\Delta x^{0},x^{1})-W(x)\right].\end{split}$$

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But this is impossible if the four increments in the brackets are independent random numbers. Therefore, it is unreasonable to see $dW(x)$ as a differential.