---
abstract: Correct quantization of free electromagnetic field is proposed
author:
- 'D.Yearchuck'
- 'Y.Yerchak'
- 'A.Alexandrov'
title: To Quantization of Free Electromagnetic Field
---

In 1873 “A Treatise on Electricity and Magnetism” by Maxwell [@Maxwell] was published, in which the discovery of the system of electrodynamics equations was reported. The equations are in fact the symmetry expressions for experimental laws, established by Faraday, and, consequently, they are mathematical mapping of experimentally founded symmetry of EM-field. It means in its turn that if some new experimental data will indicate, that symmetry of EM-field is higher, then Maxwell equations have to be generalized. That is the reason why the symmetry study of Maxwell equations is the subject of many research in field theory up to now. Heaviside [@Heaviside] in twenty years after Maxwell discovery was the first, who payed attention to the symmetry between electrical and magnnetic quantities in Maxwell equations. Mathematical formulation of given symmetry, consisting in invariance of Maxwell equations for free EM-field under the duality transformations $$\label{eq1d}
\vec {E} \rightarrow \pm\vec {H}, \vec {H} \rightarrow \mp\vec {E},$$ gave Larmor [@Larmor]. Duality transformations (\[eq1d\]) are private case of the more general dual transformations, established by Rainich [@Rainich]. Dual transformations produce oneparametric abelian group $U_1$ of chiral transformations and they are $$\label{eq2d}
\begin{split}
\raisetag{40pt}                                                    
\vec {E} \rightarrow \vec {E} cos\theta + \vec {H} sin\theta\\
\vec {H} \rightarrow \vec {H} cos\theta - \vec {E} sin\theta.
\end{split}$$ Given symmetry indicates, that both constituents $\vec {E}$ and $\vec {H}$ of EM-field are possessing equal rights, in particular they both have to consist of component with different parity. Subsequent extension of dual symmetry for the EM-field with sources leads to requirement of two type of charges. Examples of the dual symmetry display are for instance the equality of magnetic and electric energy values in LC-tank or in free electromagnetic wave. Recently concrete experimental results have been obtained concerning dual symmetry of EM-field in the matter. Two new physical phenomena - ferroelectric [@Yearchuck_Yerchak] and antiferroelectric [@Yearchuck_PL] spin wave resonances have been observed. They were predicted on the base of the model [@Yearchuck_Doklady] for the chain of electrical “spin” moments, that is intrinsic electrical moments of (quasi)particles. Especially interesting, that in [@Yearchuck_PL] was experimentally proved, that really purely imaginary electrical “spin” moment, in full correspondence with Dirac prediction [@Dirac], is responsible for the phenomenon observed. Earlier on the same samples has been registered ferromagnetic spin wave resonance, [@Ertchak_J_Physics_Condensed_Matter].

The values of splitting parameters $\mathfrak{A}^E$ and $\mathfrak{A}^H$ in ferroelectric and ferromagnetic spin wave resonance spectra allowed to find the ratio $J_{E }/J_{H}$ of exchange constants in the range of $(1.2 - 1.6)10^{4}$. Given result seems to be direct proof, that the charge, that is function, which is invariant under gauge transformations is two component function. The ratio of imagine $e_{H} \equiv g$ to real $e_{E}\equiv e $ components of complex charge is $\frac{g}{e} \sim \sqrt{J_{E }/J_{H}} \approx (1.1 - 1.3)10^{2}$. At the same time in classical and in quantum theory dual symmetry of Maxwell eqations does not take into consideration. Moreover the known solutions of Maxwell eqations do not reveal given symmetry even for free EM-field, see for instance [@Scully], although it is understandable, that the general solutions have to posseess by the same symmetry.

The aim of given work is to find the cause of symmetry difference of Maxwell eqations and their solutions and to propose correct field functions for classical and quantized EM-field. Suppose EM-field in volume rectangular cavity. Suppose also, that the field polarization is linear in z-direction. Then the vector of electrical component can be represented in the form $$E_x(z,t) = \sum_{\alpha=1}^{\infty}A_{\alpha}q_{\alpha}(t)\sin(k_{\alpha}z),$$ where $q_{\alpha}(t)$ is amplitude of $\alpha$-th normal mode of the cavity, $\alpha \in N$, $k_{\alpha} = \alpha\pi/L$, $A_{\alpha}=\sqrt{2 \nu_{\alpha}^2m_{\alpha}/(V\epsilon_0)}$, $\nu_{\alpha} = \alpha\pi c/L$, $L$ is cavity length along z-axis, $V$ is cavity volume, $m_{\alpha}$ is parameter, which is introduced to obtain the analogy with mechanical harmonic oscillator. Using the equation $$\epsilon_0\partial_t \vec{E}(z,t) = \left[ \nabla\times\vec{H}(z,t)\right]$$ we obtain in suggestion of transversal EM-field the expression for magnetic field $${H}_y(z,t) =  \sum_{\alpha=1}^{\infty}\epsilon_0\frac{A_{\alpha}}{k_{\alpha}}\frac{dq_{\alpha}}{dt}\cos(k_{\alpha}z) + H_{y0}(t),$$ where $H_{y0} = \sum_{\alpha=1}^{\infty} f_{\alpha}(t)$, $\{f_{\alpha}(t)\}$, $\alpha \in N$, is the set of arbitrary functions of the time. The partial solution is usually used, in which the function $H_{y0}(t)$ is identically zero. The field Hamiltonian $\mathcal{H}^{[1]}(t)$, corresponding given partial solution, is $$\begin{split}
&\mathcal{H}^{[1]}(t) = \frac{1}{2}\iiint\limits_{(V)}\left[\epsilon_0E_x^2(z,t)+\mu_0H_y^2(z,t)\right]dxdydz\\
&= \frac{1}{2}\sum_{\alpha=1}^{\infty}\left[m_{\alpha}\nu_{\alpha}^2q_{\alpha}^2(t) + \frac{p_{\alpha}^2(t)}{m_{\alpha}} \right],
\end{split}$$ where $$p_{\alpha} = m_{\alpha} \frac{dq_{\alpha}(t)}{dt}.$$ Then, using the equation $$\left[ \nabla\times\vec{E}\right] = -\frac{\partial \vec{B}}{\partial t} = -\mu_0 \frac{\partial \vec{H}}{\partial t}$$ it is easily to find the field functions $\{q_{\alpha}(t)\}$. They will satisfy to differential equation $$\frac{d^2q_{\alpha}(t)}{dt^2}+\frac{k_{\alpha}^2}{\mu_0\epsilon_0}q_{\alpha}(t)=0.$$ Consequently, taking into account $\mu_0\epsilon_0 = 1/c^2$, we have $$q_{\alpha}(t) = C_1e^{i\nu_{\alpha}t}+C_2e^{-i\nu_{\alpha}t}$$

Thus, real free Maxwell field equations result in well known in the theory of differential equations situation - the solutions are complex-valued functions. It means, that generally the field function for free Maxwell field produce complex space.

From general expression for the field $\vec{H}(\vec{r},t)$ $$\vec{H}(\vec{r},t) =  \left[\sum_{\alpha=1}^{\infty}A_{\alpha}\frac{\epsilon_0}{k_{\alpha}}\frac{dq_{\alpha}(t)}{dt}\cos(k_{\alpha}z) + f_{\alpha}(t)\right]\vec{e}_y$$ it is easily to obtain differential equation for $f_{\alpha}(t)$ $$\begin{split}
&\frac{d f_{\alpha}(t)}{dt} + A_{\alpha}\frac{\epsilon_0}{k_{\alpha}}\frac{\partial^2q_{\alpha}(t)}{\partial t^2}\cos(k_{\alpha}z) \\
&- \frac {1}{\mu_0} A_{\alpha}k_{\alpha}q_{\alpha}(t)cos(k_{\alpha}z) = 0.
\end{split}$$ Its solution in general case is $$f_{\alpha}(t) = \int A_{\alpha} \cos(k_{\alpha}z)\left[q_{\alpha}(t)\frac{k_{\alpha}}{\mu_0}-\frac{d^2q_{\alpha}(t)}{dt^2}\frac{\epsilon_0}{k_{\alpha}}\right]
dt +C_{\alpha}$$ Then we have another solution of Maxwell equations $$\vec{H}(\vec{r},t) = \frac{1}{\mu_0}\left\{\sum_{\alpha=1}^{\infty}k_{\alpha}A_{\alpha} \cos(k_{\alpha}z) q_{\alpha}'(t)\right\}\vec{e}_y,$$ $$\vec{E}(\vec{r},t) = \left\{\sum_{\alpha=1}^{\infty}A_{\alpha}\frac{dq_{\alpha}'(t)}{dt}\sin(k_{\alpha}z)\right\}\vec{e}_x,$$ where $q_{\alpha}'(t) = \int q_{\alpha}(t) dt + C_{\alpha}'$ Then Hamiltonian $\mathcal{H}^{[2]}(t)$ is $$\mathcal{H}^{[2]}(t) = \frac{1}{2}\sum_{\alpha=1}^{\infty}\left[m_{\alpha} \nu_{\alpha}^4 q{'}_{\alpha}^2(t) + {m_{\alpha}\nu_{\alpha}^2 (\frac{dq_{\alpha}'(t)}{dt})^2} \right].$$ Let us introduce new variables $$\begin{split}
&q{''}_{\alpha}(t) = \nu_{\alpha}q{'}_{\alpha}(t) \\
&p{''}_{\alpha}(t) = m_{\alpha}\nu_{\alpha}\frac{dq_{\alpha}'(t)}{dt}
\end{split}$$ Then $$\mathcal{H}^{[2]}(t) = 
\frac{1}{2}\sum_{\alpha=1}^{\infty}\left[m_{\alpha}\nu_{\alpha}^2q{''}_{\alpha}^2(t) + \frac{p{''}_{\alpha}^2(t)}{m_{\alpha}} \right].$$ We use further the standard procedure of field quantization. So for the first partial solution we have $$\begin{split}
&\left[\hat {p}_{\alpha}(t) , \hat {q}_{\beta}(t)\right] = i\hbar\delta_{{\alpha}\beta}\\
&\left[\hat {q}_{\alpha}(t) , \hat {q}_{\beta}(t)\right] = \left[\hat {p}_{\alpha}(t) , \hat {p}_{\beta}(t)\right] = 0,
\end{split}$$ where $\alpha, \beta \in N$. Introducing the operators $\hat{a}_{\alpha}(t)$ and $ \hat{a}^{+}_{\alpha}(t)$ $$\begin{split}
&\hat{a}_{\alpha}(t) = \frac{1}{ \sqrt{ \frac{1}{2} \hbar  m_{\alpha} \nu_{\alpha}}} \left[ m_{\alpha} \nu_{\alpha}\hat {q}_{\alpha}(t) + i \hat {p}_{\alpha}(t)\right]\\
&\hat{a}^{+}_{\alpha}(t) = \frac{1}{ \sqrt{ \frac{1}{2} \hbar  m_{\alpha} \nu_{\alpha}}} \left[ m_{\alpha} \nu_{\alpha}\hat {q}_{\alpha}(t) - i \hat {p}_{\alpha}(t)\right],
\end{split}$$ we have for the operators of canonical variables $$\begin{split}
&\hat {q}_{\alpha}(t) = \sqrt{\frac{\hbar}{2 m_{\alpha} \nu_{\alpha}}} \left[\hat{a}^{+}_{\alpha}(t) + \hat{a}_{\alpha}(t)\right]\\
&\hat {p}_{\alpha}(t) = i \sqrt{\frac{\hbar m_{\alpha} \nu_{\alpha}}{2}} \left[\hat{a}^{+}_{\alpha}(t) - \hat{a}_{\alpha}(t)\right]. 
\end{split}$$ Then field function operators are $$\hat{\vec{E}}(\vec{r},t) = \{\sum_{\alpha=1}^{\infty} \sqrt{\frac{\hbar \nu_{\alpha}}{V\epsilon_0}} \left[\hat{a}^{+}_{\alpha}(t) + \hat{a}_{\alpha}(t)\right] sin(k_{\alpha} z)\} \vec{e}_x,$$

$$\hat{\vec{H}}(\vec{r},t) = ic\epsilon_0 \{\sum_{\alpha=1}^{\infty} \sqrt{\frac{\hbar \nu_{\alpha}}{V\epsilon_0}} \left[\hat{a}^{+}_{\alpha}(t) - \hat{a}_{\alpha}(t)\right] cos(k_{\alpha} z)\} \vec{e}_y,$$

For the second partial solution, corresponding to Hamiltonian $\mathcal{H}^{[2]}(t)$ we have $$\begin{split}
&\left[\hat{p}{''}_{\alpha}(t) , \hat {q}{''}_{\beta}(t)\right] = i\hbar\delta_{{\alpha}\beta}\\
&\left[\hat {q}{''}_{\alpha}(t) , \hat {q}{''}_{\beta}(t)\right] = \left[\hat {p}{''}_{\alpha}(t) , \hat {p}{''}_{\beta}(t)\right] = 0,
\end{split}$$ $\alpha, \beta \in N$. The operators $\hat{a}{''}_{\alpha}(t)$, $\hat{a}{''}^{+}_{\alpha}(t)$ are introduced analogously $$\begin{split}
&\hat{a}{''}_{\alpha}(t) = \frac{1}{ \sqrt{ \frac{1}{2} \hbar  m_{\alpha} \nu_{\alpha}}} \left[ m_{\alpha} \nu_{\alpha}\hat {q}{''}_{\alpha}(t) + i \hat {p}{''}_{\alpha}(t)\right]\\
&\hat{a}{''}^{+}_{\alpha}(t) = \frac{1}{ \sqrt{ \frac{1}{2} \hbar  m_{\alpha} \nu_{\alpha}}} \left[ m_{\alpha} \nu_{\alpha}\hat {q}{''}_{\alpha}(t) - i \hat {p}{''}_{\alpha}(t)\right]
\end{split}$$ Relationships for canonical variables are $$\begin{split}
&\hat {q}{''}_{\alpha}(t) = \sqrt{\frac{\hbar}{2 m_{\alpha} \nu_{\alpha}}} \left[\hat{a}{''}^{+}_{\alpha}(t) + \hat{a}{''}_{\alpha}(t)\right]\\
&\hat {p}{''}_{\alpha}(t) = i \sqrt{\frac{\hbar m_{\alpha} \nu_{\alpha}}{2}} \left[\hat{a}{''}^{+}_{\alpha}(t) - \hat{a}{''}_{\alpha}(t)\right] 
\end{split}$$ For the field function operators we obtain $$\hat{\vec{E}}^{[2]}(\vec{r},t) = i \{\sum_{\alpha=1}^{\infty} \sqrt{\frac{\hbar \nu_{\alpha}}{V\epsilon_0}} \left[\hat{a}{''}^{+}_{\alpha}(t) - \hat{a}{''}_{\alpha}(t)\right] sin(k_{\alpha} z)\} \vec{e}_x,$$ $$\begin{split}
\hat{\vec{H}}^{[2]}(\vec{r},t) = \frac{1}{\mu_0 c}  \{\sum_{\alpha=1}^{\infty} \sqrt{\frac{\hbar \nu_{\alpha}}{V\epsilon_0}} \left[\hat{a}{''}^{+}_{\alpha}(t) + \hat{a}{''}_{\alpha}(t)\right] cos(k_{\alpha} z)\} \vec{e}_y,
\end{split}$$ Let us designate $\sqrt{\frac{\hbar \nu_{\alpha}}{V\epsilon_0}} = E_0$. In accordance with definition of complex quantities we have $$(\vec{E}(\vec{r},t), \vec{E}^{[2]}(\vec{r},t)) \rightarrow \vec{E}(\vec{r},t) +  i \vec{E}^{[2]}(\vec{r},t) = \vec{E}(\vec{r},t).$$ Consequently, correct field operators for quantized EM-field are $$\begin{split}
\hat{\vec{E}}(\vec{r},t) =  \{\sum_{\alpha=1}^{\infty} E_0 \{\left[\hat{a}^{+}_{\alpha}(t) + \hat{a}_{\alpha}(t)\right]\\ + \left[\hat{a}{''}_{\alpha}(t) - \hat{a}{''}^{+}_{\alpha}(t)\right]\} sin(k_{\alpha} z)\} \vec{e}_x,
\end{split}$$ and $$(\vec{H}^{[2]}(\vec{r},t), \vec{H}(\vec{r},t)) \rightarrow \vec{H}^{[2]}(\vec{r},t) +  i \vec(\vec{r},t) = \vec{H}(\vec{r},t)$$ $$\begin{split}
\hat{\vec{H}}(\vec{r},t) =  \{\sum_{\alpha=1}^{\infty} E_0 \{\frac{1}{\mu_0 c} \left[\hat{a}{''}_{\alpha}(t) + \hat{a}{''}^{+}_{\alpha}(t)\right]\\ + c \epsilon_0 \left[\hat{a}_{\alpha}(t) - \hat{a}^{+}_{\alpha}(t)\right]\} cos(k_{\alpha} z) + C_{\alpha}\hat{e}\} \vec{e}_y,
\end{split}$$

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