Let consider the Dirac equation for an electron in the field of an electromagnetic wave

$$(\gamma ^{\mu }(\hat{p} _{\mu }+eA_{\mu })-m)\psi= (-i\gamma ^{\mu }\partial _{\mu }+e\gamma ^{\mu }A_{\mu }-m)\psi=0$$

Under our assumptions the field of a plane wave with wave $4$-vector $k$ (with $k^2 = k^{\mu }k_{\mu } =0$) depends only on the combination $\phi= kx = k^{\mu }x_{\mu }$, so that the $4$-potential is

$$A^{\mu }= A^{\mu }(\phi)$$

and satisfies the Lorenz gauge condition

$$\partial _{\mu}A^{\mu } = k _{\mu}{A^{\mu }}' =0$$

In order to find a solution the equation was squared

$$ [\partial _{\mu }\partial ^{\mu }-2ie(A_{\mu }\partial ^{\mu }) + e^2A _{\mu }A ^{\mu }-m^2 -ie\hat{k}\hat{A}]\psi=0$$

with $\hat{a}:= \gamma ^{\mu } a _{\mu }$.

One makes an ansatz

$$\psi = e^{-ip ^{\mu }x _{\mu }}F(\phi)$$

where $p^{\mu }$ is a constant $4$-vector.

Adding to $p^{\mu }$ a constant multiple of $k^{\mu }$ we can suppose $p^2 = m^2$.

Volkov solved the problem exactly by

$$\psi_p = [1+\frac{e}{2(kp)}\hat{k}\hat{A}]\frac{u}{\sqrt{2p_0}}e^{iS}$$


$$S = -px - \int^{kx} _0 [\frac{e}{(kp)}(pA)-\frac{e^2}{2(kp)}A^2]d\phi$$

(Remark: $kp = k^{\mu }p_{\mu }$); Sources: Berestetkii's "Relativistic Quantum Theory" (§40, page 122) or Miroslav Pardy's "Volkov solution for an electron in the two wave fields".

Obviously $\psi_p$ solves the squared Dirac equation, but tht's not clear why it also solves Dirac

$$(\gamma ^{\mu }(\hat{p} _{\mu }+eA_{\mu })-m)\psi=0$$

Inserting $\psi_p$ into to equation I get using some identities for Gamma martices

$$[\hat{p}+\hat{k}[\frac{e}{(kp)}(pA)-\frac{e^2}{2(kp)}A^2]- e\hat{A}+m]\psi_p$$

but it's not clear why $(\hat{p}+\hat{k}[\frac{e}{(kp)}(pA)-\frac{e^2}{2(kp)}A^2]- e\hat{A}+m)$ vanishes.

  • $\begingroup$ It is "Lorenz" condition, after Ludvig Valentin Lorenz. en.wikipedia.org/wiki/Ludvig_Lorenz. $\endgroup$ – my2cts May 18 '18 at 19:10
  • $\begingroup$ It should perhaps solve $(-i\gamma ^{\mu }(\partial _{\mu }-ieA^\mu)-m)\psi=0$. $\endgroup$ – my2cts May 19 '18 at 20:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.