**Authors:** Masataka Ohta

Consider four-dimensional Hilbert space H over ℂ as a direct product of two two-dimensional Hilbert spaces over ℂ, in which two binary quantum states are represented, respectively, that is, |00 = (1, 0, 1, 0), |10> = (0, 1, 1, 0), |01> = (1, 0, 0, 1) and |11> = (0, 1, 0, 1). Then, |00>+|11> = |01> + |10> = (1, 1, 1, 1). Existence of such linear dependency is obvious considering three binary quantum states in six dimensional Hilbert space, because there are eight combinations of 0 and 1 in a six dimensional space. Moreover, though it may be surprising that basis set {|00>. |10>, |01>. |11>} is not enough to cover H, considering degree of freedom of quantum state spaces represented in four and two dimensional Hilbert spaces over ℂ counted by ℝ are 7 and 3, respectively, and 7-3*2=1, it is also obvious. The natural basis set of H is {(1, 0, 0, 0), (0, 1, 0, 0), (0, 0, 1, 0), (0, 0, 0, 1)}. Considering practical communication with two binary quantum channels, (1, 0, 0, 0), which is a valid quantum state even traditionally, should mean that the first two-dimensional Hilbert space represent |0> is sent and the second one represent no photons sent, that is, vacuum . Moreover, (0, 0, 0, 0) should also represent a valid quantum state that no photons are sent in either channel, that is, total vacuum. Violation of Bell’s inequality not by quantum entanglement is discussed in a separate paper.

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[v1] 2018-02-14 09:11:03

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