No, the state \(|11\rangle \otimes |11\rangle\) is not an entangled state; it's actually a separable or product state. This means that it can be written as a tensor product of the individual states of each qubit, and the state of each qubit can be described independently of the other.
In a truly entangled state, like the Bell states, it becomes impossible to write the joint state as a simple tensor product of individual qubit states. This results in the condition where measuring one qubit will immediately give you information about the other, regardless of the distance between them.
In the state \(|11\rangle \otimes |11\rangle\), each qubit is already in a well-defined state of its own and measuring one doesn't provide any information about the other. This is unlike entangled states where the whole point is that the qubits are not in well-defined individual states but are correlated in such a way that the state of one depends on the state of the other.
To summarize, \(|11\rangle \otimes |11\rangle\) is not an entangled state, and the state of each qubit can be described independently of the others.
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