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To distinguish between a coherent and de-cohered stage of the same system what experiments can provide the answer? The term Experiment is used here in the Bohr-Einstein-debate sense, a realizable physically sensible procedure, not necessarily practical in current labs.

Assumptions: Suppose the system initially is in a true superposition state |ψ⟩ = 1/√2(|0⟩+|1⟩), thus its density matrix initially has coherences (off diagonal) terms.

Also assume that If we decide to switch-on an external auxiliary apparatus, and bring it into interaction with our system then, due to the interaction with that auxiliary apparatus its density matrix becomes \begin{bmatrix}0.5 & 0 \\ 0 & 0.5\end{bmatrix} We'll call that new stage 'the system has decohered'.

The question is - what experimental test can determine whether the system has decohered or not.

(*As the model-system any convenient physical 2- state model can be used i.e. spins, energy levels etc.)

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So basically you want to distinguish the state $| + \rangle \langle + | $ from the dephased state $\frac{1}{2}(| 0 \rangle \langle 0 | + | 1 \rangle \langle 1 | ) = \frac{\mathbb{I}}{2}$.

Here's a simple experiment: apply a Hadamard to both states and then measure in the $\sigma_{z}$ basis. For the ``true superposition'', this transforms it into the state $| 0 \rangle \langle 0 | $ and so we get the output $0$ with probability 1 (when measured in the $\sigma_{z}$ basis). The mixed state on the other hand is unitarily invariant and therefore yields probabilities $(\frac{1}{2}, \frac{1}{2})$ in any basis (inclusing the $\sigma_{z}$ basis).

Update: This process can be generalized to other states and bases -- the idea is to transform to a basis where the coherence in the pure state manifests itself (while the incoherent state transforms into something else; allowing them to be distinguished).

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  • $\begingroup$ @MusashiK Algebraically, you want to know the trace of \rho^2. Where \rho is the density matrix. For a pure state the value will be 1, for a mixed state it will be less than one. For mixed state in this case, \rho^2 = 1/4 * (|0><0| + |1><1|) . The trace is 1/2. $\endgroup$
    – QurakNerd
    Commented Jul 2, 2020 at 19:29
  • $\begingroup$ @QurakNerd Is it possible to formulate Experiment that measures $Tr(\rho^2)$ ? $\endgroup$
    – MusashiK
    Commented Jul 2, 2020 at 19:52
  • $\begingroup$ @keisuke.akira Thank you very much for the answer. You have provided two methods - a specific for the concrete mixture, and a general suggestion.Your solution of the specific case utilized the specific property of THIS particular decohered -dephased st. - namely that being a multiple of unit matrix it is invariant. Following your suggestion - dephased states are (only) diagonal in some basis, so what is the common property that would DISTINGUISH them algebraically? I want to stress, that I ask as an appreciation of your solution, and suggestion (not in any other sense) $\endgroup$
    – MusashiK
    Commented Jul 2, 2020 at 19:55
  • $\begingroup$ #1: I'll give two answers: the general way to distinguish two states and the specific case where one of them is decohered. In the most general case, one would need to do quantum state tomography -- that is, measure the state in many different bases and then (experimentally) reconstruct the states to find that they are indeed different. For example, given two qubit states $\rho, \sigma$, to distinguish them, you'd need to measure them in the bases $\sigma_{x}, \sigma_{y}, \sigma_{z}$ and then you can distinguish any two different states. $\endgroup$ Commented Jul 3, 2020 at 8:41
  • $\begingroup$ #2: In the specific case where you know that one of the states is decohered, things can simplify a bit (for example, in the original question). What is the difference between a dephased state and one with off-diagonal elements? This is what is called quantum coherence. Given a fixed basis (for example, the $\sigma_{z}$ basis for a single qubit), all states that are diagonal in this basis are called ``incoherent'' states, while those that are not diagonal are called coherent states. $\endgroup$ Commented Jul 3, 2020 at 8:44

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