矢量场的量子化 |
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洛伦兹规范下的场量子化#
按照正则量子化的手续，我们定义矢量场的共轭动量为 (728)#\[\begin{equation} \pi^\mu(x) = \frac{\partial {\cal L_{Maxwell+g.f.}}}{\partial \dot A_\mu} \end{equation}\]具体到分量上，有 (729)#\[\begin{equation} \pi^0(x) = - \partial_\mu A^\mu \,, \qquad \pi^i(x) = F^{i0}(x) = \partial^i A^0(x) - \partial^0 A^i(x) \end{equation}\]注意到\(A_\mu\)到共轭动量为\(\pi^\mu\)，因此等时对易关系为 (730)#\[\begin{equation} [A_\mu(t, \vec x), \pi^\nu (t, \vec y)] = i \delta_\mu^\nu \delta^3(\vec x - \vec y) \end{equation}\]或者 (731)#\[\begin{equation} [A^\mu(t, \vec x), \pi^\nu (t, \vec y)] = i \eta^{\mu\nu} \delta^3(\vec x - \vec y) \end{equation}\]其它对易关系为零。 洛伦兹规范下的运动方程 (732)#\[\begin{equation} \Box A^\mu(x) = \partial^\nu \partial_\nu A^\mu(x) = 0 \end{equation}\]与四组无质量克莱因-戈登方程一致。因此我们可以将\(A^\mu\)作模式展开为 (733)#\[\begin{equation} A^\mu(x) = \int \frac{d^3 k}{(2\pi)^3 2 \omega_k} \sum_{\lambda = 0}^3 \epsilon_\lambda^\mu(\vec k) \left[ a_\lambda(\vec k) e^{- i k \cdot x} + a_\lambda^\dagger(\vec k) e^{i k \cdot x} \right] \end{equation}\]其中\(\omega_k = k^0 = |\vec k|\)。为了保持指标平衡，我们还引入费曼规范下的极化矢量 (734)#\[\begin{equation} \epsilon_{0, \mu} = \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix} \,, \qquad \epsilon_{1, \mu} = \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix} \,, \qquad \epsilon_{2, \mu} = \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix} \,, \qquad \epsilon_{3, \mu} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix} \end{equation}\]对应的将洛伦兹指标提升后的极化矢量为 (735)#\[\begin{equation} \epsilon_{0}^\mu = \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix} \,, \qquad \epsilon_{1}^\mu = \begin{pmatrix} 0 \\ -1 \\ 0 \\ 0 \end{pmatrix} \,, \qquad \epsilon_{2}^\mu = \begin{pmatrix} 0 \\ 0 \\ -1 \\ 0 \end{pmatrix} \,, \qquad \epsilon_{3}^\mu = \begin{pmatrix} 0 \\ 0 \\ 0 \\ -1 \end{pmatrix} \end{equation}\]它们满足如下关系： (736)#\[\begin{equation} \epsilon_{\lambda}^\mu \epsilon_{\lambda'}^\nu \eta_{\mu\nu} = \eta_{\lambda\lambda'} \end{equation}\]利用这个关系可以求出矢量场产生湮灭算符的对易关系。 简单的计算可以证明，产生湮灭算符可以写为 (737)#\[\begin{align} a_{\lambda}(\vec k) = &\ i \int d^3 x \, \epsilon_\lambda^\mu(\vec k) \eta^{\lambda \lambda} e^{i kx } \overset{\leftrightarrow}{\partial}_0 A_\mu(x) \\ a_{\lambda}^\dagger(\vec k) = &\ i \int d^3 x \, A^{\mu}(x) \overset{\leftrightarrow}{\partial}_0 e^{-i kx } \epsilon_{\lambda, \mu}(\vec k)\eta^{\lambda \lambda} \end{align}\]注意此处的极化指标\(\lambda\)不作求和。利用这个关系，可以验证 (738)#\[\begin{equation} [a_\lambda(\vec k), a_{\lambda'}(\vec k')] = 0 \end{equation}\] (739)#\[\begin{equation} [a_\lambda^\dagger(\vec k), a_{\lambda'}^\dagger(\vec k')] = 0 \end{equation}\]剩余的对易关系我们通过如下的计算得到。首先从等时对易关系 (740)#\[\begin{equation} [A_\mu(t, \vec x), A_\nu (t, \vec y)] = 0 \end{equation}\]得到如下等式 (741)#\[\begin{equation} [ \partial^i A_\mu(t, \vec x), A_\nu (t, \vec y)] = 0 \end{equation}\]进一步得到下述等式： (742)#\[\begin{align} [A^0(t, \vec x), \pi^0(t, \vec x)] = &\ [A^0, - \dot A^0 + \partial_i A^i] = - [ A^0, \dot A^0] = i \delta^{(3)} (\vec x - \vec y) \\ [A^i(t, \vec x), \pi^0(t, \vec y)] = &\ 0 = [A^0, \dot A^0] \\ [A^0(t, \vec x), \pi^i(t, \vec y)] = &\ [A^0, \partial^i A^0 - \partial^0 A^i] = - [A^0, \dot A^i] = 0 \\ [A^i(t, \vec x), \pi^j(t, \vec y)] = &\ [A^i, \partial^j A^0 - \partial^0 A^j] = - [A^i, \dot A^j] = i \eta^{ij} \delta^{(3)} (\vec x - \vec y) \end{align}\]上面的四个关系又可以统一写为： (743)#\[\begin{equation} [A^\mu(t, \vec x), \dot A^\nu(t, \vec y)] = - i \eta^{\mu\nu} \delta^{(3)} (\vec x - \vec y) \end{equation}\]利用这个结果，可以求出产生湮灭算符的对易关系： (744)#\[\begin{align} [a_\lambda(\vec p), a_{\lambda'}(\vec q)] = &\ i^2 \int d^3 x\, \epsilon_\lambda^\mu(\vec p) \eta^{\lambda \lambda} \int d^3 y \, \epsilon_{\lambda'}^\nu (\vec q) \eta^{\lambda' \lambda'} \\ &\ \times e^{i px - i qy} \Big[ \dot A_\mu (x) - i p^0 A_\mu(x) , -iq^0 A_\nu(y) - \dot A_\nu(y) \Big] \\ =&\ (-1) \int d^3 x\, \epsilon_\lambda^\mu(\vec p) \eta^{\lambda \lambda} \int d^3 y \, \epsilon_{\lambda'}^\nu (\vec q) \eta^{\lambda' \lambda'} e^{i px - i qy} \\ &\ \times e^{i px - i qy} \Big( -i q^0 i \eta_{\mu \nu} - i p^0 i \eta_{\mu\nu} \Big) \delta^{(3)}(\vec x - \vec y) \\ =&\ -(2 \pi)^3 2 p^0 \eta_{\lambda \lambda'} \delta^{(3)}(\vec p - \vec q) \end{align}\]我们注意到当\(\lambda = \lambda' = 0\)时， (745)#\[\begin{equation} [a_0(\vec p), a_0^\dagger(\vec q)] = - (2 \pi)^3 2 p^0 \delta^{(3)}(\vec p - \vec q) \end{equation}\]右侧出现的负号意味着\(\lambda=0\)所对应的极化态是非物理的。例如，考虑\(a_0^\dagger\)所产生的单粒子态， (746)#\[\begin{equation} |k, 0 \rangle = a_0^\dagger(\vec k) |0\rangle \end{equation}\]这个态具有负模长： (747)#\[\begin{equation} \langle q, 0 | k, 0 \rangle = \langle 0 | a_0(\vec q) a_0^\dagger(\vec k) | 0 \rangle = - (2 \pi)^3 2 k^0 \delta^{(3)}( \vec q - \vec k) < 0 \end{equation}\]因此不能作为物理态存在。 另外，无质量矢量粒子（例如光子）仅有两个物理自由度，而\(a_1^\dagger(\vec k)\)，\(a_2^\dagger(\vec k)\)，\(a_3^\dagger(\vec k)\)可以产生三个不同的物理态，因此其中必有一个是非物理的。为了回避非物理态的贡献，可以引入Guptar-Bleuler条件：对于任意的物理态\(|\psi_T\rangle\)和\(|\phi_T \rangle\)，其中下标\(T\)代表物理态，有 (748)#\[\begin{equation} \langle \psi_T | \partial_\mu A^\mu |\phi_T \rangle = 0 \end{equation}\]将\(A^\mu\)用模式展开代入，得到 (749)#\[\begin{equation} \int \frac{d^3 k}{(2 \pi)^3 2 \omega_k} \sum_{\lambda = 0}^3 i k_\mu \epsilon_\lambda^\mu(\vec k) \langle \psi_T | a_{\lambda}(\vec k) - a_{\lambda}^\dagger(\vec k) |\phi_T \rangle = 0 \end{equation}\]或者分别写为 (750)#\[\begin{equation} \sum_{\lambda = 0}^3 k_\mu \epsilon_\lambda^\mu(\vec k) a_{\lambda}(\vec k) |\phi_T \rangle = 0 \,, \qquad \forall k \end{equation}\]以及 (751)#\[\begin{equation} \sum_{\lambda = 0}^3 k_\mu \epsilon_\lambda^\mu(\vec k) \langle \psi_T | a_{\lambda}^\dagger(\vec k) = 0 \,, \qquad \forall k \end{equation}\]为了看出Guptar-Bleuler条件确实可以排除非物理态对物理量的贡献，不妨考虑自由矢量场的哈密顿量： (752)#\[\begin{equation} H = \int d^3 x \, {\cal H}(x) = \int^3 d^3x ( \pi^\mu \dot A_\mu - {\cal L}_{Maxwell+g.f.}) \end{equation}\]通过计算可以发现： (753)#\[\begin{equation} :H: = \int \frac{d^3 k}{(2\pi)^3 2 \omega_k} \omega_k \left[ \sum_{\lambda = 1}^3 a_\lambda^\dagger(\vec k) a_\lambda(\vec k) -a_0^\dagger(\vec k) a_0(\vec k) \right] \end{equation}\]不妨设\(k^\mu = (k^0, 0, 0, k^0)\)，则Guptar-Bleuler条件可以写为 (754)#\[\begin{equation} (a_0(\vec k) + a_3(\vec k)) | \phi_T \rangle = 0\,, \qquad \langle \psi_T| (a_0^\dagger(\vec k) + a_3^\dagger(\vec k)) = 0 \end{equation}\]方括号中的量在物理态\(|\psi_T\rangle\)中的矩阵元可以写为 (755)#\[\begin{align} \langle \psi_T | \left[ \sum_{\lambda = 1}^3 a_\lambda^\dagger(\vec k) a_\lambda(\vec k) - a_3^\dagger(\vec k)a_0(\vec k) \right] | \psi_T \rangle =&\ \langle \psi_T | \left[ \sum_{\lambda = 1}^3 a_\lambda^\dagger(\vec k) a_\lambda(\vec k) - (-a_3^\dagger(\vec k)) (- a_3(\vec k)) \right] | \psi_T \rangle \\ =&\ \langle \psi_T | \left[ \sum_{\lambda = 1}^2 a_\lambda^\dagger(\vec k) a_\lambda(\vec k) \right] | \psi_T \rangle \end{align}\]而 (756)#\[\begin{equation} {\cal N}(\vec k) = \left[ \sum_{\lambda = 1}^2 a_\lambda^\dagger(\vec k) a_\lambda(\vec k) \right] \end{equation}\]可以理解为物理态中动量模式为\(\vec k\)的态数目算符。因此，\(:H:\)在物理态中确实给出物理自由度的总能量。 最后，我们求一下协变费曼规范下的矢量场传播子， (757)#\[\begin{align} \langle 0 | T \{A_\mu(x) A_\nu(y) \} | 0 \rangle = &\ \theta(x^0 - y^0) \langle 0 | A_\mu(x) A_\nu(y) | 0 \rangle + \theta(y^0 - x^0) \langle 0 | A_\nu(y) A_\mu(x) | 0 \rangle \end{align}\]而 (758)#\[\begin{align} \langle 0 | A^\mu(x) A^\nu(y) | 0 \rangle = &\ \int \frac{d^3 k}{(2\pi)^3 2 \omega_k} \int \frac{d^3 k'}{(2\pi)^3 2 \omega_{k'}} \sum_{\lambda, \lambda'} \epsilon_\lambda^\mu(\vec k) \epsilon_{\lambda'}^\nu(\vec k') \langle 0 | a_\lambda(\vec k) a_{\lambda'}^\dagger(\vec k) | 0 \rangle e^{-i kx + i k' y} \\ =&\ \int \frac{d^3 k}{(2\pi)^3 2 \omega_k} \sum_{\lambda, \lambda'} \epsilon_\lambda^\mu(\vec k) \epsilon_{\lambda'}^\nu(\vec k') (-\eta_{\lambda \lambda'}) e^{-i k(x - y)} \\ =&\ - \eta^{\mu\nu} \int \frac{d^3 k}{(2\pi)^3 2 \omega_k} e^{-i k(x - y)} \end{align}\]结合正时序和反时序部分就得到 (759)#\[\begin{equation} \langle 0 | T\{A_\mu(x) A_\nu(y) \} | 0 \rangle = \int \frac{d^4 k}{(2\pi)^4} \frac{-i \eta_{\mu\nu}}{k^2 + i \epsilon} e^{-i k \cdot (x - y)} \end{equation}\] |
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