Markov Chain expressed in Density Matrix formalism

Question

Suppose we have two states of a system where I tell you that there is a probability $p_1$ of being in state $1$, and probability $p_2$ of being in state $2$. The total state can be written as a vector in $L^1$ normed space:

$$p=\begin{pmatrix}p_1 \\ p_2 \end{pmatrix}, ||p||=p_1+p_2=1$$

If we define a transition matrix for a Markov process:

$$T=\begin{pmatrix}t_{11}&t_{12} \\ t_{21}&t_{22}\end{pmatrix}$$

Then the next state would be:

$$p'=Tp=\begin{pmatrix}t_{11}p_1+t_{12}p_2 \\ t_{21}p_1+t_{22}p_2\end{pmatrix}$$

Now my understanding of density matrices and quantum mechanics is that it should contain classical probability theory in addition to strictly quantum phenomena.

Classical probabilities in the density matrix formalism are mapped as:

$$p=\begin{pmatrix}p_1 \\ p_2 \end{pmatrix} \rightarrow \rho=\begin{pmatrix}p_1&0 \\ 0&p_2 \end{pmatrix}$$

And I want to obtain:

$$p'=\begin{pmatrix}t_{11}p_1+t_{12}p_2 \\ t_{21}p_1+t_{22}p_2\end{pmatrix} \rightarrow \rho'=\begin{pmatrix}t_{11}p_1+t_{12}p_2&0 \\ 0&t_{21}p_1+t_{22}p_2\end{pmatrix}$$

My attempt:

Define an operator $U$ such that:

$$\rho'=U\rho U^\dagger$$ $$\implies \begin{pmatrix}t_{11}p_1+t_{12}p_2&0 \\ 0&t_{21}p_1+t_{22}p_2\end{pmatrix}=\begin{pmatrix}u_{11}&u_{12} \\ u_{21}&u_{22}\end{pmatrix}\begin{pmatrix}p_1&0 \\ 0&p_2\end{pmatrix}\begin{pmatrix}u_{11}^*&u_{21}^* \\ u_{12}^*&u_{22}^*\end{pmatrix}$$

$$=\begin{pmatrix}|u_{11}|^2p_1+|u_{12}|^2p_2&u_{11}u_{21}^*p_1+u_{12}u_{22}^*p_2 \\ u_{21}u_{11}^*p_1+u_{12}^*u_{22}p_2 & |u_{21}|^2p_1+|u_{22}|^2p_2 \end{pmatrix}$$

Evidently, $|u_{ij}|^2=t_{ij}$, but the off diagonal terms aren't easily made zero, (I've wrestled with the algebra and applied all the proper normalizations of probability theory).

What would be the correct way to apply a Markov process in the density matrix formalism? It seems really basic and something that this formalism should be able to naturally handle.

Edit: Repost of : repost

Answer 1

The most general quantum evolution is a completely positive (CP) map: $$ \rho\mapsto \mathcal E(\rho) = \sum_i M_i\rho M_i^\dagger \ . $$ Here, $$ M_1=\left(\begin{matrix}\sqrt{t_{11}}&0\\0&0\end{matrix}\right)\,, \ M_2=\left(\begin{matrix}0&0\\\sqrt{t_{21}}&0\end{matrix}\right)\quad $$ $$ M_3=\left(\begin{matrix}0&\sqrt{t_{12}}\\0&0 \end{matrix}\right)\,,\ M_4=\left(\begin{matrix}0&0\\0&\sqrt{t_{22}} \end{matrix}\right)\ . $$

Answer 2

The easiest way to get rid of off diagonal elements is to measure. You could then apply some post-measurement unitaries which depend on the result, as well as some classical randomness. Clearly you need more than just a unitary to apply such a process. Instead you'll need a more general CP map, as mentioned by the other answers.

I am going to assume that $t_{11}+t_{12}=t_{21}+t_{22}=1$ in what follows, so that the Markov process is also trace preserving.

Let's say you measure in the $\{|1\rangle, |2\rangle\}$ basis, but don't look at the result. The state is then described by

$$ P_1 \, \rho \, P_1 + P_2 \, \rho \, P_2. $$

Now let's consider a different process. Suppose you applied a random process that flipped the states with probability $t_{xy}$, and did nothing with probability $t_{xx}$. The state is then

$$ t_{xx} \, \rho + t_{xy} \, X \,\rho \, X $$ where here I use $X$ to denote the unitary that performs the flip.

What you want is the process that combines the two. First measure, and then apply the random flip with probabilities that depend on the results. The map you need is then

$$ t_{11} \, P_1 \rho P_1 + t_{12} \, X P_1 \, \rho P_1 \, X \,+ t_{22} \, P_2 \rho P_2 + t_{21} \, X \, P_2 \rho P_2 \, X \, . $$

This can be expressed as as the CPTP map $$ \rho\mapsto \mathcal E(\rho) = \sum_i M_i\rho M_i^\dagger \ . $$

with $M_{11} = \sqrt{t_{11}} P_1$, $M_{12} = \sqrt{t_{12}} X P_1$, $M_{22} = \sqrt{t_{22}} P_2$ and $M_{21} = \sqrt{t_{21}} X P_2$.

Note that this expression is not unique. The interpretation of how to do it is not unique either. Measurements are not necessarily required, but it can never be as simple as just a unitary.

To see why, note that your Markov process changes the value of $\rm{tr}(\rho^2)$. This cannot be done by a single qubit unitary, because

$$ (U \rho U^{\dagger})^2 = U \rho U^{\dagger} U \rho U^{\dagger} = U \rho^2 U^{\dagger},$$

and the trace is unitary invariant. A process that changes this value requires either measurement, or interaction with an external system.

Answer 3

As mentioned by Norbert Schuch, the most general quantum operation (i.e. preserving quantum mechanical interpretation) is completely positive and trace-preserving map (CPTP). You can find more on this subject in the field of open quantum systems (these maps carry information on the interaction with the environment) - see the book: H-P. Breuer, F. Petruccione "The theory of Open quantum systems".

One reason why it's not easy (only in special cases) to express your stochastic process as unitary transformation is that the transition matrix needs not be invertible (and if it is, it is not necessarily stochastic matrix), while unitaries are always invertible and describe the legitimate quantum transformation.