**7. Compton effect initiated by two orthogonal plane waves**

The project with the two laser waves is the next goal and the future direction of the laser physics of elementary particles. The two laser beams can replace many laser beams in the thermonuclear reactor such as ITER in Cadarache near Aix-en-Provence in France. At the same time, the two-laser system can be considered in chemistry as a catalyzer which was not known before the laser physics.

We solve the Dirac equation for two different four-potentials of the plane electromagnetic waves. We specify the solutions of the Dirac equation for two orthogonal plane waves.

The modified Compton formula for the scattering of two photons on an electron is determined. The solution of the Dirac equation for two waves was found by Sen Gupta [20, 21] in the form of the Fourier series, however without immediate application. The solution of the Dirac equation for two waves with the perpendicular polarization was given by Lyulka [22-25] who described the decay of particles in two laser fields. The derivation of two-wave solution is not presented in his articles. So, we investigated the situation and presented new results.

The total four-potential *V<sup>µ</sup>* is a superposition of the potentials *A<sup>µ</sup>* and *B<sup>µ</sup>* as follows:

$$V\_{\mu} = A\_{\mu}(\varphi) + B\_{\mu}(\chi), \tag{143}$$

where *ϕ* = *kx* and *χ* = *κx* and *k* = *κ*.

The Lorentz condition gives

$$
\partial\_{\mu}V^{\mu} = 0 = k\_{\mu}\frac{\partial A^{\mu}}{\partial \varphi} + \kappa\_{\mu}\frac{\partial B^{\mu}}{\partial \chi} = k\_{\mu}A^{\mu}\_{\varphi} + \kappa\_{\mu}B^{\mu}\_{\chi\prime} \tag{144}
$$

where the subscripts *ϕ*, *χ* denote partial derivatives. Equation (144) takes a more simple form if we notice that partial differentiation with respect to *ϕ* concerns only *A<sup>µ</sup>* and partial differentiation with respect to *χ* concerns only *Bµ*. So we write instead eq. (144).

$$
\partial\_{\mu}V^{\mu} = 0 = k\_{\mu}(A^{\mu})^{\prime} + \kappa\_{\mu}(B^{\mu})^{\prime} = kA^{\prime} + \kappa B^{\prime}.\tag{145}
$$

Without loss of generality, we can write instead of eq. (145) the following one:

$$(k\_{\mu}(A^{\mu})' = 0; \quad \kappa\_{\mu}(B^{\mu})' = 0,\tag{146}$$

or,

focus is not considered here. The nonlinear Compton process was experimentally confirmed,

The present text is a continuation of the author discussion on laser acceleration [16, 17, 18],

The *δ*-form laser pulses are here considered as an idealization of the experimental situation in laser physics. Nevertheless, it was demonstrated theoretically that at the present time, the zeptosecond and sub-zeptosecond laser pulses of duration 10−<sup>21</sup> <sup>−</sup> <sup>10</sup>−<sup>22</sup> s can be realized

The project with the two laser waves is the next goal and the future direction of the laser physics of elementary particles. The two laser beams can replace many laser beams in the thermonuclear reactor such as ITER in Cadarache near Aix-en-Provence in France. At the same time, the two-laser system can be considered in chemistry as a catalyzer which was not

We solve the Dirac equation for two different four-potentials of the plane electromagnetic waves. We specify the solutions of the Dirac equation for two orthogonal plane waves.

The modified Compton formula for the scattering of two photons on an electron is determined. The solution of the Dirac equation for two waves was found by Sen Gupta [20, 21] in the form of the Fourier series, however without immediate application. The solution of the Dirac equation for two waves with the perpendicular polarization was given by Lyulka [22-25] who described the decay of particles in two laser fields. The derivation of two-wave solution is not presented in his articles. So, we investigated the situation and

The total four-potential *V<sup>µ</sup>* is a superposition of the potentials *A<sup>µ</sup>* and *B<sup>µ</sup>* as follows:

*∂A<sup>µ</sup> ∂ϕ* <sup>+</sup> *κµ*

differentiation with respect to *χ* concerns only *Bµ*. So we write instead eq. (144).

*∂B<sup>µ</sup>*

where the subscripts *ϕ*, *χ* denote partial derivatives. Equation (144) takes a more simple form if we notice that partial differentiation with respect to *ϕ* concerns only *A<sup>µ</sup>* and partial

*∂µV<sup>µ</sup>* = 0 = *kµ*(*Aµ*) + *κµ*(*Bµ*) = *kA* + *κB*

*∂χ* <sup>=</sup> *<sup>k</sup>µA<sup>µ</sup>*

*V<sup>µ</sup>* = *Aµ*(*ϕ*) + *Bµ*(*χ*), (143)

*<sup>ϕ</sup>* <sup>+</sup> *κµB<sup>µ</sup>*

*<sup>χ</sup>*, (144)

. (145)

where the Compton model of laser acceleration was proposed.

**7. Compton effect initiated by two orthogonal plane waves**

for instance, by Bulla et al. [15].

136 Graphene - New Trends and Developments

by the petawatt lasers [19].

known before the laser physics.

presented new results.

where *ϕ* = *kx* and *χ* = *κx* and *k* = *κ*.

*∂µV<sup>µ</sup>* = 0 = *k<sup>µ</sup>*

The Lorentz condition gives

$$kA = \text{const} = 0; \quad \text{\(\pi B = const = 0\)},\tag{147}$$

putting integrating constant to zero.

The electromagnetic tensor *Fµν* is expressed in the new variables as in eq. (5)

$$F\_{\mu\nu} = k\_{\mu}A\_{\nu}^{\prime} - k\_{\nu}A\_{\mu}^{\prime} + \kappa\_{\mu}B\_{\nu}^{\prime} - \kappa\_{\nu}B\_{\mu}^{\prime}.\tag{148}$$

Now, we write Dirac equation for the two potentials in the form

$$[-\partial^2 - 2ie(V\partial) + e^2V^2 - m^2 - \frac{i}{2}eF\_{\mu\nu}\sigma^{\mu\nu}]\psi = 0,\tag{149}$$

where *V* = *A* + *B*, *Fµν* is given by eq. (148) and the *σ*-term is defined as follows:

$$\frac{i}{2}eF\_{\mu\nu}\sigma^{\mu\nu} = ie(\gamma k)(\gamma A') + ie(\gamma \kappa)(\gamma B') \tag{150}$$

We assume the solution of eq. (149) in the Volkov form, or

$$
\psi = e^{-ip\chi} F(\varphi\_\prime \chi). \tag{151}
$$

After performing all operations in eq. (149), we get the partial differential equation for function *F*(*ϕ*, *χ*):

$$-2k\kappa \mathcal{F}\_{\varphi\chi} + (2ipk - 2ik\mathcal{B})\mathcal{F}\_{\Psi} + (2ip\kappa - 2ieA\kappa)\mathcal{F}\_{\chi} \quad +$$

$$(e^2(A+B)^2 - 2e(A+B)p - ie(\gamma k)(\gamma A\_{\Psi}) - ie(\gamma \kappa)(\gamma B\_{\chi}))F = 0. \tag{152}$$

Equation (152) was simplified by the author [20], putting *kκ* = 0. However, to get the Compton effect-initiated two orthogonal waves, we ignore this simplification and write eq. (152) in the following form:

$$aF\_{\varphi} + bF\_{\chi} + cF = 2k\mathfrak{x}F\_{\varphi\chi} \tag{153}$$

where

$$a = 2ipk - 2iekB; \quad b = 2ip\kappa - 2ie\kappa A \tag{154}$$

and

$$\mathcal{L} = e^2 (A+B)^2 - 2e(A+B)p - ie(\gamma k)(\gamma A') - ie(\gamma \kappa)(\gamma B') \tag{155}$$

and the term of the two partial derivations is not present because of *kκ* = 0. For the field which we specify by the conditions

$$kB = 0; \quad \kappa A = 0; \quad AB = 0,\tag{156}$$

we have

$$\begin{split} 2ip\mathbf{k}\mathbf{F}\_{\boldsymbol{\theta}} + 2ip\mathbf{x}\mathbf{F}\_{\boldsymbol{\lambda}} + (e^2A^2 + e^2B^2 - 2epA - 2epB - ie(\gamma k)(\gamma A') \\ &\quad \mathrm{ie}(\gamma \boldsymbol{\kappa})(\gamma B'))\mathbf{F} = 2\mathbf{k}\mathbf{x}\mathbf{F}\_{\boldsymbol{\theta}\boldsymbol{\chi}}. \end{split} \tag{157}$$

Now, let us put

$$F(\varphi, \chi) = X(\varphi)Y(\chi). \tag{158}$$

After insertion of eq. (158) into eq. (157) and division of the new equation by *XY*, we get the terms depending only on *ϕ* and on *χ*. Or we get

$$\begin{aligned} \left(2i(pk+ik\mathbf{x})\frac{X'}{X} + e^2A^2 - 2epA - ie(\gamma k)(\gamma A')\right) &+ \\ \left(2i(p\kappa+ik\mathbf{x})\frac{Y'}{Y} + e^2B^2 - 2epB - ie(\gamma \kappa)(\gamma B')\right) &= 0 \end{aligned} \tag{159}$$

So, there are terms dependent on *ϕ* and terms dependent on *χ* only in eq. (159). The only possibility is that they are equal to some constant *λ* and −*λ*. Then,

$$2i(pk+i\mathbf{k}\mathbf{x})X' + (e^2A^2 - 2epA - ie(\gamma k)(\gamma A'))X = \lambda X \tag{160}$$

and

*aF<sup>ϕ</sup>* + *bF<sup>χ</sup>* + *cF* = 2*kκFϕχ*, (153)

*a* = 2*ipk* − 2*iekB*; *b* = 2*ipκ* − 2*ieκA* (154)

) − *ie*(*γκ*)(*γB*

*kB* = 0; *κA* = 0; *AB* = 0, (156)

<sup>2</sup>*B*<sup>2</sup> <sup>−</sup> <sup>2</sup>*epA* <sup>−</sup> <sup>2</sup>*epB* <sup>−</sup> *ie*(*γk*)(*γ<sup>A</sup>*

*ie*(*γκ*)(*γB*

*F*(*ϕ*, *χ*) = *X*(*ϕ*)*Y*(*χ*). (158)

) 

> )

+

) (155)

) −

))*F* = 2*kκFϕχ*. (157)

= 0 (159)

where

and

we have

Now, let us put

*c* = *e*

138 Graphene - New Trends and Developments

For the field which we specify by the conditions

2*ipkF<sup>ϕ</sup>* + 2*ipκF<sup>χ</sup>* + (*e*

terms depending only on *ϕ* and on *χ*. Or we get

2*i*(*pk* + *ikκ*)

2*i*(*pκ* + *ikκ*)

*X <sup>X</sup>* <sup>+</sup> *<sup>e</sup>*

possibility is that they are equal to some constant *λ* and −*λ*. Then,

*Y <sup>Y</sup>* <sup>+</sup> *<sup>e</sup>*

<sup>2</sup>(*<sup>A</sup>* <sup>+</sup> *<sup>B</sup>*)<sup>2</sup> <sup>−</sup> <sup>2</sup>*e*(*<sup>A</sup>* <sup>+</sup> *<sup>B</sup>*)*<sup>p</sup>* <sup>−</sup> *ie*(*γk*)(*γ<sup>A</sup>*

and the term of the two partial derivations is not present because of *kκ* = 0.

<sup>2</sup>*A*<sup>2</sup> + *e*

After insertion of eq. (158) into eq. (157) and division of the new equation by *XY*, we get the

So, there are terms dependent on *ϕ* and terms dependent on *χ* only in eq. (159). The only

<sup>2</sup>*A*<sup>2</sup> <sup>−</sup> <sup>2</sup>*epA* <sup>−</sup> *ie*(*γk*)(*γ<sup>A</sup>*

<sup>2</sup>*B*<sup>2</sup> <sup>−</sup> <sup>2</sup>*epB* <sup>−</sup> *ie*(*γκ*)(*γ<sup>B</sup>*

$$2\mathbf{i}(p\mathbf{x} + i\mathbf{k}\mathbf{x})Y' + (e^2\mathbf{B}^2 - 2ep\mathbf{B} - ie(\gamma\kappa)(\gamma\mathcal{B}'))Y = -\lambda Y \tag{161}$$

We put *λ* = 0 without loss of generality. In such a way, the solution of eq. (159) is the solution of two equations only. Because the form of every equation is similar to the form of eq. (14), we can write the solution of these equations as follows:

$$X = \left[1 + \frac{e}{2(kp + ik\kappa)}(\gamma k)(\gamma A)\right] \frac{\mu}{\sqrt{2p\_0}} e^{iS\_1},\tag{162}$$

with

$$S\_1 = \int\_0^{kx} \frac{e}{(kp + ik\kappa)} \left[ (pA) - \frac{e}{2}(A)^2 \right] d\varphi. \tag{163}$$

and

$$Y = \left[1 + \frac{e}{2(\kappa p + i\mathbf{k}\kappa)}(\gamma\kappa)(\gamma B)\right] \frac{\mu}{\sqrt{2p\_0}} e^{i\mathbf{S}\_2},\tag{164}$$

with

$$S\_2 = -\int\_0^{\kappa \chi} \frac{e}{(\kappa p + ik\kappa)} \left[ (pB) - \frac{e}{2} (B)^2 \right] d\chi. \tag{165}$$

The total solution is then of the form

$$\psi\_p = \left[1 + \frac{e}{2(kp + i\mathbf{k}\cdot\mathbf{r})}(\gamma k)(\gamma A)\right] \left[1 + \frac{e}{2(\kappa p + i\mathbf{k}\cdot\mathbf{r})}(\gamma \kappa)(\gamma B)\right] \frac{u}{\sqrt{2p\_0}} e^{i(\mathcal{S}\_1(A) + \mathcal{S}\_2(B))}.\tag{166}$$

In the case of the two non-collinear laser beams, the problem was solved by Lyulka in 1974 for two linearly polarized waves [22]:

$$A = a\_1 \cos \varphi; \quad B = a\_2 \cos(\chi + \delta) \tag{167}$$

with the standard conditions for *ϕ*, *χ*, *k*, *κ*, and *δ* being the phase shift.

The two-wave Volkov solution is given by eq. (166), and the matrix elements with corresponding calculation ingredients are given by the standard method as it was shown by Lyulka [22].

It was shown in [22] that

$$q^{\mu} = p^{\mu} - \varepsilon^2 \frac{a\_1^2}{2(kp)} k^{\mu} - \varepsilon^2 \frac{a\_2^2}{2(\kappa p)} \kappa^{\mu} \tag{168}$$

and

$$m\_\*^2 = m^2 \left(1 - \frac{\varepsilon^2 a\_1^2}{m^2} - \frac{\varepsilon^2 a\_2^2}{m^2}\right) ,\tag{169}$$

which is for the massless graphene limit, the following one *m*<sup>2</sup> <sup>∗</sup>(*<sup>m</sup>* <sup>=</sup> <sup>0</sup>) = <sup>−</sup>*e*2(*a*<sup>2</sup> <sup>1</sup> <sup>+</sup> *<sup>a</sup>*<sup>2</sup> 2).

The matrix element involves the extended law of conservation. Namely:

$$k\mathbf{k} + t\mathbf{\kappa} + q = q' + \mathbf{k}' + \mathbf{\kappa}',\tag{170}$$

where the *s* and *t* are natural numbers. The last equation has natural interpretations. The photon object with momenta *sk* and *tκ* interacts with electron with momentum *q*. After interaction, the electron has a momentum *q* and two photons are emitted with momenta *k* and *κ* .

We can write eq. (170) in the equivalent form:

$$
\kappa k + q - k' = q' + \kappa' - t\kappa.\tag{171}
$$

From the squared form of the last equation and after some modification, we get the generalized equation of the double Compton process for *s* = *t* = 1:

$$\frac{1}{\omega\prime} - \frac{1}{\omega} = \frac{1}{m\_\*} (1 - \cos\Theta) + \frac{\Omega\prime - \Omega}{\omega\omega\prime} - \frac{\Omega\Omega\prime}{\omega\omega\prime m\_\*} (1 - \cos\Xi),\tag{172}$$

where Ξ is the angle between the 3-momentum of the *κ*-photon and the 3-momentum of the *κ* -photons with frequency Ω and Ω , respectively. In the situation of graphene with the massless limit, we replace the renormalized mass in eq. (172) by *m*∗(*m* = 0) = *e* <sup>−</sup>*a*<sup>2</sup> <sup>1</sup> <sup>−</sup> *<sup>a</sup>*<sup>2</sup> 2.

Let us remark that if the frequencies of the photons of the first wave substantially differ from the frequencies of the photons of the second wave, then eq. (172) can be experimentally verified by the same method as the original Compton formula. To our knowledge, formula (172) is not involved in the standard textbooks on quantum electrodynamics.
