8.3: Two-State System
( \newcommand{\kernel}{\mathrm{null}\,}\)
Consider a system in which the time-independent Hamiltonian possesses two eigenstates, denoted
H0|1⟩ =E1|1⟩, ??? H0|2⟩ =E2|2⟩. ???Suppose, for the sake of simplicity, that the diagonal matrix elements of the interaction Hamiltonian, H1 , are zero:
⟨1|H1|1⟩=⟨2|H1|2⟩=0. ???The off-diagonal matrix elements are assumed to oscillate sinusoidally at some frequency ω :
⟨1|H1|2⟩=⟨2|H1|1⟩∗=γexp(iωt), ???where γ and ω are real. Note that it is only the off-diagonal matrix elements that give rise to the effect which we are interested in--namely, transitions between states 1 and 2.
For a two-state system, Equation ??? reduces to
iℏdc1dt =γexp[+i(ω−ω21)t]c2, ??? iℏdc2dt =γexp[−i(ω−ω21)t]c1, ???where ω21=(E2−E1)/ℏ , and it is assumed that t0=0 . Equations ??? and ??? can be combined to give a second-order differential equation for the time variation of the amplitude c2 :
d2c2dt2+i(ω−ω21)dc2dt+γ2ℏ2c2=0. ???Once we have solved for c2 , we can use Equation ??? to obtain the amplitude c1 . Let us look for a solution in which the system is certain to be in state 1 at time t=0 . Thus, our initial conditions are c1???=1 and c2???=0 . It is easily demonstrated that the appropriate solutions are
c2(t)=![$ \frac{-{\rm i}\, \gamma/\hbar} {[\gamma^2/\hbar^2 + (\omega-\omeg...
.../2]\,\sin\left([\gamma^2/\hbar^2+(\omega-\omega_{21})^{\,2}/4]^{1/2}\,t\right),$](http://farside.ph.utexas.edu/teaching/qm/lectures/img1813.png)
![$ - \frac{{\rm i}\,(\omega-\omega_{21})/2 }{[\gamma^2/\hbar^2 + (\o...
...2]\,\sin\left( [\gamma^2/\hbar^2+(\omega-\omega_{21})^{\,2}/4]^{1/2}\,t\right).$](http://farside.ph.utexas.edu/teaching/qm/lectures/img1816.png)
The probability of finding the system in state 1 at time t is simply P1(t)=|c1|2 . Likewise, the probability of finding the system in state 2 at time t is P2(t)=|c2|2 . It follows that
P2(t)![$ = \frac{\gamma^2/\hbar^2}{ \gamma^2/\hbar^2 + (\omega-\omega_{21}...
..., \sin^2\left([\gamma^2/\hbar^2+ (\omega-\omega_{21})^{\,2}/4]^{1/2}\,t\right),$](http://farside.ph.utexas.edu/teaching/qm/lectures/img1820.png)
Equation ??? exhibits all the features of a classic resonance. At resonance, when the oscillation frequency of the perturbation, ω , matches the frequency ω21 , we find that
P1(t) =cos2(γt/ℏ), ??? P2(t) =sin2(γt/ℏ). ???According to the above result, the system starts off at t=0 in state 1 . After a time interval πℏ/2γ , it is certain to be in state 2. After a further time interval πℏ/2γ , it is certain to be in state 1, and so on. In other words, the system periodically flip-flops between states 1 and 2 under the influence of the time-dependent perturbation. This implies that the system alternatively absorbs and emits energy from the source of the perturbation.
The absorption-emission cycle also take place away from the resonance, when ω≠ω21 . However, the amplitude of oscillation of the coefficient c2 is reduced. This means that the maximum value of P2(t) is no longer unity, nor is the minimum value of P1(t) zero. In fact, if we plot the maximum value of P2(t) as a function of the applied frequency, ω , then we obtain a resonance curve whose maximum (unity) lies at the resonance, and whose full-width half-maximum (in frequency) is 4γ/ℏ . Thus, if the applied frequency differs from the resonant frequency by substantially more than 2γ/ℏ then the probability of the system jumping from state 1 to state 2 is very small. In other words, the time-dependent perturbation is only effective at causing transitions between states 1 and 2 if its frequency of oscillation lies in the approximate range ω21±2γ/ℏ . Clearly, the weaker the perturbation (i.e., the smaller γ becomes), the narrower the resonance.
Contributors
Richard Fitzpatrick (Professor of Physics, The University of Texas at Austin)