Resonant generation of electromagnetic modes in nonlinear electrodynamics: quantum perturbative approach

Abstract

The paper studies resonant generation of higher-order harmonics in a closed cavity in Euler-Heisenberg electrodynamics from the point of view of pure quantum field theory. We consider quantum states of the electromagnetic field in a rectangular cavity with conducting boundary conditions and calculate the cross section for the merging of three quanta of cavity modes into a single one ($3 \rightarrow 1$ process) as well as the scattering of two cavity mode quanta ($2 \rightarrow 2$ process). We show that the amplitude of the merging process vanishes for a cavity with an arbitrary aspect ratio and provide an explanation based on plane wave decomposition for cavity modes. Contrary, the scattering amplitude is nonzero for specific cavity aspect ratio. This $2 \rightarrow 2$ scattering is a crucial elementary process for the generation of a quantum of a high-order harmonics with frequency $2\omega _1 - \omega _2$ in an interaction of two coherent states of cavity modes with frequencies $\omega _1$ and $\omega _2$. For this process, we calculate the mean number of quanta in the final state in a model with dissipation, which supports the previous result of resonant higher-order harmonics generation in an effective field theory approach (Kopchinskii and Satunin in Phys Rev A 105:013508, 2022).

Graphical abstract

Appendices

Elements of calculation of the matrix element for the $3 \rightarrow 1$ merging process

In this Appendix, we provide more details for the merging matrix element calculation (12), whose expansion into the components of electromagnetic field strength operators has the following form,

$$\begin{aligned} \langle f|\textsf{S} |i\rangle&= 4i\kappa S \int \limits _{-\infty }^{+\infty }\textrm{d}t \int \limits _0^{L_x}\textrm{d}x\langle a^-_{l,3n} {:}{\textbf {E}} ^4 - 2{\textbf {B}} ^2 {\textbf {E}} ^2 + {\textbf {B}} ^4 \nonumber \\ {}&\quad + 4\beta ({\textbf {B}} {\textbf {E}} )^2{:} (a^+_{i,n})^2 a^+_{j,n}\rangle {0} \nonumber \\&= \langle {\textbf {E}} ^4\rangle - 2\langle {\textbf {B}} ^2{\textbf {E}} ^2\rangle + \langle {\textbf {B}} ^4\rangle + 4\beta \langle ({\textbf {B}} {\textbf {E}} )^2\rangle \nonumber \\&= \left\{ \langle E_y^4\rangle + 2\langle E_y^2E_z^2\rangle + \langle E_z^4\rangle \right\} ~\nonumber \\&\quad -~ 2\left\{ \langle B_y^2 E_y^2\rangle + \langle B_y^2 E_z^2\rangle + \langle B_z^2 E_y^2\rangle + \langle B_z^2 E_z^2\rangle \right\} \nonumber \\&\quad +~ \left\{ \langle B_y^4\rangle + 2\langle B_y^2B_z^2\rangle + \langle B_z^4\rangle \right\} ~\nonumber \\&\quad +~ 4\beta \left\{ \langle B_y^2 E_y^2\rangle + 2\langle B_y E_y B_z E_z\rangle + \langle B_z^2 E_z^2\rangle \right\} . \end{aligned}$$

(34)

Here, the brackets for the field strength operators are defined in the first line of (34). In total, 13 terms appear in Eq. (34). Let us present an evaluation for one of them:

$$\begin{aligned} \langle E_y^4\rangle&= 4i\kappa S \int \limits _{-\infty }^{+\infty }\textrm{d}t \int \limits _0^{L_x}\textrm{d}x {\langle a^-_{l,3n} \,{:} E_y E_y E_y E_y{:}\, a^+_{i,n} a^+_{i,n} a^+_{j,n}\rangle {0}} \\&= 4i\kappa S \int \limits _{-\infty }^{+\infty }\textrm{d}t \int \limits _0^{L_x}\textrm{d}x \,4!\, \left( +i\delta _{ly} \sqrt{\frac{\omega _{3n}}{V}} \sin (k_{3n}x) \, e^{+i\omega _{3n}t}\right) \\&\quad \times ~ \left( -i\delta _{iy} \sqrt{\frac{\omega _n}{V}} \sin (k_nx) \, e^{-i\omega _nt}\right) ^2 \nonumber \\&\quad \left( -i\delta _{jy} \sqrt{\frac{\omega _n}{V}} \sin (k_nx) \, e^{-i\omega _nt}\right) \\&= 4i\kappa S \,4!\, \delta _{iy}\delta _{jy}\delta _{ly} \, i(-i)^3 \, \sqrt{\frac{\omega _{3n}\omega _n^3}{V^4}}\\&\quad \int \limits _{-\infty }^{+\infty } e^{i(\omega _{3n}-3\omega _n)t} \textrm{d}t \int \limits _0^{L_x}\sin (k_{3n}x)\sin ^3(k_nx)\textrm{d}x \\&= 2\pi \delta (0) \, \frac{12\sqrt{3}i\pi ^2 n^2\kappa }{L_x^3 S} \, \delta _{iy}\delta _{jy}\delta _{ly}. \end{aligned}$$

Here, 4! is a combinatorial factor arising due to the identity of four $E_y$ operators. At the last step we use $\omega _n=k_n=\frac{\pi n}{L_x}$ and the similar relation for the 3n subscript.

Details of the matrix element $2 \rightarrow 2$ calculation

Let us present the calculation for the terms of the matrix element (19) in more detail. The first term $\langle ({\textbf {E}} {\textbf {E}} )^2\rangle$ reads,

$$\begin{aligned}&\langle ({\textbf {E}} {\textbf {E}} )^2\rangle : \int \textrm{d}^3\textrm{x} \langle 0|\tfrac{1}{\sqrt{2}}a^{\text {TE}}_{110}a^{\text {TE}}_{110}\,{:}{\textbf { (EE)(EE)}} {:}\,a^{\text {TM}+}_{110}a^{\text {TM}+}_{130}|0\rangle \nonumber \\&\quad =(+i)^2(-i)^2 \frac{\sqrt{\omega _{130}\omega _{011}^2\omega _{110}}}{\sqrt{2}\sqrt{2^4V^4}}\times \nonumber \\&\qquad \times \int \textrm{d}^3\textrm{x} \,8\big [\underbrace{({\varvec{\mathcal {A}}}^\text {TM}_{130}{\varvec{\mathcal {A}}}^\text {TM}_{110}) ({\varvec{\mathcal {A}}}^\text {TE}_{011}{\varvec{\mathcal {A}}}^\text {TE}_{011})}_{\int = -V/2} ~\nonumber \\&\qquad +~ 2 \underbrace{({\varvec{\mathcal {A}}}^\text {TM}_{130}{\varvec{\mathcal {A}}}^\text {TE}_{011}) ({\varvec{\mathcal {A}}}^\text {TM}_{110}{\varvec{\mathcal {A}}}^\text {TE}_{011})}_{\int = 0}\big ] = \frac{-\sqrt{\omega _{130}\omega _{011}^2\omega _{110}}}{\sqrt{2}V}. \end{aligned}$$

(35)

Here, 8 and 2 are combinatorial coefficients, for shortness we indicate the result of integration with underbraces. Thus, the second term in (35) vanishes.

Analogically, the following term $\langle ({\textbf {B}} {\textbf {B}} )^2\rangle$ reads,

$$\begin{aligned}&\langle ({\textbf {B}} {\textbf {B}} )^2\rangle : \int \textrm{d}^3\textrm{x} \langle 0|\tfrac{1}{\sqrt{2}}a^{\text {TE}}_{110}a^{\text {TE}}_{110}\,{:}{\textbf { (BB)(BB)}} {:}\,a^{\text {TM}+}_{110}a^{\text {TM}+}_{130}|0\rangle \nonumber \\&\quad =(+i)^2(-i)^2 \frac{1}{\sqrt{2}\sqrt{2^4V^4}\sqrt{\omega _{130}\omega _{011}^2\omega _{110}}}\nonumber \\&\qquad \times \int \textrm{d}^3\textrm{x} \,8\big [\underbrace{({\textrm{rot}} {\varvec{\mathcal {A}}}^\text {TM}_{130}{\textrm{rot}}{\varvec{\mathcal {A}}}^\text {TM}_{110}) ({\textrm{rot}}{\varvec{\mathcal {A}}}^\text {TE}_{011}{\textrm{rot}}{\varvec{\mathcal {A}}}^\text {TE}_{011})}_{\int = V/2\cdot \pi ^4(L_{x}^{-2} + 3 L_{y}^{-2}) (L_{y}^{-2} - L_{z}^{-2})} ~\nonumber \\&\qquad +~ 2 \underbrace{({\textrm{rot}}{\varvec{\mathcal {A}}}^\text {TM}_{130}{\textrm{rot}}{\varvec{\mathcal {A}}}^\text {TE}_{011}) ({\textrm{rot}} {\varvec{\mathcal {A}}}^\text {TM}_{110}{\textrm{rot}}{\varvec{\mathcal {A}}}^\text {TE}_{011})}_{\int = -V/2 \cdot \pi ^4 \,L_{x}^{-2} L_{z}^{-2}}\big ] \nonumber \\&\quad = \frac{\pi ^4}{\sqrt{2}V}\frac{(L_{x}^{-2} + 3 L_{y}^{-2}) (L_{y}^{-2} - L_{z}^{-2}) - 2 L_{x}^{-2} L_{z}^{-2}}{\sqrt{\omega _{130}\omega _{011}^2\omega _{110}}}\nonumber \\&\quad = \frac{1}{\sqrt{2}V} \frac{\pi ^4}{L_z^4}2r^2\frac{ 2r^2 - 3}{ \sqrt{\omega _{011}^2\omega _{110}\omega _{130}}}. \end{aligned}$$

(36)

The combinatorial coefficients 8 and 2 coincide with the previous case. However, the result of the integration differs. The first of the mixed terms $\langle ({\textbf {B}} {\textbf {B}} ) ({\textbf {E}} {\textbf {E}} )\rangle$ reads,

$$\begin{aligned}&\langle ({\textbf {B}} {\textbf {B}} ) ({\textbf {E}} {\textbf {E}} )\rangle :\int \textrm{d}^3\textrm{x} \langle 0|\tfrac{1}{\sqrt{2}}a^{\text {TE}}_{110}a^{\text {TE}}_{110}\,{:}{\textbf { (BB)(EE)}} {:}\,a^{\text {TM}+}_{110}a^{\text {TM}+}_{130}|0\rangle \nonumber \\&= \frac{1}{\sqrt{2^4V^4}} \, \int \textrm{d}^3\textrm{x}\, 4\,\nonumber \\&\quad \times \bigg [(-i)^2\sqrt{\frac{\omega _{011}^2}{\omega _{130}\omega _{110}}}\underbrace{({\textrm{rot}}{\varvec{\mathcal {A}}}^\text {TM}_{130}{\textrm{rot}}{\varvec{\mathcal {A}}}^\text {TM}_{110})({\varvec{\mathcal {A}}}^\text {TE}_{011}{\varvec{\mathcal {A}}}^\text {TE}_{011})}_{\int = -V/2 \cdot \pi ^2(L_{x}^{-2} + 3L_{y}^{-2})} ~\nonumber \\&\qquad +~ i^2\sqrt{\frac{\omega _{130}\omega _{110}}{\omega _{011}^2}}\underbrace{({\textrm{rot}}{\varvec{\mathcal {A}}}^\text {TE}_{011}{\textrm{rot}}{\varvec{\mathcal {A}}}^\text {TE}_{011}) ({\varvec{\mathcal {A}}}^\text {TM}_{130}{\varvec{\mathcal {A}}}^\text {TM}_{110})}_{\int = V/2 \cdot \pi ^2(L_{y}^{-2} - L_{z}^{-2})} ~\nonumber \\&\quad +~2i(-i)\sqrt{\frac{\omega _{110}}{\omega _{130}}}\underbrace{({\textrm{rot}}{\varvec{\mathcal {A}}}^\text {TM}_{130}{\textrm{rot}}{\varvec{\mathcal {A}}}^\text {TE}_{011}) ({\varvec{\mathcal {A}}}^\text {TM}_{110}{\varvec{\mathcal {A}}}^\text {TE}_{011})}_{\int = 0} ~\nonumber \\&\qquad + ~ 2i(-i)\sqrt{\frac{\omega _{130}}{\omega _{110}}}\underbrace{({\textrm{rot}}{\varvec{\mathcal {A}}}^\text {TM}_{110}{\textrm{rot}}{\varvec{\mathcal {A}}}^\text {TE}_{011}) ({\varvec{\mathcal {A}}}^\text {TM}_{130}{\varvec{\mathcal {A}}}^\text {TE}_{011})}_{\int = 0}\bigg ]\nonumber \\&\quad = \frac{1}{2\sqrt{2}V}\frac{\pi ^2}{L_z^2}\left( 4r^2 \sqrt{\frac{\omega _{011}^2}{\omega _{110}\omega _{130}}} - (r^2-1) \sqrt{\frac{\omega _{110}\omega _{130}}{\omega _{011}^2}} \right) . \end{aligned}$$

(37)

The last term,

$$\begin{aligned}&\langle ({\textbf {B}} {\textbf {E}} )^2 \rangle : \int \textrm{d}^3\textrm{x} \langle 0|\tfrac{1}{\sqrt{2}}a^{\text {TE}}_{110}a^{\text {TE}}_{110}\,{:}{\textbf { (BE)(BE)}} {:}\,a^{\text {TM}+}_{110}a^{\text {TM}+}_{130}|0\rangle \nonumber \\&= \frac{1}{\sqrt{2^4V^4}} \, \int \textrm{d}^3\textrm{x}\, 4\,\nonumber \\&\quad \times \bigg [(-i)^2\sqrt{\frac{\omega _{011}^2}{\omega _{130}\omega _{110}}}\underbrace{({\textrm{rot}}{\varvec{\mathcal {A}}}^\text {TM}_{130}{\varvec{\mathcal {A}}}^\text {TE}_{011})({\textrm{rot}}{\varvec{\mathcal {A}}}^\text {TM}_{110}{\varvec{\mathcal {A}}}^\text {TE}_{011})}_{\int = -3V/2 \cdot \pi ^2 L_{y}^{-2}}\nonumber \\&\qquad +~ i^2\sqrt{\frac{\omega _{130}\omega _{110}}{\omega _{011}^2}}\underbrace{({\textrm{rot}}{\varvec{\mathcal {A}}}^\text {TE}_{011}{\varvec{\mathcal {A}}}^\text {TM}_{130}) ({\textrm{rot}}{\varvec{\mathcal {A}}}^\text {TE}_{011}{\varvec{\mathcal {A}}}^\text {TM}_{110})}_{\int = V/2 \cdot \pi ^2 L_y^{-2}} \nonumber \\&\qquad +~i(-i)\sqrt{\frac{\omega _{110}}{\omega _{130}}}\underbrace{({\textrm{rot}}{\varvec{\mathcal {A}}}^\text {TM}_{130}{\varvec{\mathcal {A}}}^\text {TE}_{011}) ({\textrm{rot}}{\varvec{\mathcal {A}}}^\text {TM}_{110}{\varvec{\mathcal {A}}}^\text {TE}_{011})}_{\int = 0} ~\nonumber \\&\qquad + ~ i(-i)\sqrt{\frac{\omega _{130}}{\omega _{110}}}\underbrace{({\textrm{rot}}{\varvec{\mathcal {A}}}^\text {TM}_{110}{\varvec{\mathcal {A}}}^\text {TM}_{130}) ({\textrm{rot}}{\varvec{\mathcal {A}}}^\text {TE}_{011}{\varvec{\mathcal {A}}}^\text {TE}_{011})}_{\int = 0}\nonumber \\&\qquad +~i(-i)\sqrt{\frac{\omega _{110}}{\omega _{130}}}\underbrace{({\textrm{rot}}{\varvec{\mathcal {A}}}^\text {TM}_{130}{\varvec{\mathcal {A}}}^\text {TE}_{011}) ({\textrm{rot}} {\varvec{\mathcal {A}}}^\text {TE}_{011}{\varvec{\mathcal {A}}}^\text {TM}_{110})}_{\int = 3V/2\cdot \pi ^2 L_y^{-2}} ~\nonumber \\&\qquad + ~ i(-i)\sqrt{\frac{\omega _{130}}{\omega _{110}}}\underbrace{({\textrm{rot}}{\varvec{\mathcal {A}}}^\text {TM}_{110}{\varvec{\mathcal {A}}}^\text {TE}_{011}) ({\textrm{rot}}{\varvec{\mathcal {A}}}^\text {TE}_{011}{\varvec{\mathcal {A}}}^\text {TM}_{130})}_{\int = 0}\bigg ] \nonumber \\&\quad =\frac{1}{2\sqrt{2}V}\frac{\pi ^2}{L_z^2} r^2\left( 3\sqrt{\frac{\omega _{011}^2}{\omega _{110}\omega _{130}}} +3 \sqrt{\frac{\omega _{110}}{\omega _{130}}} - \sqrt{\frac{\omega _{130}}{\omega _{110}}}\right. \nonumber \\&\qquad \left. - \sqrt{\frac{\omega _{110}\omega _{130}}{\omega _{011}^2}} \right) . \end{aligned}$$

(38)

Finally, restoring the coefficient $4\kappa$, the result (20)–(22) is obtained.