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.)


1 Answer 1


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).

  • $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.