La sustitución de $\partial_\mu \to D_\mu \equiv \partial_\mu + ieA_\mu$ permite la introducción de las interacciones electromagnéticas

Question

Quiero mostrar que la sustitución de $\partial_u \to D_\mu \equiv \partial_\mu + ieA_\mu$, o, equivalentemente, $p_\mu \to p_\mu - eA_\mu$ permite la introducción de las interacciones electromagnéticas. Aquí $e$ es la carga eléctrica de la partícula en cuestión $($$e=-|e|$ para un electrón$)$, e $A^\mu = (\Phi, \vec{A})$ es el vector potencial. Mediante la conversión de la ecuación de Dirac en el formulario$$[\vec{\alpha} \cdot (\vec{p} - e\vec{A}) + \beta m]\psi = (E - e\Phi)\psi$$to a second-order equation, and taking the low-energy limit, show that the interaction with the electromagnetic field gives rise to a change in energy in the presence of a magnetic field $\vec{B} = \nabla \times \vec{A}$ of the form$$\Delta E = -{{e}\over{2m}}\vec{\sigma} \cdot \vec{B}$$and hence implies a value of $g=2$ for the electron's magnetic moment $\vec{\mu}$ defined in terms of its spin $S$ as$$\vec{u} = g\left({e\over{2m}}\right)\vec{S}.$$Progress so far: The Dirac equation is $$(i\gamma^\mu D_\mu - m)\psi = (i\gamma^\mu(\partial_\mu + ieA_\mu) - m)\psi = 0.$$If we take $\gamma^0 = \beta$, $\gamma^i = \beta\alpha_i$, $\partial_\mu = (\partial_i, \nabla)$, and $A_\mu = (\Phi, -{\bf Un})$ then we can write this as $$[i\beta(\partial_t + ie\Phi) + i\beta\alpha \cdot (\nabla - ie{\bf A}) - m]\psi = 0$$or, since $\beta^2 = 1$,$$[i(\partial_t + ie\Phi) + i\alpha \cdot (\nabla - ie{\bf A}) - m\beta]\psi = 0.$$he intentado algunas cosas después de esto, pero no funcionó. Puede alguien darme un paso en la dirección correcta?

Answer 1

Lo que tiene es un buen comienzo. Si hacemos la habitual asignaciones ${\partial\over{\partial t}} \to -iE$ $\nabla \to i{\bf p}$ entonces tenemos$$(E - e\Phi)\psi = (\alpha \cdot ({\bf p} - e{\bf A}) + m\beta)\psi.$$Now, pick a particular representation$$\beta = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix},\text{ }\alpha_i = \begin{pmatrix} 0 & \sigma^i \\ \sigma^i & 0\end{pmatrix}.$$It is easy to check that these give the correct anti-commutation relations. Then if we denote$$\psi = \begin{pmatrix} \chi \\ \varphi\end{pmatrix}$$and plug this into the Dirac equation we obtain$$(E - e\Phi)\begin{pmatrix} \chi \\ \varphi\end{pmatrix} = \sigma \cdot ({\bf p} - e{\bf A})\begin{pmatrix} \varphi \\ \chi\end{pmatrix} + m\begin{pmatrix} \chi \\ -\varphi\end{pmatrix}.$$If we note that the nonrelativistic energy $E'$ is related to the relativistic by $E' = E - m$ then the equation becomes$$E'\begin{pmatrix} \chi \\ \varphi\end{pmatrix} = \sigma \cdot ({\bf p} - E{\bf A})\begin{pmatrix} \varphi \\ \chi\end{pmatrix} + e\Phi\begin{pmatrix} \chi \\ \varphi\end{pmatrix} - 2m\begin{pmatrix} 0 \\ \varphi\end{pmatrix}.$$In the nonrelativistic limit $E'\ll m$ so the second component of the above equation can be written$$\varphi = {{\sigma \cdot ({\bf p} - e{\bf A})\chi}\over{2m}}.$$We can then write the first component as a second order equation:$$E'\chi = \bigg\{{1\over{2m}}\sigma \cdot ({\bf p} - e{\bf A}) \sigma \cdot ({\bf p} - e{\bf A}) + e\Phi\bigg\}\chi.$$Since $\sigma^i\sigma^j = \delta^{ij} + i\epsilon^{ijk}\sigma^k$ we have $(\sigma \cdot {\bf a})(\sigma \cdot {\bf a}) = {\bf a} \cdot {\bf b} + i\sigma \cdot ({\bf} \times {\bf b})$. So,$$\sigma \cdot ({\bf p} - e{\bf A})\sigma \cdot ({\bf p} - e{\bf A}) = ({\bf p} - \epsilon{\bf A})^2 + i\epsilon^{ijk}\sigma^k(-i\partial_i-eA_i)(-i\partial_j - eA_j)$$$$=({\bf p} - e{\bf})^2 + i\epsilon^{ijk}\sigma^k(es decir,\partial_iA_j + ieA_i\partial_j)$$$$=({\bf p} - e{\bf A})^2 - e\epsilon^{ijk}\sigma^k((\partial_iA_j) + A_j\partial_i + A_i\partial_j)$$$$=({\bf p} - e{\bf})^2 - e\epsilon^{ijk}\sigma^k(\partial_iA_j)$$$$=({\bf p} - e{\bf A})^2 - e\sigma \cdot (\nabla \times{\bf A})$$$$=({\bf p} - e{\bf})^2 - e\sigma \cdot {\bf B}.$$We get that$$E'\chi = \bigg\{{{({\bf p} - e{\bf A})^2}\over{2m}} - {{e\sigma \cdot {\bf B}}\over{2m}} + e\Phi\bigg\}\chi.$$