>>91617570you clearly don't understand quantum mechanics enough to even argue about it. The Superposition is not a state, it's the summation of all possible states. again in a two bit system that means the superposition state is 2^2, or 4.
Wikipedia: To better understand this point, consider a classical computer that operates on a three-bit register. If the exact state of the register at a given time is not known, it can be described as a probability distribution over the {\displaystyle 2^{3}=8} 2^{3}=8 different three-bit strings 000, 001, 010, 011, 100, 101, 110, and 111. If there is no uncertainty over its state, then it is in exactly one of these states with probability 1. However, if it is a probabilistic computer, then there is a possibility of it being in any one of a number of different states.
The state of a three-qubit quantum computer is similarly described by an eight-dimensional vector {\displaystyle (a_{0},a_{1},a_{2},a_{3},a_{4},a_{5},a_{6},a_{7})} {\displaystyle (a_{0},a_{1},a_{2},a_{3},a_{4},a_{5},a_{6},a_{7})}. Here, however, the coefficients {\displaystyle a_{k}} a_{k} are complex numbers, and it is the sum of the squares of the coefficients' absolute values, {\displaystyle \sum _{i}|a_{i}|^{2}} {\displaystyle \sum _{i}|a_{i}|^{2}}, that must equal 1. For each {\displaystyle k} k, the absolute value squared {\displaystyle \left|a_{k}\right|^{2}} {\displaystyle \left|a_{k}\right|^{2}} gives the probability of the system being found after a measurement in the {\displaystyle k} k-th state. However, because a complex number encodes not just a magnitude but also a direction in the complex plane, the phase difference between any two coefficients (states) represents a meaningful parameter. This is a fundamental difference between quantum computing and probabilistic classical computing.[13]
