Time Dependent Perturbation Theory) Dipan Kumar Ghosh UM-DAE Centre for Excellence in Basic Sciences Kalina, Mumbai November 15, 2018

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Quantum Mechanics-II Approximation Methods : (Time Dependent Perturbation Theory) Dipan Kumar Ghosh UM-DAE Centre for Excellence in Basic Sciences Kalina, Mumbai November 15, 2018

1 Introduction

Till now we have discussed systems where the Hamiltonian had no explicit time dependence. We will now consider situation where a time dependent perturbation V (t) is present in addition to the time independent Hamiltonian H0, the solution of the latter problem is known to us and its eigenstates are given by H0 | φni = En | φni (1) In the absence of the perturbation, if the system initially happened to be in a particular eigenstate —| ψni of H0, it would continue to be in that state for ever, apart from picking up an unimportant phase factor eiEnt/~. A time dependent perturbation would change that and the system would no longer in that eigenstate but be found in some state | ψ(t)i. Since any arbitrary state can always be expressed as a linear combination of the complete set of eigenstates {| φni} of H0, it implies that the perturbation induces transition between diﬀerent eigenstates. The time evolution operator is no longer exp iHt/~, as would be the case if H were independent of time. The question that we address is what is the probability of transition from the given initial state | ψni to diﬀerent eigenstates ψm of the Hamiltonian H0, where m 6= n ?

2 Schr¨odinger,Heisenberg and Interaction Pictures

We start with a review of Schrödingerand Heisenberg pictures, which we had come across in QM-1 course. In the Schrödingerpicture, the state vectors describing the system are time independent, their evolution being governed by Schrödingerequation

1 d | ψ (t)i i s = H | ψ (t) (2) ~ dt s formal solution of the equation is given in terms of the time evolution operator ˆ | ψs(t)i =U(t, t0i | ψs(t0)i ˆ −iH(t−t0)/~ ≡ e | ψs(t0)i (3) ˆ The operator U(t, t0) is unitary and satisfies the following identities Uˆ(t, t) = 1 ˆ † ˆ ˆ −1 U (t, t0) = U(t0, t) = U (t, t0) ˆ ˆ ˆ U(t1, t2)U(t2, t3) = U(t1, t3) (4) It may be noted that the operators corresponding to the physical observables are time independent in the Schrödinger picture. In the Heisenberg picture, on the other hand, the state vectors are time independent while the operators depend on time. One can therefore, relate the state vectors of the two pictures by the fact that the Schrödingerstate at time t can be obtained from some initial state at t = 0 by

−iHt/~ | ψs(t)i = e | ψs(0)i ≡| ψH i (5) where ψH are time independent. ˆ The Heisenberg operators AH (t) satisﬁes the following equation of motion dA (t) i H = [Aˆ , H] (6) ~ dt H where H is the Hamiltonian of the system, assumed to have no explicit time dependence. It may be observed that the expectation value of an operator is the same in both the pictures ˆ ˆ hψs(t) | As | ψs(t)i = hψh | AH (t) | ψH i (7)

ˆ iHt/~ ˆ iHt/~ where AH (t) = e Ase . Let {| φni} denote a complete set of eigenstates of the Hamiltonian H. Consider | ψH i, the Heisenberg state of the system or equivalently, the Schr¨odngerstate of the system at t = 0 ). We can write X | ψH i = cn | φni (8) n

2 The coeﬃcients cn are time independent since the left hand side of the above equation is so. The Schr¨odingerstate at time t can be written as

−iHt/~ | ψs(t)i = e | ψH i

−iHt/~ X = e cn | φni n

X −iEnt/~ = cne | φni (9) n Note that the time dependence on the right is contained only in the expo- −iEnt/~ nential term e , while the coeﬃcient cn are still time independent, a consequence of the fact that the Hamiltonian has no explicit time dependence. Suppose now, the Hamiltonian, in addition to the time independent part, which will henceforth denote by H0, has a time dependent part V (t)

H = H0 + V (t), the form of the Schr¨odinger state at time t, instead of (5) can be written as

X −iEnt/~ | ψs(t)i = cn(t)e | φni (10) n where the coefficients cn(t) now depend on time. We now define what is known as interaction picture or Dirac picture which is intermediate between the two pictures defined above. A state in the interaction picture is defined by iH0t/~ | ψI (t)i = e | ψs(t)i (11) Note that though at t = 0, the state of the system coincides with that of Schrödingerpicture, the time evolution of the state from t = 0 is governed by the time-independent Hamiltonian H0 and not by the full Hamiltonian H, as would be the case for the Schrödingerstate. Using the definition (11), we have ∂ ∂ i | ψ (t)i = −H eiH0t/~ | ψ (t)i + eiH0t/~(i ) | ψ (t)i ~∂t I 0 s ~ ∂t s iH0t/~ iH0t/~ = −H0e | ψs(t)i + e H | ψs(t)i

iH0t/~ iH0t/~ −iH0t/~ iH0t/~ = −H0e | ψs(t)i + e He e | ψs(t)i

iH0t/~ iH0t/~ −iH0t/~ = −H0e | ψs(t)i + e He | ψI (t)i

≡ VI (t) | ψI (t)i (12)

3 where, we have deﬁned

iH0t/~ −iH0t/~ VI (t) = e (H − H0)e (13) which is very similar to the way in which Heisenberg operators are connected to theSchr¨odingeroperators, with H0 replacing H in the time evolution operator. ˆ Clearly, the equation of motion for the operator AI (t) in the interaction picture is given by dA (t) i I = [Aˆ (t), H ] (14) ~ dt I 0 It may be noted that the matrix elements of the operators in both the rep- resentations (and hence by (7) in all the three pictures) remain the same,

iH0t/~ −iH0t/~ hψs(t) | As | ψs(t)i = hψI (t) | e Ase | ψI (t)i

= hψI (t) | AI (t) | ψI (t)i (15)

3 First Order Perturbation

We will now obtain a solution of equation (12) governing the time develop- ment of the wavefunction in the interaction picture, ∂ i | ψ (t)i = V (t) | ψ (t)i ~∂t I I I Let us deﬁne time evolution operator by

U(t) | ψI (t = 0)i =| ψ(t)i (16)

The operator U(t) is unitary and we have ∂ ∂ i (U(t) | ψ (0)i) = | ψ (t)i ~∂t I ∂t I = VI (t) | ψI (t)i

= VI (t)U(t) | ψ(t = 0)i (17)

Thus the equation satisﬁed by the evolution operator is ∂ i U(t) = V (t)U(t) (18) ~∂t I 4 with U(0) = 1. To ﬁrst order i perturbation, the solution of the above is

Z t i 0 0 U(t) = 1 − Vi(t )dt (19) ~ 0

Let us expand | ψI (t)i in terms of the complete eigenstates of the unperturbed Hamiltonian H 0 X | ψI (t)i = cn(t) | φni (20) n where H0 | φni = En | φni. The coeﬃcients cn in (20) are the same as in (10) corresponding to the Hamiltonian H0 , as can be seen by operating both sides of the latter by eiH0t/~,

iH0t/~ X iEnt/~ iH0t/~ e | ψs(t)i = cn(t)e e | φni n X = cn(t) | φni n In terms of these coeﬃcients, (17) can be written as

∂ X X i c (t) | φ i = V (t) c (t) | φ i ~∂t n n I m m n m

Multiplying both sides of above by hφk | and introducing a complete set of eigenstates, we have

∂ X X i c (t)hφ | φ i = c (t)hφ | V (t) | φ ihφ | φ i ~∂t n k n m k I l l m n m,l

Using the orthogonality property of the eigen states, we get

∂ X i c (t) = c (t)hφ | V (t) | φ ic (t) (21) ~∂t k m k I m m m But

iH0t/~ −iH0t/~ hφk | VI (t) | φmi = hφk | e V (t)e | φmi

i(Ek−Em)t/~ = e hφk | V (t) | φmi

i(Ek−Em)t/~ iωkmt = Vkme ≡ Vkme (22)

5 where Ek − Em ωkm = = −ωmk ~ The equation (21) then becomes

∂ X i c (t) = V eiωkmtc (t) (23) ~∂t k km m m

This equation is still exact but it leads to coupled differential equations for the coefficients ck(t). We will return to the solution of (23) a little later. Presently, let us assume that the system is initially in a particular eigenstate | φni. Thus cn(t = 0) = 1 and all other coefficients are initially zero. Thus

| ψI (t)i = U(t) | ψI (t = 0)i

= U(t) | φni i Z t = 1 − VI (t)dt | φni (24) ~ 0

The coeﬃcients cm(t) are thus given by

cm(t) = hφm | ψI (t)i (25) i Z t = − hφm | VI (t) | φnidt ~ 0 Z t i iωmnt = − e hφm | V (t) | φni (26) ~ 0

The transition probability from the initial state | φni to a state | φmi is given by 2 Pn→m = |cm| (27)

3.1 Constant perturbation If V does not depend on time but is switched on at time t = 0 for a time t, one can ﬁnd out the transition rate easily. Let us designate the initial state

6 by i and the final state by f. The coefficient cf is then given by (26), i Z t iωif t/~ cf (t) = − hφf | V | φii e dt ~ 0 i eiωift − 1 = − V if ~ iωif i 2 sin(ω t/2) ωif t/2 if = − Vif e (28) ~ ωif The transition probability to the state f is thus given by 2 2 4|Vif | 2 ωif t Pi→f = |cf | = 2 2 sin (29) ~ ωif 2 Note that the transition probability depends on (i) the matrix element of the perturbation between the initial and the final states, and, (ii) thefrequency difference between the two states, the maximum transition probability oc- curring when the two states have the same energy, i.e. perturbation neither extracts energy from the system nor supplies any to it. Taking ωif as a con- tinuous variable ω, the figure shows the variation of Pi→f as a function of the frequency difference between the initial state i and the final state f.

4 sin2(ωt/2) ω2

ω −6π/t −4π/t −2π/t 0 2π/t 4π/t 6π/t

Figure 1: Transition Probability as a function of frequency diﬀerence between initial and ﬁnal states

The ﬁgure shows that there is a central peak around ωif = 0 which has a width of 4π/t and a height of t2. The widths of the central peak are smaller

7 and height much reduced. Thus appreciable transition from the initial state takes place to only those ﬁnal states which lie within |ωif | < 2π/t. This may be considered as a reﬂection of the uncertainty principle. For large time, we have the following representation of the δ function sin2(xt) lim = πδ(x) t→∞ x2t Thus we have ω t sin2 if 2 lim = πδ(ωif t/2) = 2 πδ( ωif ) t→∞ 2 ~ ~ ωif t 4 where we have used δ(cx) = (1/|x|)δ(x). Using this, we get, the transition probability after long enough time to be

2π 2 Pi→f = t | Vif | δ(Ef − Ei) (30) ~ 3.2 Fermi’s Golden Rule in many situations (for example, in radioactive decay of atoms), there are many ﬁnal states possible. In such a case, the transition probability has to be summed over all possible ﬁnal states, keeping in mind the requirement of energy conservation. The probability expression given above (Eqn. (30)) is multiplied by a factor ρf . The reason for the presence of this ”blurring factor” in Fermi’s Golden rule is the fact that a for transition to take place, there has to exist states for the system to make a transition to. For instance, a typical nuclear decay involves release of a quantum γ. Unless there are states with energy equal to the energy of the initial state minus the energy of the gamma quantum (see below, discussion on harmonic perturbation), the decay does not occur.

2π 2 Pi→f = t | Vif | ρf (Ef )δ(Ef − Ei) (31) ~ where ρf is the density of the final states is defined by ρ(E)dE being the number of states within an energy interval dE around energy E. Summing over all possible final states, we get the transition probability to be given by Z X 2π 2 P i → f | Vif | t ρ(Ef )δ(Ei − Ef )dEf f ~

8 The integration is straightforward because of the delta function and the transition rate, i.e. probability of transition per unit time to be given by

2π 2 Wi→f = | Vif | ρ(Ei) (32) ~ 3.3 Harmonic Perturbation Let us consider a perturbation which varies sinusoidally with time. We have seen that significant transition takes place only when the time dependent perturbation has a frequency dependence such that the initial and the final states are connected by Bohr condition. It is thus desirable to impress the system with a perturbation having frequency which connects the initial and final states having energies corresponding to such energy levels. Let us consider a perturbation V (t) = V (~r)eiωt (33) where V (~r) is a hermitian operator, independent of time. In practice, we need a sine or a cosine function. However,it is convenient to use the exponential form above and at the end of the calculation we can always take real or the imaginary part to get appropriate result. The coefficient cf (t) is calculated in the first order perturbation theory as follows:

i Z t iωfit cf (t) = − dte hφf | V (t | )iφi ~ 0 i Z t i(ωfi−ω)t = − hφf | V (~r) | φii dte ~ 0 i ei(ωfi−ω)t − 1 = − hφf | V (~r) | φii ~ i(ωfi − ω)  (ωfi − ω)t  i sin i(ωfi−ω)t/2 2 = − hφf | V (~r) | φiie   (34) ~  (ωfi − ω)  2 Considering the hermitian conjugate of the perturbation V (t) ∼ eiωt will also give a similar expression with ω in the above expression being replaced by −ω. One has to add the two terms and take the square modulus of the sum to get the transition probability. However, the transition becomes appreciable only if the resonance condition is satisﬁed or nearly satisﬁed, in which case,

9 the trasition probability is given by the square modulus of only one of the terms. If the ﬁnal state has a higher energy than the initial state, i.e. if ωf = ωi + ω, the transition takes place by absorption of radiation from the perturbing source, whereas, in the reverse case ωf = ωi − ω, the system emits radiation by a process known as stimulated radiation. For the case of absorption 2 (ωfi − ω)t | V |2 sin P = fi 2 (35) i→f 2 2 ~ ωfi − ω 2 where Vfi = hφf | V (~r) | φii. The transition probability varies with time with a period 2π/(ωfi − ω), the peak in the probability occurs when ωf = ωi + ω. The resonance has a ﬁnite width ∆ω = 2π/t so that ∆ωδt = 2π, which is seen to be a consequence of the uncertainty principle. To see what happens if the perturbation is switched on for a long time, it is convenient to assume that the perturbation was switched on at t → −∞. we take the perturbation to subsist from t = −T/2 to +T/2 and then take T → ∞ limit. In this case, the calculation of cf (t) gives, changing the integration limits over time from t = −T/2 to +T/2

i Z T/2 i(ωfi−ω)t cf (t) = − Vfi lim e dt ~ T →∞ −T/2 i = − Vfi2πδ(ωfi − ω) ~ The probability of transition is given by

2 4π 2 Pi→f = | Vfi | δ(ωfi)δ(ωfi) (36) ~2 The product of two delta functions is handled as follows. We write the rate of transition

Pi→f Ri→f = lim T →∞ T 2π 1 Z T/2 2 i(ωfi−ω)t = 2 | Vfi | δ(ωfi) lim e dt (37) ~ T T →∞ −T/2

10 Because of the ﬁrst delta function, we put ωfi = ω in the above integral, which then evaluates to T and we get

2π 2 2π 2 R = | Vfi | δ(ωfi − ω) = | Vfi | δ(~ωfi − ~ω) (38) ~2 ~ The delta function ensures conservation of energy. As discussed earlier, transition may occur to a host of diﬀerent ﬁnal states. The transition rate in that case is given by Fermi’s Golden Rule

2π 2 R = | Vfi | ρ(Ef − Ei − ~ω)δ(Ef − Ei − ~ω) (39) ~ Fermi’s golden rule can also be derived by assuming that the time dependent potential is switched on slowly at t → ∞ according to V (t) = V ete−iωt (wth > 0) We look at time scales much smaller than 1/ i Z t i(ωfi−ω−i) cf (t) = − Vfi lim e tdt ~ →0 −∞ i ei(ωfi−ω−i)t = − Vfi lim ~ →0 (ωfi − ω − i) Thus 2t 2 1 2 e | cf | = 2 | Vfi | lim 2 2 ~ →0 (ωfi − ω) + The transition rate is 2t d 2 2e 2π 2 R = | cf (t) | = lim 2 2 = 2 | Vfi | δ(ωfi − ω) (40) dt →0 (ωfi − ω) + ~ which is the same as derived in (39).

4 Two State System

Let us return to the equation of motion (23). We will consider a two state system with harmonic potential which connects the two states. The unperturbed Hamiltonian is

Hu = E1 | 1ih1 | +E2 | 2ih2 | (41)

11 The perturbation term is

V (t) = V e−iωt | 2ih1 | +V ∗eiωt | 1ih2 | (42)

One can, of course, solve the problem exactly. We will, however, use this example to demonstrate how the equation of motion works. Suppose, initially only the lower level, i.e. | 1i is occupied, i.e. c1(t = 0) = 1 and C2(t = 0) = 0. The equation of motion is couples the two states

dc1 i = V eiω12tc ~ dt 12 2 iω12t = h1 | V | 2ie C2

∗ i(ω+ω12)t = V e c2

∗ i(ω−ω21)t = V e c2 (43)

Likewise, the equation of motion for c2 is given by

dc2 i = V e−i(ω−ω21)tc (44) ~ dt 1 One can convert the pair of equations (43) and (44) into a pair of uncoupled but second order diﬀerential equations for c1 and c2 by diﬀerentiating each of the equation once again and substituting from the other equation into it. For instance, using (44) we have

2 d c2 dc1 − 2 = (i )(−i)(ω − ω )e−i(ω−ω21)tc + V e−i(ω−ω21)t(i ) ~ dt2 ~ 21 1 ~ dt dc = (ω − ω )(i ) 2 + | V |2 c (45) ~ 21 ~ dt 2 where the last line of above is obtained by substituting from (43) and (44). We therefore have a second order diﬀerential equation

2 2 d c2 dc2 | V | + i(ω − ω21) + c2 = 0 (46) dt2 dt ~2

We need to solve this equation subject to the initial condition c2(t = 0) = 0. iΩt Solution is obtained by assuming a solution of the form c2 ∼ e . Substitut- ing this form in (46) we get a characteristic equation

2 2 | V | Ω + Ω(ω − ω21) − = 0 ~2 12 which has solutions r ω − ω (ω − ω )2 | V |2 Ω = − 21 ± 21 + 2 4 ~2 The general solution of the diﬀerential equation (46) is thus

v v  u 2 2 u 2 2  u α | V | u α | V | it + t −it + t −iαt/2  4 2 4 2  c2(t) = e Ae ~ + Be ~ 

where α = ω − ω21. The condition c2(t = 0) = 0 gives B = −A, as a result of which r 2 2 ! −iαt/2 α | V | c2(t) = 2iAe sin + t 4 ~2 The overall constant A can be ﬁxed by using (44) for time t = 0. This gives

dc i 2 (t = 0) = V c (t = 0) ~ dt 1

Diﬀerentiating the solution obtained in the previous step for c2(t) and putting t = 0 therein

r 2 2 dc2 α | V | V V i~ (t = 0) = 2iA + = c1(0) = dt 4 ~2 i~ i~ where we have used c1(0) = 1. This determines A. Replacing α = ω − ω21, we get the following solution for c2(t) r ! −iV (ω − ω )2 | V |2 i(ω−ω21)t/2 21 c2(t) = e sin + t r 2 2 2 (ω − ω21) | V | 4 ~ ~ + 4 ~2 The probability of ﬁnding the system at the upper level at time t is given by the square modulus of the above expression r ! V 2 (ω − ω )2 | V |2 P =| c |2= sin2 21 + t (47) 2 2 2 2 2 (ω − ω21) | V | 4 ~ ~2 + 4 ~2

13 and, of course, the probability of being in the ground state is P1 = 1 − P2. Note that the probability of ﬁnding the system in the upper level varies with an angular frequency 2Ω where r (ω − ω )2 | V |2 Ω = 21 + (48) 4 ~2

When the resonance condition ω = ω21 is satisfied, the amplitude of oscillation becomes very large. In this situation, at t = 0 (when the system is in the ground state) the system starts absorbing energy from the time dependent field, and after a time t = π~/2V = h/4V , it will be found in the excited state. The system then starts returning energy to the system to return back to the ground state. The states oscillates with a time period h/2V . The situation described above has applications to Masers. In ammonia maser, for instance, there are two eigenstates, the ground state | Si and the higher state | Ai, the energy difference between the two states lies in the microwave region. At normal temperatures, the average thermal energy being much higher than this energy difference (∼ 10−5 eV), the moecules equally populate these two states. The molecules are separated by passing them through an inhomogeneous electric field. The molecules in the state | Ai are then passed through a microwave cavity in such a way that the time spent by the molecules in the cavity is ∼ h/4V so that when the molecules emerge from the cavity, they are in the lower state, the energy released has been absorbed by the field. This is the way we obtain microwave amplifica- tion by stimulated emission of radiation.

5 Berry Phase

Adiabatic Approximation: In several situations, the Hamiltonian H changed slowly. When this condition is satisﬁed, the system has enough time to adopt to the developing Hamiltonian and the state of the system at a given instant corresponds to the Hamiltonian at that instant of time. This is in contrast to Sudden approximation where the system stays in the state of the Hamil- tonian before change as it is unable to adapt to the sudden change. In many cases, the Hamiltonian depends on a parameter λ ( or on a set of parameters, which we collectively denote by λ), where λ(t) is a slowly varying function of time. For instance, if we consider spins in a slowly varying magnetic ﬁeld, λ(t) could correspond to Bx(t),By(t) and Bz(t). In such a

14 situation, we obtain the solution of the Schr¨odingerequation by adiabatic approximation. It must be noted that the phrase adiabatic in this context, unlike in its thermodynamic parlance, has no relationship with the conservation of energy. A simple example from classical mechanics may illustrate the process.

N θ

A B

Figure 2: The path of a simple pendulum oscillating in a vertical plane.

Consider an ideal simple pendulum oscillating in a vertical plane. Sup- posing we start at the north pole N with the plane of oscillation being along NA (and of course, the local vertical direction at the location of the pendulum. If the movement of the support of the pendulum along NA is sudden, the resulting motion of the pendulum will be chaotic. However, if the support is moved slowly, the motion will still remain steady, remaining in the vertical plane containing the longitude line NA. After reaching the point A, we move the support parallel to the great circle till the point B and then back to north pole along BN, thus completing the circuit. However, the plane in which the pendulum swings is no longer the same (or parallel to) the plane of oscillation during the outward motion from N, the outbound plane and the

15 incoming plane making an angle θ with one another.This angle is of purely geometric origin, independent of the shape of the earth and it is equal to the solid angle subtended by the area NAB at the centre of the latitude circle along which AB is situated. If R be the radius of this circle, then A R2θ Ω = = = θ R2 R2 Foucault’s pendulum is also an example of rotation of plane of oscillation, except that, in that case the support is moved by the earth. Let us return to quantum mechanics. If the Hamiltonian is time independent, the evolution of the eigenstate | ψni with time is given by

−iEnt/~ | ψn(t)i = e | ψn(0)i (49) where e−iEnt/~ is a purely dynamic phase. We could have multiplied the same solution by a purely unimportant phase factor eiϕ. However, if the Hamiltonian is time dependent, we would have d i | ψ (t)i = H(t) | ψ (t)i = E (t) | ψ (t)i (50) ~dt n n n n

Here we have assumed that the Hamiltonian varies “slowly” so that En varies with time but remains an energy eigenvalue. Let us write the solution of above in a way which reduces to (49) when H does not depend on time. Our guess for a solution in this case is i R t 0 0 − 0 En(t )dt | ψn(t)i = e ~ | φn(t)i (51) where H(t) | φn(t)i = En(t) | φn(t)i Note that the solution is correct if H is time independent. However, this is not the whole story because it does not take into account the slow variation of H(t) itself. To correct for this omission, we take i R t 0 0 − En(t )dt 0 iγn(t) | ψn(t)i = e ~ | φn(t)ie (52) where the additional factor eiγn(t) must be unity (or eiϕ with ϕ being independent of time) if our previous ansatz (50) is correct. Substitute (52) into

This implies that the last two terms on the right had side of the ﬁrst step of the above equation equals zero,

∂ | φ (t)i ∂γ i n − | φ (t)i n = 0 ~ ∂t ~ n ∂t which gives Z t 0 ∂φn(t) γn(t) = i dt hφn(t) | i (54) 0 ∂t If we revert back to the parameter λ, which determines the time variation of H(t), we get Z λf ∂φn(t) γn(t) = i dλhφn(λ) | i (55) λi ∂λ

5.1 Example: Consider a particle whose spin is aligned in the direction of a magnetic ﬁeld B~ (t) which precesses along the z- direction with an angular speed ω, so that the azimuthal angle ϕ = ωt changes linearly with time.

B~ = B[ˆi sin θ cos ωt + ˆj sin ωt + kˆ cos θ]

Here θ is the angle of precession which remains constant. The magnetic ﬁeld is expressed in energy units by absorbing a factor µ. The Hamiltonian is given by

H = S~ · B~ = ~σ · B~ 2 B cosθ sin θ cos ωt − i sin θ sin ωt = ~ 2 sin θ cos ωt + i sin θ sin ωt − cos θ

17 B which has eigenvalues ±~ . The spin is initially aligned in the direction of 2 the magnetic ﬁeld. This is a two state problem of the type developed in the previous section.