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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

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  • $\begingroup$ How is this related to quantum computing? $\endgroup$ – Kiro Apr 24 '18 at 6:11
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    $\begingroup$ @Kiro It is basically asking about how to apply a given operation to a qubit. $\endgroup$ – James Wootton Apr 24 '18 at 11:27
  • $\begingroup$ @JamesWootton I see, that is fine then. $\endgroup$ – Kiro Apr 24 '18 at 12:31
  • $\begingroup$ I asked it on physics but they said to post it here... $\endgroup$ – Connor Dolan Apr 24 '18 at 12:58
  • $\begingroup$ @ConnorDolan "Them" was only one person there ... It would have fitted equally well there, they just don't like double postings, that's why it got closed. $\endgroup$ – Norbert Schuch Apr 24 '18 at 16:22
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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.

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  • $\begingroup$ Good point. I made an edit. $\endgroup$ – James Wootton Apr 24 '18 at 9:31
  • $\begingroup$ Nitpick: Shouldn't the Markov probabilities be $t_{11}+t_{21}=t_{12}+t_{22}=1$? Your answer makes sense to me, but the explanation seems a little artificial. There are natural systems where I shouldn't have to "measure" a result or "flip" a bit in order to obtain the behavior of a Markov chain. I was hoping that classical probability theory would fit in there more seamlessly. $\endgroup$ – Connor Dolan Apr 24 '18 at 13:29
  • $\begingroup$ The explanation is not unique, so you might find something you like better. I have now added a note on this. $\endgroup$ – James Wootton Apr 24 '18 at 14:31
  • $\begingroup$ @ConnorDolan If you want to know how this fits in nicely, you might want to read up on Stinespring dilations, which allow you to implement any CP map by acting with a unitary (or linear map) on a larger system. $\endgroup$ – Norbert Schuch Apr 24 '18 at 16:21
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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, $$ M_1=\left(\begin{matrix}\sqrt{t_{11}}&0\\0&0\end{matrix}\right)\,, \ M_2=\left(\begin{matrix}0&0\\\sqrt{t_{21}}&0\end{matrix}\right)\quad $$ $$ M_3=\left(\begin{matrix}0&\sqrt{t_{12}}\\0&0 \end{matrix}\right)\,,\ M_4=\left(\begin{matrix}0&0\\0&\sqrt{t_{22}} \end{matrix}\right)\ . $$

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  • $\begingroup$ Interesting. Can you explain or give a reference for the intuition of these? $\endgroup$ – Connor Dolan Apr 23 '18 at 23:52
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    $\begingroup$ I tried applying these to rho=|0><0| and didn't get agreement. Perhaps my matrix powers are failing today. Has anyone else had success? $\endgroup$ – James Wootton Apr 24 '18 at 11:40
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    $\begingroup$ @JamesWootton Fair point. I guess it does indeed have to be Kraus rank 4 - at least it feels like every entry of $T$ describes part of a random process. (Is there a proof for that?) $\endgroup$ – Norbert Schuch Apr 24 '18 at 11:45
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    $\begingroup$ @NorbertSchuch I guess there is some freedom in the channel because the original question does not specify the action for non-diagonal density operators, and we can exploit this freedom to reduce the rank. For example, a channel with two Kraus operators $N_1 = M_1 + M_4$ and $N_2 = M_2 + M_3$ does the job (although one could argue it does not really represent the classical process $T$). $\endgroup$ – John Watrous Apr 24 '18 at 19:43
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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.

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    $\begingroup$ Welcome to the Stack Exchange, and thanks for your answer. Would you be able to add more details on how OP might derive the map for the process described? $\endgroup$ – James Wootton Apr 24 '18 at 11:20
  • $\begingroup$ What do you mean by OP? It was already mentioned that there is no unique representation of this process and several of them ware already presented. $\endgroup$ – Filip Wudarski Apr 25 '18 at 11:28

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