本篇详细介绍了傍轴条件下高斯激光和拉盖尔-高斯激光的推导。介绍了拉盖尔高斯激光的轨道角动量,以及圆偏振、椭圆偏振光携带的自旋角动量。
傍轴激光及其角动量
高斯激光与高斯脉冲激光
傍轴亥姆霍兹方程
时域中高斯激光是波动方程的解,
\[ \nabla^2U-\frac{1}{c^2}\frac{\partial^2U}{\partial t^2}=0 \]
令
\[ U=U(\omega)e^{i\omega t} \]
得到频域中对应的亥姆霍兹方程
\[ \nabla^2U+k^2U=0 \tag{1}\]
其中 \(k=\omega/c\).
假设激光主要沿 \(z\) 轴方向传播,设解具有如下形式
\[ U(r,z,\omega)=u(r,z,\omega)e^{-ikz} \tag{2}\]
其中,\(r=\sqrt{x^2+y^2}\),将 Equation 2 代入式 Equation 1,得到
\[ \frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2}+\frac{\partial^2u}{\partial z^2}-2ik\frac{\partial u}{\partial z}=0 \]
我们考虑沿 \(z\) 方向缓变的场强,即 \(|\partial u/\partial z|\ll k|u|\),亦可得 \(|\partial^2 u/\partial z^2|\ll k|\partial u/\partial z|\). 忽略 \(u\) 对 \(z\) 的二阶导,将上式简化为
\[ \frac{\partial^2u}{\partial x^2} +\frac{\partial^2u}{\partial y^2} -2ik\frac{\partial u}{\partial z}=0 \tag{3}\]
此为傍轴近似下的亥姆霍兹方程,也可以在柱坐标系 \((r,\theta,z)\) 下表达为
\[ \frac{\partial^2 u}{\partial r^2}+\frac{1}{r}\frac{\partial u}{\partial r}+\frac{1}{r^2}\frac{\partial^2 u}{\partial \theta^2}-2ik\frac{\partial u}{\partial z}=0 \tag{4}\]
高斯激光
先考虑简单的情况,\(u\) 沿 \(z\) 轴具有旋转对称性,可将 Equation 4 化为
\[ \frac{\partial^2 u}{\partial r^2}+\frac{1}{r}\frac{\partial u}{\partial r}-2ik\frac{\partial u}{\partial z}=0 \tag{5}\]
假设 \(u\) 具有如下高斯形式
\[ u=Ae^{-i\frac{kr^2}{2q(z)}}e^{-ip(z)} \tag{6}\]
代入方程 Equation 5,得到
\[ -\frac{k^2r^2}{q^2}\left(1-\frac{\mathrm{d}q}{\mathrm{d}z}\right)-2k\left(\frac{i}{q}-\frac{\mathrm{d}p}{\mathrm{d}z}\right)=0 \tag{7}\]
上式对任意 \(r\) 都成立,只能
\[ 1-\frac{\mathrm{d} q}{\mathrm{~d} z}=0 \tag{8}\]
\[ \frac{i}{q}-\frac{\mathrm{d} p}{\mathrm{~d} z}=0 \tag{9}\]
解方程 Equation 8,得
\[ q(z)=z+iz_0 \tag{10}\]
这里设定了 \(q(z)\) 的初始值为 \(iz_0\),\(z_0\) 为实数。利用
\[ \frac{1}{q(z)}=\frac{1}{z\left(1+\frac{z_0^2}{z^2}\right)}-\frac{i}{z_0\left(1+\frac{z^2}{z_0^2}\right)} \]
Equation 6 中第一个 \(e\) 指数可以化为
\[ e^{-i\frac{kr^2}{2R(z)}}e^{-\frac{r^2}{\sigma^2}} \]
这里,我们定义高斯光波前的曲率半径
\[ R(z)=z\left(1+\frac{z_0^2}{z^2}\right) \] 和腰斑 \[ \sigma=\sigma_0\sqrt{1+\frac{z^2}{z_0^2}} \] 其中,\(\sigma_0\) 是束腰 \[ \sigma_0=\sqrt{\frac{2z_0}{k}}=\sqrt{\frac{\lambda z_0}{\pi}} \] 利用上式也反解出 \(z_0\)
\[ z_0=\frac{k\sigma_0^2}{2} \tag{11}\]
将 Equation 10 代入 Equation 9 得到
\[ \frac{\mathrm dp}{\mathrm d z}=\frac{1}{iz-z_0} \]
可以解得
\[ \begin{aligned} p(z) & =-i \ln \left(i z-z_0\right) \\ & =-i \ln \sqrt{z^2+z_0^2}+\arctan \left(-z / z_0\right) \\ & =i \ln \frac{\sigma_0}{z_0 \sigma}-\varphi(z) \end{aligned} \] 这里我们定义了古伊相位
\[ \varphi(z)=\arctan(z/z_0) \] 将解出的 \(p(z)\) 代回式 Equation 6 就得到了高斯形式解
\[ U(r,z,\omega)=A\frac{\sigma_0}{\sigma}e^{-\frac{r^2}{\sigma^2}}e^{-i\frac{kr^2}{2R(z)}}e^{i\varphi(z)}e^{-ikz} \tag{12}\] 注意上式已经乘上了 \(e^{-ikz}\).
高斯脉冲激光
上一小节推导了频域上傍轴亥姆霍兹方程的高斯形式解。对于单色光,要得到时域上的解,只需在 Equation 12 后乘以 \(e^{i\omega t}\) 即可。如若考虑色散,则需对 Equation 12 作傅里叶逆变换。
考虑共焦腔中的高斯激光,束腰和波数有依赖关系
\[ \sigma_0=\sqrt{L/k} \] 其中 \(L\) 是共焦腔的长度,这时 \(z_0=L/2\) 是一个与激光频率无关的常量,从而 \(q(z),\varphi(z)\) 和 \(\sigma_0/\sigma\) 与频率无关。根据真空电磁波的色散关系 \(k=\omega/c\),重写 Equation 12 得
\[ U(r,z,\omega)=A(\omega)\frac{\sigma_0}{\sigma}\exp\left\{-i\frac{\omega}{c}\left[z+\frac{r^2}{2q(z)}\right]+i\varphi(z)\right\} \tag{13}\]
对其作傅里叶逆变换
\[ \begin{aligned} & \frac{1}{\sqrt{2 \pi}} \int A(\omega) \frac{\sigma_0}{\sigma} \exp \left\{-i \frac{\omega}{c}\left[z+\frac{r^2}{2 q(z)}\right]+i \varphi(z)\right\} e^{i \omega t} \mathrm{~d} \omega \\ = & \frac{\sigma_0}{\sigma} e^{i \varphi(z)} \cdot \frac{1}{\sqrt{2 \pi}} \int A(\omega) e^{i \omega \tau} \mathrm{d} \omega \\ = & \frac{\sigma_0}{\sigma} e^{i\left[\omega_0 \tau+\varphi(z)\right]} \cdot \frac{1}{\sqrt{2 \pi}} \int A\left(\omega-\omega_0\right) e^{i\left(\omega-\omega_0\right) \tau} \mathrm{d}\left(\omega-\omega_0\right) \end{aligned} \] 其中,第一个等号定义了复数形式的时间
\[ \tau=t-\left[z+\frac{r^2}{2q(z)}\right]/c=t-\left[z+\frac{r^2}{2R(z)}\right]/c+i\frac{r^2}{\omega_0\sigma^2(z)} \tag{14}\]
\(\omega_0\) 是中心频率。于是,我们得到了高斯激光在时域中的表达式
\[ U(r,z,t)=\frac{\sigma_0}{\sigma(z)}A(\tau)e^{i[\omega_0\tau+\varphi(z)]} \tag{15}\] 其中 \(A(\tau)\) 是 \(A(\omega-\omega_0)\) 的傅里叶逆变换。
考虑一个高斯脉冲信号
\[ C(t)=C_0\exp\left(-\frac{t^2}{2t_0^2}\right)e^{i\omega_0t} \tag{16}\]
对其作傅里叶变换,得到
\[ C(\omega-\omega_0) =\frac{1}{\sqrt{2\pi}}\int C_0\exp\left(-\frac{t^2}{2t_0^2}\right)e^{i\omega_0t} e^{-i\omega t}\mathrm{d}\omega =C_0t_0\exp\left[-\frac{(\omega-\omega_0)^2}{2(1/t_0)^2}\right] \]
替换 Equation 13 中的 \(A(\omega)\) 可以得到频域中的场强分布。再对频域中的场强作傅里叶逆变换,即可得到时域中场强分布
\[ \begin{aligned} U(r, z, t) & =\frac{\sigma_0}{\sigma} e^{i\left[\omega_0 \tau+\varphi(z)\right]} \cdot \frac{1}{\sqrt{2 \pi}} \int C\left(\omega-\omega_0\right) e^{i\left(\omega-\omega_0\right) \tau} \mathrm{d}\left(\omega-\omega_0\right) \\ & =C_0 \frac{\sigma_0}{\sigma(z)} \exp \left(-\frac{\tau^2}{2 t_0^2}\right) e^{i\left[\omega_0 \tau+\varphi(z)\right]} \end{aligned} \] 其中 \(\tau\) 由 Equation 14 定义。 把 \(\tau\) 的表达式代入,得
\[ \begin{aligned} U(r, z, t)=& A_{0} \frac{\sigma_{0}}{\sigma(z)} \exp \left[-\left(t-\frac{r^{2}}{2 c R(z)}-\frac{z}{c}\right)^{2} / 2 t_{0}^{2}\right] \\ & \times \exp \left[-\frac{r^{2}}{\sigma^{2}(z)}+\frac{r^{4}}{2 \sigma^{4}(z)}\left(\frac{\delta_{0}}{\omega_{0}}\right)^{2}\right] \\ & \times \exp \left\{i \omega_{0}\left[1-\frac{r^{2}}{\sigma^{2}(z)}\left(\frac{\delta_{0}}{\omega_{0}}\right)^{2}\right] \left[t-\frac{r^{2}}{2 c R(z)}-\frac{z}{c}\right]+i\varphi(z)\right\} %\\ & \times \exp [i \varphi(z)] \end{aligned} \] 其中 \(\delta_0=1/t_0\).考察束腰处的场强,令 \(z=0\),则 \(R(z)=\infty,~\sigma(z)=\sigma_0,~\tau=t+ir^2/\omega_0\sigma_0^2,~\varphi(z)=0\),得到
\[ U(r,t)=C_0\exp\left(-\frac{t^2}{2t_0^2}\right)\exp\left[-\frac{r^2}{\sigma_0^2}+\frac{r^4}{2\sigma_0^4}\left(\frac{\delta_0}{\omega_0}\right)^2\right]\exp\left\{ i\omega_0\left[1-\frac{r^2}{\sigma_0^2}\left(\frac{\delta_0}{\omega_0}\right)^2\right]t\right\} \]
拉盖尔-高斯 (LG) 激光的推导及其角动量
LG 激光的表达式
下面在柱坐标系下求解式 Equation 4
\[ \frac{\partial^2 u}{\partial r^2}+\frac{1}{r}\frac{\partial u}{\partial r}+\frac{1}{r^2}\frac{\partial^2 u}{\partial \theta^2}-2ik\frac{\partial u}{\partial z}=0 \]
令
\[ u=f(r,z)\Theta(\theta) \]
分离变量后得到
\[ \frac{r^2}{f} \frac{\partial^2 f}{\partial r^2}+\frac{r}{f} \frac{\partial f}{\partial r}-2 i k \frac{r^2}{f} \frac{\partial f}{\partial z}=\ell^2 \tag{17}\]
\[ \Theta^{\prime \prime}+\ell^2 \Theta=0 \tag{18}\]
方程 Equation 18 有解
\[ \Theta(\theta)=e^{-i\ell\theta} \]
由周期性边界条件可知,\(\ell\) 只能取整数。 设
\[ f(r,z)=Ag(r,z)e^{-i\frac{kr^2}{2q(z)}}e^{-ip(z)} \tag{19}\] 其中 \(e^{-i\frac{kr^2}{2q(z)}}e^{-ip(z)}\) 是高斯激光的解。将 Equation 19 代入方程 Equation 17,得到
\[ -\frac{k^2r^2}{q^2}\left(1-\frac{\mathrm{d}q}{\mathrm{d}z}\right)-2k\left(\frac{i}{q}-\frac{\mathrm{d}p}{\mathrm{d}z}\right)+\frac{1}{g}\left[\frac{\partial^2 g}{\partial r^2}+\left(\frac{1}{r}-\frac{2ikr}{q}\right)\frac{\partial g}{\partial r}-2ik\frac{\partial g}{\partial z}\right]=\frac{\ell^2}{r^2} \]
由方程 Equation 7 可知上式前两项之和为0. 方程变为
\[ \frac{1}{g}\left[\frac{\partial^2 g}{\partial r^2}+\left(\frac{1}{r}-\frac{2ikr}{q}\right)\frac{\partial g}{\partial r}-2ik\frac{\partial g}{\partial z}\right]=\frac{\ell^2}{r^2} \tag{20}\]
定义变量\(\xi\)
\[ \xi=\frac{2r^2}{\sigma(z)^2}=\frac{2r^2}{\sigma_0^2\left(1+\frac{z^2}{z_0^2}\right)} \]
令
\[ g(r,z)=h(\xi,z) \] 则有
\[ \begin{aligned} & \frac{\partial g}{\partial r}=\frac{\partial h}{\partial \xi} \frac{\partial \xi}{\partial r}=\frac{4 r}{\sigma^2} \frac{\partial h}{\partial \xi} \\ & \frac{\partial^2 g}{\partial r^2}=\frac{4}{\sigma^2} \frac{\partial h}{\partial \xi}+\frac{16 r^2}{\sigma^4} \frac{\partial^2 h}{\partial \xi^2} \\ & \frac{\partial g}{\partial z}=\frac{\partial h}{\partial z}+\frac{\partial h}{\partial \xi} \frac{\partial \xi}{\partial z}=\frac{\partial^2 h}{\partial z^2}-\frac{4 z \xi}{\sigma^2 z_0 k} \frac{\partial h}{\partial \xi} \end{aligned} \]
利用这些式子以及 \(z_0\) 与 \(\sigma_0\) 的关系 Equation 11 可以将 \(g\) 的偏微分方程 Equation 20 改成 \(h\) 的偏微分方程,化简得
\[ \xi \frac{\partial^2 h}{\partial \xi^2}+(1-\xi) \frac{\partial h}{\partial \xi}-\frac{h \ell^2}{4 \xi}=\frac{i k \sigma^2}{4} \frac{\partial h}{\partial z} \]
令
\[ h(\xi,z)=\Xi(\xi)Z(z) \]
分离变量得
\[ \frac{\xi}{\Xi} \frac{\partial^2 \Xi}{\partial \xi^2}+\frac{(1-\xi)}{\Xi} \frac{\partial \Xi}{\partial \xi}-\frac{\ell^2}{4 \xi}=\frac{i k \sigma^2}{4 Z} \frac{\partial Z}{\partial z}=-\left(\frac{|\ell|}{2}+p\right) \] 其中 \(p\) 为常数,得到两个方程
\[ \frac{\xi}{\Xi} \frac{\partial^2 \Xi}{\partial \xi^2}+\frac{(1-\xi)}{\Xi} \frac{\partial \Xi}{\partial \xi}-\frac{|\ell|^2}{4 \xi}+\left(\frac{|\ell|}{2}+p\right)=0 \tag{21}\]
\[ \frac{i k \sigma^2}{4 Z} \frac{\partial Z}{\partial z}+\left(\frac{|\ell|}{2}+p\right)=0 \tag{22}\]
进一步令 \[ \Xi=(\sqrt{\xi})^{|\ell|}L(\xi) \]
代入方程 Equation 21,得到
\[ \xi \frac{\partial^2 L}{\partial \xi^2}+(|\ell|+1-\xi) \frac{\partial L}{\partial \xi}+p L=0 \] 这是广义拉盖尔方程,它的解是广义拉盖尔多项式
\[ L(\xi)=L_p^{|\ell|}(\xi) \]
于是,我们解得\(\Xi\)
\[ \Xi=(\sqrt{\xi})^{|\ell|}L_p^{|\ell|}(\xi) \]
容易验证,方程 Equation 22 具有解
\[ Z=e^{iN\arctan(z/z_0)} \] 其中
\[ N=|\ell|+2p \]
至此,我们得到了完整的柱坐标系傍轴亥姆霍兹方程的解
\[ \begin{aligned} u_{p, \ell}(r, \theta, z) & =A \Xi(\xi) Z(z) \Theta(\theta) e^{-i \frac{k r^2}{2 q(z)}} e^{-i p(z)} \\ & =A(\sqrt{\xi})^{|\ell|} L_p^{|\ell|}(\xi) e^{-i \frac{k r^2}{2 q(z)}} e^{-i p(z)} e^{i N \arctan \left(z / z_0\right)} e^{-i \ell \theta} \\ & =A \frac{\sigma_0}{\sigma}\left(\frac{\sqrt{2} r}{\sigma(z)}\right)^{|\ell|} L_p^{|\ell|}\left(\frac{2 r^2}{\sigma(z)^2}\right) e^{-\frac{r^2}{\sigma^2}} e^{-i \frac{k r^2}{2 R(z)}} e^{i \varphi(z)} e^{-i \ell \theta} \end{aligned} \] 这里,古伊相位的定义是 \[ \varphi(z)=(N+1)\arctan(z/z_0) \]
激光的轨道角动量
假设一线偏振激光,其矢势为
\[ \boldsymbol{A}=\hat{\boldsymbol x}u(x,y,z)e^{i\omega t-ikz} \]
对于单色光,标势可以由矢势求出 \[ \phi=\frac{i}{\omega \mu_0 \epsilon_0} \nabla \cdot \boldsymbol{A}=\frac{i}{\omega \mu_0 \epsilon_0} \frac{\partial u}{\partial x} e^{i \omega t-i k z} \]
分别求出电场和磁场
\[ \boldsymbol{B}=\nabla \times \boldsymbol{A}=-i k\left[\hat{\boldsymbol{y}} u-\frac{i}{k} \hat{\boldsymbol{z}} \frac{\partial u}{\partial y}\right] e^{i \omega t-i k z} \]
因为 \(\left|\frac{\partial u}{\partial z}\right| \ll|k u|\) 而在结果中舍去了和 \(\frac{\partial u}{\partial z}\) 有关的项。 \[ \boldsymbol{E}=-\frac{\partial \boldsymbol{A}}{\partial t}-\nabla \phi=-i \omega\left[\hat{\boldsymbol{x}} u-\frac{i}{k} \hat{\boldsymbol{z}} \frac{\partial u}{\partial x}\right] e^{i \omega t-i k z} \] 其中,计算 \(\nabla\phi\) 时,由于 \(|\nabla u|\ll|ku|\) 而舍去了和 \(\nabla u\) 有关的项。 由此可以计算动量密度(时间平均)
\[ \begin{aligned} \mathcal{P} & =\epsilon_0\langle\Re(\boldsymbol{E}) \times \Re(\boldsymbol{B})\rangle=\frac{\epsilon_0}{2}\left(\boldsymbol{E}^* \times \boldsymbol{B}+\boldsymbol{E} \times \boldsymbol{B}^*\right) \\ & =i \omega \frac{\epsilon_0}{2}\left[u^*\left(\hat{\boldsymbol{x}} \frac{\partial u}{\partial x}+\hat{\boldsymbol{y}} \frac{\partial u}{\partial y}\right)-u\left(\hat{\boldsymbol{x}} \frac{\partial u^*}{\partial x}+\hat{\boldsymbol{y}} \frac{\partial u^*}{\partial y}\right)\right]+\omega k \epsilon_0|u|^2 \hat{\boldsymbol{z}} \\ & =i \omega \frac{\epsilon_0}{2}\left(u^* \nabla u-u \nabla u^*\right)+\omega k \epsilon_0|u|^2 \hat{\boldsymbol{z}} \end{aligned} \]
假设 \(u\) 对 \(\theta\) 的依赖关系只体显在 \(\exp{(-i\ell \theta)}\) \[ u(x,y,z)=u(r,\theta,z)=u_0(r,z)e^{-i\ell \theta} \] 则 \[ \nabla u=e^{-i\ell \theta}\nabla u_0-\frac{i\ell}{r}\hat{\boldsymbol \theta}u_0e^{-i\ell \theta} \] 代入刚才计算的动量密度,得到 \[ \mathcal{P}=i\omega \frac{\epsilon_0}{2}(u_0^*\nabla u_0-u_0\nabla u_0^*) +\frac{\omega\epsilon_0\ell}{r}|u_0|^2\hat{\boldsymbol{\theta}} +\omega k \epsilon_0|u_0|^2\hat{\boldsymbol z} \] 利用傍轴近似 \(|\partial u/\partial z|\ll k|u|\), 可以忽略 \((u_0^*\nabla u_0-u_0\nabla u_0^*)\) 中 \(\hat{\boldsymbol{z}}\) 方向的量。可以把动量密度写为 \[ \mathcal{P}=i \omega \frac{\epsilon_0}{2}\left(u_0^* \frac{\partial u_0}{\partial r}-u_0 \frac{\partial u_0^*}{\partial r}\right) \hat{\boldsymbol{r}}+\frac{\omega \epsilon_0 \ell}{r}\left|u_0\right|^2 \hat{\boldsymbol{\theta}}+\omega k \epsilon_0\left|u_0\right|^2 \hat{\boldsymbol{z}} \] 利用 \[ \mathcal{L}=\boldsymbol{r} \times \mathcal{P}=(r \hat{\boldsymbol{r}}+z \hat{\boldsymbol{z}}) \times \mathcal{P} \] 可以计算得角动量密度
\[ \begin{aligned} \mathcal{L} & =\mathcal{L}_r(r, z) \hat{\boldsymbol{r}}+\mathcal{L}_\theta(r, z) \hat{\boldsymbol{\theta}}+\mathcal{L}_z(r, z) \hat{\boldsymbol{z}} \\ & =-\omega \epsilon_0 \ell \frac{z}{r}\left|u_0\right|^2 \hat{\boldsymbol{r}}+\omega \epsilon_0\left[\frac{i z}{2}\left(u_0^* \frac{\partial u_0}{\partial r}-u_0 \frac{\partial u_0^*}{\partial r}\right)-r k\left|u_0\right|^2\right] \hat{\boldsymbol{\theta}}+\omega \epsilon_0 \ell\left|u_0\right|^2 \hat{\boldsymbol{z}} \end{aligned} \]
激光场的角动量是角动量密度在激光所在区域进行积分 \[ \boldsymbol{L}=\int\mathcal{L}\mathrm{d}V=\int\mathrm{d}z\int_0^{\infty}r\mathrm{d}r\int_0^{2\pi}\mathrm{d}\theta~\mathcal{L} \] 按分量来看
\[ \begin{aligned} \int \mathcal{L}_r(r, z) \hat{\boldsymbol{r}}(\theta) \mathrm{d} \boldsymbol{r} & =\int \mathrm{d} z \int_0^{\infty} r \mathrm{~d} r \int_0^{2 \pi} \hat{\boldsymbol{r}}(\theta) \mathrm{d} \theta \mathcal{L}_r(r, z) \\ & =\int \mathrm{d} z \int_0^{\infty} r \mathrm{~d} r\left\{\int_0^\pi \hat{\boldsymbol{r}}(\theta) \mathrm{d} \theta+\int_\pi^{2 \pi} \hat{\boldsymbol{r}}(\theta) \mathrm{d} \theta\right\} \mathcal{L}_r(r, z) \\ & =\int \mathrm{d} z \int_0^{\infty} r \mathrm{~d} r\left\{\int_0^\pi \hat{\boldsymbol{r}}(\theta) \mathrm{d} \theta+\int_0^\pi \hat{\boldsymbol{r}}(\theta+\pi) \mathrm{d} \theta\right\} \mathcal{L}_r(r, z) \\ & =0 \end{aligned} \]
最后一步用到了 \(\hat{\boldsymbol{r}}(\theta)=-\hat{\boldsymbol{r}}(\theta+\pi)\). 同理,由于 \(\hat{\boldsymbol{\theta}}(\theta)=-\hat{\boldsymbol{\theta}}(\theta+\pi)\), 角向的积分也为0. 于是 \[ \boldsymbol{L}=L_z \hat{\boldsymbol{z}}=\hat{\boldsymbol{z}} \omega \epsilon_0 \ell \int\left|u_0\right|^2 \mathrm{~d} V \] 同样地,对于动量,有 \[ \boldsymbol{P}=P_z \hat{\boldsymbol{z}}=\hat{\boldsymbol{z}} \omega k \epsilon_0 \int\left|u_0\right|^2 \mathrm{~d} V \]
激光中单个光子的平均角动量是激光场的角动量比上光子数,即 \[ l_z=\frac{\boldsymbol{L}}{c\boldsymbol{P}/\hbar\omega}=\frac{\omega\epsilon_0\ell}{c\omega k \epsilon_0/\hbar\omega}=\ell \hbar \]
因为在求解的时候假设激光是线偏振的,这样的激光没有自旋角动量,所以\(l_z\)计算的是光子轨道角动量。值得注意的是,推得这个结果只用到了傍轴近拟和假设 \(u(r,\theta,z)=u_0(r,z)e^{-i\ell\theta}\),没有对 \(u_0\) 的具体形式作任何要求。我们可以得到结论:如果傍轴近似下,激光场对于 \(\theta\)的依赖只取决于 \(\exp{(-i\ell\theta)}\), 则该激光场具有轨道角动量,且单个光子的平均轨道角动量大小为 \(\ell\hbar\).
激光的自旋角动量与总角动量
上面的推导假定了 \(\boldsymbol{A}\) 的线偏振的,如果 \(\boldsymbol{A}\) 具有圆偏振,矢势设为 \[ \boldsymbol{A}=\frac{1}{\sqrt{2}}\hat{\boldsymbol{x}}ue^{i\omega t-ikz}+\frac{1}{\sqrt{2}}\hat{\boldsymbol{y}}ue^{i\sigma_z\frac{\pi}{2}+i\omega t-ikz} \] 其中 \(\sigma_z=\pm 1\) 对应右旋圆偏光和左旋圆偏光,\(\sigma_z=0\) 对应线偏光。下在求动量密度 \[ \left(\boldsymbol{E}^* \times \boldsymbol{B}+\boldsymbol{E} \times \boldsymbol{B}^*\right)_{n e w}=\frac{1}{2} i \omega \frac{\partial|u|^2}{\partial y} \hat{\boldsymbol{x}}\left(e^{i \sigma_z \frac{\pi}{2}}-e^{-i \sigma_z \frac{\pi}{2}}\right)+\frac{1}{2} i \omega \frac{\partial|u|^2}{\partial x} \hat{\boldsymbol{y}}\left(e^{-i \sigma_z \frac{\pi}{2}}-e^{i \sigma_z \frac{\pi}{2}}\right) \] 其中,脚标 \(new\) 的意思是相对于线偏振的结果多出了的项,利用 \[ \sin\sigma_z\frac{\pi}{2}=\sigma_z \tag{23}\] 得到 \[ \left(\boldsymbol{E}^* \times \boldsymbol{B}+\boldsymbol{E} \times \boldsymbol{B}^*\right)_{n e w}=\omega \sigma_z\left(\hat{\boldsymbol{y}} \frac{\partial}{\partial x}-\hat{\boldsymbol{x}} \frac{\partial}{\partial y}\right)|u|^2=\omega \sigma_z \frac{\partial|u|^2}{\partial r} \hat{\boldsymbol{\theta}} \] 上式第二个等号用到了条件 \(u(r,\theta,z)=u_0(r,z)e^{-i\ell\theta}\). 于是,我们得到圆偏振激光的动量密度
\[ \begin{aligned} \mathcal{P} & =i \omega \frac{\epsilon_0}{2}\left(u^* \nabla u-u \nabla u^*\right)+\sigma_z \omega \frac{\epsilon_0}{2} \frac{\partial|u|^2}{\partial r} \hat{\boldsymbol{\theta}}+\omega k \epsilon_0|u|^2 \hat{\boldsymbol{z}} \\ & =i \omega \frac{\epsilon_0}{2}\left(u_0^* \nabla u_0-u_0 \nabla u_0^*\right)+\left(\frac{\omega \epsilon_0 \ell}{r}\left|u_0\right|^2+\sigma_z \omega \frac{\epsilon_0}{2} \frac{\partial\left|u_0\right|^2}{\partial r}\right) \hat{\boldsymbol{\theta}}+\omega k \epsilon_0\left|u_0\right|^2 \hat{\boldsymbol{z}} \\ & =i \omega \frac{\epsilon_0}{2}\left(u_0^* \frac{\partial u_0}{\partial r}-u_0 \frac{\partial u_0^*}{\partial r}\right) \hat{\boldsymbol{r}}+\left(\frac{\omega \epsilon_0 \ell}{r}\left|u_0\right|^2+\sigma_z \omega \frac{\epsilon_0}{2} \frac{\partial\left|u_0\right|^2}{\partial r}\right) \hat{\boldsymbol{\theta}}+\omega k \epsilon_0\left|u_0\right|^2 \hat{\boldsymbol{z}} \end{aligned} \]
与线偏振的情况相同,动量密度的体积分只有 \(\hat{\boldsymbol{z}}\) 分量,角动量密度也如此,因而我们下面只关注它们的 \(\hat{\boldsymbol{z}}\) 分量。角动量密度的 \(\hat{\boldsymbol{z}}\) 分量为 \[ \mathcal{J}_z \hat{\boldsymbol{z}}=r \hat{\boldsymbol{r}} \times \mathcal{P}_\theta \hat{\boldsymbol{\theta}}=\left(\omega \epsilon_0 \ell\left|u_0\right|^2+\sigma_z \omega \frac{\epsilon_0}{2} r \frac{\partial\left|u_0\right|^2}{\partial r}\right) \hat{\boldsymbol{z}} \tag{24}\]
其中第一项即 \(\mathcal{L}_z\), 是激光的轨道角动量,第二项是由于光的圆偏振而导致的角动量,可认为它是光的自旋角动量。
考虑到 \[ \begin{aligned} \int_0^{\infty} r \mathrm{~d} r \frac{1}{2} r \frac{\partial\left|u_0\right|^2}{\partial r} & =\left.\frac{1}{2} r^2\left|u_0\right|^2\right|_{r=0} ^{r \rightarrow \infty}-\int_0^{\infty} r \mathrm{~d} r\left|u_0\right|^2 \\ & =-\int_0^{\infty} r \mathrm{~d} r\left|u_0\right|^2 \end{aligned} \]
这里用到了 \(r^2|u_0|^2\to 0(r\to\infty)\), 这对LG激光是成立的。 于是,我们可以求得激光场的总角动量 \[ \begin{aligned} \boldsymbol{J}=J_z \hat{\boldsymbol{z}} & =\hat{\boldsymbol{z}} \omega \epsilon_0 \int \mathrm{d} z \int_0^{2 \pi} \mathrm{d} \theta \int_0^{\infty} r \mathrm{~d} r\left(\ell\left|u_0\right|^2+\frac{\sigma_z}{2} r \frac{\partial\left|u_0\right|^2}{\partial r}\right) \\ & =\hat{\boldsymbol{z}} \omega \epsilon_0 \int \mathrm{d} z \int_0^{2 \pi} \mathrm{d} \theta \int_0^{\infty} r \mathrm{~d} r\left(\ell-\sigma_z\right)\left|u_0\right|^2 \\ & =\hat{\boldsymbol{z}} \omega \epsilon_0\left(\ell-\sigma_z\right) \int\left|u_0\right|^2 \mathrm{~d} V \end{aligned} \]
对动量密度积分得到激光场的总动量 \[ \boldsymbol{P}=P_z \hat{\boldsymbol{z}}=\hat{\boldsymbol{z}} \omega k \epsilon_0 \int\left|u_0\right|^2 \mathrm{~d} V \] 从而可以计算单个光子的平均总角动量 \[ j_z=\frac{\boldsymbol{J}}{c\boldsymbol{P}/\hbar\omega}=\frac{\omega\epsilon_0(\ell-\sigma_z)}{c\omega k \epsilon_0/\hbar\omega}=(\ell-\sigma_z) \hbar \] 如果令 \(\sigma_z=+1\) 为左旋圆偏振光,\(\sigma_z=-1\) 为右旋圆偏振光,则1 \[ j_z=(\ell+\sigma_z) \hbar \]
椭圆偏振光的自旋角动量
根据上一节的推导,傍轴圆偏振光 \[ \boldsymbol{A}=\left(\hat{\boldsymbol{x}}+\hat{\boldsymbol{y}} e^{-i \sigma_z \frac{\pi}{2}}\right) e^{i \theta} \tag{25}\] 具有角动量 \(\sigma_z\hbar\). 这里 \(\theta=\omega t-kz\)为简写,\(\sigma_z=1\) 为左旋圆偏光,\(\sigma_z=-1\) 为右旋圆偏光。
当 \(\sigma_z\) 取 \([-1,1]\) 之间的非整数时,Equation 25 表示的的是椭圆偏振光。因为此时不再满足 \(\sin\sigma_z\frac{\pi}{2}=\sigma_z\),激光的角动量应为 \(\hbar\sin\sigma_z\frac{\pi}{2}\). 下面讨论椭圆偏振光如何分解成左旋和右旋圆偏振光的叠加。
假设 \[ \hat{\boldsymbol{x}}+\hat{\boldsymbol{y}} e^{-i \sigma_z \frac{\pi}{2}}=A\left(\hat{\boldsymbol{x}}+\hat{\boldsymbol{y}} e^{-i \frac{\pi}{2}}\right)+B\left(\hat{\boldsymbol{x}}+\hat{\boldsymbol{y}} e^{i \frac{\pi}{2}}\right) \] 可以解得 \[ \left\{\begin{array}{l} A=\frac{1}{2}\left(1-e^{-i \sigma_z \frac{\pi}{2}}\right) \\ B=\frac{1}{2}\left(1+e^{i \sigma_z \frac{\pi}{2}}\right) \end{array}\right. \] 叠加的振幅是复数,这是因为按 Equation 25 定义的椭圆偏振光的长轴不在坐标上,而是与坐标轴成45度角。
假设在新坐标系下,椭圆偏振光具有形式 \[ \boldsymbol{A}=\left(\hat{\boldsymbol{x}}^{\prime} a+\hat{\boldsymbol{y}}^{\prime} b e^{-i \frac{\pi}{2}}\right) e^{i \theta^{\prime}} \tag{26}\] 或写成参数方程 \[ \left\{\begin{array}{l} x^{\prime}=a \cos \theta^{\prime} \\ y^{\prime}=b \cos \left(\theta^{\prime}-\frac{\pi}{2}\right)=b \sin \theta^{\prime} \end{array}\right. \tag{27}\] 这里 \(a\) 和 \(b\) 代表椭圆的半长轴或半短轴。上面两式的写法已经假定了光的偏振形式为左旋椭偏光,为了具有更一般的含义,可以设定 \(b\) 具有符号。\(b>0\) 代表左旋椭偏光,\(b<0\) 代表右旋椭偏光。
原坐标系下椭偏光参数方程为 \[ \left\{\begin{array}{l} x=\cos \theta \\ y=\cos \left(\theta-\sigma_z \frac{\pi}{2}\right) \end{array}\right. \tag{28}\]
椭圆 Equation 28 是由椭圆 Equation 27 转过45角得到的。对椭圆 Equation 27 作转动操作,得到
\[ \begin{aligned} & \frac{1}{\sqrt{2}}\left[\begin{array}{cc} 1 & -1 \\ 1 & 1 \end{array}\right]\left[\begin{array}{l} a \cos \theta^{\prime} \\ b \sin \theta^{\prime} \end{array}\right]=\frac{1}{\sqrt{2}}\left[\begin{array}{l} a \cos \theta^{\prime}-b \sin \theta^{\prime} \\ a \cos \theta^{\prime}+b \sin \theta^{\prime} \end{array}\right] \\ & =\sqrt{\frac{a^2+b^2}{2}}\left[\begin{array}{l} \cos \left(\theta^{\prime}+\phi\right) \\ \sin \left(\theta^{\prime}+\phi^{\prime}\right) \end{array}\right]=\sqrt{\frac{a^2+b^2}{2}}\left[\begin{array}{c} \cos \theta \\ \sin \left(\theta+\phi^{\prime}-\phi\right) \end{array}\right] \end{aligned} \tag{29}\]
其中,\(\theta=\theta'+\phi\),并且 \[ \cos\phi=\frac{a}{\sqrt{a^2+b^2}},~\sin\phi=\frac{b}{\sqrt{a^2+b^2}},~\cos\phi'=\frac{a}{\sqrt{a^2+b^2}},~\sin\phi'=-\frac{b}{\sqrt{a^2+b^2}} \]
式 Equation 29 得到的椭圆应与式 Equation 28 为同一个椭圆,可以得到
\[ a^2+b^2=2 \tag{30}\]
\[ \cos\sigma_z\frac{\pi}{2}=\cos(\phi-\phi')=\frac{a^2-b^2}{2} \]
\[ \sin\sigma_z\frac\pi2=\sin(\phi-\phi')=ab \tag{31}\] 从而求得 \[ \left\{\begin{array}{l} a^2=1+\cos \sigma_z \frac{\pi}{2} \\ b^2=1-\cos \sigma_z \frac{\pi}{2} \end{array}\right. \]
因为 \(a>0\),\(b\) 的符号与 \(\sin\sigma_z\frac\pi2\) 相同,对于 \(\sigma_z\in[-1,1]\),有 \[ \left\{\begin{array}{l} a=\sqrt{1+\cos \sigma_z \frac{\pi}{2}} \\ b=\sqrt{1-\cos \sigma_z \frac{\pi}{2}} \cdot \operatorname{sign}\left(\sigma_z\right) \end{array}\right. \tag{32}\]
这样,我们完成了对椭圆偏振光坐标的转换。对于 Equation 26 决定的椭偏光,将其分解成圆偏振光的叠加,可令 \[ a \hat{\boldsymbol{x}}^{\prime}+b \hat{\boldsymbol{y}}^{\prime} e^{-i \frac{\pi}{2}}=A\left(\hat{\boldsymbol{x}}^{\prime}+\hat{\boldsymbol{y}}^{\prime} e^{-i \frac{\pi}{2}}\right)+B\left(\hat{\boldsymbol{x}}^{\prime}+\hat{\boldsymbol{y}}^{\prime} e^{i \frac{\pi}{2}}\right) \] 可以解得 \[ \left\{\begin{array}{l} A=\frac{1}{2}(a+b) \\ B=\frac{1}{2}(a-b) \end{array}\right. \tag{33}\] 于是,我们得到如下分解
\[ \hat{\boldsymbol{x}}+\hat{\boldsymbol{y}} e^{-i \sigma_z \frac{\pi}{2}}=\frac{1}{2}(a+b)\left(\hat{\boldsymbol{x}}^{\prime}+\hat{\boldsymbol{y}}^{\prime} e^{-i \frac{\pi}{2}}\right)+\frac{1}{2}(a-b)\left(\hat{\boldsymbol{x}}^{\prime}+\hat{\boldsymbol{y}}^{\prime} e^{i \frac{\pi}{2}}\right) \] 其中,\(a,b\) 由式 Equation 32 决定。可以看到,分解的振幅为实数。
当然,我们更关注的是任意按振幅为 \(A,B\) 叠加的左旋和右旋圆偏振光的角动量是多少。为此,用 Equation 33 反表示出 \(a,b\),即
\[ \left\{\begin{array}{l} a=A+B \\ b=A-B \end{array}\right. \] 将其代入 Equation 30 和 Equation 31 可以得到 \[ A^2+B^2=1 \] 及 \[ \sin\sigma_z\frac\pi2=A^2-B^2 \] 于是有以下结论:任意椭偏光可以分解成两个圆偏光的叠加 \(\hat{\boldsymbol{x}}+\hat{\boldsymbol{y}} e^{-i \sigma_z \frac{\pi}{2}}=A\left(\hat{\boldsymbol{x}}^{\prime}+\hat{\boldsymbol{y}}^{\prime} e^{-i \frac{\pi}{2}}\right)+B\left(\hat{\boldsymbol{x}}^{\prime}+\hat{\boldsymbol{y}}^{\prime} e^{i \frac{\pi}{2}}\right)\),其中 \(A,B\) 为实数,满足归一化条件 \(A^2+B^2=1\),该椭偏光具有大小为 \(\hbar\sin\sigma_z\frac\pi2=\hbar(A^2-B^2)\) 的角动量。
References
Footnotes
若按此定义,则 Equation 24 的第二项会与文献 (Allen et al. 1992) 中式(10)差一负号。↩︎