In Example 4.3:
(a) If you measured the component of spin angular momentum along the x direction, at time t, what is the probability that you would get +ћ/2?
(b) Same question, but for the y component.
(c) Same, for the z component.

Figures 4.14

Transcribed Image Text:

Example 4.3 Larmor precession: Imagine a particle of spin 1/2 at rest in a uniform magnetic field, which points in the z-direction: B = Bok The Hamiltonian (Equation 4.158) is у Bol (9) The eigenstates of Hare the same as those of S.: H=-y BoS; = X+₁ with energy E+ = -(y Boh) /2, X-. with energy E-=+(y Boh)/2. ax =Hx. ar can be expressed in terms of the stationary states: ih The energy is lowest when the dipole moment is parallel to the field-just as it would be classically. Since the Hamiltonian is time independent, the general solution to the time-dependent Schrödinger equation, X(1) = ax+e-¹₁/h +by_e-1E-1/h = - (be-ty Bot 12) The constants a and bare determined by the initial conditions: - (8). x(0) = X(t) = (Sx) = X(t) SxX(t) = (cos (a/2)e-ty But/2 sin (a/2)¹y the/2) / (1 o) (sin (1/2)e-ty Box 12) (of course, la² + b² = 1). With no essential loss of generality I'll write a = cos (a/2) and b = sin (a/2), where a is a fixed angle whose physical significance will appear in a moment. Thus ħ = sin a cos (y Bor). 2 Similarly, cos (a/2) ely Bal/2 sin (a/2) e-ly Bot/2). To get a feel for what is happening here, let's calculate the expectation value of S, as a function of time: (S₂) = x(1) S,x() = and (S₂) = x(1) S₂x (1) = h 2 ħ (4.159) 2 (4.160) sin a sin (y Bot). (4.161) cos a.. (4.162) (4.163) (4.164) (4.165) (4.166) Thus (S) is tilted at frequency w = y Bo. (4.167) just as it would classically (see Figure 4.14). No surprise here-Ehrenfest's theorem (in the form derived in Problem 4.23) guarantees that (S) evolves according to the classical laws. But it's nice to see how this works out in a specific context. constant angle a to the z axis, and precesses about the field at the Larmor