Chapter 9 Practice Problems

OK so in physics, a delta function is the thing where the area under it is 1,

and something like this is what we call the Dirac delta function.

The problem tells us to do it with a rectangle, so let’s cook one up ourselves with area 1:

$$t\quad <\quad -\epsilon \quad :\quad \text{before perturbation is turned on},\quad \to \quad C_{a}\quad =\quad 1\quad ,\quad C_{b}\quad =\quad 0\\ t\quad >\quad +\epsilon \quad :\quad \text{since perturbation is turned off},\quad \to \quad C_{a}\quad =\quad const\quad =\quad a\quad ,\quad C_{b}\quad =\quad const\quad =\quad a$$$$-\epsilon \quad <\quad t\quad <\quad +\epsilon \quad \text{when perturbation is running,} \\ \text{the}\quad \begin{cases}{\dot {C_{a}}\quad =\quad -\frac {i}{\hbar}H'_{ab}e^{-iw_{0}t}\cdot C_{b}}\\{\quad}\\{\dot {C_{b}}\quad =\quad -\frac {i}{\hbar}H'_{ba}e^{-iw_{0}t}\cdot C_{a}}\end{cases}\quad \text{that we derived}$$

Alright — let’s crunch the H’ values!

$$H'_{ab}\quad =\quad \left< \psi_{a}|H'|\psi_{b} \right> \quad =\quad \left< \psi_{a}|U\delta (t)|\psi_{b} \right> \quad =\quad \int _{-\infty}^{\infty}{\quad}\psi_{a}U\delta (t)\psi_{b}\quad =\quad \frac {1}{2\epsilon}U_{ab}$$

The value of the delta function!!! We used 1/2ε!!!!

OK so now, for real this time, let’s solve this thing.

I’m going to follow the exact same logical playbook we used for Prob 9.2!!!

(http://gdpresent.blog.me/220580074156) Right here.

Quantum Mechanics I Studied #38. Time-dependent Perturbation

………………. ……haah~ T_T T_T T_T T_T in chapter 6 we did time-independent perturbation theory, and from there…

blog.naver.com

Here we go!!!!

$$\dot {C_{b}}\quad =\quad -\frac {i}{\hbar}\frac {U_{ba}}{2\epsilon}e^{iw_{0}t}C_{a}\\ \text{differentiating both sides again with respect to}\quad t,\\ \ddot {C_{b}}\quad =\quad -\frac {i}{\hbar}\frac {U_{ba}}{2\epsilon}\left( iw_{0}e^{iw_{0}t}C_{a}\quad +\quad e^{iw_{0}t}\dot {C_{a}} \right) \\ =\quad iw_{0}\left( -\frac {i}{\hbar}\frac {U_{ba}}{2\epsilon}e^{iw_{0}t}C_{a} \right) \quad -\frac {i}{\hbar}\frac {U_{ba}}{2\epsilon}\quad e^{iw_{0}t}\dot {C_{a}} \\ \text{the red part is}\quad \dot {C_{b}}\quad \text{and the blue part, sub it back in,}\\ =\quad iw_{0}\dot {C_{b}}\quad -\frac {i}{\hbar}\frac {U_{ba}}{2\epsilon}\quad e^{iw_{0}t}\cdot \left( -\frac {i}{\hbar}\frac {U_{ba}}{2\epsilon}\quad e^{-iw_{0}t}C_{b} \right) \\ =\quad iw_{0}\dot {C_{b}}\quad -\frac {1}{\hbar^{2}}\frac {\left| U_{ba} \right|^{2}}{\left( 2\epsilon \right)^{2}}\quad C_{b} \\ \text{and since we said}\quad U_{ba}\quad \equiv \quad \alpha, \\ \ddot {C_{b}}\quad =\quad iw_{0}\dot {C_{b}}\quad -\frac {1}{\hbar^{2}}\frac {\alpha^{2}}{\left( 2\epsilon \right)^{2}}\quad C_{b}\\ \ddot {C_{b}}\quad -\quad iw_{0}\dot {C_{b}}\quad +\frac {1}{\hbar^{2}}\frac {\alpha^{2}}{\left( 2\epsilon \right)^{2}}\quad C_{b}\quad =\quad 0$$$$\left( D^{2}\quad -\quad iw_{0}D\quad +\frac {\alpha^{2}}{\left( 2\hbar \epsilon \right)^{2}} \right) \quad C_{b}\quad =\quad 0\\ D\quad =\quad \frac {iw_{0}\quad \pm \quad \sqrt {-w_{0}-4\frac {\alpha^{2}}{\left( 2\hbar \epsilon \right)^{2}}}}{2}\quad \\ =\quad \frac {iw_{0}\quad \pm \quad i\sqrt {w_{0}\quad +\quad \frac {\alpha^{2}}{\left( \hbar \epsilon \right)^{2}}}}{2}\quad \\ =\quad \frac {i}{2}\left( w_{0}\quad \pm \quad \sqrt {w_{0}\quad +\quad \left( \frac {\alpha}{\hbar \epsilon} \right)^{2}} \right) \\ \text{let}\quad \sqrt {w_{0}\quad +\quad \left( \frac {\alpha}{\hbar \epsilon} \right)^{2}}\quad \equiv \quad w\quad !$$$$\text{so the general solution for}\quad C_{b}(t)\quad \text{is}\\ C_{b}(t)\quad =\quad Ae^{i\frac {\left( w_{0}+w \right)}{2}t}\quad +\quad Be^{i\frac {\left( w_{0}-w \right)}{2}t}\\ \quad =\quad e^{i\frac {w_{0}}{2}t}\left( Ae^{i\frac {w}{2}t}\quad +\quad Be^{-i\frac {w}{2}t} \right)$$$$\text{Since}\quad C_{b}(-\epsilon )\quad =\quad 0,\\ e^{-i\frac {w_{0}}{2}\epsilon}\left( Ae^{-i\frac {w}{2}\epsilon}\quad +\quad Be^{i\frac {w}{2}\epsilon} \right) \quad =\quad 0 \\ \text{i.e.,}\quad Ae^{-i\frac {w}{2}\epsilon}\quad =\quad -Be^{i\frac {w}{2}\epsilon}\\ B\quad =\quad -Ae^{-iw\epsilon} \\ C_{b}(t)\quad =\quad e^{i\frac {w_{0}}{2}t}\left( Ae^{i\frac {w}{2}t}\quad +\quad -Ae^{-iw\epsilon}e^{-i\frac {w}{2}t} \right) \\ =\quad Ae^{i\frac {w_{0}}{2}t}\left( e^{i\frac {w}{2}t}\quad +\quad -e^{-iw\left( \epsilon +\frac {t}{2} \right)} \right)$$$$\text{Now back to the very top,} \\ \dot {C_{b}}\quad =\quad -\frac {i}{\hbar}\frac {U_{ba}}{2\epsilon}e^{iw_{0}t}C_{a}\quad =\quad -\frac {i}{\hbar}\frac {\alpha}{2\epsilon}e^{iw_{0}t}C_{a}\\ \text{and this has to equal the derivative of the general solution we just got,} \\ \text{so let's differentiate the general solution w.r.t.}\quad t\\ C_{b}(t)\quad =\quad Ae^{i\frac {w_{0}}{2}t}\left( e^{i\frac {w}{2}t}\quad -e^{-iw\left( \epsilon +\frac {t}{2} \right)} \right) \quad \text{differentiate w.r.t.}\quad t\quad \text{go go go}\\ \dot {C_{b}}(t)\quad =\quad A\left[ i\frac {w_{0}}{2}e^{i\frac {w_{0}}{2}t}\left( e^{i\frac {w}{2}t}\quad -e^{-iw\left( \epsilon +\frac {t}{2} \right)} \right) \quad +\quad e^{i\frac {w_{0}}{2}t}\left( i\frac {w}{2}e^{i\frac {w}{2}t}\quad +\quad i\frac {w}{2}e^{-iw\left( \epsilon +\frac {t}{2} \right)} \right) \right] \\ =\quad A\left[ i\frac {w_{0}}{2}e^{i\frac {w_{0}}{2}t}\left( e^{i\frac {w}{2}t}\quad -e^{-iw\left( \epsilon +\frac {t}{2} \right)} \right) +\quad i\frac {w}{2}e^{i\frac {w_{0}}{2}t}\left( e^{i\frac {w}{2}t}\quad +\quad e^{-iw\left( \epsilon +\frac {t}{2} \right)} \right) \right] \\ =\quad Ae^{i\frac {w_{0}}{2}t}\left[ i\frac {w_{0}}{2}\left( e^{i\frac {w}{2}t}\quad -e^{-iw\left( \epsilon +\frac {t}{2} \right)} \right) +\quad i\frac {w}{2}\left( e^{i\frac {w}{2}t}\quad +\quad e^{-iw\left( \epsilon +\frac {t}{2} \right)} \right) \right] \\ =\quad Ae^{i\frac {w_{0}}{2}t}\frac {i}{2}\left[ w_{0}e^{i\frac {w}{2}t}-\quad w_{0}e^{-iw\left( \epsilon +\frac {t}{2} \right)}\quad +\quad we^{i\frac {w}{2}t}\quad +\quad we^{-iw\left( \epsilon +\frac {t}{2} \right)} \right] \\ \quad \therefore \quad \dot {C_{b}}(t)\quad =\quad \frac {i}{2}Ae^{i\frac {w_{0}}{2}t}\left[ \left( w\quad +\quad w_{0} \right) e^{i\frac {w}{2}t}+\quad \left( w-w_{0} \right) e^{-iw\left( \epsilon +\frac {t}{2} \right)} \right]$$$$\dot {C_{b}}\quad =\quad -\frac {i}{\hbar}\frac {\alpha}{2\epsilon}e^{iw_{0}t}C_{a}\\ \dot {C_{b}}\quad =\quad \frac {i}{2}Ae^{i\frac {w_{0}}{2}t}\left[ \left( w\quad +\quad w_{0} \right) e^{i\frac {w}{2}t}+\quad \left( w-w_{0} \right) e^{-iw\left( \epsilon +\frac {t}{2} \right)} \right] \\ \text{so,}\quad C_{a}\quad =\quad \frac {\frac {i}{2}Ae^{i\frac {w_{0}}{2}t}}{-\frac {i}{\hbar}\frac {\alpha}{2\epsilon}e^{iw_{0}t}}\left[ \left( w\quad +\quad w_{0} \right) e^{i\frac {w}{2}t}+\quad \left( w-w_{0} \right) e^{-iw\left( \epsilon +\frac {t}{2} \right)} \right] \\ =\quad -\frac {A\hbar \epsilon}{\alpha}e^{-i\frac {w_{0}}{2}t}\left[ \left( w\quad +\quad w_{0} \right) e^{i\frac {w}{2}t}+\quad \left( w-w_{0} \right) e^{-iw\left( \epsilon +\frac {t}{2} \right)} \right]$$$$\text{Since}\quad C_{a}(-\epsilon )\quad =\quad 1\quad \text{has to hold,}\\ C_{a}(-\epsilon )\quad =\quad -\frac {A\hbar \epsilon}{\alpha}e^{i\frac {w_{0}}{2}\epsilon}\left[ \left( w\quad +\quad w_{0} \right) e^{-i\frac {w}{2}\epsilon}+\quad \left( w-w_{0} \right) e^{-iw\left( \epsilon -\frac {\epsilon}{2} \right)} \right] \\ =\quad -\frac {A\hbar \epsilon}{\alpha}2we^{i\frac {w_{0}-w}{2}\epsilon}\quad \\ =\quad 1 \\ \therefore \quad A\quad =\quad \frac {-\alpha}{2\hbar \epsilon w}e^{-i\frac {w_{0}-w}{2}\epsilon}$$$$\text{So,}\\ C_{a}(t)\quad =\quad \frac {1}{2w}e^{i\frac {w}{2}\epsilon}e^{-i\frac {w_{0}}{2}\left( \epsilon +t \right)}\left\{\left( w+w_{0} \right) e^{i\frac {w}{2}t}\quad +\quad \left( w-w_{0} \right) e^{-iw\left( \epsilon +\frac {t}{2} \right)} \right\} \\ =\quad \frac {1}{2w}e^{-i\frac {w_{0}}{2}\left( \epsilon +t \right)}\left\{\left( w+w_{0} \right) e^{i\frac {w}{2}\left( \epsilon +t \right)}\quad +\quad \left( w-w_{0} \right) e^{-iw\left( \epsilon +t \right)} \right\} \\ =\quad \frac {1}{2w}e^{-i\frac {w_{0}}{2}\left( \epsilon +t \right)}\left\{w\left( 2\frac {e^{i\frac {w}{2}\left( \epsilon +t \right)}+e^{-i\frac {w}{2}\left( \epsilon +t \right)}}{2} \right) \quad +\quad w_{0}\left( 2i\frac {e^{i\frac {w}{2}\left( \epsilon +t \right)}-e^{-i\frac {w}{2}\left( \epsilon +t \right)}}{2i} \right) \right\} \\ =\quad \frac {1}{2w}e^{-i\frac {w_{0}}{2}\left( \epsilon +t \right)}\left\{2wcos\left( \frac {w}{2}\left( \epsilon +t \right) \right) \quad +\quad i2w_{0}sin\left( \frac {w}{2}\left( \epsilon +t \right) \right) \right\} \\ =\quad e^{-i\frac {w_{0}}{2}\left( \epsilon +t \right)}\left\{cos\left( \frac {w}{2}\left( \epsilon +t \right) \right) \quad +\quad i\frac {w_{0}}{w}sin\left( \frac {w}{2}\left( \epsilon +t \right) \right) \right\} \\ \therefore \quad C_{a}(t)\quad =\quad e^{-i\frac {w_{0}}{2}\left( \epsilon +t \right)}\left\{cos\left( \frac {w}{2}\left( \epsilon +t \right) \right) \quad +\quad i\frac {w_{0}}{w}sin\left( \frac {w}{2}\left( \epsilon +t \right) \right) \right\}$$$$\text{And since}\quad C_{b}(t)\quad =\quad Ae^{i\frac {w_{0}}{2}t}\left( e^{i\frac {w}{2}t}\quad -\quad e^{-iw\left( \epsilon +\frac {t}{2} \right)} \right) \quad \text{was what we had,}\\ C_{b}(t)\quad =\quad \left( -\frac {\alpha}{2\hbar \epsilon w}e^{i\frac {w-w_{0}}{2}\epsilon} \right) e^{i\frac {w_{0}}{2}t}\left( e^{i\frac {w}{2}t}\quad -\quad e^{-iw\left( \epsilon +\frac {t}{2} \right)} \right) \\ =\quad -\frac {\alpha}{2\hbar \epsilon w}e^{i\frac {w}{2}\epsilon}e^{i\frac {w_{0}}{2}\left( t-\epsilon \right)}\left( e^{i\frac {w}{2}t}\quad -\quad e^{-iw\left( \epsilon +\frac {t}{2} \right)} \right) \\ =\quad -\frac {\alpha}{2\hbar \epsilon w}e^{i\frac {w}{2}\left( t-\epsilon \right)}\left( e^{i\frac {w}{2}\left( \epsilon +t \right)}\quad -\quad e^{-iw\left( \epsilon +t \right)} \right) \\ =\quad -\frac {\alpha}{2\hbar \epsilon w}e^{i\frac {w}{2}\left( t-\epsilon \right)}2i\frac {\left( e^{i\frac {w}{2}\left( \epsilon +t \right)}\quad -\quad e^{-iw\left( \epsilon +t \right)} \right)}{2i}\\ =\quad -\frac {\alpha}{2\hbar \epsilon w}e^{i\frac {w}{2}\left( t-\epsilon \right)}sin\left( \frac {w}{2}\left( \epsilon +t \right) \right) \\ C_{b}(t)\quad =\quad -\frac {\alpha}{2\hbar \epsilon w}e^{i\frac {w}{2}\left( t-\epsilon \right)}sin\left( \frac {w}{2}\left( \epsilon +t \right) \right)$$$$C_{a}(\epsilon )\quad =\quad a\quad =\quad e^{-i\frac {w_{0}}{2}\left( \epsilon +\epsilon \right)}\left\{cosw\epsilon \quad +\quad i\frac {w_{0}}{w}sinw\epsilon \right\} e^{-iw_{0}\epsilon}\left\{cosw\epsilon \quad +\quad i\frac {w_{0}}{w}sinw\epsilon \right\} \\ C_{b}(\epsilon )\quad =\quad b\quad =\quad -\frac {i\alpha}{\hbar \epsilon w}sinw\epsilon \\ w\quad \text{was, from earlier,}\quad w\quad =\quad \sqrt {w_{0}^{2}+\left( \frac {\alpha}{\epsilon \hbar} \right)^{2}} \\ \text{which means}\quad w\epsilon \quad =\quad \sqrt {\left( \epsilon w_{0} \right)^{2}+\left( \frac {\alpha}{\hbar} \right)^{2}} \\ \text{so if}\quad \epsilon \to 0,\quad \text{then}\quad w\epsilon \quad \to \quad \left| \frac {\alpha}{\hbar} \right|$$$$\lim _{\epsilon \to 0}{b}\quad =\quad -\frac {i\alpha}{\hbar \frac {\left| \alpha \right|}{\hbar}}sin\frac {\left| \alpha \right|}{\hbar}\quad =\quad i\frac {\alpha}{\left| \alpha \right|}sin\frac {\left| \alpha \right|}{\hbar}\\ \lim _{\epsilon \to 0}{a}\quad =\quad cos\left( \frac {\left| \alpha \right|}{\hbar} \right) \quad +\quad 0 \\ \text{so,}\quad P_{a\to b}\quad =\quad \left| b \right|^{2}\quad =\quad sin^{2}\frac {\left| \alpha \right|}{\hbar} \\ \text{and,}\quad \left| C_{a}(t) \right|^{2}\quad +\quad \left| C_{b}(t) \right|^{2}\quad =\quad cos^{2}\frac {w\left( \epsilon +t \right)}{2}\quad +\quad \frac {w_{0}}{w}sin^{2}\frac {w\left( \epsilon +t \right)}{2}\quad +\quad \left( \frac {\alpha}{\hbar \epsilon w} \right)^{2}sin^{2}\frac {w\left( \epsilon +t \right)}{2}\\ =\quad cos^{2}\frac {w\left( \epsilon +t \right)}{2}\quad +\quad \left( \frac {w_{0}}{w}+\left( \frac {\alpha}{\hbar \epsilon w} \right)^{2} \right) sin^{2}\frac {w\left( \epsilon +t \right)}{2}\quad \\ =\quad cos^{2}\frac {w\left( \epsilon +t \right)}{2}\quad +\quad sin^{2}\frac {w\left( \epsilon +t \right)}{2}\quad \\ =\quad 1$$

How clean is that?!

Same playbook as when we first went through the concept —

plug the given Hamiltonian and wave function into the time-dependent Schrödinger equation and solve:

$$H\psi \quad =\quad i\hbar \frac {\partial \psi}{\partial t}\\ \left( H^{0}\quad +\quad H'(t) \right) \sum {\quad}C_{n}(t)\psi_{n}e^{-i\frac {E_{n}}{\hbar}t}\quad =\quad i\hbar \frac {\partial}{\partial t}\sum {\quad}C_{n}(t)\psi_{n}e^{-i\frac {E_{n}}{\hbar}t} \\ \sum {\quad}C_{n}(t)\left[ H^{0}\psi_{n} \right] e^{-i\frac {E_{n}}{\hbar}t}\quad +\quad \sum {\quad}C_{n}(t)\left[ H'(t)\psi_{n} \right] e^{-i\frac {E_{n}}{\hbar}t}\quad =\quad i\hbar \sum {\quad}\dot {C_{n}}(t)\psi_{n}e^{-i\frac {E_{n}}{\hbar}t}\quad +\quad \sum {\quad}C_{n}(t)E_{n}\psi_{n}e^{-i\frac {E_{n}}{\hbar}t} \\ \text{the red ones go splat — dead, gone.}\\ \sum {\quad}C_{n}(t)\left[ H'(t)\psi_{n} \right] e^{-i\frac {E_{n}}{\hbar}t}\quad =\quad i\hbar \sum {\quad}\dot {C_{n}}(t)\psi_{n}e^{-i\frac {E_{n}}{\hbar}t}$$$$\text{Back when it was a two-level system,}\\ \text{we multiplied by}\quad <\quad \psi_{1}\quad |,\quad <\quad \psi_{2}\quad |\quad \text{one at a time,} \\ \text{but here we want to hit both sides with}\quad <\quad \psi_{m}\quad |.$$$$\text{Multiplying both sides by}\quad <\quad \psi_{m}\quad |,\\ \sum _{n}^{\quad}{\quad}C_{n}(t)\left< \psi_{m}|H'(t)\psi_{n} \right> e^{-i\frac {E_{n}}{\hbar}t}\quad =\quad i\hbar \sum _{n}^{\quad}{\quad}\dot {C_{n}}(t)\left< \psi_{m}|\psi_{n} \right> e^{-i\frac {E_{n}}{\hbar}t}\\ \text{The red ones all die except when the indices match.}\quad \text{Walk through}\quad n=1,2,3,4,...,\quad \text{only the}\quad n=m\quad \text{term survives.} \\ \sum _{n}^{\quad}{\quad}C_{n}(t)\left< \psi_{m}|H'(t)\psi_{n} \right> e^{-i\frac {E_{n}}{\hbar}t}\quad =\quad i\hbar \dot {C_{m}}(t)e^{-i\frac {E_{m}}{\hbar}t} \\ \therefore \quad \dot {C_{m}}(t)\quad =\quad \frac {1}{i\hbar}e^{i\frac {E_{m}}{\hbar}t}\sum _{n}^{\quad}{\quad}C_{n}(t)\left< \psi_{m}|H'(t)\psi_{n} \right> e^{-i\frac {E_{n}}{\hbar}t}$$

$$\dot {C_{m}}(t)\quad =\quad \frac {1}{i\hbar}e^{i\frac {E_{m}}{\hbar}t}\sum _{n}^{\quad}{\quad}C_{n}(t)\left< \psi_{m}|H'(t)\psi_{n} \right> e^{-i\frac {E_{n}}{\hbar}t}\\ \text{first, sub}\quad N\quad \text{in for}\quad m\quad \text{go go go}\\ \dot {C_{N}}(t)\quad =\quad \frac {1}{i\hbar}\sum _{n}^{\quad}{\quad}C_{n}(t)H'_{Nn}e^{i\frac {E_{N}-E_{n}}{\hbar}t} \\ \text{0th order is}\quad C_{N}(0)\quad =\quad 1,\quad \text{and every other}\quad C_{n}(0)\quad \text{is zero}\\ (\text{because if there were no perturbation, it'd just cruise on as is!}) \\ \dot {C_{N}}(t)\quad =\quad \frac {1}{i\hbar}C_{N}(t)H'_{NN}e^{i\frac {E_{N}-E_{N}}{\hbar}t}\quad =\quad \frac {1}{i\hbar}C_{N}(t)H'_{NN}$$

I really should label which order of perturbation I’m on… T_T T_T

$$\dot {C_{N}}^{(1)}(t)\quad =\quad \frac {1}{i\hbar}H'_{NN}\quad \text{integrate both sides go go go}\\ \dot {C_{N}}^{(1)}(t)\quad =\quad \frac {1}{i\hbar}\int _{0}^{t}{H'_{NN}(t')dt'}\quad +\quad constant \\ \text{the}\quad constant\quad \text{is just the 0th order, so it's}\quad 1!\\ \dot {C_{N}}^{(1)}(t)\quad =\quad 1-\frac {i}{\hbar}\int _{0}^{t}{H'_{NN}(t')dt'}$$$$\text{Also,} \\ \dot {C_{m}}(t)\quad =\quad \frac {1}{i\hbar}\sum _{n}^{\quad}{\quad}C_{n}(t)H'_{mn}e^{i\frac {E_{m}-E_{n}}{\hbar}t},\quad \text{as}\quad n\quad \text{runs}\quad 1,3,4,5,...,\quad \text{every}\quad n\neq N\quad \text{term zeroes out}\\ \dot {C_{m}}(t)\quad =\quad \frac {1}{i\hbar}C_{N}(t)H'_{mN}e^{i\frac {E_{m}-E_{N}}{\hbar}t}\quad \text{and 0th-order}\quad C_{N}(t)\quad =1,\quad \text{so,}\\ \quad =\quad -\frac {i}{\hbar}H'_{mN}e^{i\frac {E_{m}-E_{N}}{\hbar}t}\\ \therefore \quad \dot {C_{m}}^{(1)}(t)\quad =\quad -\frac {i}{\hbar}\int _{0}^{t}{\quad}H'_{mN}e^{i\frac {E_{m}-E_{N}}{\hbar}t'}dt'$$

$$\dot {C_{m}}^{(1)}(t)\quad =\quad -\frac {i}{\hbar}\int _{0}^{t}{\quad}H'_{mN}e^{i\frac {E_{m}-E_{N}}{\hbar}t'}dt'\\ \dot {C_{M}}^{(1)}(t)\quad =\quad -\frac {i}{\hbar}\int _{0}^{t}{\quad}H'_{MN}e^{i\frac {E_{M}-E_{N}}{\hbar}t'}dt'\quad \text{and the red thing is a constant,}\\ \\ =\quad -\frac {i}{\hbar}H'_{MN}\int _{0}^{t}{\quad}e^{i\frac {E_{M}-E_{N}}{\hbar}t'}dt'\quad =\quad -\frac {i}{\hbar}H'_{MN}\left[ \frac {1}{i\frac {E_{M}-E_{N}}{\hbar}}e^{i\frac {E_{M}-E_{N}}{\hbar}t'} \right]_{0}^{t}\\ =\quad -H'_{MN}\left( \frac {e^{i\frac {E_{M}-E_{N}}{\hbar}t}-1}{E_{M}-E_{N}} \right) \quad =\quad -H'_{MN}\left( \frac {e^{i\frac {E_{M}-E_{N}}{\hbar}t}-1}{E_{M}-E_{N}} \right) \frac {e^{-i\frac {E_{M}-E_{N}}{2\hbar}}}{e^{-i\frac {E_{M}-E_{N}}{2\hbar}}}\\ =\quad -\frac {H'_{MN}}{E_{M}-E_{N}}e^{i\frac {E_{M}-E_{N}}{2\hbar}}\left( e^{i\frac {E_{M}-E_{N}}{2\hbar}t}-e^{-i\frac {E_{M}-E_{N}}{2\hbar}t} \right) \\ =\quad -\frac {H'_{MN}}{E_{M}-E_{N}}e^{i\frac {E_{M}-E_{N}}{2\hbar}}\left( 2i\frac {e^{i\frac {E_{M}-E_{N}}{2\hbar}t}-e^{-i\frac {E_{M}-E_{N}}{2\hbar}t}}{2i} \right) \\ =\quad -2i\frac {H'_{MN}}{E_{M}-E_{N}}e^{i\frac {E_{M}-E_{N}}{2\hbar}}sin\left( \frac {E_{M}-E_{N}}{2\hbar}t \right) \\ \therefore \quad \dot {C_{M}}^{(1)}(t)\quad =\quad -2i\frac {H'_{MN}}{E_{M}-E_{N}}e^{i\frac {E_{M}-E_{N}}{2\hbar}}sin\left( \frac {E_{M}-E_{N}}{2\hbar}t \right)$$$$\therefore \quad \left| \dot {C_{M}}^{(1)}(t) \right|^{2}\quad =\quad 4\frac {\left| H'_{MN} \right|^{2}}{\left( E_{M}-E_{N} \right)^{2}}sin^{2}\left( \frac {E_{M}-E_{N}}{2\hbar}t \right)$$

$$\text{In the infinite potential well,}\\ \psi_{n}\quad =\quad \sqrt {\frac {2}{a}}sin\frac {n\pi}{a}x\\ E_{n}\quad =\quad \frac {\pi^{2}\hbar^{2}}{2ma^{2}}n^{2}$$$$\text{It says the thing started in the ground state, so}\\ C_{n}^{(1)}(t)\quad =\quad 1\quad \text{I guess,}\quad \text{but we're asked for the probability of transitioning to}\quad \psi_{2}$$$$\text{The}\\ \quad \left| \dot {C_{M}}^{(1)}(t) \right|^{2}\quad =\quad 4\frac {\left| H'_{MN} \right|^{2}}{\left( E_{M}-E_{N} \right)^{2}}sin^{2}\left( \frac {E_{M}-E_{N}}{2\hbar}t \right) \\ \text{we packaged up above is gonna be super handy!!!}$$$$\left| \dot {C_{2}}^{(1)}(t) \right|^{2}\quad =\quad 4\frac {\left| H'_{12} \right|^{2}}{\left( E_{M}-E_{N} \right)^{2}}sin^{2}\left( \frac {E_{2}-E_{1}}{2\hbar}t \right)$$$$E_{2}-E_{1}\quad =\quad \frac {\pi^{2}\hbar^{2}}{2ma^{2}}\left( 4-1 \right) \quad =\quad \frac {3\pi^{2}\hbar^{2}}{2ma^{2}}$$$$H'_{12}\quad =\quad \left< \psi_{1}|H'|\psi_{2} \right> \\ =\quad \int _{0}^{\frac {a}{2}}{\quad}\sqrt {\frac {2}{a}}sin\left( \frac {\pi}{a}x \right) V_{0}\sqrt {\frac {2}{a}}sin\left( \frac {2\pi}{a}x \right) dx\\ \\ =\quad \frac {2V_{0}}{a}\int _{0}^{\frac {a}{2}}{\quad}sin\left( \frac {\pi}{a}x \right) sin\left( \frac {2\pi}{a}x \right) dx\quad \\ =\quad \frac {2V_{0}}{a}\int _{0}^{\frac {a}{2}}{\quad}sin\left( \frac {\pi}{a}x \right) \left\{2sin\left( \frac {\pi}{a}x \right) cos\left( \frac {\pi}{a}x \right) \right\} dx\\ =\quad \frac {4V_{0}}{a}\int _{0}^{\frac {a}{2}}{\quad}sin^{2}\left( \frac {\pi}{a}x \right) cos\left( \frac {\pi}{a}x \right) dx$$$$*\quad u\text{-sub time}\\ sin\left( \frac {\pi}{a}x \right) \quad =\quad t\\ \frac {\pi}{a}cos\left( \frac {\pi}{a}x \right) dx\quad =\quad dt\\ cos\left( \frac {\pi}{a}x \right) dx\quad =\quad \frac {a}{\pi}dt$$$$=\quad \frac {4V_{0}}{a}\int _{0}^{\frac {a}{2}}{\quad}2sin^{2}\left( \frac {\pi}{a}x \right) cos\left( \frac {\pi}{a}x \right) dx\\ =\quad \frac {4V_{0}}{a}\int _{0}^{1}{\quad}\frac {a}{\pi}t^{2}dt\\ =\quad \frac {4V_{0}}{\pi}\left[ \frac {1}{3}t^{3} \right]_{0}^{1}\quad =\quad \frac {4V_{0}}{3\pi}$$$$\text{So,}\\ P_{1\to 2}(t)\quad =\left| \quad C_{2}^{(1)}(t) \right|^{2}\\ =\quad 4\frac {\left( \frac {4V_{0}}{3\pi} \right)^{2}}{\left( \frac {3\pi^{2}\hbar^{2}}{2ma^{2}} \right)^{2}}sin^{2}\left( \frac {\frac {3\pi^{2}\hbar^{2}}{2ma^{2}}}{2\hbar}t \right) \quad =\quad \left( \frac {16ma^{2}V_{0}}{9\pi^{3}\hbar^{2}} \right)^{2}sin^{2}\left( \frac {3\pi^{2}\hbar}{4ma^{2}}t \right) \\ \\ \quad =\quad \left[ \frac {16ma^{2}V_{0}}{9\pi^{3}\hbar^{2}}sin\left( \frac {3\pi^{2}\hbar}{4ma^{2}}t \right) \right]^{2}$$

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Originally written in Korean on my Naver blog (2015-12). Translated to English for gdpark.blog.