**6. Compton effect initiated by a laser pulse**

Let us start with the classical theory of interaction of particle with an impulsive force. We idealize the impulsive force by the dirac *δ*-function. Newton's second law for the interaction of a massive particle with mass *m* with an impulsive force *Pδ*(*t*) is as follows:

$$m\frac{d^2\mathbf{x}}{dt^2} = P\delta(t),\tag{75}$$

where *P* is some constant.

*s* 1 *<sup>ω</sup>* <sup>−</sup> <sup>1</sup>

*<sup>ω</sup>* <sup>=</sup> *<sup>s</sup>*

*m*∗(*m* = 0)

of the renormalized mass and the occurrence of index of refraction.

limiting case with *m* → 0 has the appropriate angle limit *θ* → 0.

Using relation *m*<sup>2</sup>

form:

or equivalently

the graphene sheet:

*s* 1 *<sup>ω</sup>* <sup>−</sup> <sup>1</sup>

124 Graphene - New Trends and Developments

experimental verification, namely:

which is a modification of the original equation for the Compton process

1 *<sup>ω</sup>* <sup>−</sup> <sup>1</sup>

*<sup>ω</sup>* <sup>=</sup> *<sup>s</sup> m*∗

*<sup>ω</sup>* <sup>=</sup> <sup>1</sup>

<sup>∗</sup>(*<sup>m</sup>* <sup>=</sup> <sup>0</sup>) = <sup>−</sup>*e*2*a*2, we get eq. (66) for the situation of the Compton effect in

*e*

(<sup>1</sup> <sup>−</sup> *<sup>n</sup>*<sup>2</sup> cos *<sup>θ</sup>*) = *<sup>s</sup>*

So, we see that the last Compton formula differs from the original one only by the existence

We know that the original Compton formula can be written in the form suitable for the

*h*¯ *mc*

which was used by Compton for the verification of the quantum nature of light [1]. The

If we consider the Compton process in dielectric, then the last formula goes to the following

*h*¯

sin<sup>2</sup> *<sup>θ</sup>* 2

∆*λ* = 4*π*

∆*λ* = 2*π*

It is evident that relation *λ* − *λ* ≥ 0 follows from eq. (1). However, if we put

1

(<sup>1</sup> <sup>−</sup> *<sup>n</sup>*<sup>2</sup> cos *<sup>θ</sup>*), (66)

*<sup>m</sup>* (<sup>1</sup> <sup>−</sup> cos *<sup>θ</sup>*). (67)

√−*a*<sup>2</sup> (<sup>1</sup> <sup>−</sup> *<sup>n</sup>*<sup>2</sup> cos *<sup>θ</sup>*), (68)

, (69)

*mc* (<sup>1</sup> <sup>−</sup> *<sup>n</sup>*<sup>2</sup> cos *<sup>θ</sup>*). (70)

<sup>1</sup> <sup>−</sup> *<sup>n</sup>*<sup>2</sup> cos *<sup>θ</sup>* <sup>≤</sup> 0, (71)

*<sup>n</sup>*<sup>2</sup> <sup>≤</sup> cos *<sup>θ</sup>* <sup>≤</sup> 1, (72)

Using the Laplace transform on the last equation, with

$$\int\_0^\infty e^{-st} x(t) dt = X(s),\tag{76}$$

$$\int\_0^\infty e^{-st} \ddot{\mathbf{x}}(t) dt = s^2 X(s) - sx(0) - s\dot{x}(0) \tag{77}$$

$$\int\_0^\infty e^{-st} \delta(t) dt = 1,\tag{78}$$

we obtain

$$
\hbar \mathbf{s} \mathbf{s}^2 \mathbf{X}(\mathbf{s}) - \hbar \mathbf{s} \mathbf{x}(\mathbf{0}) - \boldsymbol{m} \dot{\mathbf{x}}(\mathbf{0}) = \mathbf{P}.\tag{79}
$$

For a particle starting from rest with *x*˙(0) = 0, *x*(0) = 0, we get

$$X(\mathbf{s}) = \frac{P}{m\mathbf{s}^2},\tag{80}$$

and using the inverse Laplace transform, we obtain

$$x(t) = \frac{P}{m}t\tag{81}$$

and

$$
\dot{\mathbf{x}}(t) = \frac{P}{m}.\tag{82}
$$

Let us remark that if we express *δ*-function by the relation *δ*(*t*) = *H*˙ (*t*), *H* being defined as a step function, then from eq. (75) follows *x*˙(*t*) = *P*/*m*, immediately. The physical meaning of the quantity *P* can be deduced from equation *F* = *Pδ*(*t*). After *t*-integration, we have *Fdt* = *mv* = *P*, where *m* is mass of a body and *v* its final velocity (with *v*(0) = 0). It means that the value of *P* can be determined a posteriori and then this value can be used in more complex equations than eq. (75). Of course it is necessary to suppose that *δ*-form of the impulsive force is adequate approximation of the experimental situation.

If we consider the *δ*-form electromagnetic pulse, then we can write

$$F\_{\mu\nu} = a\_{\mu\nu} \delta(\varphi). \tag{83}$$

where *ϕ* = *kx* = *ωt* − **kx**. In order to obtain the electromagnetic impulsive force in this form, it is necessary to define the four-potential in the following form:

$$A\_{\mu} = a\_{\mu} H(\varphi),\tag{84}$$

where function *H* is the Heaviside unit step function defined by the relation

$$H(\varphi) = \begin{cases} 0, & \varphi < 0 \\ 1, & \varphi \ge 0 \end{cases}.\tag{85}$$

If we define the four-potential by eq. (85), then the electromagnetic tensor with impulsive force is of the form

$$F\_{\mu\nu} = \partial\_{\mu}a\_{\nu} - \partial\_{\nu}a\_{\mu} = (k\_{\mu}a\_{\nu} - k\_{\nu}a\_{\mu})\delta(\varphi) = a\_{\mu\nu}\delta(\varphi). \tag{86}$$

To find motion of an electron in the *δ*-form electromagnetic force, we must solve immediately the Lorentz equation, or, to solve Lorentz equation in general with four-potential *A<sup>µ</sup>* = *aµA*(*ϕ*) and then replace the four-potential by the eta-function. Following Meyer [13], we apply his method and then replace *Aµ*(*ϕ*) by *aµH*(*ϕ*) in the final result.

The Lorentz equation reads:

$$\frac{dp\_{\mu}}{d\tau} = \frac{e}{m} F\_{\mu\nu} p^{\nu} = \frac{e}{m} (k\_{\mu} a \cdot p - a\_{\mu} k \cdot p) A'(\varphi),\tag{87}$$

where the prime denotes derivation with regard to *ϕ*, *τ* is proper time, and *p<sup>µ</sup>* = *m*(*dxµ*/*dτ*). After multiplication of the last equation by *kµ*, we get with regard to the Lorentz condition <sup>0</sup> <sup>=</sup> *∂µA<sup>µ</sup>* <sup>=</sup> *<sup>a</sup>µ∂µA*(*ϕ*) = *<sup>k</sup>µaµ<sup>A</sup>* or *<sup>k</sup>* · *<sup>a</sup>* <sup>=</sup> 0 and *<sup>k</sup>*<sup>2</sup> <sup>=</sup> 0 the following equation:

$$\frac{d(k \cdot p)}{d\tau} = 0\tag{88}$$

and it means that *k* · *p* is a constant of the motion and it can be defined by the initial conditions, for instance, at time *τ* = 0. If we put *pµ*(*τ* = 0) = *p*<sup>0</sup> *<sup>µ</sup>*, then we can write *<sup>k</sup>* · *<sup>p</sup>* <sup>=</sup> *<sup>k</sup>* · *<sup>p</sup>*0. At this moment, we have

$$k \cdot p = \frac{mk \cdot d\mathbf{x}}{d\tau} = m\frac{d\boldsymbol{\varrho}}{d\tau} \,\tag{89}$$

or

 ∞ 0 *e*

<sup>−</sup>*stx*¨(*t*)*dt* = *s*

 ∞ 0 *e*

*<sup>X</sup>*(*s*) = *<sup>P</sup>*

*<sup>x</sup>*(*t*) = *<sup>P</sup> m*

*<sup>x</sup>*˙(*t*) = *<sup>P</sup>*

Let us remark that if we express *δ*-function by the relation *δ*(*t*) = *H*˙ (*t*), *H* being defined as a step function, then from eq. (75) follows *x*˙(*t*) = *P*/*m*, immediately. The physical meaning of the quantity *P* can be deduced from equation *F* = *Pδ*(*t*). After *t*-integration, we have *Fdt* = *mv* = *P*, where *m* is mass of a body and *v* its final velocity (with *v*(0) = 0). It means that the value of *P* can be determined a posteriori and then this value can be used in more complex equations than eq. (75). Of course it is necessary to suppose that *δ*-form of the

impulsive force is adequate approximation of the experimental situation.

If we consider the *δ*-form electromagnetic pulse, then we can write

 ∞ 0 *e*

For a particle starting from rest with *x*˙(0) = 0, *x*(0) = 0, we get

and using the inverse Laplace transform, we obtain

we obtain

126 Graphene - New Trends and Developments

and

<sup>−</sup>*stx*(*t*)*dt* = *X*(*s*), (76)

<sup>2</sup>*X*(*s*) <sup>−</sup> *sx*(0) <sup>−</sup> *sx*˙(0), (77)

<sup>−</sup>*stδ*(*t*)*dt* = 1, (78)

*ms*<sup>2</sup> , (80)

*t* (81)

*<sup>m</sup>*. (82)

*Fµν* = *aµνδ*(*ϕ*). (83)

*ms*2*X*(*s*) <sup>−</sup> *msx*(0) <sup>−</sup> *mx*˙(0) = *<sup>P</sup>*. (79)

$$\frac{d\boldsymbol{\varrho}}{d\boldsymbol{\tau}} = \frac{\boldsymbol{k} \cdot \boldsymbol{p}^0}{m}.\tag{90}$$

So, using the last equation and relation *d*/*dτ* = (*d*/*ϕ*)*dϕ*/*dτ*, we can write eq. (87) in the form

$$\frac{dp\_{\mu}}{d\varphi} = \frac{e}{k \cdot p^{0}} (k\_{\mu}a \cdot p - a\_{\mu}k \cdot p^{0})A'(\varphi) \tag{91}$$

giving (after multiplication by *aµ*)

$$\frac{d(a \cdot p)}{d\varphi} = -ea^2 A'\tag{92}$$

or

$$a \cdot p = a \cdot p^0 - e a^2 A. \tag{93}$$

Substituting the last formula into (91), we get

$$\frac{dp\_{\mu}}{d\varphi} = e \left( a\_{\mu} - \frac{k\_{\mu}a \cdot p^{0}}{k \cdot p^{0}} \right) \frac{dA}{d\varphi} - \frac{e^{2}a^{2}}{2k \cdot p^{0}} \frac{d(A^{2})}{d\varphi} k\_{\mu}. \tag{94}$$

This equation can be immediately integrated to give the resulting momentum in the form

$$p\_{\mu} = p\_{\mu}^{0} - e \left( A\_{\mu} - \frac{A^{\nu} p\_{\nu}^{0} k\_{\mu}}{k \cdot p^{0}} \right) - \frac{e^{2} A^{\nu} A\_{\nu} k\_{\mu}}{2k \cdot p^{0}}.\tag{95}$$

Now, if we put into this formula the four-potential (84) of the impulsive force, then for *ϕ >* 0 when *H >* 1, we get

$$p\_{\mu} = p\_{\mu}^{0} - e \left( a\_{\mu} - \frac{a^{\nu} p\_{\nu}^{0} k\_{\mu}}{k \cdot p^{0}} \right) - \frac{e^{2} a^{\nu} a\_{\nu} k\_{\mu}}{2k \cdot p^{0}}.\tag{96}$$

The last equation can be used to determine the magnitude of *aµ*. It can be evidently expressed as the number of *k*-photons in electromagnetic momentum. For *ϕ <* 0, it is *H* = 0 and, therefore, *p<sup>µ</sup>* = *p*<sup>0</sup> *µ*

It is still necessary to say what is the practical realization of the *δ*-form potential. We know from the Fourier analysis that the Dirac *δ*-function can be expressed by integral in the following form:

$$\delta(\varphi) = \frac{1}{\pi} \int\_0^\infty \cos(s\varphi) ds. \tag{97}$$

So, the *δ*-potential can be realized as the continual superposition of the harmonic waves. In case it will not be possible to realize it experimentally, we can approximate the integral formula by the summation formula as follows:

$$\delta(\varphi) \approx \frac{1}{\pi} \sum\_{0}^{\infty} \cos(s\varphi). \tag{98}$$

Now, the quantum mechanical problem is to find solution of the Dirac equation with the *δ*-form four-potential (84) and, from this solution, determine the quantum motion of the charged particle under this potential. After insertion of Ψ*<sup>p</sup>* and Ψ¯ *<sup>p</sup>* into the current density *j <sup>µ</sup>* = Ψ<sup>∗</sup> *<sup>p</sup>γµ*)Ψ*p*), we have

$$j^{\mu} = \frac{1}{p\_0} \left\{ p^{\mu} - eA^{\mu} + k^{\mu} \left( \frac{e(pA)}{(kp)} - \frac{e^2A^2}{2(kp)} \right) \right\}. \tag{99}$$

which is evidently related to eq. (23).

The so-called kinetic momentum corresponding to *j <sup>µ</sup>* is as follows:

$$J^{\mu} = \Psi\_p^\*(p^{\mu} - eA^{\mu})\Psi\_p = \bar{\Psi}\_p\gamma^0(p^{\mu} - eA^{\mu})\Psi\_p =$$

$$\left\{ p^{\mu} - eA^{\mu} + k^{\mu} \left( \frac{e(pA)}{(kp)} - \frac{e^2A^2}{2(kp)} \right) \right\} + k^{\mu}\frac{ie}{8(kp)p\_0}F\_{a\beta}(u^\*\sigma^{a\beta}u)\_{\,\,\,\,} \tag{100}$$

where

*dϕ <sup>d</sup><sup>τ</sup>* <sup>=</sup> *<sup>k</sup>* · *<sup>p</sup>*<sup>0</sup>

*d*(*a* · *p*)

*dp<sup>µ</sup> <sup>d</sup><sup>ϕ</sup>* <sup>=</sup> *<sup>e</sup>*

giving (after multiplication by *aµ*)

128 Graphene - New Trends and Developments

Substituting the last formula into (91), we get

*dp<sup>µ</sup> <sup>d</sup><sup>ϕ</sup>* <sup>=</sup> *<sup>e</sup>* *p<sup>µ</sup>* = *p*<sup>0</sup>

*p<sup>µ</sup>* = *p*<sup>0</sup>

*<sup>µ</sup>* − *e*

*<sup>µ</sup>* − *e*

*<sup>a</sup><sup>µ</sup>* <sup>−</sup> *<sup>k</sup>µ<sup>a</sup>* · *<sup>p</sup>*<sup>0</sup> *<sup>k</sup>* · *<sup>p</sup>*<sup>0</sup>

 *dA <sup>d</sup><sup>ϕ</sup>* <sup>−</sup> *<sup>e</sup>*2*a*<sup>2</sup> <sup>2</sup>*<sup>k</sup>* · *<sup>p</sup>*<sup>0</sup>

This equation can be immediately integrated to give the resulting momentum in the form

*<sup>A</sup><sup>µ</sup>* <sup>−</sup> *<sup>A</sup><sup>ν</sup> <sup>p</sup>*<sup>0</sup>

Now, if we put into this formula the four-potential (84) of the impulsive force, then for *ϕ >* 0

*<sup>a</sup><sup>µ</sup>* <sup>−</sup> *<sup>a</sup><sup>ν</sup> <sup>p</sup>*<sup>0</sup>

The last equation can be used to determine the magnitude of *aµ*. It can be evidently expressed as the number of *k*-photons in electromagnetic momentum. For *ϕ <* 0, it is *H* = 0 and,

*<sup>ν</sup>k<sup>µ</sup> <sup>k</sup>* · *<sup>p</sup>*<sup>0</sup>

*<sup>ν</sup>k<sup>µ</sup> <sup>k</sup>* · *<sup>p</sup>*<sup>0</sup>

form

or

when *H >* 1, we get

therefore, *p<sup>µ</sup>* = *p*<sup>0</sup>

*µ*

So, using the last equation and relation *d*/*dτ* = (*d*/*ϕ*)*dϕ*/*dτ*, we can write eq. (87) in the

*<sup>k</sup>* · *<sup>p</sup>*<sup>0</sup> (*kµ<sup>a</sup>* · *<sup>p</sup>* <sup>−</sup> *<sup>a</sup>µ<sup>k</sup>* · *<sup>p</sup>*0)*<sup>A</sup>*

*<sup>m</sup>* . (90)

*<sup>d</sup><sup>ϕ</sup>* <sup>=</sup> <sup>−</sup>*ea*2*<sup>A</sup>* (92)

*<sup>a</sup>* · *<sup>p</sup>* <sup>=</sup> *<sup>a</sup>* · *<sup>p</sup>*<sup>0</sup> <sup>−</sup> *ea*2*A*. (93)

*d*(*A*2) *dϕ*

<sup>−</sup> *<sup>e</sup>*2*AνAνk<sup>µ</sup>*

<sup>−</sup> *<sup>e</sup>*2*aνaνk<sup>µ</sup>*

(*ϕ*) (91)

*kµ*. (94)

<sup>2</sup>*<sup>k</sup>* · *<sup>p</sup>*<sup>0</sup> . (95)

<sup>2</sup>*<sup>k</sup>* · *<sup>p</sup>*<sup>0</sup> . (96)

$$
\sigma^{\mathfrak{a}\mathfrak{f}} = \frac{1}{2} (\gamma^{\mathfrak{a}} \gamma^{\mathfrak{f}} - \gamma^{\mathfrak{f}} \gamma^{\mathfrak{a}}).\tag{101}
$$

Now, we express the four-potential by the step function. In this case, the kinetic momentum contains the tensor *Fµν* involving *δ*-function. It means that there is a singularity at point *ϕ* = 0. This singularity plays no role in the situation for *ϕ >* 0 because, in this case, the *δ*-function is zero. Then, the kinetic momentum is the same as *j <sup>µ</sup>*. The emission of photons by electron in the delta-pulse force follows from the matrix element *M* corresponding to the emission of photon by electron in the electromagnetic field [12]:

$$M = -ie^2 \int d^4x \bar{\Psi}\_{p'}(\gamma e^{\prime \ast}) \Psi\_p \frac{e^{ik^\prime \ge}}{\sqrt{2\omega^\prime}},\tag{102}$$

where Ψ*<sup>p</sup>* is the wave function of an electron before interaction with a pulse and Ψ*<sup>p</sup>* is the wave function of an electron after interaction and emission of photon with components *k<sup>µ</sup>* = (*ω* , **k** ). The symbol *e*∗ is the four-polarization vector of emitted photon.

If we write Volkov wave function Ψ*<sup>p</sup>* in the form (18), then, for the impulsive vector potential (84), we have

$$S = -p\mathbf{x} - \left[e\frac{ap}{kp} - \frac{e^2}{2kp}a^2\right]\boldsymbol{\varrho}, \quad \mathbf{R} = \left[1 + \frac{e}{2kp}(\gamma k)(\gamma a)H(\boldsymbol{\varrho})\right].\tag{103}$$

So, we get the matrix element in the form

$$M = g \int d^4x \bar{\Psi}\_{p'} O \Psi\_p \frac{e^{ik'x}}{\sqrt{2\omega'}},\tag{104}$$

where *<sup>O</sup>* <sup>=</sup> *<sup>γ</sup>e*∗, *<sup>g</sup>* <sup>=</sup> <sup>−</sup>*ie*<sup>2</sup> in case of the electromagnetic interaction and

$$\Psi\_{p'} = \frac{i\overline{l}}{\sqrt{2p'\_0}} \mathcal{R}(p') e^{-iS(p')}. \tag{105}$$

In such a way, using the above definitions, we write the matrix element in the form

$$M = \frac{\mathcal{g}}{\sqrt{2\omega'}} \frac{1}{2p\_0' 2p\_0} \int d^4 \mathbf{x} \bar{\mathcal{R}}(p') \mathcal{O} \mathcal{R}(p) e^{-i\mathcal{S}(p') + i\mathcal{S}(p)} e^{i\mathcal{k}' \mathbf{x}}.\tag{106}$$

The quantity *R*¯(*p* ) follows immediately from eq. (103), namely:

$$\bar{\mathcal{R}}' = \overline{\left[1 + \frac{e}{2kp'} (\gamma k)(\gamma a)H(\varphi)\right]} = \left[1 + \frac{e}{2kp'} (\gamma a)(\gamma k)H(\varphi)\right].\tag{107}$$

Using

$$-i\mathbf{S}(p') + i\mathbf{S}(p) = i(p'-p) + i(a'-a)\boldsymbol{\varrho},\tag{108}$$

where

$$\alpha = \left( e \frac{ap}{kp} - \frac{e^2}{2} \frac{a^2}{kp} \right), \quad \alpha' = \left( e \frac{ap'}{kp'} - \frac{e^2}{2} \frac{a^2}{kp'} \right), \tag{109}$$

we get

where Ψ*<sup>p</sup>* is the wave function of an electron before interaction with a pulse and Ψ*<sup>p</sup>* is the wave function of an electron after interaction and emission of photon with components

). The symbol *e*∗ is the four-polarization vector of emitted photon. If we write Volkov wave function Ψ*<sup>p</sup>* in the form (18), then, for the impulsive vector potential

*ϕ*, *R* =

*d*4*x*Ψ¯ *<sup>p</sup>O*Ψ*<sup>p</sup>*

*R*¯(*p* )*e* −*iS*(*p* )

 1 + *e*

> *eik x* √ 2*ω*

<sup>2</sup>*kp* (*γk*)(*γa*)*H*(*ϕ*)

, (104)

. (105)

*<sup>x</sup>*. (106)

. (107)

, (109)

. (103)

*k<sup>µ</sup>* = (*ω*

(84), we have

, **k**

130 Graphene - New Trends and Developments

*S* = −*px* −

So, we get the matrix element in the form

*<sup>M</sup>* <sup>=</sup> *<sup>g</sup>* √ 2*ω*

*R*¯ = 1 +

The quantity *R*¯(*p*

Using

where

 *e ap kp* <sup>−</sup> *<sup>e</sup>*<sup>2</sup> <sup>2</sup>*kp <sup>a</sup>*<sup>2</sup> 

> *M* = *g*

where *<sup>O</sup>* <sup>=</sup> *<sup>γ</sup>e*∗, *<sup>g</sup>* <sup>=</sup> <sup>−</sup>*ie*<sup>2</sup> in case of the electromagnetic interaction and

1 2*p* <sup>0</sup>2*p*<sup>0</sup>

<sup>2</sup>*kp* (*γk*)(*γa*)*H*(*ϕ*)

*e*

− *iS*(*p*

*α* = *e ap kp* <sup>−</sup> *<sup>e</sup>*<sup>2</sup> 2 *a*2 *kp* 

<sup>Ψ</sup>¯ *<sup>p</sup>* <sup>=</sup> *<sup>u</sup>*¯

) follows immediately from eq. (103), namely:

2*p* 0

*d*4*xR*¯(*p*

 = 1 +

, *α* =

 *e ap kp* <sup>−</sup> *<sup>e</sup>*<sup>2</sup> 2 *a*2 *kp* 

)*OR*(*p*)*e*

−*iS*(*p*

*e*

)+*iS*(*p*) *e ik*

<sup>2</sup>*kp* (*γa*)(*γk*)*H*(*ϕ*)

) + *iS*(*p*) = *i*(*p* − *p*) + *i*(*α* − *α*)*ϕ*, (108)

In such a way, using the above definitions, we write the matrix element in the form

$$M = \frac{\mathcal{g}}{\sqrt{2\omega'}} \frac{1}{2p\_0' 2p\_0} \int d^4 \mathbf{x} i\bar{\mathcal{R}}(p') \mathcal{O} \mathcal{R}(p) \mu e^{i(p'-p)\mathbf{x}} e^{i(a'-a)\mathbf{q}} e^{i\mathbf{k'} \mathbf{x}}.\tag{110}$$

Putting

$$\bar{R}(p')OR(p)e^{i(a'-a)\varphi} = \frac{1}{2\pi} \int\_{-\infty}^{\infty} D(s)e^{-is\varphi}ds\tag{111}$$

with the inverse transform

$$D(s) = \int\_{-\infty}^{\infty} d\varphi e^{j s \varphi} \bar{R}(p') O R(p) e^{j(a'-a)},\tag{112}$$

we get after *x*-integration

$$M = \frac{\mathcal{g}}{\sqrt{2\omega'}} \frac{1}{\sqrt{2p\_0' 2p\_0}} \int ds \bar{u}(p') D(s)u(p) \delta^{(4)}(ks + p - k' - p'). \tag{113}$$

We see from the presence of the *δ*-function in eq. (113) that during the process of the interaction of electron with the laser pulse, the energy-momentum conservation law holds good

$$sk + p = k' + p'.\tag{114}$$

The last equation has physical meaning for *s* = 1, 2, . . . *N*, *N* being a positive integer. *s* = 1 means the conservation of energy momentum of the one-photon Compton process and *s* = 2 of the two-photon Compton process and *s* = *N* means the multiphoton interaction with *N* photons of laser pulse with electron. The multiphoton interaction is nonlinear and differs from the situation where electron scatters twice or more as it traverses the laser focus. The analogical situation is valid for the photoelectric equation.

Now, let us determine *D*(*s*). With regard to the mathematical relation *H*2(*ϕ*) = *H*(*ϕ*), we can put

$$
\bar{R}(p')OR(p) = A + BH(\varphi) \tag{115}
$$

where

$$A = \gamma e^{\prime \ast} \tag{116a}$$

and

$$B = \frac{e}{2(kp)}(\gamma e'^\ast)(\gamma k)(\gamma a) + \frac{e}{2(kp')}(\gamma a)(\gamma k)(\gamma e'^\ast) +$$

$$\frac{e^2}{4(kp)(kp')}(\gamma a)(\gamma k)(\gamma e'^\ast)(\gamma k)(\gamma a). \tag{116b}$$

In such a way,

$$D(s) = A \int\_{-\infty}^{\infty} e^{i(a'-a+s)\cdot\varphi} d\varphi + B \int\_{-\infty}^{\infty} e^{i(a'-a+s)\cdot\varphi} H(\varphi) d\varphi = $$
 
$$(2\pi)A\delta(a'-a+s) + (2\pi i)B\frac{1}{a'-a+s} \tag{117}$$

as a consequence of the integral

$$\int\_0^\infty e^{-\varepsilon \mathbf{x}} \sin m\mathbf{x} d\mathbf{x} = \frac{m}{\varepsilon^2 + m^2} \tag{118}$$

for *m* → 0.

The total probability of the emission of photons during the interaction of the laser pulse with electron is as follows:

$$W = \int \frac{1}{2} \sum\_{\text{spin.polar.}} \frac{|M|^2}{VT} \frac{d^3 p' d^3 k'}{(2\pi)^6}. \tag{119}$$

There are two substantial mathematical operations for the evaluation of *W*. The one step is to use, after performing |*M*| 2, the following mathematical identity:

$$(2\pi)^8 \delta^{(4)}(sk + p - p' - k') \delta^{(4)}(p + s'k - p' - k') = $$

$$(2\pi)^4 VT \frac{\delta(s - s')}{\delta(0)} \delta^{(4)}(p + sk - p' - k'), \tag{120}$$

where *V* is the space volume and *T* is time interval, and the second step is the determination of Trace, because according to the quantum electrodynamics of a spin, it is possible to show that [12]

$$\frac{1}{2} \sum\_{\text{spin.polar.}} |M|^2 = \frac{1}{2} \text{Tr} \left\{ (\gamma p' + m) D(\gamma p + m) \gamma^0 D^+ \gamma^0 \right\},\tag{121}$$

where in our case, quantity *D* is given by eq. (117).

In order to determine *Tr* or spur of the combinations of *γ*-matrix, it is suitable to know some relations. For instance:

$$\text{Tr}(a\gamma)(b\gamma) = 4ab, \quad \text{Tr}(a\gamma)(b\gamma)(c\gamma) = 0,\tag{122}$$

$$\operatorname{Tr}(a\gamma)(b\gamma)(c\gamma)(d\gamma) = 4\left[ (ab)(cd) - (ac)(bd) + (ad)(bc) \right].\tag{123}$$

Then,

*A* = *γe*

<sup>∗</sup>)(*γk*)(*γa*) + *<sup>e</sup>*

<sup>−</sup>*α*+*s*)*ϕdϕ* + *B*

(*γa*)(*γk*)(*γe*

2(*kp*)

 ∞ −∞ *e i*(*α*

<sup>−</sup>*ε<sup>x</sup>* sin *mxdx* <sup>=</sup> *<sup>m</sup>*

The total probability of the emission of photons during the interaction of the laser pulse with

There are two substantial mathematical operations for the evaluation of *W*. The one step is

2, the following mathematical identity:

)*δ*(4)

) *<sup>δ</sup>*(0) *<sup>δ</sup>*(4)

(*p* + *s*  *k* − *p* − *k*

(*p* + *sk* − *p* − *k*

) =


*d*<sup>3</sup> *p d*3*k*

(2*π*)*Aδ*(*<sup>α</sup>* <sup>−</sup> *<sup>α</sup>* <sup>+</sup> *<sup>s</sup>*)+(2*πi*)*<sup>B</sup>* <sup>1</sup>

(*γa*)(*γk*)(*γe*

<sup>−</sup>*α*+*s*)*ϕH*(*ϕ*)*dϕ* =

*<sup>B</sup>* <sup>=</sup> *<sup>e</sup>* 2(*kp*)

*D*(*s*) = *A*

as a consequence of the integral

132 Graphene - New Trends and Developments

 ∞ −∞ *e i*(*α*

(*γe*

*e*2 4(*kp*)(*kp*)

> ∞ 0 *e*

*W* =

(2*π*)8*δ*(4)

1

(*sk* + *p* − *p* − *k*

(2*π*)4*VT <sup>δ</sup>*(*<sup>s</sup>* <sup>−</sup> *<sup>s</sup>*

<sup>2</sup> ∑ *spin*.*polar*.

and

In such a way,

for *m* → 0.

electron is as follows:

to use, after performing |*M*|

∗ (116a)

∗)+

∗)(*γk*)(*γa*). (116b)

*<sup>α</sup>* <sup>−</sup> *<sup>α</sup>* <sup>+</sup> *<sup>s</sup>* (117)

*<sup>ε</sup>*<sup>2</sup> <sup>+</sup> *<sup>m</sup>*<sup>2</sup> (118)

(2*π*)<sup>6</sup> . (119)

), (120)

$$\text{Tr}\left[ (\gamma p' + m)D(\gamma p + m)\bar{D} \right] = \text{S}\_1 + \text{S}\_2 + \text{S}\_3 + \text{S}\_4; \quad \bar{D} = \gamma^0 D^+ \gamma^0 \tag{124}$$

with

$$S\_1 = \text{Tr}[\gamma p'D\gamma p\bar{D}] \tag{125}$$

$$S\_2 = \text{Tr}[mD\gamma p\bar{D}]\tag{126}$$

$$S\_3 = \text{Tr}[m\gamma p'D\bar{D}]\tag{127}$$

$$S\_4 = \text{Tr}[m^2D\vec{D}]\_\prime \tag{128}$$

where *D* is given by eq. (117) and *A* and *B* are given by eqs. (116b) and (116b).

Now, it is evident that the total calculation is complex and involves many algebraic operations with *γ*-matrices and *δ*-functions. At this moment, we restrict the calculations to the most simple approximation where we replace the term in brackets in eq. (18) by unit, and so we write instead of eq. (18)

$$\Psi\_p \sim \frac{u}{\sqrt{2p\_0}} e^{iS} \tag{129}$$

which is usually used in similar form for the nonrelativistic calculations as it is discussed in [12]. Then,

$$M = \frac{g}{\sqrt{2\omega'}} \frac{1}{2p\_0' 2p\_0} \int d\mathbf{x}'^4 \overline{u}(\gamma e'^\*) u e^{i(p'-p)\mathbf{x}} e^{i(\mathbf{a}'-\mathbf{a})\cdot\mathbf{p}} e^{i\mathbf{k}'\mathbf{x}} = $$
 
$$\frac{g}{\sqrt{2\omega'}} \frac{1}{2p\_0' 2p\_0} \overline{u}(\gamma e'^\*) u \delta^{(4)}(lk + p - p' - k'), \tag{130}$$

where

$$l = \mathfrak{a} - \mathfrak{a}'.\tag{131}$$

In this simplified situation, *ROR* ¯ reduces to *A* = *γe*∗. Then, using relation *γ*¯*<sup>µ</sup>* = *γ<sup>µ</sup>* with a consequence *A*¯ = *A* and relations (122) and (123), we get

$$\mathcal{S}\_1 = \text{Tr}[\gamma p' A \gamma p \bar{A}] = 4 \left[ (p' \varepsilon'^\*)(p \varepsilon'^\*) - (p p')(\varepsilon'^\* \varepsilon'^\*) + (p' \varepsilon'^\*)(p \varepsilon'^\*) \right] \tag{132}$$

$$S\_2 = \text{Tr}[\pi A \gamma p \bar{A}] = 0\tag{133}$$

$$S\_3 = \text{Tr}[m\gamma p'A\bar{A}] = 0\tag{134}$$

$$S\_4 = \text{Tr}[m^2 A \bar{A}] = 4m^2(e^{\prime \ast} e^{\prime \ast}).\tag{135}$$

At this moment, we can write probability of the process *W* in the form:

$$W = \int \frac{1}{2} \sum\_{\text{spin. polar.}} \frac{|M|^2}{VT} \frac{d^3 p' d^3 k'}{(2\pi)^6} =$$

$$\int \frac{d^3 p' d^3 k'}{(2\pi)^6} \frac{1}{2} (\mathbf{S}\_1 + \mathbf{S}\_2 + \mathbf{S}\_3 + \mathbf{S}\_4) \frac{1}{(2\pi)^2} \delta^{(4)}(lk + p - p' - k') =$$

$$\int \frac{d^3 p' d^3 k'}{(2\pi)^6} \frac{1}{2} \delta^{(4)}(lk + p - p' - k') 4 \left\{ (p' e'^\*)(p \epsilon^{\prime \*}) + (e'^\* e'^\*)(m^2 - (p p')) \right\}. \tag{136}$$

The presence of the *δ*-function in the last formula is expression of the conservation law *lk* + *p* = *k* + *p* , which we write in the form

$$lk + p - k' = p'.\tag{137}$$

If we introduce the angle Θ between **k** and **k** , then, with |**k**| = *ω* and |**k** | = *ω* , we get from the squared eq. (137) in the rest system of electron, where *p* = (*m*, 0), the following equation:

$$l\frac{1}{\omega'} - \frac{1}{\omega} = \frac{l}{m}(1 - \cos\Theta); \quad l = \alpha - \alpha',\tag{138}$$

which is a modification of the original equation for the Compton process

which is usually used in similar form for the nonrelativistic calculations as it is discussed in

*u*¯(*γe*

*l* = *α* − *α*

In this simplified situation, *ROR* ¯ reduces to *A* = *γe*∗. Then, using relation *γ*¯*<sup>µ</sup>* = *γ<sup>µ</sup>* with a

<sup>∗</sup>)(*pe*∗) − (*pp*

<sup>∗</sup>)*uei*(*<sup>p</sup>*

<sup>∗</sup>)*uδ*(4)

<sup>−</sup>*p*)*xe i*(*α* <sup>−</sup>*α*)*ϕe ik <sup>x</sup>* =

)(*e* ∗*e*

∗*e*

(*lk* + *p* − *p* − *k*

∗)+(*p e* ∗)(*pe*∗)

*S*<sup>2</sup> = Tr[*mAγpA*¯] = 0 (133)

*AA*¯] = 0 (134)


(*lk* + *p* − *p* − *k*

<sup>∗</sup>)(*m*<sup>2</sup> <sup>−</sup> (*pp*

*d*<sup>3</sup> *p d*3*k* (2*π*)<sup>6</sup> <sup>=</sup>

) =

)) 

. (136)

∗). (135)

. (131)

), (130)

(132)

*dx*4*u*¯(*γe*

1 2*p* <sup>0</sup>2*p*<sup>0</sup>

[12]. Then,

where

*<sup>M</sup>* <sup>=</sup> *<sup>g</sup>* √ 2*ω*

134 Graphene - New Trends and Developments

*S*<sup>1</sup> = Tr[*γp*

1 2*p* <sup>0</sup>2*p*<sup>0</sup>

consequence *A*¯ = *A* and relations (122) and (123), we get

*AγpA*¯] = 4

 (*p e*

At this moment, we can write probability of the process *W* in the form:

*d*<sup>3</sup> *p*

*d*<sup>3</sup> *p*

*lk* + *p* = *k* + *p*

*d*3*k* (2*π*)<sup>6</sup>

1 2 *δ*(4)

*d*3*k* (2*π*)<sup>6</sup>

, which we write in the form

1 2

(*lk* + *p* − *p* − *k*

*S*<sup>3</sup> = Tr[*mγp*

*S*<sup>4</sup> = Tr[*m*2*AA*¯] = 4*m*2(*e*

*W* =

(*S*<sup>1</sup> <sup>+</sup> *<sup>S</sup>*<sup>2</sup> <sup>+</sup> *<sup>S</sup>*<sup>3</sup> <sup>+</sup> *<sup>S</sup>*4) <sup>1</sup>

The presence of the *δ*-function in the last formula is expression of the conservation law

)4 (*p e* 1

<sup>2</sup> ∑ *spin*. *polar*.

(2*π*)<sup>2</sup> *<sup>δ</sup>*(4)

∗*e*

∗)(*pe*∗)+(*e*

*g* √ 2*ω*

$$\frac{1}{\omega'} - \frac{1}{\omega} = \frac{1}{m}(1 - \cos\Theta). \tag{139}$$

We observe that the basic difference between single-photon interaction and *δ*-pulse interaction is the factor *l* = *α* − *α* .

We know that the last formula of the original Compton effect can be written in the form suitable for the experimental verification, namely:

$$
\Delta\lambda = 4\pi \frac{\hbar}{mc} \sin^2 \frac{\Theta}{2},
\tag{140}
$$

which was applied Compton for the verification of the quantum nature of light.

We can express eq. (138) in a new form. From equation *lk* + *p* = *k* + *p* , we get after multiplying it by *k* in the rest frame of electron

$$kp'=\omega m-\omega\omega'(1-\cos\Theta)\tag{141}$$

Then, *l* in eq. (138) is given by the formula (*a* ≡ (*v*, **w**)):

$$l = \frac{2evm - e^2a^2}{2\omega m} - \frac{2eap' - e^2a^2}{2\omega \left[m - \omega'(1 - \cos\Theta)\right]}.\tag{142}$$

This equation (138) can be experimentally verified by the similar methods which were used by Compton for the verification of his formula. However, it seems that the interaction of the photonic pulse substantially differs from the interaction of a single photon with electron.

The equation *lk* + *p* = *k* + *p* is the symbolic expression of the nonlinear Compton effect in which several photons are absorbed at a single point, but only single high-energy photon is emitted. The second process where electron scatters twice or more as it traverses the laser focus is not considered here. The nonlinear Compton process was experimentally confirmed, for instance, by Bulla et al. [15].

The present text is a continuation of the author discussion on laser acceleration [16, 17, 18], where the Compton model of laser acceleration was proposed.

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 by the petawatt lasers [19].
