equ

I Like This

Figure 1: 嘤嘤嘤

$$S = \int_{t_1}^{t_2} L(q, \dot{q}, t) dt$$

$$(1)$$

The formula 式. 1 declares the definition of action.

Table 1: 今晚加鸡腿

A	B	C
1	2	3

$$jb \sin$$

$$(2)$$

式. 3

$$no as$$

$$(3)$$

$$\psi(x)=\left\{ \begin{aligned} &\frac{D}{\sqrt{|p(x)|}}\exp \left( -\frac{1}{\hbar}\int_{x_2}^{x}|p(x')|dx' \right), \quad (x > x_2) &(a)\\ &\frac{2D}{\sqrt{p(x)}}\sin \left( \frac{1}{\hbar}\int_{x}^{x_2}p(x')dx' + \frac{\pi}{4} \right), \quad (x_1 < x < x_2) &(b)\\ &\frac{D}{\sqrt{|p(x)|}}\left[ 2\cos\theta \cdot \exp \left( \frac{1}{\hbar}\int_{x}^{x_1}|p(x')|dx' \right) \right. \\ &\left. +\sin\theta \cdot \exp \left( -\frac{1}{\hbar}\int_{x}^{x_1}|p(x')|dx' \right) \right], \quad (x < x_1) &(c)\\ \end{aligned} \right.$$

$$(4)$$

do you like 式. 4(a)?

去找喵星人

$$\psi(x)=\left\{ \begin{aligned} &\frac{D}{\sqrt{|p(x)|}}\exp \left( -\frac{1}{\hbar}\int_{x_2}^{x}|p(x')|dx' \right), \quad (x > x_2) &(a)\\ &\frac{2D}{\sqrt{p(x)}}\sin \left( \frac{1}{\hbar}\int_{x}^{x_2}p(x')dx' + \frac{\pi}{4} \right), \quad (x_1 < x < x_2) &(b)\\ &\frac{D}{\sqrt{|p(x)|}}\left[ 2\cos\theta \cdot \exp \left( \frac{1}{\hbar}\int_{x}^{x_1}|p(x')|dx' \right) \right. \\ &\left. +\sin\theta \cdot \exp \left( -\frac{1}{\hbar}\int_{x}^{x_1}|p(x')|dx' \right) \right], \quad (x < x_1) &(c)\\ \end{aligned} \right.$$

$$(5)$$

式. 1, 3, 4

fig. 1 吗？

no ass

Warning lala

This document is protected by ...

Warning Header

Welcome to Hexo!