Klein-Gordon场量子化

Klein-Gordon方程为

$$\left(-\partial^2+\omega_0^2\right)\psi=0\tag{1}$$

在之前的文章中已经给出对易关系

$$[\phi(x),\phi(y)]=c_+\int\frac{\mathrm{d}^3\vec k}{(2\pi)^3}\frac1{2\omega_{\vec k}}\left[e^{-i\omega_{\vec k}t+i\vec k\cdot\vec x}-e^{i\omega_{\vec k}t+i\vec k\cdot\vec x}\right]\tag{2}$$

其中

$$\omega_{\vec k}=\sqrt{\vec k^2+\omega_0^2}\tag{3}$$

我们将在其他地方证明

$$[\phi(x),\phi(y)]=c_+\mathrm{sign}(x^0-y^0)\left[i\omega_0\theta(\tau^2)\frac{J_1(\omega_0\tau)}{4\pi\tau}-\frac i{2\pi}\delta(\tau^2)\right]\tag{4}$$

其中$\tau=\sqrt{-(x-y)^2}$。我们将从此处继续完成Klein-Gordon场的量子化。

计算$c_+$

因为场算符是厄米的，所以（第二个等号用到厄米性）

$$[\phi(x),\phi(y)]^*=[\phi^\dagger(y),\phi^\dagger(x)]=[\phi(y),\phi(x)]=-[\phi(x),\phi(y)]$$

最后一步仅仅用到对易子的定义。所以

$$C^*(x)=-C(x)$$

根据(4)可知$c_+\in\mathbb R$。又因为场算符可以任意伸缩，只需要伸缩量为实数以保持厄米性，所以可以另$c_+=\pm1$。

接下来要计算到底是哪一个，工具是利用哈密顿量有下界。首先我们有

$$[\phi(t,\vec x),\phi(t,\vec y)]\equiv 0\tag{5}$$

当$\vec x=\vec y$时是显然的，否则是因为类空。进一步可以计算

$$\begin{aligned} &[\phi(t,\vec x),\dot\phi(t,\vec y)]\\ =&\lim_{\epsilon\rightarrow0}\left[\phi(t,\vec x),\frac1\epsilon(\phi(t+\epsilon,\vec y)-\phi(t,\vec y))\right]\\ =&\lim_{\epsilon\rightarrow0}\frac1\epsilon[\phi(t,\vec x),\phi(t+\epsilon,\vec y)]\\ =&\lim_{\epsilon\rightarrow0}\frac{c_+}\epsilon\mathrm{sign}(-\epsilon)\left[i\omega_0\theta(\epsilon^2-|\vec x-\vec y|^2)\frac{J_1(\omega_0\sqrt{\epsilon^2-|\vec x-\vec y|^2})}{4\pi\sqrt{\epsilon^2-|\vec x-\vec y|^2}}-\frac i{2\pi}\delta(\epsilon^2-|\vec x-\vec y|^2)\right] \end{aligned}\tag{6}$$

当$\vec x-\vec y\neq\vec 0$时，只需$|\epsilon|<|\vec x-\vec y|$即可得到$\theta,\delta$函数均为$0$，所以$[\phi(t,\vec x),\dot\phi(t,\vec y)]\propto\delta(\vec x-\vec y)$。

(6)式第一项中的$J_1(x)=\frac x2-\frac{x^3}{16}+o(x^5)$，所以对$\vec x$积分后领头阶为$\epsilon^3$，在趋于零的时候贡献为$0$。所以对$\vec x$全空间积分为

$$\begin{aligned} \int[\phi(t,\vec x),\dot\phi(t,\vec y)]\mathrm{d}^3\vec x=&\lim_{\epsilon\rightarrow0}-\frac{ic_+}{2\pi\epsilon}\mathrm{sign}(-\epsilon)\int\mathrm{d}^3\vec x\ \delta(\epsilon^2-|\vec x-\vec y|^2)\\ =&\lim_{\epsilon\rightarrow0}-\frac{ic_+}{2\pi\epsilon}\mathrm{sign}(-\epsilon)\int_0^\infty 4\pi r^2\mathrm{d}r\delta(\epsilon^2-r^2)\\ =&\lim_{\epsilon\rightarrow0}-\frac{ic_+}{2\pi\epsilon}\mathrm{sign}(-\epsilon)\int_0^\infty 4\pi r^2\mathrm{d}r\frac{1}{2|\epsilon|}\delta(\epsilon-r)\\ =&ic_+ \end{aligned}$$

最后一步是注意到符号函数。因此

$$\left\{\begin{array}{l} [\phi(t,\vec x),\dot\phi(t,\vec y)]=ic_+\delta(\vec x-\vec y)\\ [\phi(t,\vec x),\phi(t,\vec y)]=0 \end{array}\right.\tag{8}$$

根据Heisenberg绘景下的运动方程

$$i\dot\phi(x)=[\phi(x),\hat H]\Rightarrow i\ddot\phi(x)=[\dot\phi(x),\hat x]\tag{9}$$

再根据Klein-Gordon方程

$$(-\partial^2+\omega_0^2)\phi=0\Rightarrow[\dot\phi(x),\hat H]=i(\nabla^2\phi-\omega_0^2\phi)\tag{10}$$

利用(9)“推出”符号左边的等式猜测$\hat H$中有$\frac{1}{2c_+}\int\mathrm{d}^3\vec y\ \dot\phi^2(t,\vec y)$项，再利用(10)“推出”符号右边的等式猜测

$$\hat H=\frac{1}{2c_+}\int\mathrm{d}^3\vec x\left[\dot\phi^2+(\nabla\phi)^2+\omega_0^2\phi^2\right]\tag{11}$$

验证(10)式如下:

$$\begin{aligned} &[\dot\phi(x),\hat H]=\frac1{2c_+}\int\mathrm{d}^3\vec y\ [\dot\phi(x),(\nabla_{\vec y}\phi(\vec y))^2+\omega_0^2\phi^2(\vec y)]\\ =&\frac1{2c_+}\int\mathrm{d}^3\vec y\ [\dot\phi(x),(\nabla_{\vec y}\phi(\vec y))^2]+\frac1{2c_+}\int\mathrm{d}^3\vec y\ [\dot\phi(x),\omega_0^2\phi^2(\vec y)]\\ =&\frac1{2c_+}\int\mathrm{d}^3\vec y\ [\dot\phi(x),(\nabla_{\vec y}\phi(\vec y))^2]+\frac1{c_+}\int\mathrm{d}^3\vec y\ [-ic_+\delta(\vec x-\vec y)]\omega^2_0\phi(\vec y)\\ =&\frac1{2c_+}\int\mathrm{d}^3\vec y\ [\dot\phi(x),(\nabla_{\vec y}\phi(\vec y))^2]-i\omega^2_0\phi(\vec x)\\ =&\frac1{2c_+}\int\mathrm{d}^3\vec y\ \nabla_{\vec y}\phi(\vec y) [\dot\phi(x),\nabla_{\vec y}\phi(\vec y)]\\ &+\frac1{2c_+}\int\mathrm{d}^3\vec y\ [\dot\phi(x),\nabla_{\vec y}\phi(\vec y)]\nabla_{\vec y}\phi(\vec y)-i\omega_0^2\phi(\vec x)\\ =&\frac1{c_+}\int\mathrm{d}^3\vec y\ \nabla_{\vec y}\phi(\vec y)\nabla_{\vec y}[-ic_+\delta(\vec x-\vec y)]-i\omega_0^2\phi(\vec x)\\ =&-i\int\mathrm{d}^3\vec y\ \nabla_{\vec y}\phi(\vec y)\nabla_{\vec y}\delta(\vec x-\vec y)-i\omega_0^2\phi(\vec x)\\ =&i\int\mathrm{d}^3\vec y\ \delta(\vec x-\vec y)\nabla^2_{\vec y}\phi(\vec y)-i\omega_0^2\phi(\vec x)\\ =&i\nabla^2_{\vec x}\phi(\vec x)-i\omega_0^2\phi(\vec x)\\ \end{aligned}$$

根据哈密顿有有下界可知$c_+=1$，因为积分内明显是正定的。

还有一个小问题，这里没计算$[\dot\phi(t,\vec x),\dot\phi(t,\vec y)]$
不过可以将(11)式$\dot\phi\rightarrow\pi$认为是逻辑起点，正如大多数教材都从拉格朗日量出发。

哈密顿量的四矢量（4-动量）

(9)式写成四矢量的形式为下式的$0$分量

$$i\partial^\mu\phi(x)=[P^\mu,\phi(x)],i\partial^\mu\dot\phi(x)=[P^\mu,\dot\phi(x)]\tag{12}$$

猜测

$$\vec P=-\int\mathrm{d}^3\vec y\ \dot\phi(\vec y)\nabla_{\vec y}\phi(\vec y)\tag{13}$$

验证(12)左式

$$\begin{aligned} &[P^\mathrm i,\phi(x)]=-\int\mathrm d^3\vec y\ [\dot\phi(\vec y),\phi(x)]\partial^\mathrm i_{\vec y}\phi(\vec y)\\ =&-\int\mathrm d^3\vec y\ [-i\delta(\vec x-\vec y)]\partial^\mathrm i_{\vec y}\phi(\vec y)\\ =&i\partial^\mathrm i\phi(\vec x) \end{aligned}$$

二次量子化第一步：变换到动量空间

定义

$$\left\{\begin{aligned} &\tilde\phi(k)=\int\mathrm d^4x\ e^{-ik\cdot x}\phi(x)\\ &\phi(x)=\int\frac{\mathrm d^4k}{(2\pi)^4}\ e^{ik\cdot x}\tilde\phi(k)\\ \end{aligned}\right.\tag{14}$$

则Klein-Gordon方程变为

$$(k^2+\omega_0^2)\tilde\phi(k)=0\tag{15}$$

厄米性条件变为

$$\tilde\phi^\dagger(k)=\int\mathrm d^4x\ e^{ik\cdot x}\phi^\dagger(x)=\tilde\phi(-k)\tag{16}$$

根据(15)式，满足Lorentz不变性的非平庸解为

$$\tilde\phi(k)=2\pi\delta(k^2+\omega_0^2)\left[a(\vec k)\theta(k^0)+a^\dagger(-\vec k)\theta(-k^0)\right]\tag{17}$$

此处仅假设了$k^0>0$；至于$k^0<0$时，利用了(16)式$\tilde\phi(k)=\tilde\phi^\dagger(-k)=2\pi\delta(k^2+\omega_0^2)a^\dagger(-\vec k)$

$$\begin{aligned}&[H,\tilde\phi(k)]=\int\mathrm d^4x\ e^{-ik\cdot x}[H,\phi(x)]\\ =&\int\mathrm d^4x\ e^{-ik\cdot x}[-i\dot\phi(x)]\\ =&-i\int\mathrm d^3\vec x\ e^{-i\vec k\cdot\vec x}\int\mathrm dt\ e^{i\omega_{\vec k}t}\dot\phi(x)\\ =&i\int\mathrm d^3\vec x\ e^{-i\vec k\cdot\vec x}\int\mathrm{d}t\ i\omega_{\vec k}\phi(x)e^{i\omega_{\vec k}t}=-\omega_{\vec k}\tilde\phi(k)\end{aligned}$$

所以$H\tilde\phi(k)|E\rangle=(E-\omega_{\vec k})\tilde\phi(k)|E\rangle$，即$k^0=\omega_{\vec k}>0$处$\tilde\phi(k)$为湮灭算符，$k^0<0$处为产生算符

对易关系为

$$\begin{aligned} &[\tilde\phi(k),\tilde\phi(l)]=\int\mathrm{d}^4x\mathrm{d}^4y\ e^{-ik\cdot x-il\cdot y}[\phi(x),\phi(y)]\\ =&\int\mathrm{d}^4x\mathrm{d}^4y\ e^{-ik\cdot x-il\cdot y}\int\frac{\mathrm{d}k'}{(2\pi)^4}\ e^{ik'\cdot(x-y)}2\pi\delta(k'^2+\omega_0^2)\mathrm{sign}(k'^0)\\ =&(2\pi)^4\int\mathrm{d}k'\ \delta(k'-k)\delta(k'+l)2\pi\delta(k'^2+\omega_0^2)\mathrm{sign}(k'^0)\\ =&(2\pi)^4\delta(k+l)2\pi\delta(k^2+\omega_0^2)\mathrm{sign}(k^0)\\ \end{aligned}\tag{18}$$

二次量子化：探索$a,a^\dagger$的物理意义

利用(17)式可将场算符写为

$$\phi(x)=\int\frac{\mathrm d^4k}{(2\pi)^4}\ e^{ik\cdot x}2\pi\delta(k^2+\omega_0^2)\left[a(\vec k)\theta(k^0)+a^\dagger(-\vec k)\theta(-k^0)\right]$$

同样对$k^0$积分后为

$$\begin{aligned}\phi(x)=&\int\frac{\mathrm d^3\vec k}{(2\pi)^32\omega_{\vec k}}\ e^{i\vec k\cdot\vec x}\left[a(\vec k)e^{-i\omega_{\vec k}t}+a^\dagger(-\vec k)e^{i\omega_{\vec k}t}\right]\\ =&\int\frac{\mathrm d^3\vec k}{(2\pi)^32\omega_{\vec k}}\ \left[a(\vec k)e^{ik\cdot x}+a^\dagger(\vec k)e^{-ik\cdot x}\right]\end{aligned}\tag{19}$$

为了反解出$a(\vec k)$，我们计算

$$\left[\omega_{\vec k}+i\frac{\partial}{\partial t}\right]e^{i\xi k\cdot x}=\left[\omega_{\vec k}+\xi k^0\right]e^{i\xi k\cdot x}=(1+\xi)\omega_{\vec k}e^{ik\cdot x}$$

所以作用在$\phi(x)$上后$a^\dagger$的贡献消失，$a$的贡献与分母恰好约掉，即

$$\left[\omega_{\vec k}+i\frac{\partial}{\partial t}\right]\phi(x)=\int\frac{\mathrm d^3\vec k}{(2\pi)^3}a(\vec k)e^{ik\cdot x}$$

所以两边同时乘上$e^{i\omega_{\vec k}t}$后做傅里叶变换

$$\begin{aligned}a(\vec k)=&\int\mathrm{d}^3\vec k\ e^{i\omega_{\vec k}t-i\vec k\cdot\vec x}\left[\omega_{\vec k}+i\frac{\partial}{\partial t}\right]\phi(x)\\ =&\int\mathrm{d}^3\vec k\ e^{-ik\cdot x}\left[\omega_{\vec k}+i\frac{\partial}{\partial t}\right]\phi(x)\end{aligned}\tag{20}$$

另一方面，(17)式对$k^0$积分得

$$\int_0^\infty\mathrm{d}k^0\ \tilde\phi(k)=\frac{2\pi}{2\omega_{\vec k}}a(\vec k)$$

同理可得

$$a^\dagger(\vec k)=\frac{\omega_{\vec k}}\pi\int_0^\infty\mathrm dk^0\tilde\phi(-k)$$

所以对易关系为

$$\left\{\begin{aligned}\left[a(\vec k),a(\vec l)\right]&=\frac{\omega_{\vec k}\omega_{\vec l}}{\pi^2}\int_0^\infty\mathrm{d}k^0\mathrm{d}l^0\ [\tilde\phi(k),\tilde\phi(l)]=0\\ [a^\dagger(\vec k),a^\dagger(\vec l)]&=\frac{\omega_{\vec k}\omega_{\vec l}}{\pi^2}\int_0^\infty\mathrm{d}k^0\mathrm{d}l^0\ [\tilde\phi(-k),\tilde\phi(-l)]=0\\ [a(\vec k),a^\dagger(\vec l)]&=\frac{\omega_{\vec k}\omega_{\vec l}}{\pi^2}\int_0^\infty\mathrm{d}k^0\mathrm{d}l^0\ [\tilde\phi(k),\tilde\phi(-l)]\\ &=\frac{\omega_{\vec k}\omega_{\vec l}}{\pi^2}\int_0^\infty\mathrm{d}k^0\mathrm{d}l^0\ (2\pi)^4\delta(k-l)2\pi\delta(k^2+\omega_0^2)\mathrm{sign}(k^0)\\ &=\frac{\omega_{\vec k}\omega_{\vec l}}{\pi^2}\int_0^\infty\mathrm{d}k^0\ (2\pi)^4\delta(\vec k-\vec l)2\pi\delta(k^2+\omega_0^2)\\ &=\frac{\omega_{\vec k}\omega_{\vec l}}{\pi^2}(2\pi)^4\delta(\vec k-\vec l)2\pi\frac{1}{2\omega_{\vec k}}=(2\pi)^32\omega_{\vec k}\delta(\vec k-\vec l) \end{aligned}\right.\tag{21}$$

至此二次量子化完毕，只需升降算符的对易关系满足(21)式，且场算符为

$$\tilde\phi(k)=\int\frac{\mathrm{d}^3\vec k}{(2\pi)^32\omega_{\vec k}}\left[a(\vec k)e^{ik\cdot x}+a^\dagger(\vec k)e^{-ik\cdot x}\right]\tag{22}$$

4-动量的二次量子化

进一步，4-动量(11)式和(13)式用升降算符表示（即将(19)式代入）。为此，先计算其导数$\dot\phi,\nabla\phi$。简记$\frac{\mathrm d^3\vec k}{(2\pi)^32\omega_{\vec k}}=\tilde{\mathrm dk}$

$$\left\{\begin{aligned} \phi(x)&=\int\tilde{\mathrm dk}\ \left[a(\vec k)e^{ik\cdot x}+a^\dagger(\vec k)e^{-ik\cdot x}\right]\\ \dot\phi(x)&=\int\tilde{\mathrm dk}\ (-i\omega_{\vec k})\left[a(\vec k)e^{ik\cdot x}-a^\dagger(\vec k)e^{-ik\cdot x}\right]\\ \nabla\phi(x)&=\int\tilde{\mathrm dk}\ i\vec k\left[a(\vec k)e^{ik\cdot x}-a^\dagger(\vec k)e^{-ik\cdot x}\right]\\ \end{aligned}\right.\tag{23}$$

4-动量的空间部分仅有一项，先行计算

$$\begin{aligned} &\vec P=-\int\mathrm{d}^3\vec x\int\tilde{\mathrm dk}\tilde{\mathrm dk'}\ \omega_{\vec k}\vec k' \left[a(\vec k)e^{ik\cdot x}-a^\dagger(\vec k)e^{-ik\cdot x}\right] \left[a(\vec k')e^{ik'\cdot x}-a^\dagger(\vec k')e^{-ik'\cdot x}\right]\\ =&-\int\mathrm{d}^3\vec x\int\tilde{\mathrm dk}\tilde{\mathrm dk'}\ \omega_{\vec k}\vec k' \left[a(\vec k)a(\vec k')e^{i(k+k')\cdot x}+a^\dagger(\vec k)a^\dagger(\vec k')e^{-i(k+k')\cdot x}\right]\\ &+\int\mathrm{d}^3\vec x\int\tilde{\mathrm dk}\tilde{\mathrm dk'}\ \omega_{\vec k}\vec k' \left[a(\vec k)a^\dagger(\vec k')e^{i(k-k')\cdot x}+a^\dagger(\vec k)a(\vec k')e^{-i(k-k')\cdot x}\right]\\ =&-\int\tilde{\mathrm dk}\tilde{\mathrm dk'}\ (2\pi)^3\delta(\vec k+\vec k')\omega_{\vec k}\vec k' \left[a(\vec k)a(\vec k')e^{-i(\omega_{\vec k}+\omega_{\vec k'})t}+a^\dagger(\vec k)a^\dagger(\vec k')e^{i(\omega_{\vec k}+\omega_{\vec k'})t}\right]\\ &+\int\mathrm{d}^3\vec x\int\tilde{\mathrm dk}\tilde{\mathrm dk'}\ \omega_{\vec k}\vec k' \left[a(\vec k)a^\dagger(\vec k')e^{i(k-k')\cdot x}+a^\dagger(\vec k)a(\vec k')e^{-i(k-k')\cdot x}\right]\\ =&-\frac12\int\tilde{\mathrm dk'}\ \vec k' \left[a(-\vec k')a(\vec k')e^{-i2\omega_{\vec k'}t}+a^\dagger(-\vec k')a^\dagger(\vec k')e^{i2\omega_{\vec k'}t}\right]\\ &+\int\mathrm{d}^3\vec x\int\tilde{\mathrm dk}\tilde{\mathrm dk'}\ \omega_{\vec k}\vec k' \left[a(\vec k)a^\dagger(\vec k')e^{i(k-k')\cdot x}+a^\dagger(\vec k)a(\vec k')e^{-i(k-k')\cdot x}\right]\\ =&-\frac12\int\tilde{\mathrm dk'}\ \vec k' \left[a(-\vec k')a(\vec k')e^{-i2\omega_{\vec k'}t}+a^\dagger(-\vec k')a^\dagger(\vec k')e^{i2\omega_{\vec k'}t}\right]\\ &+\frac12\int\tilde{\mathrm dk'}\ \vec k' \left[a(\vec k')a^\dagger(\vec k')+a^\dagger(\vec k')a(\vec k')\right]\\ \end{aligned}\tag{24}$$

最后一个等号是对第二行做了和第一行一样的操作。因为湮灭算符之间对易，产生算符之间也对易，所以第一行被积函数是奇函数，进而结果为$0$。第二行利用对易关系得到$\int\mathrm d^3\vec k\ \vec k\delta(0)$，也根据奇偶性消去这项。最终

$$\vec P=\int\tilde{\mathrm dk}\ \vec ka^\dagger(\vec k)a(\vec k)\tag{25}$$

能量部分也类似计算得到

$$\begin{aligned} H=&\frac12\int\frac{\tilde{\mathrm{d}k}}{2\omega_{\vec k}}(-\omega_{\vec k}^2+\vec k^2+\omega_0^2)\left[a(-\vec k)a(\vec k)e^{-i2\omega_{\vec k}t}+a^\dagger(-\vec k)a^\dagger(\vec k)e^{i2\omega_{\vec k}t}\right]\\ &+\frac12\int\frac{\tilde{\mathrm{d}k}}{2\omega_{\vec k}}(\omega_{\vec k}^2+\vec k^2+m^2)\left[a(\vec k)a^\dagger(\vec k)+a^\dagger(\vec k)a(\vec k)\right] \end{aligned}\tag{26}$$

用$\omega_{\vec k}^2=\vec k^2+\omega_0^2$即可将第一项清零并将第二项化简，结果为

$$H=\int\tilde{\mathrm dk}\ \omega_{\vec k}a^\dagger(\vec k)a(\vec k)+\int\mathrm{d}^3\vec k\ \omega_{\vec k}\delta(0)\tag{27}$$

第二项是自能发散，往往可以直接置为$0$，因为我们并不关心它。