A qubit is in state α|0+β|1. Given that α=(√3)/8, choose all possible values of β fromthe given options.(1-√3)/8(1+√3)/8-(√5)/8(√5)/8

The state of a qubit is given by α|0⟩ + β|1⟩, where α and β are complex numbers. However, these numbers are not arbitrary, they must satisfy the normalization condition, which states that |α|^2 + |β|^2 = 1.

Given that α = (√3)/8, we can substitute this into the normalization condition to find the possible values of β.

|α|^2 + |β|^2 = 1 |((√3)/8)|^2 + |β|^2 = 1 3/64 + |β|^2 = 1 |β|^2 = 1 - 3/64 |β|^2 = 61/64

Since β is a complex number, it can have two possible values, +√(61/64) and -√(61/64).

Let's simplify these:

+√(61/64) = (√61)/8 = 7.81024967591/8 = 0.97628095948875 -√(61/64) = - (√61)/8 = -7.81024967591/8 = -0.97628095948875

So, the possible values of β are not in the given options.