# Field in the presence of a background electromagnetic field

#### Wledig

Problem 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:

$$-\gamma^{\mu}\gamma^{\nu}D_{\mu} D_{\nu} - im\gamma^{\mu} D_{\mu} + im\gamma^{\nu}D_{\nu} - m^2 =$$
$$-\gamma^{\mu}\gamma^{\nu}(\partial_{\mu} + ieA_{\mu})(\partial_{\nu}-m) - im \gamma^{\mu}(\partial_{\mu}+ieA_{\mu})+im\gamma^{\nu}(\partial_{\nu}+ieA_{\nu}) - m^2 =$$
$$\gamma^{\mu}\gamma^{\nu}(\partial_{\mu} \partial_{\nu} - \partial_{\mu} m + ieA_{\mu} \partial_{\nu} - ieA_{\mu} m) - im\gamma^{\mu} \partial_{\mu} + emA_{\mu}\gamma^{\mu} + im\gamma^{\nu}\partial_{\nu}-em\gamma^{\nu} A_{\nu} - m^2$$

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.

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#### Wledig

Just noticed a bunch of typos at my attempted solution, I apologize for that. Here's my new attempt, if we focus on the first term:
$$-\gamma^\mu \gamma^\nu D_\mu D_\nu = -\dfrac{1}{2}( \{\gamma^\mu, \gamma^\nu \} + [\gamma^\mu,\gamma^\nu]) D_\mu D_\nu =$$
$$-\dfrac{1}{2}(2\eta^{\mu \nu} + [\gamma^\mu, \gamma^\nu]) D_\mu D_\nu =$$

$$\underline{-\eta^{\mu \nu}}(\underline{\partial_\mu \partial_\nu} + \partial_\mu i e A_\nu + ieA_\mu \partial_\nu - eA_\mu A_\nu) - \dfrac{1}{2}[\gamma^\mu, \gamma^\nu] D_\mu D_\nu$$

Now we can recover the Klein-Gordon equation combining the terms underlined up there with the $-m^2$ in:

$$-\gamma^{\mu}\gamma^{\nu}D_{\mu} D_{\nu} - im\gamma^{\mu} D_{\mu} + im\gamma^{\nu}D_{\nu} - m^2$$

I still need some time to reavaluate the additional term though...

Last edited:

#### Wledig

Ended up with an additional term of the form:
$$c = -\eta^{\mu \nu} (\partial_\mu i e A_\nu + ie A_\mu \partial_\nu - eA_\mu \partial_\nu) - \dfrac{1}{2}[\gamma^\mu, \gamma^\nu]D_\mu D_\nu - im\gamma^\mu D_\mu + im \gamma^\nu D_\nu$$

I'll probably have to fiddle with the indices in order to make $[D_\mu , D_\nu]$ appear, I'm just not sure how to go about doing this...

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