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I'd like to revisit a question I posted earlier. While I now know what coherence means (and how it can be defined) I still struggle to give an intuitive explanation that goes along with the word "decoherence". Including I want to look at this from a discrete point of view - so only by using qubits. This makes the definition of "coherence" much harder, as it's not really possible to talk about interference of probability distributions. My goal is to explain why it is called "decoherence" with the "boundaries" I just explained.

In the end, the explanation of coherence should arrive at a point where it links to (or where coherence even is equal to?) the property of states to be in superpositions of (classically distinct) states - because those are the ones that "vanish" during the decoherence process.

I tried to use Baumgratz, Cramer, Plenio's definition (see equation 1 here) of incoherent states to characterize coherence, but the considered article doesn't mention decoherence which makes me step back from it.

How is coherence defined in a (plain) discrete context? As the ability of systems to have superposition states? As this is clearly basis-dependent, is this even a proper definition?

Qmechanic
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manuel459
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1 Answers1

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Imagine a generic pure state$^\dagger$, which is a superposition of some basis states \begin{equation} |\Psi\rangle = \sum_{k=1}^N c_k | k \rangle \end{equation} To be concrete, we can take the basis states $|k\rangle$ to be the computational basis with $N$ qubits. For example, for $N=2$, $|k\rangle$ would be a member of the set $\{|00\rangle, |01\rangle, |10\rangle, |11\rangle\}$.

Now, let's write the complex coefficients as a real amplitude and phase: $c_k = a_k e^{i \varphi_k}$. So \begin{equation} |\Psi\rangle = \sum_{k=1}^N a_k e^{i \varphi_k} |k\rangle \end{equation} Now consider the probability for the state to be $|m\rangle$ after collapse (here $|m\rangle$ is an eigenstate of some observable which isn't diagonal in the computational basis) \begin{eqnarray} P_m = |\langle m | \Psi \rangle|^2 &=& \sum_{k=1}^N \sum_{k'=1}^N \langle m | k \rangle \langle k' | m \rangle a_k a_{k'} e^{i(\varphi_k - \varphi_{k'})} \\ &=& \sum_{k=1}^N a_k^2 \langle m | k\rangle + \sum_{k=1}^N \sum_{k'=1, k'\neq k}^N a_k a_{k'} e^{i (\varphi_k - \varphi_{k'})} \langle m | k' \rangle \langle k | m\rangle \end{eqnarray} If the phases $\varphi_k$ are uncorrelated, then you would expect the second term above to tend to cancel, particularly if $N$ is large. This is essentially because you are adding complex numbers with roughly the same amplitude but with random phases. In the limit where the cancellation is perfect, then the probability becomes \begin{equation} P_m = \sum_{k=1}^N a_k^2 \langle m | k\rangle \end{equation} This cancellation is the key idea of decoherence. The probabilities for different basis states simply add, as they would in normal probability theory.

On the other hand, a coherent state is one where the phases are correlated, For example, if $\varphi_k=\varphi_{k'}$ for all $k, k'$, then there will not be any cancellation in the second term at all. The interference term, which only appears in quantum mechanics and not ordinary probability theory, is crucial for correctly calculating the probabilities in this case. The computational advantage of quantum computing comes from arranging these correlations in a smart way.

Therefore I would not say that coherence is not equivalent to the superposition principle. I would say that coherence and decoherence are descriptions of different types of superposition, depending on the phase coherence of the expansion coefficients.


$^\dagger$ Of course you can also formulate this for mixed states in terms of density matrices, but I personally find it useful to understand the pure state case first.

Andrew
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  • Thank you for this answer. I learned a lot from it. The only thing I don't get is how the measurement process you described (with collapse and the cancellation of the interference terms in the probability) is comparable to the decoherence process. There no collapse occurs, one just switches to another point of view (the view of an observer only capable to access a subsystem) by tracing out the environment. Also, the subsystem has no state $|\psi\rangle$ but is in an entangled state with the environment. – manuel459 May 04 '22 at 13:24
  • @manuel459 I'll try to add some more details later, but very briefly: You can think of the whole system (environment+qubits) as being in a pure state if you want, in which case my answer applies with very large $N$ (so the cancellation is very good). You can also phrase it in density matrix language, the idea is that the off diagonal terms in the density matrix tend to cancel if the phases aren't correlated, for essentially the same reason as in the example I gave. After tracing over the environment variables, you'll be in a mixed state with a very close to diagonal density matrix. – Andrew May 04 '22 at 13:46
  • Given one thinks of the whole system (environment+qubits). Why does your explanation need a collapse to make the coherences dissapear? I always thought this could also be seen by just looking at the reduced density matrix? – manuel459 May 04 '22 at 14:12
  • It might be completely wrong but after studying your explanation thouroughly i come to the conclution that the interaction with the environment somehow changes the phase correlation... is that correct? – manuel459 May 04 '22 at 15:05
  • @manuel459 That's a good way to look at it. The coupling with the environment causes the phases of the qubits to become correlated with the phases of the environment variables, instead of with each other (since there are so many more environment variables). Then when you trace over the environment variables, you get a density matrix that is very close to diagonal. – Andrew May 04 '22 at 16:10