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By H. S. W. Massey

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But the angles q q1 and q1 q are equal, and by the assumptions of our lemma, it then follows that 2w (q) = w (q1 ) . There is a line L q , perpendicular to pq at a point r , passing through q , and a right triangle that can be constructed from r to the apex q2 to q, with the line r pq as hypotenuse. To satisfy Pythagoras’s theorem, we see that the distance pr is λ1 . pq2 is unity (this line is orthogonal to qp). The distance qq2 is 1 + λ, and the distance rq2 is 1 + λ1 . Finally, q r is λ − λ1 .

1 Introduction Now we will turn to some elementary and familiar examples of quantum mechanics to remind us of matters which will be used in the subsequent discussions. , 1977; Jordan, 1986; Liboff, 1998). 2 Harmonic oscillator The Hamiltonian operator is 1 2 pˆ + ω2 qˆ 2 = Hˆ † . Hˆ = 2 The classical equations of motion are ∂H dq = =p dt ∂p ∂H dp =− = −ω2 q. 2) In quantum mechanics, q, ˆ pˆ = i h¯ . 3) The “hat” denotes operator. The time-dependent Heisenberg equations are of the same form as the classical counterpart: d q(t) ˆ = p(t) ˆ dt d pˆ (t) = −ω2 qˆ (t) .

47) and σ 21. 48) = 0. 49) ≡ |−1 . The α state is spin positive (m s = +1) along the “3” direction, and β spin down (m s = −1). Generally, |ψ = aα + bβ = a |+1 + b |−1 a + b2 = 1. 3 Spin one-half and two-level atoms 29 the familiar Pauli matrices. Continuing, we find 0 1 0 0 σ+ = 0 0 1 0 , σ− = . 53) σ + |−1 = |+1 σ − |+1 = |−1 σ − |− = 0. ” Recall that σ and I form a complete set of 2 × 2 matrices. e. ρ= 1 [a0 I + r · σ ] . 54) The coefficients may be written a0 = Trρ ri = Trρσ i . The above operators have been written in the Schrödinger picture.

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