If I have an entangled state shared between two parties Alice and Bob $$\frac{1}{\sqrt{2}}|00\rangle+\frac{1}{\sqrt{2}}|11\rangle....(1)$$ then the reduced density operator of Alice's side is $\frac{I}{2}$, that is Alice's qubit is in mixed state. But being in a mixed state is a classical phenomena. By classical phenomena I mean when we write density operator for mixed state that can be in either of $\{|\psi_i\rangle\}$ states with probability $p_i$, we know that system is exactly one of the states but just don't know which .Such uncertainties are due to defects in device etc is a classical phenomena. So is entanglement due to the reason Alice's/Bob's reduced density operator is mixed ( which is a classical phenomena if I am not wrong )?
Sorry if I could not put this more properly in words.


In short: No, your reasoning is backwards. The reduced density matrix is mixed because of entanglement and not vise versa.

If two systems A and B are in an entangled state, we can no longer separate the state description and write down individual state vectors for system A and B, i.e:

$$ |\psi\rangle_A \otimes |\psi\rangle_A $$

Remember that a state is just a mathematical object which allows to predict the expectation values for any observable and the probabilities for any measurement outcome. The reduced state for system A (or B) is the state which allows one to calculate the value of any observable which acts solely on system A, in other words, for any measurement of the properties of system A which ignores system B. The reduced state is obtained by an operation called partial trace. After performing the partial trace, you end up with a mixed state, as you yourself has said. This is because some of the information of the composite system is lost during the partial trace (well, we are "tracing it out"!) and hence we end up with a mixed state as our final answer.

So really, entanglement is not a classical phenomenon, instead, the reduced density matrix of either Alice or Bob is a mixed state for the reasons described above.


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