• #### Problem 2, solution

(a)  The z-component of B is constant, while the component of B perpendicular to the z-axis rotates ccw about the z-axis with frequency w.

(b)  H(t)=-gS×B(t)=-gB0Sz-gB1(coswt Sx+sinwt Sy)=w0Sz+w1(coswt Sx+sinwt Sy)

Let U(R) be the rotation operator for a ccw rotation R(f) about the z-axis.

U(R)=exp(-(i/h)Szf)

|y(t)>=U(t,0)|y(0)>,  U(R)|y(t)>=U(R)U(t,0)|y(0)>

U(R) can be viewed as an operator that rotates every state vector ccw through an angle f or an operator that changes the basis vectors and therefore rotates the coordinate system cw through an angle f.

Let f=-wt, then U(R) rotates the coordinate system ccw with angular frequency w.   The coordinate system then rotates with the magnetic field.

Consider an infinitesimal rotation.

U(R)|y(dt)> = (I+(i/h)Szwdt)|y(dt)> = (I+(i/h)Szwdt)(I-(i/h)Hdt)|y(0)>

U(R)|y(dt)> = (I+(i/h)Szwdt)(I-(i/h)(w0Sz+w1(coswt Sx+sinwt Sy))dt)| +>

U(R)|y(dt)> = (I+(i/h)((w-w0)Sz-w1Sxcoswt-w1Sysinwt)dt)| +>

S is a vector observable, its components transform under rotation R as V’=R-1V.

For a ccw rotation R(f) we have Vx’=Vxcosf+Vysinf.

Therefore Sx’=Sxcoswt+Sysinwt is the x-component of the observable S in a frame that rotates ccw with angular frequency w about the z-axis, i.e. in a frame that rotates with the magnetic field.

U(R)|y(dt)>=|y'(dt)>=(I+(i/h)((w-w0)Sz-w1Sx’)dt)|+>

|y'(dt)>=(I+(i/h)((w-w0)Sz’-w1Sx’)dt)|y’(0)>

|y'> denotes the state vector in a frame rotating ccw about the z-axis with angular frequency w.

In this frame the evolution operator is U(t,0)=exp(-(i/h)((w-w0)Sz’-w1Sx’)t).

The Hamiltonian therefore is H’=(w-w0)Sz’-w1Sx’.

The matrix of the Hamiltonian is

in the {|+>, |->} basis.

The eigenfunctions and eigenvalues are

E+=+(h/2)(Dw2+w12)1/2,   |y'+>=cos(q/2)|+>+sin(q/2)|->,

E-=-(h/2)(Dw2+w12)1/2,   |y'->=-sin(q/2)|+>+cos(q/2)|->,

where tanq=-w1/Dw,  sinw=-w1/(Dw2+w12)1/2.

Therefore:

|y’(0)> = |+> = cos(q/2)|y'+>-sin(q/2)|y'->,

|y’(t)> = cos(q/2)exp(-(i/h)E+t)|y'+>-sin(q/2)exp((i/h)E+t)|y'->

= cos(E+t/h)[cos(q/2)|y'+>-sin(q/2)|y'->]-i sin(E+t/h)[cos(q/2)|y'+>+sin(q/2)|y'->]

= cos(E+t/h)|+>-i sin(E+t/h)[cos(q)|+>+sin(q)|->]=b+|+>+b-|->

To find the eigenfunctions in the lab frame we use U-1(R)|y’(t)> = |y(t)>.

exp(-(i/h)Szwt)|y’(t)> = exp(-iwt/2) b+|+>+exp(i/wt/2) b-|-> = |y(t)>.

We now have:

a+(t)= exp(-iwt/2)b+ = exp(-iwt/2)(cos(E+t/h)-i sin(E+t/h)cos(q))

a-(t)= exp(iwt/2)b- = -i exp(iwt/2)sin(E+t/h)sin(q)

Special case:

If w=w0, the Dw=0 and E+=(h/2)w1,  tanq=-w1/Dw=- ¥, q=-p/2, cos(q)=0, sin(q)=-1.

Then

a+(t)= exp(-iwt/2) b+ = exp(-iwt/2)cos((w1/2)t),

a-(t)= exp(iwt/2) b- = i exp(iwt/2)sin((w1/2)t).

(c)  P+-(t)=|a-(t)|2= sin2(E+t/h)sin2(q)=w12/(Dw2+w12) sin2((Dw2+w12)1/2t/2)

For the special case w=w0 P+-(t)= sin2((w1/2)t).  Then P+-(t)= 1 if  (w1/2)t=np/2, n=odd.

(d)  |+> and |-> are both excited states and both decay with probability 1/t per unit time.  If n atoms are excited at t=0, then at time t n exp(t/t) are still in an excited state.  The atoms that are in an excited state flip between |+> and |->.

If an atom is excited at t=0, then the probability of finding it in the state |-> at time t is exp(t/t) P+-(t).  Assume n atoms are excited per unit time. As t goes to infinity we find the total number of atoms in the |-> state by summing up the contributions from all prior time intervals, i.e. by calculating the integral  .

Use .

We multiply the number of atoms thus obtained by their probability of decay 1/t to find the number N which decay from the state |-> per unit time.

N=(n/2)w12/(Dw2+w12+(1/t)2)

N plotted versus Dw is a Lorentz curve with half width L=(w12+(1/t)2)1/2.

Dw=w-w0=w+gB0.