# Path Integrals

$$\Psi(x'',t)=\int dx' K(x'',t,x',t_0)\Psi(x',t_0)$$, $$K(x'',t,x',t_0)=\langle x_N|U(t,t_0)|x_0\rangle$$.

$$K(x_n,t_n,x_0,t_0)=\lim_{N\to\infty}\left(\frac{m}{2\pi i\hbar \varepsilon}\right)^{N/2}\int_{\infty}^\infty dx_1\cdots\int_{\infty}^\infty dx_{N-1}\exp\frac{i}{\hbar}\sum_{n=1}^N\varepsilon\sum\frac{m(x_n-x_{n-1})^2}{2\varepsilon^2} = \int_{x_0}^{x_N}\mathcal{D}[x(t)]\exp\frac{i}{\hbar}\int_{t_0}^t\mathcal{L}_{classical}dt$$

## Free Particle

$$t_n-t_{n-1}=\varepsilon$$ $$\mathcal{L}=\frac{1}{2}m\dot{x}^2, S=\int_{t_0}^{t_N}\mathcal{L}dt = \sum_{n=1}^N\frac{1}{2}m\left(\frac{x_n-x_{n-1}}{t_n-t_{n-1}}\right)^2(t_n-t_{n-1})$$ Thus, $$K(x_N,t_N,x_0,t_0)=\int_{x_0}^{x_N}\mathcal{D}[x(t)]\exp\frac{i}{\hbar}S$$ Hence, $$\mathcal{D}[x(t)] = \left(\frac{m}{2\pi i\hbar \varepsilon}\right)^{N/2}\int_{\infty}^\infty dx_1\cdots\int_{\infty}^\infty dx_{N-1}$$.

### Single Step

$$\Psi(x'',t)=\int_\mathbb{R}K\Psi(x',t_0)$$ $$K(x_0+\eta,t_0+\varepsilon;x_0,t_0) = \left(\frac{m}{2\pi i\hbar \varepsilon}\right)^{1/2}\exp\frac{i}{\hbar}\frac{m\eta^2}{2\varepsilon}$$ Let $$x=x_0+\eta, t=t_0+\varepsilon$$. Then, $$\Psi(x,t)=\left(\frac{m}{2\pi i\hbar \varepsilon}\right)^{1/2}\int_\mathbb{R}d\eta \Psi(x_0,t_0) \exp\frac{i}{\hbar}\frac{m\eta^2}{2\varepsilon}=\left(\frac{m}{2\pi i\hbar \varepsilon}\right)^{1/2}\int_\mathbb{R}d\eta \Psi(x-\eta,t_0) \exp\frac{i}{\hbar}\frac{m\eta^2}{2\varepsilon}=\left(\frac{m}{2\pi i\hbar \varepsilon}\right)^{1/2}\int_\mathbb{R}d\eta \left[\Psi(x,t_0)-\eta\frac{\partial Psi}{\partial x}+\frac{1}{2}\eta^2\frac{\partial^2\Psi}{\partial x^2}\right] \exp\frac{i}{\hbar}\frac{m\eta^2}{2\varepsilon}=\left(\frac{m}{2\pi i\hbar \varepsilon}\right)^{1/2}\left[\Psi(x,t_0)\left(\frac{2\pi i\hbar \varepsilon}{m}\right)^{1/2} + 0 + \frac{\hbar\varepsilon}{2im}\left(\frac{2\pi i\hbar\varepsilon}{m}\right)^{1/2}\frac{\partial^2 \Psi}{h\partial x^2}\right] = \Psi(x,t_0)+\frac{\hbar i}{2m}\frac{\partial^2 \Psi}{\partial x^2}\varepsilon$$

So, $$\Psi(x,t_0+\varepsilon)-\Psi(x,t_0)=\frac{\partial \Psi}{\partial t}\varepsilon = \frac{i\hbar\varepsilon}{2m}\Psi''$$ hence $$i\hbar\dot{\Psi}=\frac{p^2}{2m}\Psi$$

## Usefulness

Used in Field Theory but not in typical particle QM.

Created: 2024-05-30 Thu 21:18

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