Need help with qn5 (qn4 is just for reference) At time t = 0, suppose that the w

Question

Need help with qn5 (qn4 is just for reference)

At time t = 0, suppose that the wavefunction of the spin of an electron is given by (psi_z (s_z = 1/2, t = 0) psi_z (s_z = -1/2,t = 0)) = (1 0) Under an external magnetic field in the x circ direction, it can be shown that the wavefunction evolves in time according to the equation (psi_z (1/2, t) psi_z (-1/2, t)) = (cos (omega t) -j sin(omega t) -j sin (omega t) cos (omega t)) (psi_z (1/2,0) psi_z (-1/2,0)) where omega is a constant depending on the magnitude of the magnetic field. Assuming that omega > 0, what is the minimum time needed for the spin to flip to a definite value of s_z = 1/2, i.e., what is the minimum t such that |psi_z (1/2, t)|^2 = 0, |psi_z (-1/2, t)|^2 = 1? (A) t = 0 (B) t = pi/(2 omega) (C) t = pi/omega (D) t = 3 pi/(2 omega) (E) t = 2 pi/omega (Time evolution, incompatible observables) Consider the same setup described in Q4. If I measure the y component of the spin at a time t, what is the probability distribution of my measurement outcome? (A) P(s_y = 1/2, t) = 1/2, P(s_y = -1/2, t) = 1/2 (B) P(s_y = 1/2, t) = 1, P(s_y = -1/2, t) = 0 (C) P(s_y = 1/2, t) = 0, P(s_y = -1/2, t) = 1 (D) P(s_y = 1/2, t) = cos^2 (omega t), P(s_y = -1/2, t) = sin^2 (omega t) (E) P(s_y = 1/2, t) = [1 - sin(2 omega t)]/2, P(s_y = -1/2, t) = [1 + sin (2 omega t)]1/2

Explanation / Answer

Ans:- As per Question no. 4 the time revolution we are taken the t =0 & t =1 in wave function of spin electron we are given the external magnetic field x direction. is a magnitude of magnetic field. so as per magnetic dipole moments the charge particle is minimum time =0 and max time is 1 so our time is t = 3 pie /2 as relation of dipole moment

= (q/2m) S

Assume is dipole moment

q (Charge electro or particle)

m ( mass)

S (angular momentum)

arising of the spinning.

Psy( sy = 1/2 ) = gq /2m S      Psy( sy = 1/2 = (gq h /2mh) S                                                   (1).

Here qh/(2m) incompatible observables units of dipole moment.

For an electron g = x and so the particle or electron charge is -ve

so we get Sz= -1/2 = e = 2B /h

The dipole moment are ant parallel of angular momentum

If the particle is insert in a magnitude magnetic field · B the Hamiltonian spin system

HS = · B = B· S = (BxSˆx + BySˆy + BzSˆz).                                                                                                               (2)

know we consider a magnetic spin pointing at time=0 to time =1 along the direction special angles by

(0,0) : (0 0)

P (sy = 1/2 ,t) |, 0) = cos |+)+ sin ei0 |)

sp given HS = B Sˆz + [ [ 1+sin 2t] /2

HS|±)= Bn

Then |,t) = e = e (cos)+ sin e(i0 |)

        P (sy = 1/2 ,t) =cos2(t)= cos e(iHSt/n|+)+ sin ei0eiHSt/n)

                      cos2(t)= |,t) = e cos |+)+ sin e +iBt/2i(0Bt)                                                          (3)

so overall phase relevant to the spin state of recognize as the spin state related to the vector jn(time)

defined by angles (t)= ( 0 , 1)

(t)= 0 Bt .

so that our main important fact electromagnetic theory those theory we need to be solve a magnetic field of dipole moment so our torque experiences given by .

= × B.

so our the familiar mechanics equation are related to the different rate of change of angular momentum are same or equal to the torque

dS = B× S.

Hamiltonian spin of a general spin magnitude of magnetic field

HS = · B = B· S = L · S.

so that our this related to g = p(E)2 +2 and so that the energy eigen probability distribution of outcome are

EE(E)= E0 + p0 + E3 +3

EG(E)= E0 p1 + E3 +3                                                                                                      (4)

so as equation to be find (1),(2),(3) & (4) is related to the probability distribution of spin momentum so our Ans is (E) . those could be find the all four relation.

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