- #1
Wledig
- 69
- 1
- Homework Statement
- A field obeying the Dirac equation in the presence of a background electromagnetic field also obeys the second-order equation:$$(i\gamma^{\mu}D_{\mu}+m)(i\gamma^{\nu}D_{\nu}-m)\Psi = 0$$ Where ##D_{\mu} = (\partial_{\mu} +ieA_{\mu})## Simplify this equation by using the identity $$\gamma^{\mu}\gamma^{\nu} = \dfrac{1}{2}\{\gamma^{\mu},\gamma^{\nu}\} + \dfrac{1}{2}[\gamma^{\mu} \gamma^{\nu}]$$
a) Show that it reduces to the Klein-Gordon equation plus one extra term.
b) Simplify the new term by proving the identity $$[D_{\mu},D_{\nu}] = +ieF_{\mu \nu}$$ . Using the explicit form of the ##\gamma^{\mu}## matrices evaluate this term in a background magnetic field for which ##F_ij = \epsilon_{ijk}B^k## and ##F_{0i} = 0##.
c) Act the resulting equation on ##\Psi = \begin{pmatrix}
\xi \\
0
\end{pmatrix}
e^{-imt}##. Show that to first order in B, the energy of the state is shifted by an term of the form of ##\Delta E = \mu \cdot B##. In the expression for ##\mu##, identify g = 2.
- Relevant Equations
- Klein Gordon equation: ##(\partial^2 + m^2)\phi(x) = 0##
Dirac equation solution he is referring to:
##\Psi = \begin{pmatrix}
\xi \\
0
\end{pmatrix}
e^{-imt}##
Magnetic moment: ##\mu = \dfrac{-geS}{2m}##
Attempt at a solution:
$$ -\gamma^{\mu}\gamma^{\nu}D_{\mu} D_{\nu} - im\gamma^{\mu} D_{\mu} + im\gamma^{\nu}D_{\nu} - m^2 =$$
$$ -\gamma^{\mu}\gamma^{\nu}(\partial_{\mu} + ieA_{\mu})(\partial_{\nu}-m) - I am \gamma^{\mu}(\partial_{\mu}+ieA_{\mu})+im\gamma^{\nu}(\partial_{\nu}+ieA_{\nu}) - m^2 =$$
$$ \gamma^{\mu}\gamma^{\nu}(\partial_{\mu} \partial_{\nu} - \partial_{\mu} m + ieA_{\mu} \partial_{\nu} - ieA_{\mu} m) - im\gamma^{\mu} \partial_{\mu} + emA_{\mu}\gamma^{\mu} + im\gamma^{\nu}\partial_{\nu}-em\gamma^{\nu} A_{\nu} - m^2$$
Rearranging the terms we can recover the Klein-Gordon equation with this additional term c:
$$ -\gamma^{\mu}\gamma^{\nu}\partial_\mu \partial_nu + im\gamma^{\nu}\partial_\nu - im\gamma^{\mu}\partial_\mu - m^2 + c$$
$$ -\eta^{\mu \nu} \partial_\mu \partial_\nu - m^2 + c$$
$$ c = \gamma^{\mu}\gamma^{\nu}ieA_{\mu}\partial_{\nu} + \gamma^{\mu}\gamma^{\nu}iemA_{\mu} + eA_{\mu}\gamma^{\mu} - m\gamma^{\nu}eA_{\nu}$$
I don't see a way to incorporate the identity he suggested and still recover the Klein Gordon equation.
$$ -\gamma^{\mu}\gamma^{\nu}D_{\mu} D_{\nu} - im\gamma^{\mu} D_{\mu} + im\gamma^{\nu}D_{\nu} - m^2 =$$
$$ -\gamma^{\mu}\gamma^{\nu}(\partial_{\mu} + ieA_{\mu})(\partial_{\nu}-m) - I am \gamma^{\mu}(\partial_{\mu}+ieA_{\mu})+im\gamma^{\nu}(\partial_{\nu}+ieA_{\nu}) - m^2 =$$
$$ \gamma^{\mu}\gamma^{\nu}(\partial_{\mu} \partial_{\nu} - \partial_{\mu} m + ieA_{\mu} \partial_{\nu} - ieA_{\mu} m) - im\gamma^{\mu} \partial_{\mu} + emA_{\mu}\gamma^{\mu} + im\gamma^{\nu}\partial_{\nu}-em\gamma^{\nu} A_{\nu} - m^2$$
Rearranging the terms we can recover the Klein-Gordon equation with this additional term c:
$$ -\gamma^{\mu}\gamma^{\nu}\partial_\mu \partial_nu + im\gamma^{\nu}\partial_\nu - im\gamma^{\mu}\partial_\mu - m^2 + c$$
$$ -\eta^{\mu \nu} \partial_\mu \partial_\nu - m^2 + c$$
$$ c = \gamma^{\mu}\gamma^{\nu}ieA_{\mu}\partial_{\nu} + \gamma^{\mu}\gamma^{\nu}iemA_{\mu} + eA_{\mu}\gamma^{\mu} - m\gamma^{\nu}eA_{\nu}$$
I don't see a way to incorporate the identity he suggested and still recover the Klein Gordon equation.