Question

Q7R.2 Suppose we have a box that emits electrons in a definite but unknown spin state | ψ). If we send electrons

Answer 1

The electron are in the state 1 phi) the probability p((s_0 = h/2 Vertical-bar 4)) = 3/4 a(so, h/2)) + b(so h/2 Vertical-bar b) = squareroot 3/4 also a(s-x = h/2 Vertical-bar a) + b(s_x^+/b) = roo0.93 if p((s_0 + a)) = cos 62 (alpha - 120) p((s_0^+ /b)) = sin 62 (1206 degree - alpha) and p((s-c^+ /a)) = cos^2 (alpha) p((s_x6 + b)) = sin^2 (alpha) so we need to find alpha replacing (30 in (1) , (2), {a cos(alpha - 1206 degree) + b sin(120 - alpha) = squareroot 3/4 a cos alpha + b sin alpha = squareroot 0.93 {a (squareroot 3/2 sin alpha - cos alpha/2) + b(squareroot 3/2 cos alpha + sin alpha/2) = roo3/2 a cos alpha + b sin alph = 0.964 a^2 + b^2 = 1 we have three equations and three unknowns to be solved using b = squareroot 1 - a^2 (a squareroot 3/2 + squareroot 1 - a^2/2) sin alpha + (squareroot 3(1 - a^2)/2 - a/2) cos alpha = squareroot 3/2 but from (5) & (4) a = 0.6064 b = 0.7951 alpha = 0.6515 radian = 37.32 degree