Let us first recall how to reverse a classical diffusion process.

Reverse of Classical Diffusion Process

Let $f:\RR^d\times [0,T]\to \RR^d$ and $\sigma:\RR^m\times [0,T]\to \RR^d$, let the $\RR^d$-valued stochastic process $Y=(Y_s)_{s\in [0,T]}$ be the solution to the SDE

where $B_s$ is a standard $m$-dimensional Brownian motion. Assume that $Y$ has density $p_Y$, which satisfies the Fokker-Planck equation given by

where $D:=\frac{1}{2}\sigma\sigma^T$.

Now consider the reverse time stochastic process \((\overleftarrow{Y}_s)_{s\in [0,T]}\) satisfying

for every $t\in [0,T]$. We ask for a SDE description of this process.

Using the Fokker-Planck equation of $Y_s$, we have

The negative divergence prohibits us from directly viewing the above equation as a Fokker-Planck equation. However, we can write

In this way, once we know $\nabla\log \overleftarrow{p}_Y$, we can cpnvert the negative divergence term to the drift term.

As a special case, if we choose to maintain the original diffusion term, then we obtain the following SDE:

where

Reversing Lindblad Equation

The quantum version of Fokker-Planck equation, also named (Markovian) Lindblad equation, is the following:

where

Here $[A,B]=AB-BA$ is the commutator and ${A,B}=AB+BA$ is the anticommutator. Note that the Hamiltonian $H$ is Hermitian. Here we assume that $H$ and $L_i$’s are not time dependent.

To reverse a Lindblad operator $L$, exactly similar to what we have done in standard diffusion model, what we need to do is solve this equation for $L_b$ and $H_b$, where $H_b$ is Hermitian:

One way to solve for $L_b$ and $H_b$ goes like this. First notice that $\rho$ is Hermitian. Thus it has an eigenvalue decomposition under a suitable set of basis, which I denote by $\rho=\sum_i\lambda_i\ket{i}\bra{i}$ following physicists convention.

Expanding $[H_b,\rho]$ under this basis gives \(\bra{i}[H_b,\rho] \ket{j}= (\lambda_j-\lambda_i)\bra{i} H_b \ket{j}.\) Plugging this expansion into the equation we want to solve, we find that the condition that $H_b$ is Hermitian imposes that \(\cD[L_b](\rho)+\cD[L](\rho)\) should be Hermitian. Moreover, the diagonal elements of \(\cD[L_b](\rho)+\cD[L](\rho)\) should be zero for $H_b$ to be solvable.

Now recall that \(\cD[L](\rho):=L\rho L^\dagger -\frac{1}{2} L^\dagger L\rho -\frac{1}{2} \rho L^\dagger L.\) This is already Hermitian, but the diagonal elements are not all zero.

Now the magic part begins. Consider \(L_b=\rho^{\frac{1}{2}}L^\dagger \rho^{-\frac{1}{2}}.\) Then

Summing and noticing that for Hermitian $A$, operator of the form

is Hermitian with zero diagonal elements.

In summary, the reverse Lindblad’s equation can be expressed by

where

Complex Fokker-Planck Equation

Let $f:\CC^d\times [0,T]\to \CC^d$ and $\sigma:\CC^m\times [0,T]\to \CC^d$, let the $\CC^d$-valued stochastic process $Z=(Z_s)_{s\in [0,T]}$ be the solution to the SDE

where $B_s$ is a standard $m$-dimensional Brownian motion. Assume that $Z$ has density $p_Z$, which satisfies the Fokker-Planck equation given by

Stochastic Unraveling

Unraveling means a stochastic wave-equation that recovers the evolution of density matrices $ρ(t)$ in expectation. More specifically, it looks for a stochastic process $X_t(\omega)$ on the Hilbert space $\CC^d$, such that the expectation $ρ(t):=\EE X_t(\omega)X_t^\dagger (\omega)$ solves the Lindblad equation.

Formally, a density matrix is an expectation of pure states $\ket{\phi}\bra{\phi}$. We seek a SDE that describes the evolution of each $\ket{\phi}$:

where $\xi_j$ are independent complex Brownian motions.

Using Ito calculus,

Noting that

we obtain

Matching this with the Lindblad equation, we require

Qubit Channel

For an arbitary density matrix $\rho$, we have the formula

where

The lindblad equation for quantum qubit channel can be written as

or in the Lindblad’s form as

At $t=0$, $\rho(0)$ has an eigenvalue decomposition $\rho(0)=U\Lambda(0) U^\dagger$, where $U$ is unitary and $\Lambda(0)$ is diagonal. As the qubit channel admits the unitary freedom, we can focus on the case where $U=I$, and the result can be easily generalized to arbitary $U$. Say

where $\lambda_i(0)\geq 0$ for $i\in [2]$ and $\lambda_1(0)+\lambda_2(0)=1$.

At time $t$, solving the Lindblad equation yields

The reversed Hamiltonian can be expressed by

and the reversed Lindblad operators can be expressed by

For general $U$, the reversed Lindblad operators can be obtained by applying the transformation $X \mapsto U^\dagger X U$. Note that the operator $Z$ can actually be dropped.

$d$-dimensional channel

For $d$-dimensional density matrix, notice that $I = \tr(\rho)I$ can be expressed as an operator sum.

Questions

For quantum state preparation, we already know the $\Lambda(0)$ we want and we want to prepare this from the state $I/2$, we do not need to learn anything, just apply the reversed Lindblad operators.

Maybe I have some misunderstanding here for we have discussed to “learn” something via quantum process tomography just like diffusion models.

How to physically implement a given (time-dependent) Lindblad operator ? To prepare a pure state, the reversed Lindblad operators seems to blow up.

For $d$-dimensional channel, it seems thaat $d^2$ Lindblad operators are necessary, which suffers from the curse of dimensionality.

Lindblad Equation as CTMC

Let $L_{mn}=\sqrt{\gamma_{mn}}\ket{n}\bra{m}$ with $\gamma_{mn}\geq 0$.

Then we have

Let $H\propto I$, then the Lindblad equation is equal to

for diagonal terms and

for $i\neq j$.

Notice that we obtain the master equation for diagonal terms.

Reverse

Let us investigate what we obtain if we set $\rho$ to be diagonal so that the evolution is purely classical. For simplicity, we shorten $\rho_{ii}$ by $\rho_i$ in this section. The eigenvalues $\lambda_i$ are exactly $\rho_i$.

The reverse Hamiltonian is

where

is proportional to $\delta_{ij}$, hence the reverse Hamiltonian is equal to zero. The reverse Lindbladian is

Thus the reverse equation is

where