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Note: Cross-posted on Physics SE.


It's a standard piece of quantum information theory that noise can be helpful in augmenting compatibility of quantum observables. For example given a qubit state $\rho$ and the two Pauli observables $\sigma_x$ and $\sigma_y$, it's known that they are incompatible meaning that there is no POVM $\lbrace M_{x, y} \rbrace$ giving two outcome $x, y$ that has $\sigma_x$ and $\sigma_y$ as marginals. This means that by calling $\lbrace E_x \rbrace$ and $\lbrace L_y \rbrace$ the effects of the measurement along $x$ and $y$ respectively we have: $$E_x = \sum_{y} M_{x, y} \quad L_y = \sum_{x} M_{x, y}$$ If a depolarizing noise is acting one the state: $\rho \rightarrow \Phi_{\lambda} \left( \rho \right)$ then the two measurement may become compatible. Indeed in the Heisenberg picture the POVMs are $\Phi^{\dagger} \left( \sigma_x \right)$ and $\Phi^{\dagger} \left( \sigma_y \right)$, this two admit realization as marginals of a joint POVM. Now the question: what are the practical application of this trick? Noise is always present in any process and this says that this very level of noise that you cannot get rid of is at least acting to improve compatibility and you can find a measurement that is better suited to extract information about the system rather than measuring $\Phi^{\dagger} \left( \sigma_x \right)$ and $\Phi^{\dagger} \left( \sigma_y \right)$. But given that you can control the map $\Phi_{\lambda}$ why would you add noise just to improve compatibility (other than for proving it can be done). If you had $N$ qubits you could just as well divide the sample in $\frac{N}{2}$ and $\frac{N}{2}$ and perform different noiseless measurement on the two bunches to extract information (of course this doesn't work if you had just 1 qubit, let alone this case). It doesn't seem to be convinient to spoil the state (and lose information) just to enhance compatibility. Is it so or am I missing something? Can added "noise" (a generic quantum map) be actually useful for retrieving information?

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