NQED（2）自由电磁场的量子化
2020-3-9 10:20

NQED（2）自由电磁场的量子化

1. 自由电磁场的量子化

本文中, 我们将采用一种推广的张量记号. 经典场被看作是一种广义的向量. 时空坐标x和场的内部指标将合在一起构成这些向量的复合指标. 因此, 电磁场在时空点x的逆变和协变分量将被记作${{A}^{x\mu }}$和${A_{x\mu }}$. 我们将用广义的度规来升降这些复合指标. 例如:

${{A}^{x\mu }}=\int{dy}\cdot {{g}^{x\mu ,y\nu }}{{A}_{y\nu }}$,

${{A}_{y\nu }}=\int{dx}\cdot {{g}_{y\nu .x\mu }}{{A}^{x\mu }}$                         (1.1)

${{g}^{x\mu ,y\nu }}={{\delta }^{xy}}{{g}^{\mu \nu }}=\delta (x-y){{g}^{\mu \nu }}$

${{g}_{y\nu .x\mu }}={{\delta }_{yx}}{{g}_{\nu \mu }}=\delta (x-y){{g}_{\nu \mu }}$        (1.2)

${\Psi = \Psi ({A^{x\mu }})}$                          (1.3)

${\Psi = \Psi (A)}$.                                (1.4)

${S_\gamma } = \int {dx} \cdot {A_{x\mu }}( - \frac{1}{2}{\hat k^2}){A^{x\mu }}$                  (1.5)

${\Psi _{0\gamma }} = \exp [\frac{i}{2}{S_{\gamma F}}],$

${S_{\gamma F}} = - \frac{1}{2}\int {dx} \cdot {A_{x\mu }}({{\hat k}^2} + i\hat \varepsilon ){A^{x\mu }}.$                  (1.6)

${{\hat A}^{( \pm )x\mu }} = \frac{{{A^{x\mu }}}}{2} \pm {D_F}{^{x\mu }_{y\nu }} \cdot \frac{1}{i}\frac{\delta }{{\delta {A_{y\nu }}}}$

$= \frac{{{A^{x\mu }}}}{2} \mp \frac{1}{{\hat k + i\varepsilon }}\frac{1}{i}\frac{\delta }{{\delta {A_{x\mu }}}},$

${{\hat A}^{( \pm )}}_{y\nu } = \frac{{{A_{y\nu }}}}{2} \pm {D_F}{^{x\mu }_{y\nu }} \cdot \frac{1}{i}\frac{\delta }{{\delta {A^{x\mu }}}}$

$= {g_{x\mu ,y\nu }}{{\hat A}^{( \pm )x\mu }},.$      (1.7)

${{\hat A}^{( - )x\mu }} \cdot {\Psi _{0\gamma }} = 0,$

${{\hat A}^{( - )}}_{x\mu } \cdot {\Psi _{0\gamma }} = 0.$           (1.8)

$[{{\hat A}^{( - )x\mu }},{{\hat A}^{( + )}}_{y\nu }] = i{D_F}{^{x\mu }_{y\nu }},$

$[{{\hat A}^{( - )x\mu }},{{\hat A}^{( + )y\nu }}] = i{g^{y\nu ,z\lambda }}{D_F}{^{x\mu }_{z\lambda }} = i{D_F}^{x\mu ,y\nu },$

$[{{\hat A}^{( - )}}_{x\mu },{{\hat A}^{( + )}}_{y\nu }] = i{g_{x\mu ,z\lambda }}{D_F}{^{z\lambda }_{y\nu }} = i{D_{Fx\mu ,y\nu }}$            (1.9)

${A^{x\mu }} = {{\hat A}^{( + )x\mu }} + {{\hat A}^{( - )x\mu }},$

${A_{x\mu }} = {{\hat A}^{( + )}}_{x\mu } + {{\hat A}^{( - )}}_{x\mu }.$ (1.10)

${\hat k^2}{\hat A^{( - )x\mu }} \cdot \Psi (A) = 0$                (1.11)

${\hat k^2}{A^{x\mu }} \cdot \Psi (A) = 0,$                     (1.12)

$({\Psi _1},{\Psi _2}) = ({\Psi _2},{\Psi _1}),$

$(a{\Psi _1},{\Psi _2}) = a({\Psi _1},{\Psi _2}),$                  (1.13)

$(\Psi ,\Psi ) = 1.$                               (1.14)

$({\Psi _{0\gamma }},{\Psi _{0\gamma }}) = 1.$              (1.15)

如果对任意两个波泛函${\Psi _1},{\Psi _2}$  都有等式:

$(\hat F{\Psi _1},{\Psi _2}) = ({\Psi _1},\hat G{\Psi _2}).$            (1.16)

$({\Psi _1},{\Psi _2}) = \int {DA} \cdot {\Psi _1}{\Psi _2}$            (1.17)

$\Psi \hat F = {\hat F^T}\Psi .$                            (1.18)

${\left( {{{\hat A}^{( + )x\mu }}} \right)^T} = {{\hat A}^{( - )x\mu }}.,$

${\left( {{{\hat A}^{( + )}}_{x\mu }} \right)^T} = {{\hat A}^{( - )}}_{x\mu ,}$              (1.19)

${\Psi _{0\gamma }} \cdot {{\hat A}^{( + )x\mu }} = 0,$

${\Psi _{0\gamma }} \cdot {{\hat A}^{( + )}}_{x\mu } = 0,$               (1.20)

$\Psi \cdot [{\hat k^2}{\hat A^{( + )x\mu }}] = 0.$                                    (1.21)

第四个原理是关于测量结果的原理. 我们假定, 一个可观察量F的多次测量的平均值是

$< F{ > _\Psi } = \frac{{\int {DA} \cdot \Psi \hat F\Psi }}{{\int {DA} \cdot \Psi \Psi }}$     (1.22)

为了能得到量子场的能量动量守恒定律, 我们假定量子场的拉格朗日密度算符是

${\hat L_\gamma }(x) = - 2{\hat A^{( + )}}_{x\lambda }{\hat k^2}{\hat A^{( - )x\lambda }}$,                      (1.23)

${\hat T^{\mu \nu }}_\gamma (x) = - 4{\hat A^{( + )}}_{x\lambda }{\hat k^\nu }{\hat k^\mu }{\hat A^{( - )x\lambda }} - {g^{\mu \nu }}{\hat L_\gamma }(x)$  .             (1.24)

${\partial _\nu }\int {DA \cdot \Psi {{\hat T}^{\mu \nu }}_\gamma (x)\Psi } = 0.$                           (1.25)

${A^{k\mu }} = u_x^k{A^{x\mu }},$

${A_{k\mu }} = u_k^x{A_{x\mu }},$                                (1.26)

$u_x^k = {(2\pi )^{ - 2}}{e^{ + ikx}},$

$u_k^x = {(2\pi )^{ - 2}}{e^{ - ikx}}.$                            (1.27)

${A^{k\mu *}} = {({A^{k\mu }})^*},$

$u_{x*}^{k*} = {(u_x^k)^*}.$                          (1.28)

${A^{k\mu *}} = u_{x*}^{k*}{A^{x\mu *}},$                          (1.29)

${A_{k\mu }} = {g_{k\mu ,k'\nu *}}{A^{k'\nu *}} = {g_{kk'*}}{g_{\mu \nu *}}{A^{k'\nu *}},$

${g_{kk'*}} = \delta (k - k'),$

${g_{\mu \nu *}} = {g_{\mu \nu }},$

${A^{k\mu }} = {g^{k\mu ,k'\nu *}}{A_{k'\nu *}} = {g^{kk'*}}{g^{\mu \nu *}}{A_{k'\nu *}},$

${g^{kk'*}} = \delta (k - k'),$

${g^{\mu \nu *}} = {g^{\mu \nu }}.$                   (1.30)

${D_F}{^{k\mu }_{k'\nu }} = {\delta ^k}_{k'}{\delta ^\mu }_\nu {D_F}({k^2}),$

${D_F}({k^2}) = - \frac{1}{{{k^2} + i\varepsilon ({k^2})}}.$            (1.31)

${D_F}{^{x\mu }_{y\nu }} = u_k^x{D_F}{^{k\mu }_{k'\nu }}u_y^{k'}.$          (1.32)

${{\hat A}^{( \pm )k\mu }} = u_x^k{{\hat A}^{( \pm )x\mu }}$

$= \frac{{{A^{k\mu }}}}{2} \pm {D_F}({k^2})\frac{1}{i}\frac{\delta }{{\delta {A_{k\mu }}}},$

${{\hat A}^{( \pm )k\mu *}} = u_{x*}^{k*}{{\hat A}^{( \pm )x\mu *}}$

$= \frac{{{A^{k\mu *}}}}{2} \pm {D_F}({k^2})\frac{1}{i}\frac{\delta }{{\delta {A_{k\mu *}}}}$       (1.33)

${{\hat A}^{( \pm )}}_{k\mu } = \frac{{{A_{k\mu }}}}{2} \pm {D_F}({k^2})\frac{1}{i}\frac{\delta }{{\delta {A^{k\mu }}}}$

$= {g_{k\mu ,k'\nu *}}{{\hat A}^{( \pm )k'\nu *}},$

${{\hat A}^{( \pm )}}_{k\mu *} = \frac{{{A_{k\mu *}}}}{2} \pm {D_F}({k^2})\frac{1}{i}\frac{\delta }{{\delta {A^{k\mu *}}}}$

$= {g_{k\mu *,k'\nu }}{{\hat A}^{( \pm )k'\nu }}$   (1.34)

$[{{\hat A}^{( - )k\mu }},{{\hat A}^{( + )}}_{k'\nu }] = i{D_F}({k^2}){\delta ^\mu }_\nu {\delta ^k}_{k'},$

$[{{\hat A}^{( - )k\mu }},{{\hat A}^{( + )k'\nu *}}] = i{D_F}({k^2}){g^{k\mu ,k'\nu *}},$

$[{{\hat A}^{( - )}}_{k\mu },{{\hat A}^{( + )}}_{k'\nu *}] = i{D_F}({k^2}){g_{k\mu ,k'\nu *}}.$                     (1.35)

${D_F}{^{k\mu }_{x\nu }} = u_x^{k'}{D_F}{^{k\mu }_{k'\nu }} = u_x^k{\delta ^\mu }_\nu {D_F}({k^2}),$

${D_F}{^{k\mu *}_{x\nu }} = u_{x*}^{k*}{\delta ^\mu }_\nu {D_F}({k^2}),$      (1.36)

$[{{\hat A}^{( - )k\mu }},{{\hat A}^{( + )}}_{x\nu }] = i{D_F}{^{k\mu }_{x\nu }},$

$[{{\hat A}^{( - )k\mu *}},{{\hat A}^{( + )}}_{x\nu }] = i{D_F}{^{k\mu *}_{x\nu }},$     (1.37)

${k^2}{\hat A^{( - )k\mu }} \cdot \Psi = 0,$                        (1.38)

${{\hat A}^{( - )k\mu }}{\Psi _{0\gamma }} = {\Psi _{0\gamma }}{{\hat A}^{( + )k\mu }} = 0,$

${{\hat A}^{( - )}}_{k\mu }{\Psi _{0\gamma }} = {\Psi _{0\gamma }}{{\hat A}^{( + )}}_{k\mu } = 0,$

${{\hat A}^{( - )k\mu }} + {{\hat A}^{( + )k\mu }} = {A^{k\mu }},$

${{\hat A}^{( - )}}_{k\mu } + {{\hat A}^{( - )}}_{k\mu } = {A_{k\mu }}.$      (1.39)

${\Psi _{k\rho \theta }} = ({e^{ + i\theta }}{\rho _\mu }{A^{k\mu }} + {e^{ - i\theta }}{\rho _{\mu *}}{A^{k\mu *}}) \cdot {\Psi _{0\gamma }}$                   (1.40)

${\rho ^\mu }{\rho _\mu } = - 1$                                (1.41)

$\int {DA} \cdot {\Psi _{k\rho \theta }}{\Psi _{k\rho \theta '}} = - 2i{D_F}({k^2}){\delta ^4}(0)\cos (\theta - \theta ').$                 (1.42)

${\partial _\mu }{\hat A^{( - )x\mu }} \cdot \Psi = 0.$                (1.43)

${\rho ^\mu }{k_\mu } = 0.$                                    (1.44)

在动量表示中, 量子场的能量动量张量密度算符可取作:

${\hat T^{\mu \nu }}_\gamma (k) = - 4{\hat A^{( + )}}_{k\lambda }{k^\nu }{k^\mu }{\hat A^{( - )k\lambda }} - {g^{\mu \nu }}{\hat L_\gamma }(k)$                   (1.45)

${\hat L_\gamma }(k) = - 2{\hat A^{( + )}}_{k\lambda }{k^2}{\hat A^{( - )k\lambda }}$    .                         (1.46)

$\int {DA} \cdot {\Psi _{k\rho \theta }} \cdot \int {dk'} \cdot {{\hat T}^{0\mu }}_\gamma (k'){\Psi _{k\rho \theta }}$

$= - 8{k^0}{k^\mu }{D_F}({k^2}){D_F}({k^2}){\delta ^4}(0)$

$= - 2i{D_F}({k^2}){\delta ^4}(0) \cdot 2\pi \delta (0) \cdot {k^\mu }$   。(1.47)