Mathematical Physics by Carl Bender

list

• https://www.youtube.com/watch?v=LYNOGk3ZjFM&list=PLB24487A8C5EDC02A

lecture 1 perturbation method

• solve $x^5+x=1$

method 1

• try $x^5+\epsilon x=1$
• find $x(\epsilon)$ satisfying $x(\epsilon)^5+\epsilon x(\epsilon)=1$
• answer

$$x(\epsilon)=1-\frac{\epsilon }{5}-\frac{\epsilon ^2}{25}-\frac{\epsilon ^3}{125}+\frac{21 \epsilon ^5}{15625}+\frac{78 \epsilon ^6}{78125}+\cdots$$

• Setting $\epsilon=1$ gives numerical value $0.75\cdots$

weak coupling approach

• use the similar idea to Feynman diagrams
• try $\epsilon x^5+ x=1$
• we get

$$ x(\epsilon)=1-\epsilon +5 \epsilon ^2-35 \epsilon ^3+285 \epsilon ^4-2530 \epsilon ^5+23751 \epsilon ^6+\cdots $$

• can we get a meaningful number out of this?
• yes, for example, Pade summation can be used

asymptotics

• $f\sim g\, \quad (x\to x_0)$ iff $$\lim_{x\to x_0}\frac{f(x)}{g(x)}=1$$
• apply the method of dominant balance to $\epsilon x^5+ x=1$
• $x^4\sim -1/\epsilon\, \quad (\epsilon \to 0)$ and thus

$$x\sim \frac{\omega}{\epsilon^{1/4}}\, \quad (\epsilon \to 0)$$ where $\omega^4=-1$

• this is the first order approximation and we can have more terms

lecture 2

• 틀:수학노트
• $y+Q(x)y=0$ Schrodinger equation
• this is a very hard problem to solve
• consider a perturbed equation $y+\epsilon Q(x)y=0$ so that its unperturbed equation is $y=0$ with initial conditions $y(0)=\alpha, y'(0)=\beta$
• take the formal solution $y(x)=\sum_{n}a_n(x)\epsilon^n$ where $a_0(x)=\alpha+\beta x$
• for it to be a solution, it should satisfy

$$ a_n''(x)=-Q(x)a_{n-1}(x) $$ for each $n>0$

• thus we get

$$ a_n(x)=-\int_0^x\int_0^t Q(s)a_{n-1}(s)\,ds\,dt $$

computational resource

• https://docs.google.com/file/d/0B8XXo8Tve1cxTmdLemI1MS1XQzQ/edit

books

• Bender, Carl M., and Steven A. Orszag. 1999. Advanced Mathematical Methods for Scientists and Engineers I: Asymptotic Methods and Perturbation Theory. Springer.