Phy5646/homeworkintimeperturbation

Problem in Time Dependent Perturbation theory: Magnetic Resonance

Consider the Hamiltonian$H=H_{0}+H_{1}(t)$

$H_{0}=E_{1}|1\rangle \langle 1|+E_{2}|2\rangle \langle 2|$

$H_{1}(t)=\gamma e^{i\omega t}|1\rangle \langle 2|+\gamma e^{-i\omega t}|2\rangle \langle 1|$

where $E_{2}>E_{1}$, and $\gamma$ and $\omega$ are real and positive. At the time $t=0$ assume thatthe lower energy level is populated, i.e. the probability for the level 1 is one and the one for level 2 is zero.

(a) Assuming that the wavefuction of the system is given by

$\psi =a_{1}(t)e^{-iE_{1}t/\hbar }|1\rangle +a_{2}(t)e^{-iE_{2}t/\hbar }|2\rangle$

(b) Solve the coupled differential equations obtained in (a). For this purporse reduce the coupled equations to a single second order differential equation for $a_{1}$. The solutions are of the form $a_{j}(t)=C_{j}e^{i\lambda _{+}t}+D_{j}e^{i\lambda _{-}t},\ j=1,2$. Obtain the frequencies$\lambda _{+}$ and $\lambda _{-}$.

(c) Determine the coefficients $C_{1}$, $C_{2}$, $C_{3}$ and $D_{4}$ using the initial conditions spedified above. Note that the coefficients are not all independent($a_{1}$ and $a_{2}$ satisfy differential equations).

(d) Obtain the time-dependent probabilities of finding the system in level 1 and in level 2.

(e) Consider the amplitude of the probability of finding the system in state 2 as a function of $\omega$. What is the resonance condition? Obtain the full width at half maximum of the resonance.

Solution:

(a)

$i\hbar {\frac {\partial \Psi }{\partial t}}=H\Psi$

$(i\hbar {\dot {a}}_{1}e^{-iE_{1}t/\hbar }+E_{1}a_{1}e^{-iE_{1}t/\hbar })|1\rangle +(i\hbar {\dot {a}}_{2}e^{-iE_{2}t/\hbar }+E_{2}a_{2}e^{-iE_{2}t/\hbar })|2\rangle =a_{1}e^{-iE_{1}t/\hbar }(E_{1}|1\rangle +\gamma e^{-i\omega t}|2\rangle )+a_{2}e^{-iE_{2}t/\hbar }(E_{2}|2\rangle +\gamma e^{i\omega t}|1\rangle )$

$i\hbar {\dot {a}}_{1}=a_{2}e^{i(E_{1}-E_{2})t/\hbar }\gamma e^{i\omega t}=\gamma e^{i(\omega -\omega _{21})t}a_{2}$

$i\hbar {\dot {a}}_{2}=a_{1}e^{i(E_{1}-E_{2})t/\hbar }\gamma e^{-i\omega t}=\gamma e^{-i(\omega -\omega _{21})t}a_{1}$

$i\hbar e^{-i(\omega -\omega _{21})t}{\dot {a}}_{1}=\gamma a_{2}$

$(\omega -\omega _{21})\hbar e^{-i(\omega -\omega _{21})t}{\dot {a}}_{1}+i\hbar e^{-i(\omega -\omega _{21})t}{\ddot {a}}_{1}=\gamma {\dot {a}}_{2}={\frac {\gamma ^{2}}{i\hbar }}e^{-i(\omega -\omega _{21})t}a_{1}$

$-\hbar ^{2}{\ddot {a}}_{1}+i\hbar ^{2}(\omega -\omega _{21}){\dot {a}}_{1}-\gamma ^{2}a_{1}=0<math><math>a\sim e^{i\lambda t}$

$\hbar ^{2}\lambda ^{2}-\hbar ^{2}\lambda (\omega -\omega _{21})-\gamma ^{2}=0$

$\lambda _{\pm }={\frac {1}{2}}(\omega -\omega _{21})\pm {\frac {1}{2}}{\sqrt {(\omega -\omega _{21})^{2}+4{\frac {\gamma ^{2}}{\hbar ^{2}}}}}$

$a_{1}(t)=C_{1}e^{i\lambda _{+}t}+B_{1}e^{i\lambda _{-}t}$

$a_{2}(t)=C_{2}e^{i\lambda _{+}t}+B_{2}e^{i\lambda _{-}t}$

$a_{2}(0)=0$

$C_{2}=-B_{2}$

$a_{2}(t)=2iC_{2}e^{i(\omega -\omega _{21})t/2}\sin[t{\sqrt {{\frac {(\omega -\omega _{21})^{2}}{4}}+{\frac {\gamma ^{2}}{\hbar ^{2}}}}}]$

$a_{1}(t)={\frac {i\hbar }{\gamma }}e^{i(\omega -\omega _{21})t}{\dot {a}}_{2}=-{\frac {2\hbar }{\gamma }}C_{2}\{i{\frac {\omega -\omega _{21}}{2}}\sin[t{\sqrt {{\frac {(\omega -\omega _{21})^{2}}{4}}+{\frac {\gamma ^{2}}{\hbar ^{2}}}}}]+{\sqrt {{\frac {(\omega -\omega _{21})^{2}}{4}}+{\frac {\gamma ^{2}}{\hbar ^{2}}}}}\cos[t{\sqrt {{\frac {(\omega -\omega _{21})^{2}}{4}}+{\frac {\gamma ^{2}}{\hbar ^{2}}}}}]\}e^{i{\frac {3}{2}}(\omega -\omega _{21})t}$

$a_{1}(0)=1$

$C_{2}=-{\frac {\gamma }{2\hbar {\sqrt {{\frac {(\omega -\omega _{21})^{2}}{4}}+{\frac {\gamma ^{2}}{\hbar ^{2}}}}}}}$

$|a_{2}(t)|^{2}={\frac {(\gamma /\hbar )^{2}}{{\frac {1}{4}}(\omega -\omega _{21})^{2}+{\frac {\gamma ^{2}}{\hbar ^{2}}}}}\sin ^{2}[t{\sqrt {{\frac {(\omega -\omega _{21})^{2}}{4}}+{\frac {\gamma ^{2}}{\hbar ^{2}}}}}]$

$|a_{1}(t)|^{2}={\frac {1}{{\frac {1}{4}}(\omega -\omega _{21})^{2}+{\frac {\gamma ^{2}}{\hbar ^{2}}}}}\{[{\frac {(\omega -\omega _{21})^{2}}{4}}+{\frac {\gamma ^{2}}{\hbar ^{2}}}]\cos ^{2}[t{\sqrt {{\frac {(\omega -\omega _{21})^{2}}{4}}+{\frac {\gamma ^{2}}{\hbar ^{2}}}}}]+{\frac {(\omega -\omega _{21})^{2}}{4}}\sin ^{2}[t{\sqrt {{\frac {(\omega -\omega _{21})^{2}}{4}}+{\frac {\gamma ^{2}}{\hbar ^{2}}}}}]\}=1-{\frac {\gamma ^{2}/\hbar ^{2}}{{\frac {(\omega -\omega _{21})^{2}}{4}}+{\frac {\gamma ^{2}}{\hbar ^{2}}}}}\sin ^{2}[t{\sqrt {{\frac {(\omega -\omega _{21})^{2}}{4}}+{\frac {\gamma ^{2}}{\hbar ^{2}}}}}]=1-|a_{2}(t)|^{2}$

$F(\omega )={\frac {(2\gamma /\hbar )^{2}}{(\omega -\omega _{21})^{2}+(2\gamma /\hbar )^{2}}}$

$\omega =\omega _{21}={\frac {E_{2}-E_{1}}{\hbar }}$

$4\gamma /\hbar$