---
abstract: |
    The quantization of the SU(2)$\times $U(1) gauge-symmetric electroweak theory is performed in the Hamiltonian path-integral formalism. In this quantization, we start from the Lagrangian given in the unitary gauge in which the unphysical Goldstone fields are absent, but the unphysical longitudinal components of the gauge fields still exist. In order to eliminate the longitudinal components, it is necessary to introduce the Lorentz gauge conditions as constraints. These constraints may be incorporated into the Lagrangian by the Lagrange undetermined multiplier method. In this way, it is found that every component of a four-dimensional vector potential has a conjugate counterpart. Thus, a Lorentz-covariant quantization in the Hamiltonian path-integral formalism can be well accomplished and leads to a result which is the same as given by the Faddeev-Popov approach of quantization.

    PACS:

    :   11.15.-qy; 11.10.Gh; 12.20.-m

address: |
    Center for Theoretical Physics, Department of Physics, Jilin\
    University,Changchun 130023, People’s Republic of China
author:
- 'Jun-Chen Su'
title: 'Quantization of The Electroweak Theory in The Hamiltonian Path-Integral Formalism'
---

In our preceding paper$^{[1]}$, the quantum electroweak theory without involving the Goldstone bosons$^{[2,3]}$ was established starting from the Lagrangian given in the unitary gauge by the Faddeev-Popov approach$^{[4]}$ in the Lagrangian path-integral formalism. The quantum theory given in the $%
\alpha $-gauge shows good renormalizability. The unitarity is ensured by the limiting procedure: $\alpha \rightarrow \infty $ by which the results calculated in the $\alpha $-gauge will be converted to the physical ones as should be obtained in the unitary gauge. An important feature of the theory is that the theory established is still of SU(2)$\times $U(1) gauge symmetry. To check this point, in this paper, the quantization of the theory will be carried out in the Hamiltonian path-integral formalism$^{[5-7]}$. This quantization, unlike the quantization by the Faddeev-Popov approach, does not concern the gauge transformation and the gauge-invariance of the Lagrangian chosen for the quantization.

For simplicity, we limit ourself to discuss the electroweak interaction system for one generation of leptons. the Lagrangian is$^{[2,3,8]}$ $${\cal L}={\cal L}_g+{\cal L}_f+{\cal L}_\phi  \eqnum{1}$$ where $${\cal L}_g=-\frac 14F^{\alpha \mu \nu }F_{\mu \nu }^\alpha  \eqnum{2}$$ $${\cal L}_f=\overline{L}i\gamma ^\mu D_\mu L+\overline{l}_Ri\gamma ^\mu D_\mu
l_R  \eqnum{3}$$ and $${\cal L}_\phi =\frac 12(D^\mu \phi )^{+}(D_\mu \phi )-\frac 12\mu ^2\phi
^{+}\phi -\frac 14\lambda (\phi ^{+}\phi )^2-\frac{f_l}{\sqrt{2}}(\overline{L%
}\phi l_R+\overline{l}_R\phi ^{+}L)  \eqnum{4}$$ are respectively the parts of the Lagrangian for the gauge boson, lepton and scalar particle fields. In the above, $$F_{\mu \nu }^\alpha =\partial _\mu A_\nu ^\alpha -\partial _\nu A_\mu
^\alpha +g\varepsilon ^{\alpha \beta \gamma }A_\mu ^\beta A_\nu ^\gamma 
\eqnum{5}$$ where $\alpha =0,1,2,3,$ A$_\mu ^i=$ (A$_\mu ^a,B_\mu ),$A$_\mu ^a$ and $%
B_\mu $ are the SU(2) and U(1) gauge fields respectively and the Levi-Civita tensor is defined by $$\varepsilon ^{\alpha \beta \gamma }=\{ 
\begin{array}{c}
\epsilon ^{abc}\text{ , if }\alpha ,\beta ,\gamma =a,b,c=1,2\text{ or }3; \\ 
0\text{ , if }\alpha ,\beta \text{ and/or }\gamma =0
\end{array}
\eqnum{6}$$ $$L=\left( 
\begin{array}{c}
\nu _L \\ 
l_L
\end{array}
\right)  \eqnum{7}$$ here $\nu _L$ represents the left handed neutrino field and $l_L$ the left-handed lepton field, $l_R$ is the right-handed lepton field , $$D_\mu =\partial _\mu -ig\frac{\tau ^a}2A_\mu ^a-ig^{\prime }\frac Y2B_\mu 
\eqnum{8}$$ is the covariant derivative in which $\frac{\tau ^a}2$ and $\frac Y2$ are the generators of SU(2)$_T$ and U(1)$_Y$ gauge groups, respectively, and $$\phi (x)=\left( 
\begin{array}{c}
0 \\ 
H(x)+v
\end{array}
\right)  \eqnum{9}$$ is the scalar field doublet, $g,g^{\prime },\mu ^2,\lambda $ and $f_l$ are the coupling constants with a relation $v=\sqrt{-\mu ^2/\lambda }$.

In above Lagrangian, still exist the unphysical longitudinal parts of the gauge fields. They may completely be eliminated by introducing the Lorentz conditions $$\chi ^\alpha \equiv \partial ^\mu A_\mu ^\alpha =0  \eqnum{10}$$

According to the general procedure of dealing with constrained systems $%
^{[5,6]}$, the Lorentz conditions, as constraints, may be incorporated into the Lagrangian by the Lagrange multiplier method $${\cal L}_\lambda ={\cal L}+\lambda ^\alpha \partial ^\mu A_\mu ^\alpha 
\eqnum{11}$$ where ${\cal L}$ is the Lagrangian shown in Eqs.(1)-(4) and $\lambda ^\alpha 
$ are the Lagrange multipliers. The above Lagrangian will suitably be chosen to be the starting point of performing the Hamiltonian path-integral quantization. The advantage of the Lagrangian is that it provides each component of a vector potential a canonically conjugate counterpart. In fact, according to the usual definition of the canonical momentum (density) of a field variable, it is found $$\Pi _\mu ^\alpha =\frac{\partial {\cal L}_\lambda }{\partial \stackrel{%
\bullet }{A}^{\alpha \mu }}=F_{\mu 0}^\alpha +\lambda ^\alpha \delta _{\mu
0}=\{ 
\begin{array}{c}
F_{k0}^\alpha \equiv E_k^\alpha \text{ , if }\mu =k=1,2,3 \\ 
\lambda ^\alpha \equiv -E_0^\alpha \text{ , if }\mu =0
\end{array}
\eqnum{12}$$ here, as we see, the Lagrange multipliers act as the time-components of the conjugate momenta. With the conjugate momenta defined above and the conjugate momenta for fermion fields and the Higgs field which are defined as $$\begin{aligned}
\Pi _l &=&\frac{\partial {\cal L}_\lambda }{\delta \stackrel{\bullet }{l}}%
=il^{+}\text{ , }\overline{\Pi }_l=\frac{\partial {\cal L}_\lambda }{\delta 
\stackrel{\bullet }{\stackrel{\_}{l}}}=0  \nonumber \\
\Pi _\nu &=&\frac{\partial {\cal L}_\lambda }{\delta \stackrel{\bullet }{\nu 
}}=i\nu _L^{+}\text{ , }\overline{\Pi }_\nu =\frac{\partial {\cal L}_\lambda 
}{\delta \stackrel{\bullet }{\stackrel{\_}{\nu }}}=0  \eqnum{13}\end{aligned}$$ and $$\Pi _H=\frac{\partial {\cal L}_\lambda }{\delta \stackrel{\bullet }{H}}=%
\stackrel{\bullet }{H}  \eqnum{14}$$ according to the standard procedure, the Lagrangian in Eq.(11) may readily be recast in the first-order form $${\cal L}_\lambda =E^{\alpha \mu }\stackrel{\bullet }{A}_\mu ^\alpha +\Pi _l%
\stackrel{\bullet }{l}+\Pi _\nu \stackrel{\bullet }{\nu }+\Pi _H\stackrel{%
\bullet }{H}-{\cal H}+A_0^\alpha \varphi ^\alpha -E_0^\alpha \chi ^\alpha 
\eqnum{15}$$ where $${\cal H}={\cal H}_g+{\cal H}_f+{\cal H}_H  \eqnum{16}$$ is the Hamiltonian (density) in which $${\cal H}_g=\frac 12[(E_k^\alpha )^2+(B_k^\alpha )^2]  \eqnum{17}$$ here $$B_k^\alpha =-\frac 12\varepsilon _{ijk}F_{ij}^\alpha  \eqnum{18}$$ $${\cal H}_f=-i\overline{\nu }_L\gamma ^k\partial _k\nu _L-i\overline{l}\gamma
^k\partial _kl-\frac 12\overline{L}\gamma ^k(g\tau ^aA_k^a-g^{\prime
}B_k)L+g^{\prime }\overline{l}_R\gamma ^kB_kl_R  \eqnum{19}$$ and $$\begin{aligned}
{\cal H}_H &=&\frac 12\Pi _H^2+\frac 12\left( \nabla H\right) ^2+\frac 12\mu
^2\left( H+v\right) ^2+\frac \lambda 4\left( H+v\right) ^4  \nonumber \\
&&\ \ \ \ \ \ +\frac 14\{g^2[(A_k^1)^2+(A_k^2)^2]+(g^{\prime
}B_k-gA_k^3)^2\}\left( H+v\right) ^2  \eqnum{20} \\
&&\ \ \ \ \ \ +\frac{f_l}{\sqrt{2}}\overline{l}l\left( H+v\right)  \nonumber\end{aligned}$$ $$\varphi ^\alpha =\varphi _g^\alpha +\varphi _f^\alpha +\varphi _H^\alpha 
\eqnum{21}$$ in which $$\varphi _g^\alpha =\partial ^\mu E_\mu ^\alpha -g\varepsilon ^{\alpha \beta
\gamma }E_k^\beta A^{\gamma k}  \eqnum{22}$$ $$\varphi _f^\alpha =\{ 
\begin{array}{c}
\frac 12g\overline{L}\gamma _0\tau ^aL\text{ , if }\alpha =a=1,2,3; \\ 
-\frac 12g^{\prime }(\overline{L}\gamma _0L+2\overline{l}_R\gamma _0l_R)%
\text{ , if }\alpha =0
\end{array}
\eqnum{23}$$ and $$\varphi _H^\alpha =\{ 
\begin{array}{c}
\frac 14g(gA_0^a-\delta ^{a3}g^{\prime }B_0)\left( H+v\right) ^2\text{ , if }%
\alpha =a=1,2,3; \\ 
\frac 14g^{\prime }(g^{\prime }B_0-gA_0^3)\left( H+v\right) ^2\text{ , if }%
\alpha =0
\end{array}
\eqnum{24}$$ and $\chi ^\alpha $ was defined in Eq.(10).

From the structure of the Lagrangian in Eq.(15), it is clearly seen that the last two terms in Eq.(15) are actually given by incorporating the constraint conditions $$\varphi ^\alpha =0  \eqnum{25}$$ and that in Eq.(10) into the Lagrangian by the Lagrange multiplier method. These constraint conditions may be derived from the stationary condition of the action $S_\lambda =\int d^4x{\cal L}_\lambda (x)$. From this derivation, one may also obtain equations of motion (see Appendix A) in which there are time-derivatives of the dynamical field variables $A_k^a,B_k,F_{k0}^a=E_k^a$ and $B_{k0}=E_k^0$; whereas in Eqs.(10) and (25) there are no such derivatives. Therefore, Eqs.(10) and (25) can only be identified with the constraint equations. Since $\partial ^\mu E_\mu ^\alpha $ =$\partial ^\mu
E_{L\mu }^\alpha $ and $\partial ^\mu A_\mu ^\alpha =\partial ^\mu A_{L\mu
}^\alpha $ where $E_{L\mu }^\alpha $ and $A_{L\mu }^\alpha $ are the longitudinal parts of the canonical variables $E_\mu ^\alpha $ and $A_\mu
^\alpha $ respectively, we see, the conditions in Eq.(26) and (27) are responsible respectively for constraining the unphysical longitudinal parts of the canonical variables $E_{L\mu }^\alpha $ and $A_{L\mu }^\alpha $ . Therefore, only the transverse parts of the variables, $E_{T\mu }^\alpha $ and $A_{T\mu }^\alpha ,$ can be viewed as independent dynamical field variables. Because each of the transverse vectors $E_{T\mu }^\alpha $ and $%
A_{T\mu }^\alpha $ contains three independent components, they are sufficient to describe the polarization states of the massive gauge fields.

Let us turn to the solutions of the equations (10) and (25). The solution of equation (10), as is well-known, is $$A_{L\mu }^\alpha =0  \eqnum{26}$$ For the equation (25) with the $\varphi ^\alpha $ being represented in Eqs.(21)-(24), we would like to note that the function $\varphi _g^\alpha $ in Eq.(22) can also be written in the form of Lorentz-covariance $$\varphi _g^\alpha =\partial ^\mu E_\mu ^\alpha -g\varepsilon ^{\alpha \beta
\gamma }E_\mu ^\beta A^{\gamma \mu }  \eqnum{27}$$ This is because the added term $g\varepsilon ^{\alpha \beta \gamma
}E_0^\beta A_0^\gamma $ in the above gives a vanishing contribution to the Lagrangian (see the term $A_0^\alpha \varphi ^\alpha $ in Eq.(15)) due to the identity $\varepsilon ^{\alpha \beta \gamma }A_0^\alpha A_0^\gamma
\equiv 0$. On substituting Eq.(26) into Eq.(25) and noticing that the longitudinal vector $E_{L\mu }^\alpha $ can always be represented as $$E_{L\mu }^\alpha =\partial _\mu Q^\alpha  \eqnum{28}$$ where $Q^a$ is a scalar function, the equation in Eq.(25) may be written in the form$^{[4,5]}$ $$K^{\alpha \beta }(x)Q^\beta (x)=R^\alpha (x)  \eqnum{29}$$ where $$K^{\alpha \beta }=\delta ^{\alpha \beta }\Box -g\varepsilon ^{\alpha \beta
\gamma }A_T^{\gamma \mu }\partial _\mu  \eqnum{30}$$ and $$R^\alpha =g\varepsilon ^{\alpha \beta \gamma }E_{T\mu }^\beta A_T^{\gamma
\mu }-\varphi _f^\alpha (A_{0T}^\alpha )-\varphi _H^\alpha (A_{0T}^\alpha ) 
\eqnum{31}$$ here $\varphi _f^\alpha (A_{0T}^\alpha )$ and $\varphi _H^\alpha
(A_{oT}^\alpha )$ are defined in Eqs.(23) and (24) with the $A_0^\alpha $ being replaced by $A_{0T}^\alpha $. With the aid of the Green’s function $%
\triangle ^{\alpha \beta }(x-y)$ (the ghost particle propagator) which satisfies the equation $$K^{\alpha \gamma }(x)\triangle ^{\gamma \beta }(x-y)=\delta ^{\alpha \beta
}\delta ^4(x-y)  \eqnum{32}$$ the solution of equation (29) is found to be $$Q^\alpha (x)=\int d^4y\triangle ^{\alpha \beta }(x-y)R^\beta (y)  \eqnum{33}$$ Inserting this result into Eq.(28), we get $$E_{L\mu }^\alpha =E_{L\mu }^\alpha (A_{T\mu }^\alpha ,E_{T\mu }^\alpha
,\cdots )  \eqnum{34}$$ which is a functional of the independent field variables $A_{T\mu }^\alpha
,E_{T\mu }^\alpha $ and others.

Now, we are ready to carry out the quantization in the Hamiltonian path-integral formalism. In accordance with the basic idea of path-integral quantization$^{[5-7]}$, we are allowed to directly write out an exact generating functional of Green’s functions by making use of the Hamiltonian which is expressed in terms of the independent field variables $$Z\left[ J\right] =\frac 1N\int D\left( \Pi ^{*},\Phi ^{*}\right) \exp
\{i\int d^4x(\Pi ^{*}\stackrel{}{\cdot \stackrel{\bullet }{\Phi ^{*}}}-{\cal %
H}^{*}\left( \Pi ^{*},\Phi ^{*}\right) +J^{*}\cdot \Phi ^{*})\}  \eqnum{35}$$ where $\Phi ^{*}$ and $\Pi ^{*}$stand for all the independent variables $%
(A_{T\mu }^\alpha ,\nu ,l,H)$ and the conjugate ones $(E_{T\mu }^\alpha ,\Pi
_\nu ,\Pi _l,\Pi _H)$ respectively, $J$ denotes the external sources and $%
{\cal H}^{*}(\Pi ^{*},\Phi ^{*})$ is the Hamiltonian defined by$^{[6]}$ $${\cal H}^{*}(\Pi ^{*},\Phi ^{*})={\cal H}(\Pi ,\Phi )|_{A_{L\mu }^\alpha
=0,E_{L\mu }^\alpha =E_{L\mu }^\alpha (\Pi ^{*},\Phi ^{*})}  \eqnum{36}$$ This Hamiltonian, as it stands, has a complicated functional structure which is not convenient for establishing the perturbation theory. Therefore, one still expects to represent the generating functional through the full vectors $A_\mu ^\alpha $ and $E_\mu ^\alpha $. For this purpose, it is necessary to introduce the delta-functionals into the generating functional like this$^{[6]}$ $$\begin{aligned}
Z\left[ J\right] &=&\frac 1N\int D\left( \Pi ,\Phi \right) \delta \left[
A_L\right] \delta [E_L-E_L(\Pi ^{*},\Phi ^{*})]  \nonumber \\
&&\ \ \ \times \exp \{i\int d^4x[\Pi \cdot \Phi \stackrel{.}{-{\cal H}\left(
\Pi ,\Phi \right) }+J\cdot \Phi ]\}  \eqnum{37}\end{aligned}$$ where $\Phi =(A_\mu ^\alpha ,\nu ,l,H)$ and $\Pi =(E_\mu ^\alpha ,\Pi _\nu
,\Pi _l,\Pi _H)$.

The delta-functionals in Eq.(37) can be expressed as a useful form as follows$^{[6]}$(see appendix B) $$\delta \left[ A_L\right] \delta \left[ E_L-E_L\left( \Pi ^{*},\Phi
^{*}\right) \right] =\det M[A]\delta [\varphi ]\delta [\chi ]  \eqnum{38}$$ where $\delta [\varphi ]$ and $\delta [\chi ]$ represent the constraint conditions in Eq.(10) and (25) and M\[A\] is a matrix whose elements are given by the following Poisson bracket$^{[6]}$ $$\begin{aligned}
M^{\alpha \beta }(x,y) &=&\{\varphi ^\alpha (x),\chi ^\beta (y)\}  \nonumber
\\
&=&\int d^4z\{\frac{\delta \varphi ^\alpha (x)}{\delta A_\mu ^\gamma (z)}%
\frac{\delta \chi ^\beta (y)}{\delta E^{\gamma \mu }(z)}-\frac{\delta
\varphi ^\alpha (x)}{\delta E_\mu ^\gamma (z)}\frac{\delta \chi ^\beta (y)}{%
\delta A^{\gamma \mu }(z)}\}  \eqnum{39}\end{aligned}$$ These matrix elements are easily evaluated by using the expressions denoted in Eqs.(10), (21), (27), (23) and (24). The result is $$M^{\alpha \beta }(x,y)=\partial _x^\mu \left[ D_\mu ^{\alpha \beta
}(x)\delta ^4(x-y)\right]  \eqnum{40}$$ where $$D_\mu ^{\alpha \beta }(x)=\delta ^{\alpha \beta }\partial _\mu
^x-g\varepsilon ^{\alpha \beta \gamma }A_{\mu ^{}}^\gamma (x)  \eqnum{41}$$

Upon inserting the relation in Eq.(38) into Eq.(37) and employing the Fourier representation for $\delta [\varphi ]$$$\delta [\varphi ]=\int D(\frac \eta {2\pi })e^{i\int d^4x\eta ^\alpha
(x)\varphi ^\alpha (x)}  \eqnum{42}$$ we have $$\begin{aligned}
Z\left[ J\right] &=&\frac 1N\int D\left( \Pi ,\Phi \right) D(\frac{d\eta }{%
2\pi })\det M[A]\delta [\chi ]  \nonumber \\
&&\ \ \ \ \times \exp \{i\int d^4x(\Pi \cdot \stackrel{}{\stackrel{\bullet }{%
\Phi }+\eta ^\alpha }\varphi ^\alpha -{\cal H}+J\cdot \Phi )\}  \eqnum{43}\end{aligned}$$ For later convenience, the $E_0^\alpha $-dependent terms will be extracted from the first two terms in the above exponent and thus Eq.(45) will be rewritten as $$\begin{aligned}
Z\left[ J\right] &=&\frac 1N\int D(\Pi ^{\prime },\Phi ^{\prime })D\left(
E_0,A_0\right) D(\frac \eta {2\pi })\det M\left[ A\right] \delta [\chi ] 
\nonumber \\
&&\ \ \ \ \times \exp \{i\int d^4x[E_0^\alpha (\stackrel{\bullet }{A}%
_0^\alpha -\stackrel{\bullet }{\eta }^\alpha )+\Pi ^{\prime }\cdot \stackrel{%
\bullet }{\Phi ^{\prime }}+\eta ^\alpha \overline{\varphi }^\alpha -{\cal H}%
+J\cdot \Phi ]\}  \eqnum{44}\end{aligned}$$ where $$\Pi ^{\prime }\cdot \stackrel{\bullet }{\Phi ^{\prime }}=E^{\alpha k}%
\stackrel{\bullet }{A}_k^\alpha +\Pi _\nu \stackrel{\bullet }{\nu }+\Pi _l%
\stackrel{\bullet }{l}+\Pi _H\stackrel{\bullet }{H}  \eqnum{45}$$ and $$\eta ^\alpha \stackrel{}{\overline{\varphi }^\alpha }=\eta ^\alpha (\partial
^kE_k^\alpha +g\varepsilon ^{\alpha \beta \gamma }A_k^\beta E^{\gamma
k}+\varphi _f^\alpha +\varphi _H^\alpha )  \eqnum{46}$$ The integral over $E_0^\alpha $ in Eq.(44) gives the delta-functional $$\delta [\stackrel{\bullet }{A}_0^\alpha -\stackrel{\bullet }{\eta }^\alpha
]=\det \left| \partial _0\right| ^{-1}\delta [\eta ^\alpha -A_0^\alpha ] 
\eqnum{47}$$ The determinant in the above, as a constant, may be put in the normalization constant N. The delta-functional $\delta [\eta ^\alpha -A_0^\alpha ]$ will be used to perform the integration over $\eta ^\alpha $ in Eq.(44). After these manipulations, we get $$\begin{aligned}
Z\left[ J\right] &=&\frac 1N\int D\left( \Pi ^{\prime },\Phi ^{\prime
}\right) D\left( A_0\right) \det M[A]\delta [\chi ]  \nonumber \\
&&\ \ \ \ \times \exp \{i\int d^4x[\Pi ^{\prime }\cdot \stackrel{\bullet }{%
\Phi ^{\prime }}+A_0^\alpha \overline{\varphi }^\alpha -{\cal H}+J\cdot \Phi
]\}  \eqnum{48}\end{aligned}$$ In the above expression, the integrals over $E_k^\alpha $ and $\Pi _H$ are of Gaussion type and hence are easily calculated, giving $$\int D\left( E_k^\alpha \right) e^{-i\int d^4x\left[ \frac 12\left(
E_k^\alpha \right) ^2+E_k^\alpha F_{ok}^\alpha \right] }=e^{i\int d^4x\frac 1%
2\left( F_{k0}^\alpha \right) ^2}  \eqnum{49}$$ $$\int D\left( \Pi _H\right) e^{-i\int d^4x\left[ \frac 12\Pi _H^2-\Pi _H%
\stackrel{\bullet }{H}\right] }=e^{i\int d^4x\frac 12\stackrel{\bullet }{H}%
^2}  \eqnum{50}$$ For the integrals over $\Pi _\nu $ and $\Pi _l$, the integration variables $%
\Pi _\nu $ and $\Pi _l$ will be changed to $\overline{\nu }$ and $\overline{l%
\text{.}}$ The Jacobian caused by this change, as a constant, may be put in the constant N. On substituting Eqs.(49) and (50) in Eq.(48), it is easy to see that in the functional integral thus obtained, except for the external source terms, the sum of the other terms in the exponent just give the original Lagrangian shown in Eqs.(1)-(9). Thus, we obtain $$Z[J]=\frac 1N\int D\left( \Psi ^{\prime }\right) \det M\delta [\chi
]e^{i\int d^4x[{\cal L}+J\cdot \Psi ^{\prime }]}  \eqnum{51}$$ where $\Psi ^{\prime }=\left( \overline{l},l,\overline{\nu },\nu ,A_\mu
^\alpha ,H\right) $. When making use of the familiar expression of the determinant$^{[4]}$ $$\begin{aligned}
\det M &=&\int D\left( \overline{C},C\right) e^{i\int d^4xd^4y\overline{C}%
^\alpha (x)M^{\alpha \beta }(x,y)C^\beta (y)}  \nonumber \\
\ &=&\int D\left( \overline{C},C\right) e^{i\int d^4x\overline{C}^\alpha
\partial ^\mu (D_\mu ^{\alpha \beta }C^\beta )}  \eqnum{52}\end{aligned}$$ and the Fresnel representation for the delta-functional $$\delta [\partial ^\mu A_\mu ^{}]=\lim_{\alpha \rightarrow 0}C[\alpha ]\exp
\{-\frac i{2\alpha }\int d^4x\left( \partial ^\mu A_\mu ^\alpha \right) ^2\}
\eqnum{53}$$ where $C[\alpha ]$ is a constant, we arrive at $$Z[J]=\frac 1N\int D(\Psi )\exp \{i\int d^4x({\cal L}_{eff}+J\cdot \Psi )\} 
\eqnum{54}$$ where $${\cal L}_{eff}={\cal L}-\frac 1{2\alpha }(\partial ^\mu A_\mu ^\alpha )^2+%
\overline{C}^\alpha \partial ^\mu (D_\mu ^{\alpha \beta }C^\beta ) 
\eqnum{55}$$ is the effective Lagrangian and the limit $\alpha \rightarrow 0$ is implied in Eq.(56). With the definition of particle fields listed below$^{[2,3,8]}$ $$W_\mu ^{\pm }=\frac 1{\sqrt{2}}(A_\mu ^1\mp iA_\mu ^2)  \eqnum{57}$$ $$\left( 
\begin{array}{c}
Z_\mu \\ 
A_\mu
\end{array}
\right) =\left( 
\begin{array}{c}
\cos \theta _w-\sin \theta _w \\ 
\sin \theta _w\text{ }\cos \theta _w
\end{array}
\right) \left( 
\begin{array}{c}
A_\mu ^3 \\ 
B_\mu
\end{array}
\right)  \eqnum{57}$$ where $\theta _w$ is the Weinberg angle $$C^{\pm }=\frac 1{\sqrt{2}}\left( C^1\mp iC^2\right) \text{ , }\overline{C}%
^{\pm }=\frac 1{\sqrt{2}}(\overline{C}^1\mp i\overline{C}^2)  \eqnum{58}$$ $$\left( 
\begin{array}{c}
C_z \\ 
C_\gamma
\end{array}
\right) =\left( 
\begin{array}{c}
\cos \theta _w-\sin \theta _w \\ 
\sin \theta _w\text{ }\cos \theta _w
\end{array}
\right) \left( 
\begin{array}{c}
C^3 \\ 
C^0
\end{array}
\right)  \eqnum{59}$$ and $$\left( 
\begin{array}{c}
\overline{C}_z \\ 
\overline{C}_\gamma
\end{array}
\right) =\left( 
\begin{array}{c}
\cos \theta _w-\sin \theta _w \\ 
\sin \theta _w\text{ }\cos \theta _w
\end{array}
\right) \left( 
\begin{array}{c}
\overline{C}^3 \\ 
\overline{C}^0
\end{array}
\right)  \eqnum{60}$$ the effective Lagrangian in Eq.(54) can be rewritten in the form $${\cal L}_{eff}={\cal L}_g+{\cal L}_f+{\cal L}_H+{\cal L}_{gf}+{\cal L}_{gh} 
\eqnum{61}$$ where $$\begin{aligned}
{\cal L}_g &=&-\frac 12W_{\mu \nu }^{+}W^{-\mu \nu }-\frac 14[Z^{\mu \nu
}Z_{\mu \nu }+A^{\mu \nu }A_{\mu \nu }]+M_w^2W_\mu ^{+}W^{-\mu }+\frac 12%
M_z^2Z^\mu Z_\mu  \nonumber \\
&&\ +ig[(W_{\mu \nu }^{+}W^{-\mu }-W_{\mu \nu }^{-}W^{+\mu })\left( \sin
\theta _wA^\nu +\cos \theta _wZ^\nu \right) +W_\mu ^{+}W_\nu ^{-}\left( \sin
\theta _wA^{\mu \nu }+\cos \theta _wZ^{\mu \nu }\right) ]  \nonumber \\
&&\ +g^2\{W_\mu ^{+}W_\nu ^{-}\left( \sin \theta _wA^\mu +\cos \theta
_wZ^\mu \right) (\sin \theta _wA^\nu +\cos \theta _wZ^\nu )-W_\mu
^{+}W^{-\mu }\left( \sin \theta _wA_\nu +\cos \theta _wZ_\nu \right) ^2 
\nonumber \\
&&\ +\frac 12[(W_\mu ^{+})^2\left( W_\nu ^{-}\right) ^2-(W_\mu ^{+}W^{-\mu
})^2]\}  \eqnum{62}\end{aligned}$$ here $$\begin{aligned}
W_\mu ^{\pm } &=&(\partial _\mu W_\mu ^{\pm }-\partial _\nu W_\mu ^{\pm
}),Z_{\mu \nu }=\partial _\mu Z_\nu -\partial _\nu Z_\mu ,  \nonumber \\
A_{\mu \nu } &=&\partial _\mu A_\nu -\partial _\nu A_\mu  \eqnum{63}\end{aligned}$$ and $$M_w=\frac 12gv,M_Z=M_w/\cos \theta _w  \eqnum{64}$$ $$\begin{aligned}
{\cal L}_f &=&\overline{\nu }i\gamma ^\mu \frac 12(1-\gamma _5)\partial _\mu
\nu +\overline{l}(i\gamma ^\mu \partial _\mu -m_l)l+\frac g{\sqrt{2}}(j_\mu
^{-}W^{+\mu }+j_\mu ^{+}W^{-\mu })-ej_\mu ^{em}A^\mu  \nonumber \\
&&\ +\frac e{2\sin 2\theta _w}j_\mu ^0Z^\mu  \eqnum{65}\end{aligned}$$ in which $$j_\mu ^{+}=\overline{l}\gamma _\mu \frac 12(1-\gamma _5)\nu =(j_\mu ^{-})^{+}
\eqnum{66}$$ $$j_\mu ^{em}=\overline{l}\gamma _\mu l  \eqnum{67}$$ $$j_\mu ^0=\overline{\nu }\gamma _\mu (1-\gamma _5)\nu -\overline{l}\gamma
_\mu (1-\gamma _5)l+4\sin ^2\theta _wj_\mu ^{em}  \eqnum{68}$$ and $$m_l=\frac 1{\sqrt{2}}f_lv  \eqnum{69}$$ $${\cal L}_H=\frac 12\left( \partial ^\mu H\right) ^2-\frac 12m_H^2H^2+\frac g4%
(W^{+\mu }W_\mu ^{-}+\frac 1{2\cos \theta _w}Z^\mu Z_\mu )\left(
H^2+2vH\right) -\frac{f_l}{\sqrt{2}}\overline{l}lH-\lambda vH^3-\frac \lambda
4H^4  \eqnum{70}$$ $${\cal L}_{gf}=-\frac 1\alpha \partial ^\mu W_\mu ^{+}\partial ^\nu W_\nu
^{-}-\frac 1{2\alpha }(\partial ^\mu Z_\mu )^2-\frac 1{2\alpha }(\partial
^\mu A_\mu )^2  \eqnum{71}$$ and $$\begin{aligned}
{\cal L}_{gh} &=&\overline{C}^{-}\Box C^{+}+\overline{C}^{+}\Box C^{-}+%
\overline{C}_z\Box C_z+\overline{C}_\gamma \Box C_\gamma  \nonumber \\
&&\ -ig\{(\partial ^\mu \overline{C}^{+}C^{-}-\partial ^\mu C^{-}C^{+})(\cos
\theta _wZ_\mu +\sin \theta _wA_\mu )  \nonumber \\
&&\ +(\partial ^\mu \overline{C}^{-}W_\mu ^{+}-\partial ^\mu \overline{C}%
^{+}W_\mu ^{-})\left( \cos \theta _wC_z+\sin \theta _wC_\gamma \right) 
\nonumber \\
&&\ +\left( \cos \theta _w\partial ^\mu \overline{C}_z+\sin \theta
_w\partial ^\mu \overline{C}_\gamma \right) (C^{+}W_\mu ^{-}-C^{-}W_\mu
^{+})\}  \eqnum{72}\end{aligned}$$ The external source terms in Eq.(56) are defined by $$\begin{aligned}
J\cdot \Psi &=&J_\mu ^{-}W^{+\mu }+J_\mu ^{+}W^{-\mu }+J_\mu ^zZ^\mu +J_\mu
^\gamma A^\mu +JH+\overline{\xi }_ll  \nonumber \\
&&\ +\overline{l}\xi _l+\overline{\xi }_\nu \nu +\overline{\nu }\xi _\nu +%
\overline{\eta }^{+}C^{-}+\overline{C}^{+}\eta ^{-}++\overline{\eta }%
^{-}C^{+}+\overline{C}^{-}\eta ^{+}+\overline{\eta }_zC_z  \nonumber \\
&&\ +\overline{C}_z\eta _z+\overline{\eta }_\gamma C_\gamma +\overline{C}%
_\gamma \eta _\gamma  \eqnum{73}\end{aligned}$$

The effective Lagrangian shown above is exactly the same as the one given in the Landau gauge which was obtained in our preceding paper$^{[1]}$. In the paper, the quantization of the electroweak theory without involving Goldstone bosons was performed in the Lagrangian path-integral formalism by the Faddeev-Popov’s approach and/or the Lagrange multiplier method and the quantum theory was given in the general $\alpha $-gauge . For the quantization carried out in the Lagrangian path-integral formalism, it is necessary to consider the SU(2)$\times $U(1) gauge-invariance of the Lagrangian given in the unitary gauge and utilize the SU(2)$\times $U(1) gauge transformations. Nevertheless, in the quantization performed in the Hamiltonian path-integral formalism, as one has seen in this paper , we do not need to consider any gauge transformation and the gauge- invariance of the Lagrangian used. The same result obtained by the both quantizations indicates that the electroweak theory without the Goldstone bosons surely has the original SU(2)$\times $U(1) gauge symmetry. This feature of the theory is natural because the theory can be written out from the ordinary $%
R_\alpha -$ gauge theory$^{[9,10]}$ by making the Higgs transformation to the original Lagrangian and striking off the Goldstone fields from the gauge-fixing terms and the ghost terms in the effective Lagrangian. This procedure was justified in our preceding paper.

**Acknowledgment**
==================

The author wishes to thank professor Shi-Shu Wu for useful discussions. This project was supported in part by National Natural Science Foundation of China and The Research Fund for the Doctoral Program of Higher Education.

**Appendix A: equations of Motion**
===================================

To help understanding the nature of the constraint conditions in Eqs.(10) and (25), we briefly sketch the derivation of equations of motion for the gauge fields. In order to get first order equations, the Lagrangian in Eq.(2) will be recast in the first order form $${\cal L}_g=\frac 12F^{a\mu \nu }F_{\mu \nu }^\alpha -\frac 12F^{a\mu \nu
}(\partial _\mu A_\nu ^a-\partial _\nu A_\mu ^a+g\epsilon ^{abc}A_\mu
^bA_\nu ^c)+\frac 14B^{\mu \nu }B_{\mu \nu }-\frac 12B^{\mu \nu }(\partial
_\mu B_\nu -\partial _\nu B_\mu )  \eqnum{A.1}$$ where the variables $F_{\mu \nu }^\alpha ,A_\mu ^\alpha ,B_{\mu \nu }$ and $%
B_\mu $ are all treated as independent. Then, from the stationary condition of the action $S_\lambda =\int d^4x{\cal L}_\lambda $ where ${\cal L}%
_\lambda $ was defined in Eqs.(1), (A.1), (3) and (4), it is not difficult to derive the following equations $$\partial ^\nu F_{\mu \nu }^a=g\epsilon ^{abc}F_{\mu \nu }^bA^{c\nu }+\frac 12%
gL\gamma _\mu \tau ^aL+\frac 14g^2\phi ^{+}\phi A_\mu ^a+\frac 14gg^{\prime
}\phi ^{+}\tau ^a\phi B_\mu -\partial _\mu \lambda ^a  \eqnum{A.2}$$ $$\partial ^\nu B_{\mu \nu }=-\frac 12g^{\prime }(L\gamma _\mu L+2\overline{l}%
_R\gamma _\mu l_R)+\frac 14g^{\prime 2}\phi ^{+}\phi B_\mu +\frac 14%
gg^{\prime }\phi ^{+}\tau ^a\phi A_\mu ^a-\partial _\mu \lambda ^0 
\eqnum{A.3}$$ $$F_{\mu \nu }^a=\partial _\mu A_\nu ^a-\partial _\nu A_\mu ^a+g\epsilon
^{abc}A_\mu ^bA_\nu ^c  \eqnum{A.4}$$ $$B_{\mu \nu }=\partial _\mu B_\nu -\partial _\nu B_\mu  \eqnum{A.5}$$ and the Lorentz condition written in Eq.(10) as well as the equations for the fermions and the Higgs particle which we do not list here for brevity.

Setting $\mu \left( \nu \right) =0$ and $k=1,2,3$, the above equations will be separately written as follows $$\partial _0A_k^a=\partial _kA_0^a-F_{k0}^a+g\epsilon ^{abc}A_k^bA_0^c 
\eqnum{A.6}$$ $$\partial _0B_k=\partial _kB_0-B_{k0}  \eqnum{A.7}$$ $$\partial _0F_{k0}^a=\partial ^lF_{lk}^a+g\epsilon
^{abc}(F_{k0}^bA_0^c+F_{kl}^bA^{cl})+\frac 12g\overline{L}\gamma _k\tau ^aL+%
\frac 14g^2\phi ^{+}\phi A_k^a+\frac 14gg^{\prime }\phi ^{+}\tau ^a\phi
B_k-\partial _k\lambda ^a  \eqnum{A.8}$$ $$\partial _0B_{k0}=\partial ^lB_{lk}-\frac 12g^{\prime }\left( \overline{L}%
\gamma _kL+2\overline{l}_R\gamma _kl_R\right) +\frac 14gg^{\prime }\phi
^{+}\tau ^a\phi A_k^a+\frac 14g^{\prime 2}\phi ^{+}\phi B_k-\partial
_k\lambda ^0  \eqnum{A.9}$$ $$F_{kl}^a=\partial _kA_l^a-\partial _lA_k^a+g\epsilon ^{abc}A_k^bA_l^c 
\eqnum{A.10}$$ $$B_{kl}=\partial _kB_l-\partial _lB_k  \eqnum{A.11}$$ $$\partial ^kF_{k0}^a=g\epsilon ^{abc}F_{k0}^bA^{ck}-\frac 12g\overline{L}%
\gamma _0\tau ^aL-\frac 14g^2\phi ^{+}\phi A_0^a-\frac 14gg^{\prime }\phi
^{+}\tau ^a\phi B_0+\partial _0\lambda ^a  \eqnum{A.12}$$ $$\partial ^kB_{k0}=\frac 12g^{\prime }\left( \overline{L}\gamma _0L+2%
\overline{l}_R\gamma _0l_R\right) -\frac 14gg^{\prime }\phi ^{+}\tau ^a\phi
A_0^a-\frac 14g^{\prime 2}\phi ^{+}\phi B_0+\partial _0\lambda ^0 
\eqnum{A.13}$$ As we see, in Eqs.(A.6)-(A.9) there are the time-derivatives of the field variables $A_k^a,B_k,F_{k0}^a=E_k^a$ and $B_{k0}=E_k^0$, whereas in Eqs.(A.10)-(A.13), there are no such derivatives. Therefore, Eqs.(A.6)-(A.9) act as the equations of motion, while, Eqs.(A.10)-(A.13) can only be identified with the constraint equations. With the definitions given in Eqs.(12) and (21-(24), we see that Eqs.(A.12) and (A.13) are just combined to give the constraint equation written in Eq.(25).

**Appendix B:The Delta-Functional Representation of Constraint Equations**
==========================================================================

In this appendix, we take an example to prove the relation shown in Eq.(38). Suppose we have two equations $$\varphi _1(x,y)=0  \eqnum{B.1}$$ $$\varphi _2(x,y)=0  \eqnum{B.2}$$ whose solutions are assumed to be ($x_s,y_s)$. Let us evaluate the integral $$I=\int dxdyf(x,y)\delta [\varphi _1(x,y)]\delta [\varphi _2(x,y)] 
\eqnum{B.3}$$ where $f(x,y)$ is an arbitrary integrable function. This integral may be easily calculated by making the change of the integration variables $$\varphi _1(x,y)=u_1\text{ , }\varphi _2(x,y)=u_2  \eqnum{B.4}$$ Correspondingly, the integration measure will be changed to $$dxdy=\det \left( \frac{\partial (\varphi _1,\varphi _2)}{\partial (x,y)}%
\right) ^{-1}du_1du_2  \eqnum{B.5}$$ Substituting Eqs.(B.4) and (B.5) in Eq.(B.3), we have $$\begin{aligned}
I &=&\int du_1du_2\det \left( \frac{\partial (\varphi _1,\varphi _2)}{%
\partial (x,y)}\right) ^{-1}\delta \left( u_1\right) \delta \left(
u_2\right) f(x(u_1,u_2),y(u_1,u_2))  \nonumber \\
&=&\det \left( \frac{\partial (\varphi _1,\varphi _2)}{\partial (x,y)}%
\right) ^{-1}f(x(u_1,u_2),y(u_1,u_2))_{|u_1=u_2=0}  \eqnum{B.6}\end{aligned}$$ Noticing $$x\left( u_1,u_2\right) _{|u_1=u_2=0}=x_s\text{ , }y\left( u_1,u_2\right)
_{\mid u_1=u_2=0}=y_s  \eqnum{B.7}$$ we can write $$I=\det \left( \frac{\partial (\varphi _1,\varphi _2)}{\partial (x,y)}\right)
_{|x=x_s,y=y_s}^{-1}f(x_s,y_s)=\int dxdyf(x,y)\det \left( \frac{\partial
(\varphi _1,\varphi _2)}{\partial (x,y)}\right) ^{-1}\delta \left(
x-x_s\right) \delta (y-y_s)  \eqnum{B.8}$$ In comparison of this expression with that denoted in Eq.(B.3), it is clear to see that $$\delta \left( x-x_s\right) \delta (y-y_s)=\det \left( \frac{\partial
(\varphi _1,\varphi _2)}{\partial (x,y)}\right) \delta (\varphi _1)\delta
(\varphi _2)  \eqnum{B.9}$$ For delta-functionals, certainly, we have the same relation, just as shown in Eq.(38)

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