# Why is entanglement so important if it introduces dependent information?

I still don't understand, why is entanglement such a crucial property of a quantum algorithm. If I understand it, it means that the information between different qubits is somehow correlated, which means redundant, at least in statistics.

• Correlations are not redundant in statistics. To measure anything, one must use correlations to establish connections between the measurement devices and the actual parameters being measured. Without correlations, you could never use the readon on a thermometer to tell you anything about the temperature. The whole point of entanglement is how to use correlations in a clever way. What is redundant is measuring two separate variables that are 100% correlated (which may be useful for checking results, but I agree it is technically redundant). Commented Dec 16, 2021 at 21:52

If you consider a classical random process, like the flipping of a coin (in which "heads" is labeled by $$|0\rangle$$ and "tails" is labeled by $$|1\rangle$$), then spinning that coin can in many ways simulate a quantum superposition like $$\left(\frac{1}{\sqrt{2}}|0\rangle + \frac{1}{\sqrt{2}}|1\rangle\right)$$ since there's a 50%, or $$|\frac{1}{\sqrt{2}}|^2$$, probability of getting "heads" or "tails" (a quantum superposition can be more general than that, and even have imaginary numbers for the coefficients, but even then it's possible to model their current state using classical statistics).
Entanglement is more complicated than that, in a fundamental way. It means that if we have two systems which can each exist in states $$|0\rangle$$ or $$|1\rangle$$, then you cannot write the state as $$|0\rangle|b\rangle$$ or $$|1\rangle|b\rangle$$ for any state $$|b\rangle$$ even if $$|b\rangle$$ is in a superposition. This can be modelled by a classical computer, but in general, the cost would be exponential in the number of members in the entangled state.