标量场理论

标量场是指场变量与参考系无关的场。在参考系变换下

$$\left\{\begin{array}{ll} x\rightarrow\bar x=\Lambda x+a & (\bar x^\mu=\Lambda^\mu_{\ \ \nu} x^\nu+a^\mu)\\ \phi\rightarrow\phi & (\bar\phi(\bar x)=\bar\phi(\bar x(x))=\phi(x)) \end{array}\right.$$

其中$\Lambda$是恰当正时Lorentz变换。假设$a=0$，则

$$\bar\phi(\bar x)=\phi(x)=\phi(\Lambda^{-1}\bar x)$$

对于无穷小变换$\Lambda^\mu_{\ \ \nu}=\delta^\mu_{\ \ \nu}+\theta^\mu_{\ \ \nu}$，其逆变换为$(\Lambda^{-1})^\mu_{\ \ \nu}=\delta^\mu_{\ \ \nu}-\theta^\mu_{\ \ \nu}$

$$\begin{aligned} \delta\phi(x)&\equiv\bar\phi(x)-\phi(x)=\phi(\Lambda^{-1}x)-\phi(x)\\ &=-\theta^\mu_{\ \ \nu}x^\nu\frac{\partial}{\partial x^\mu}\phi(x)\equiv-\theta_{\mu\nu}x^\nu\partial^\nu\phi(x) \end{aligned}$$

接下来会多次用到$\theta$的反对称性，证明思路如下：
根据间隔不变性有

$$\mathrm ds^2=g_{\mu\nu}x^\mu x^\nu=g_{\mu\nu}\Lambda^\mu_{\ \ \alpha}\Lambda^\nu_{\ \ \beta}x^\alpha x^\beta$$

因此$g_{\mu\nu}\Lambda^\mu_{\ \ \alpha}\Lambda^\nu_{\ \ \beta}=g_{\alpha\beta}$，将无穷小变换代入并略去$\theta$的高阶项得

$$g_{\alpha\nu}\theta^\nu_{\ \ \beta}+g_{\mu\beta}\theta^\mu_{\ \ \alpha}=\theta_{\alpha\beta}+\theta_{\beta\alpha}=0$$

对其进行进一步变换（第三个等号用到$\theta$反对称）

$$\begin{aligned} \delta\phi(x)&=-\theta_{\mu\nu}x^\nu\partial^\mu\phi(x)\\ &=-\frac12\left(\theta_{\mu\nu}x^\nu\partial^\mu+\theta_{\mu\nu}x^\nu\partial^\mu\right)\phi(x)\\ &=-\frac12\left(\theta_{\mu\nu}x^\nu\partial^\mu-\theta_{\nu\mu}x^\nu\partial^\mu\right)\phi(x)\\ &=-\frac12\left(\theta_{\mu\nu}x^\nu\partial^\mu-\theta_{\mu\nu}x^\mu\partial^\nu\right)\phi(x) \end{aligned}$$

定义线性变换

$$\mathscr L^{\mu\nu}=-i\left(x^\nu\partial^\mu-x^\mu\partial^\nu\right)$$

$\color{red}{\mathbf{注意}}$ 这个线性变换不是做在态矢量的Hilbert空间上，而是做在场算符组成的空间上
结论为：

$$\begin{aligned} \delta\phi(x)=&\frac i2\theta_{\mu\nu}\mathscr L^{\mu\nu}\phi(x)\\ =&i\left[\theta_{01}\mathscr L^{01}+\theta_{02}\mathscr L^{02}+\theta_{03}\mathscr L^{03}\right.\\ &+\theta_{12}\mathscr L^{12}+\theta_{23}\mathscr L^{23}+\theta_{31}\mathscr L^{31}\left.\right]\phi(x) \end{aligned}$$

容易看出，$\mathscr L^{12}$对应$z$方向角动量，同理可得其他六项的意义。
所以$\phi$与自旋无关，即对应无自旋粒子。

正则量子化

Klein-Gordon方程为

$$\left(\frac{\partial^2}{\partial t^2}-\nabla^2+\omega_0^2\right)\Psi=0$$

或写作

$$(-\partial^2+\omega_0^2)\Psi=0\tag{1.1}$$

我们假设场算符满足（反）对易关系，即

$$\phi(x)\phi(y)-\sigma\phi(y)\phi(x)=C(x-y)\tag{1.2}$$

其中$\sigma=\pm1$。为了确定具体是正是负，我们考虑

$$C(-x)=\phi(0)\phi(x)-\sigma\phi(x)\phi(0)=-\sigma C(x)\tag{1.3}$$

因此$\sigma$的正负性与$C(x)$的奇偶性相关。取$y=0$，将$(-\partial^2+\omega_0^2)$作用在(1.2)式上得

$$0=(-\partial^2+\omega_0^2)C(x)\tag{1.4}$$

为了求解$C(x)$，做变换

$$\left\{\begin{aligned} \tilde C(k)&=\int\mathrm{d}^4xe^{-ik\cdot x}C(x)\\ C(x)&=\int\frac{\mathrm{d}^4k}{(2\pi)^4}e^{ik\cdot x}\tilde C(k) \end{aligned}\tag{1.5}\right.$$

$$k\cdot x=g_{\mu\nu}k^\mu x^\nu=-k^0t+\vec k\cdot\vec x$$

因此(1.4)化为

$$0=\int\frac{\mathrm{d}^4k}{(2\pi)^4}(k^2+\omega_0^2)e^{ik\cdot x}\tilde C(k)\tag{1.6}$$

这说明除了$k^2+\omega_0^2=0$的地方以外，$\tilde C(k)\equiv0$。
根据Lorentz不变性

$$\begin{aligned}\tilde C(\Lambda k)=&\int\mathrm{d}^4(\Lambda^{-1}x)e^{-i(\Lambda k)\cdot(\Lambda^{-1}x)}C(\Lambda^{-1}x)\\ =&\int\mathrm{d}^4(\Lambda^{-1}x)e^{-ik\cdot x}C(\Lambda^{-1}x)\\ =&\int\mathrm{d}^4xe^{-ik\cdot x}C(x)=\tilde C(k)\end{aligned}\tag{1.7}$$

因此$\tilde C(k)$在$k_0=\sqrt{\vec k^2+\omega_0^2}\equiv\omega_{\vec k}$上取值均相等，在$k_0=-\omega_{\vec k}$上取值也均相等。

$$\tilde C(k)=c_+(2\pi)\delta(k^2+\omega_0^2)\theta(k^0)+c_-(2\pi)\delta(k^2+\omega_0^2)\theta(-k^0)\tag{1.8}$$

之所以选择$\delta(k^2+\omega_0^2)$而非$\delta(k^0\pm\omega_{\vec k})$是为了满足Lorentz不变性的要求。而两者之间有联系

$$\delta(k^2+\omega_0^2)=\frac{1}{2\omega_{\vec k}}\left(\delta(k^0+\omega_{\vec k})+\delta(k^0-\omega_{\vec k})\right)\tag{1.9}$$

将(1.8)(1.9)代入(1.5)变换回坐标表象得

$$C(x)=\int\frac{\mathrm{d}^3\vec k}{(2\pi)^3}\frac{1}{2\omega_{\vec k}}\left[c_+e^{-i\omega_{\vec k}t+i\vec k\cdot\vec x}+c_-e^{i\omega_{\vec k}t+i\vec k\cdot\vec x}\right]\tag{1.10}$$

取类空坐标$C(t=0,\vec x)=0$，即得到$c_++c_-=0$，于是

$$C(x)=c_0\int\frac{\mathrm{d}^3\vec k}{(2\pi)^3}\frac{1}{2\omega_{\vec k}}\left[e^{-i\omega_{\vec k}t+i\vec k\cdot\vec x}-e^{i\omega_{\vec k}t+i\vec k\cdot\vec x}\right]\tag{1.11}$$

因此

$$C(-x)=c_0\int\frac{\mathrm{d}^3\vec k}{(2\pi)^3}\frac{1}{2\omega_{\vec k}}\left[e^{i\omega_{\vec k}t-i\vec k\cdot\vec x}-e^{-i\omega_{\vec k}t-i\vec k\cdot\vec x}\right]=-C(x)\tag{1.12}$$

第二个等号取了$\vec k\rightarrow-\vec k$。根据(1.3)可知$\sigma=1$，即场算符对易关系为

$$[\phi(x),\phi(0)]=c_0\int\frac{\mathrm{d}^3\vec k}{(2\pi)^3}\frac{1}{2\omega_{\vec k}}\left[e^{-i\omega_{\vec k}t+i\vec k\cdot\vec x}-e^{i\omega_{\vec k}t+i\vec k\cdot\vec x}\right]\tag{1.13}$$

未完待续