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\title{Griffith QM Time Dependent Perturbation Theory CheatSheet}
\begin{document}
\raggedright
\footnotesize
\begin{center}
\Large{\textbf{Griffith QM Time Dependent Perturbation Theory CheatSheet (UCB 137B)}} \\
\end{center}
\begin{multicols}{3}
\setlength{\premulticols}{1pt}
\setlength{\postmulticols}{1pt}
\setlength{\multicolsep}{1pt}
\setlength{\columnsep}{2pt}
\section{TIPT}
\begin{align*}
H = H_0 + H' \\
E_n = E_n^0 + E_n^1 \\
|\psi_n\rangle = |\psi_n^0\rangle + |\psi_n^1\rangle \\
E_n^1 = \langle \psi_n^0 | H' | \psi_n^0 \rangle \\
|\psi_n^1\rangle = \sum_{m\neq n} \frac{\langle \psi_m^0 | H' | \psi_n^0 \rangle}{E_n^0-E_m^0} | \psi_m^0\rangle
\end{align*}
\subsection{Degenerate Case}
\begin{align*}
\text{Degenerate space: } \{|i\rangle\} \to E\\
\quad W_{ab} = \langle a | H' | b \rangle \text{ Non-Diagonal} \\
\text{Eigenvalue and Eivenvectors} \to E_n^1, |\hat{i}\rangle
\end{align*}
\section{Variational Method}
\begin{align*}
\langle H \rangle(\lambda) &=
\frac{\langle \psi(x,\lambda)|H|\psi(x,\lambda) \rangle}{\langle \psi(x,\lambda)|\psi(x,\lambda) \rangle} \\
\langle H \rangle(\lambda) &\geq E_{.g.s} \\
\frac{d}{d\lambda} \langle H \rangle(\lambda_0) = 0 &\Rightarrow
\langle H \rangle(\lambda_0) \approx E_{.g.s}
\end{align*}
\section{WKB Method}
\begin{align*}
\frac{d^2 \psi(x)}{dx^2} &= -k^2(x) \psi(x)& \\ k(x)&=\frac1\hbar \sqrt{2m(E-V(x)} \\
\phi(x) &= \int^x k(x) dx \\
\psi(x) &= \frac1{\sqrt{k(x)}} (C_+ e^{i \phi(x)} + C_- e^{-i \phi(x)})\\
&= \frac1{\sqrt{k(x)}} (C_1 \sin{\phi(x)} + C_2 \cos{\phi(x)})
\end{align*}
\subsection{Energy Level}
\begin{align*}
\int_{R_{classical}} k(x) dx = n\pi \\
\text{one $\infty$ wall} \quad n \to n-1/4 \\
\text{No $\infty$ wall} \quad n \to n-1/2
\end{align*}
\subsection{Tunneling}
\begin{align*}
T = e^{-2\gamma} \\
\gamma = \int_{R_{forbidden}} k(x) dx
\end{align*}
\section{TDPT}
\begin{align*}
H = H_0 + V(t) \\
\text{Eigenstate of $H_0$: } |n\rangle, E_n\\
\text{transition: } |i\rangle \to | f\rangle \\
V_{f i}(t) = \langle f | V(t) | i \rangle \\
\omega_{f i} = (E_f-E_i)/\hbar \\
c_f(T) = \frac{-i}{\hbar} \int_0^T V_{f i}(t) e^{-i \omega_{f i} t} dt
\end{align*}
\subsection{Constant Perturbation}
\begin{align*}
V(t) = \begin{cases}
V, & 0 \leq t \leq T \\
0, & \text{otherwise}
\end{cases} \\
V_{fi}(t)= constant \\
P_{i\to f}(t)= |c_f(t)|^2 = 4 \frac{|V_{fi}|^2}{\hbar^2} \frac{\sin^2(\omega_{fi})t/2}{\omega_{fi}^2} \\
\omega_{fi} \to 0 \quad \text{(degenerate states):} \\
|c_f(t)|^2 = \frac{|V_{fi}|^2}{\hbar^2} t^2
\end{align*}
\subsection{Absorption}
\begin{align*}
V(t) = V \sin(\omega t) \\
P_{i\to f}(t) = \frac{|V_{fi}|^2}{\hbar^2} \frac{\sin^2((\omega_{fi}-\omega)t/2)}{(\omega_{fi}-\omega)^2}
\end{align*}
\subsection{Simulated Emission}
\begin{align*}
E_i > E_f,\quad \omega_{fi}<0 \\
P_{i\to f}(t) = \frac{|V_{fi}|^2}{\hbar^2} \frac{\sin^2((\omega_{fi}+\omega)t/2)}{(\omega_{fi}+\omega)^2}
\end{align*}
\subsection{Fermi Golden Rule}
\begin{align*}
E_i \to E_f\text{ (continuous states)} \\
P_{i\to f} =
\frac{2\pi}{\hbar} |\langle f | V| i\rangle |^2 \rho(E_f) t \\
\end{align*}
\subsection{Selection Rule}
For spherical symmetric potential:
\begin{align*}
\langle n',l',m'| \vec{r} | n,l,m \rangle &\neq 0 \text{ when:} \\
\Delta l &= \pm 1 \text{ and:}\\
\Delta l &= \pm 1 \text{ or } 0
\end{align*}
\section{Scattering}
\begin{align*}
\psi(r,\theta) &= e^{ikz} + f(\theta) \frac{e^{ikr}}{r}, \text{ for large r} \\
k &= \frac{\sqrt{2mE}}\hbar \\
\frac{d \sigma}{d \Omega} &= |f(\theta)|^2 \\
\sigma &= \int d\omega \frac{d \sigma}{d \Omega}
\end{align*}
\subsection{Born Approximation}
\begin{align*}
f(\theta) = -\frac{m}{2\pi \hbar^2} \int V(\vec{r}) e^{i(\vec{k}'-\vec{k})\cdot \vec{r}} d^3\vec{r}
\intertext{Low Energy:}
f(\theta) = -\frac{m}{2\pi \hbar^2} \int V(\vec{r}) d^3\vec{r}
\intertext{Spherical symmetric:}
f(\theta) = -\frac{2m}{\hbar^2 \kappa} \int_0^\infty r V(r) \sin(\kappa r) dr \\
\kappa = 2k\sin(\theta/2)
\end{align*}
\subsubsection{Yukawa Potential}
\begin{align*}
V(r) = V_0 \frac{e^{-r/R}}{r} \\
f(\theta) = -\frac{2mV_0 R^2}{\hbar^2} \frac{1}{1+4k^2R^2\sin^2(\theta/2)} \\
\sigma = (\frac{2mV_0R^2}{\hbar^2})^2 \frac{4\pi}{1+4k^2R^2}
\end{align*}
\subsubsection{Rutherford Scattering}
Let $V_0 = q_1q_2/4\pi \epsilon_0$, $R=\infty$:
\begin{align*}
f(\theta) = -\frac{2mq_1q_2}{4\pi\epsilon_0\hbar^2\kappa^2}
\end{align*}
\subsection{Partial Waves}
\begin{align*}
f(\theta) &= \frac1k \sum_{i=0}^\infty (2l+1)e^{i \delta_l} \sin(\delta_l) P_l(\cos(\theta)) \\
\sigma &= \frac{4\pi}{k^2}\sum_{l=0}^\infty (2l+1) \sin^2(\delta_l)
\end{align*}
\subsubsection{Optical Theorem}
\begin{equation*}
Im[f(0)] = \frac{k \sigma}{4\pi}
\end{equation*}
\subsubsection{Hard Ball}
\begin{align*}
\delta_l = \tan^{-1}(\frac{j_l(ka)}{\eta_l(ka)}) \\
ka << 1 \to \sigma = 4\pi a^2
\end{align*}
\section{Useful Models}
\subsection{Density of States}
\begin{align*}
E &= \hbar^2 k^2 /2m \\
dN &= \frac{L^3}{(2\pi)^3} d^3k = \frac{L^3}{(2\pi)^3} d\Omega dk \\
dN &= \frac{L^3}{(2\pi)^3} 4\pi \frac{m}{\hbar^2 k} dE \\
\rho(E) &= \frac{dN}{dE} = \frac{L^3}{2\pi^2} \frac{mk}{\hbar^2}
\end{align*}
\subsection{infinite square well}
\begin{align*}
H(x) = \frac{p^2}{2m} + \begin{cases}
0, & 0\leq x\leq a \\
\infty, & \text{otherwise}
\end{cases} \\
E_n = \frac1{2m} (\frac{n \pi \hbar}{a})^2 \\
\psi_n = \sqrt{\frac{2}{a}} \sin(\frac{n\pi x}{a}) e^{-iE_nt/\hbar}
\end{align*}
\subsection{Harmonic Oscillator}
\begin{align*}
H(x) &= \frac{p^2}{2m} + \frac12 m\omega^2x^2 \\
E_n &= (n+1/2)\hbar \omega \\
\psi_n(x) &= \frac1{\sqrt{2^n n!}} (\frac{m\omega}{\pi \hbar})^{1/4} e^{-\zeta^2/2} H_n(\zeta) \\
\zeta &= \sqrt{\frac{m\omega}{\hbar}x}
\end{align*}
\subsection{Virial Theorem}
\begin{align*}
2 \langle T \rangle = \langle \vec{r} \cdot \nabla V \rangle
\quad \text{(3D)}\\
2 \langle T \rangle = \langle x \frac{dV}{dx} \rangle \quad \text{(1D)}\\
2 \langle T \rangle = n \langle V \rangle \quad (V\propto r^n) \\
\langle T\rangle = -E_n, \quad \langle V \rangle = 2 E_n \quad \text{(hydrogen)}\\
\langle T\rangle = \langle V \rangle = E_n/2 \quad \text{(harmonic oscillator)}\\
\end{align*}
\section{Math}
\subsection{Legendre Polynomials}
Domain: $(-1,1)$ \\
Even, Odd, Even, Odd ...
\begin{align*}
P_0(x) &= 1 \\
P_1(x) &= x \\
P_2(x) &= \frac12 (3x^2-1) \\
P_3(x) &= \frac12 (5x^3 - 3x)
\end{align*}
\subsection{Hankel Functions}
Solution to Radial Shrodinger Equation:
\begin{align*}
-\frac{\hbar^2}{2m} \frac{1}{r^2} \frac{\partial}{\partial r} (r^2 R_{El}) + [V(r) + \frac{\hbar^2 l (l+1)}{2m r^2}]R_{El} = E R_{El} \\
V = 0 \to R_{El} = j_l(kr) \\
V \neq 0 \to R_{El} = j_l(kr+\delta_l) \\
r \to \infty \Rightarrow R_{El} = \frac{\sin(kr-l\pi/2+\delta_l(E))}{kr}
\end{align*}
When $kr >> 1$
\begin{align*}
j_l(kr) &\to \frac{\sin{kr-l\pi/2}}{kr} \\
\eta_l(kr) &\to \frac{-\cos{kr-l\pi/2}}{kr} \\
h_l(kr) &\to \frac{e^{i(kr-l\pi/2)}}{ikr} \\
h_l^*(kr) &\to \frac{e^{-i(kr-l\pi/2)}}{-ikr} \\
j_l(kr) &= \frac{1}{2} (h_l(kr)+h^*(kr))
\end{align*}
\subsection{Hermite Polynomials}
Domain: $(-\infty,\infty)$ \\
Even, Odd, Even, Odd ...
\begin{align*}
H_0(x) &= 1 \\
H_1(x) &= 2x \\
H_2(x) &= 4x^2-2 \\
H_3(x) &= 8x^3-12x
\end{align*}
\subsection{Spherical Harmonics}
\begin{align*}
|l,m \rangle &= Y_l^m(\theta, \phi) \\
Y_0^0(\theta,\phi) &= \frac12 \frac1{\sqrt{\pi}} \\
Y_1^0(\theta,\phi) &= \frac12 \sqrt{\frac3\pi} \cos{\theta} \\
Y_1^{-1}(\theta,\phi) &= \frac12 \sqrt{\frac{3}{2\pi}} \sin{\theta} e^{-i \phi} \\
Y_1^{-1}(\theta,\phi) &= -\frac12 \sqrt{\frac{3}{2\pi}} \sin{\theta} e^{i \phi}
\end{align*}
\subsection{Green's Function}
For a Linear Operator $\hat{D}_x$
\begin{align*}
\text{Homogeneous solution: }\hat{D}_x \psi_0(x) = 0 \\
\text{Hard Problem: }\hat{D}_x \psi(x) = f(x) \\
\text{Simple Problem: }\hat{D}_x G(x,x') = \delta(x-x') \\
\psi(x) = \psi_0(x) + \int_{\text{f Domain}} G(x,x') f(x') dx'
\end{align*}
\subsection{Some Integrals}
\begin{align*}
\Gamma(n+1) &= n! \\
\Gamma(z+1) &= z \Gamma(z) \\
\int_0^\infty x^n e^{-ax} dx &= \frac{n!}{a^{n+1}} \\
\int_0^\infty e^{-ax^b} dx &= a^{-1/b} \Gamma(1/b+1) \\
\int_0^\infty e^{-ax} \sin{bx} dx &= \frac{b}{a^2+b^2} \\
\int_0^\infty e^{-ax} \cos{bx} dx &= \frac{a}{a^2+b^2} \\
\int_{-\infty}^\infty e^{-ax^2+bx} dx &= \sqrt{\frac{\pi}{a}} e^{\frac{b^2}{4a}} \\
\int_0^\infty e^{-ax^2}x^n dx &= I_n(a)\\
I_0=\frac12 \sqrt{\frac{\pi}{a}}, I_1&=\frac{1}{2a}, I_2=\frac1{4a} \sqrt{\frac{\pi}{a}}, I_3=\frac{1}{2a^2}
\end{align*}
\end{multicols}
\end{document}