In this entry we will analyse that a fermion in a plane wave background, typically used to model lasers, presents similar equations compared to Very Special Relativity (VSR), which is a model to describe the existence of neutrino mass is we assume that the nature symmetry is not Lorentz, but a subgroup of it, usually SIM(2)
Let us start with the typical Dirac equation for an electron with mass $M$ in an external electromagnetic field:
$$\left[ \left( i {\partial \!\!\!/} – e {A\!\!\!/} \right) – M \right]\psi = 0 \tag{1}$$
and we will consider that the external field is a plane wave. Hence, $A_{\mu} = A_{\mu} (k \cdot x)$. Acting with the operator $\left[ \left( i {\partial \!\!\!/} – e {A\!\!\!/} \right) + M \right]$ in the left, using the anticommutation relation of the gamma matrices: $\gamma^{\mu} \gamma^{\nu} = 2 \eta^{\mu \nu} – \gamma^{\nu} \gamma^{\mu}$, we have that
$[- \partial^2 – i e \gamma^{\mu} \gamma^{\nu} (\partial_{\mu} A_{\nu}) – 2 i
e A \cdot \partial + e^2 A^2 – M^2] \psi = 0$
and since $A_{\nu} = A_{\nu} (k \cdot x)$, therefore, $\partial_{\mu} A_{\nu}=A’_{\nu} k_{\mu}$, where the prime stands for a derivative respect to $A$. Hence,
$$\left[ – \partial^2 – i e {k\!\!\!/} {A\!\!\!/}’ – 2 i e A \cdot \partial + e^2
A^2 – M^2 \right] \psi = 0 \tag{2}$$
Using as ansatz $\psi = e^{- i p \cdot x} g (k \cdot x)$, $k^2 = 0$ since it is a electromagnetic wave-vector, and $p^2 = M^2$ for a fermion, we have
$$2 i k \cdot p g’ (k \cdot x) – i e {k\!\!\!/}{A\!\!\!/}’ g (k \cdot x) – 2
e A \cdot p g (k \cdot x) – 2 i e A \cdot k g’ (k \cdot x) + e^2 A^2 g
(k \cdot x) = 0$$
If $A$ satisfies the Lorenz gauge $k \cdot A = 0$:
$$2 i k \cdotp g’ (k \cdot x) + \left[ – i e {k\!\!\!/}{A\!\!\!/}’ – 2 e A
\cdot p + e^2 A^2 \right] g (k \cdot x) = 0$$
$$g’ (k \cdot x) = \frac{1}{2 k \cdot p} \left[ i (e^2 A^2 – 2 e A \cdot
p) + e {k\!\!\!/}{A\!\!\!/}’ \right] g (k \cdot x) \tag{3}$$
Equation (3) is of the kind $g’ (\varphi) = R (\varphi) g (\varphi)$ whit solution $g (\varphi) = c_1 \exp \left[ \int^{\varphi}_0 d \bar{\varphi} R (\bar{\varphi}) \right]$ (More precisely for our case: $g’ (\varphi) = S’ (\varphi) g (\varphi)$ and the solution is $g (\varphi) = c_2\exp [S (\varphi)]$). Using it, we have
$$g (k \cdot x) = C \exp \left[ i \int^{k \cdot x}_0 d \bar{\varphi} \left(
\frac{e^2 A^2 – 2 e A \cdot p}{2 k \cdot p} \right) + e \frac{{k\!\!\!/}
{A\!\!\!/}}{2 k \cdot p} \right] \tag{4}$$
Notice that
$$\exp \left[ e \frac{{k\!\!\!/} {A\!\!\!/}}{2 k \cdot p} \right] = 1+e \frac{\not{k} \not{A}}{2 k \cdot p} + e^2 \frac{{k\!\!\!/} {A\!\!\!/} {k\!\!\!/} {A\!\!\!/}}{4 (k \cdot p)^2} + \cdots +$$
Using the anticommutation relation of the gamma matrices, that $k \cdot A = 0$ since we assume that the field satisfies the Lorenz gauge, and ${k\!\!\!/} {k\!\!\!/} = k^2 = 0$ we have that the higher powers vanishes and we are left with
$$\exp \left[ e
\frac{{k\!\!\!/} {A\!\!\!/}}{2 k \cdot p} \right] = 1 + e \frac{{k\!\!\!/} {A\!\!\!/}}{2 k \cdot p}$$.
Hence,
$$g (k \cdot x) = C \left( 1 + e \frac{{k\!\!\!/} {A\!\!\!/}}{2 k \cdot
p} \right) \exp \left[ i \int^{k \cdot x}_0 d \bar{\varphi} \left( \frac{e^2
A^2 – 2 e A \cdot p}{2 k \cdot p} \right) \right] \tag{5}$$
It allows us to define the state $\psi$ as
$$\psi = \exp [- i S_p (x)] \Gamma_p (k \cdot x) u_{p, s} \tag{6}$$
where
$$S_p (x) = p \cdot x + \int^{k \cdot x}_0 d \bar{\varphi} \left(
\frac{2 e A \cdot p – e^2 A^2}{2 k \cdot p} \right)$$
$$\Gamma_p (k \cdot x) = \left( 1 + e \frac{{k\!\!\!/} {A\!\!\!/}}{2 k \cdot p} \right)$$
Considering the possibility to start with $p$ with opposite sign in our ansatz, the most general way to write the state is
$$\psi = \exp [- i S_p (x)] \Gamma_p (k \cdot x) u_{p, s} + \exp [- i S_{- p}
(x)] \Gamma_{- p} (k \cdot x) v_{p, s} \tag{7}$$
where we observe that if $A \rightarrow 0$: $\psi = e^{- i p \cdot x} u_{p, s} + e^{i p
\cdot x} v_{p, s}$ which is the free solution. This solution has been known for long time, from 1935, due to Volkov
We come back to the Dirac equation (1) and we use $\psi$ explicitly as in equation (6) and after some algebra we have
$$[i \gamma^{\mu} \partial_{\mu} + \gamma^{\mu} \partial_{\mu} S_p (x) – e
\gamma^{\mu} A_{\mu} – M] [\Gamma_p (k \cdot x) u_{p, s}] = 0$$
We use that
$$\frac{d}{d x} \int^{b (x)}_{a (x)} f (x, t) d t = f (x, b (x)) \frac{d b}{d x} – f (x, a (x)) \frac{d a}{d x} + \int^{b (x)}_{a (x)}
\frac{\partial}{\partial x} f (x, t) d t$$
to obtain
$$\partial_{\mu} S_p (x) = \partial_{\mu} \left( p \cdot x + \int^{k \cdot
x}_0 d \bar{\varphi} \left( \frac{2 e A \cdot p – e^2 A^2}{2 k \cdot p} \right) \right) = p_{\mu} + \left( \frac{2 e A \cdot p – e^2 A^2}{2 k \cdot p} \right) k_{\mu}$$
With this,
$$i {k\!\!\!/} \Gamma_p’ (k \cdot x) + \left( {q\!\!\!/} – M \right) \Gamma_p (k
\cdot x) = 0 \tag{8}$$
where we have defined the momentum
$$q_{\mu} = p_{\mu} – e A_{\mu} + \left( \frac{2 e A \cdot p – e^2 A^2}{2 k \cdot p} \right) k_{\mu} \tag{9}$$.
Deriving $\Gamma_p$ we have $\Gamma_p’ (k \cdot x) = – e \frac{{k\!\!\!/}{A\!\!\!/}’ (k \cdot
x)}{2 k \cdot p}$, therefore,
$${k\!\!\!/} \Gamma_p’ (k \cdot x) = 0$$
Hence,
$$\left( {q\!\!\!/} – M \right) \Gamma_p (k \cdot x) = 0 \tag{10}$$
Now averaging the momentum $q_{\mu}$:
$$\langle q_{\mu} \rangle = p_{\mu} – \frac{\langle e^2 A^2 \rangle}{2}
\frac{k_{\mu}}{k \cdot p}$$
where we are considering that in average only the amplitude of $A$ (i.e. $A^2$) is
constant. Any other term containing $A$ vanishes in average. Then, averaging $\psi$ in (7), we get
$$\langle \psi \rangle = \langle \exp [- i S_p (x)] \Gamma_p (k \cdot x)
\rangle u_{p, s} + \langle \exp [- i S_{- p} (x)] \Gamma_{- p} (k \cdot x)
\rangle v_{p, s}$$
In parallel,
$$\langle \exp [- i S_p (x)] \Gamma_p (k \cdot x) \rangle = \exp \left[ – i
\left( p \cdot x – \int^{k \cdot x}_0 d \bar{\varphi} \left( \frac{e^2
\langle A^2 \rangle}{2 k \cdot p} \right) \right) \right]$$
Since we are considering $\langle A^2 \rangle$ constant, we take it out of the integral and $\int^{k \cdot x}_0 d
\bar{\varphi} \left( \frac{e^2 \langle A^2 \rangle}{2 k \cdot p} \right)
\approx \frac{e^2 \langle A^2 \rangle}{2 k \cdot p} k \cdot x$. Therefore,
$$\langle \exp [- i S_p (x)] \Gamma_p (k \cdot x) \rangle \approx \exp \left[ i \left( p – \frac{\langle e^2 A^2 \rangle}{2 k \cdot p} k \right) \cdot
x \right] = e^{- i \langle q \rangle \cdot x}$$
Hence,
$$\langle \psi \rangle = e^{- i \langle q \rangle \cdot x} u_{p, s} + e^{i
\langle q \rangle \cdot x} v_{p, s}$$
With this, we average in equation (1) and we have
$$\left[ {p\!\!\!/} – \frac{\langle e^2 A^2 \rangle}{2} \frac{{k\!\!\!/}}{k \cdot
p} – M \right] \langle \psi \rangle = e \left\langle {A\!\!\!/} \psi
\right\rangle \tag{9}$$
After a long computation we can show that
$$e \left\langle {A\!\!\!/} \psi \right\rangle = – e^2 \frac{{k\!\!\!/} \langle A^2 \rangle}{2 k \cdot p} e^{- i \langle q\rangle \cdot x} u_{p, s}$$
Hence,
$$\left[ {p\!\!\!/} – M \right] \langle \psi \rangle = 0 \tag{10}$$
Notice that this description is equivalent to start from an equation
$$\left[ i{\partial \!\!\!/} – i \frac{\langle e^2 A^2 \rangle}{2} \frac{{k\!\!\!/}}{k \cdot
\partial} – M \right] \langle \psi \rangle = 0$$
for the average fermion field.
If we define $\langle e^2 A^2 \rangle = – m^2$ we have
$$\left[ i {\partial \!\!\!/}+ i \frac{m^2}{2} \frac{{k\!\!\!/}}{k \cdot \partial} – M \right] \langle \psi \rangle = 0 \tag{11}$$
which corresponds to the fermionic VSR equation, as we can see as example in equation (7) of a previous work