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I attempt to understand is it possible whether to calculate the next expression by means of mathematica: $\varepsilon^{\mu\nu\rho\sigma}tr(\{\partial_{\mu}L_{\nu},L_{\rho}\}(\partial_{\sigma}H)H^{\dagger})$?

where $\mu,\nu,\rho,\sigma=0,1,2,3$; $\{,\}$- anti-communicator; $L_{\mu}=\tau^{a}L^{a}_{\mu}=\tau^{0}L^{0}_{\mu}+\tau^{i} L^{i}_{\mu}$ ; $a=0,1,2,3$; $\tau^{0}$- unity matrix; $\tau^{i}$ - Pauli matrix; $L_{\mu}=L_{\mu}(x^{0},x^{1},x^{2},x^{3})$

and: $H=1+i\frac{\tau^{i}\pi^{i}}{2f}-\frac{\tau^{i}\pi^{i}\tau ^{j}\pi^{j}}{8f^{2}}$; $f$ - constant; $\pi^{i}=\pi^{i}(x^{0},x^{1},x^{2},x^{3})$;

finally $ \varepsilon^{\mu\nu\rho\sigma} $- Levi-Civita symbol; $\dagger$-Hermitian conjugation.

I write $\varepsilon^{\mu\nu\rho\sigma}tr(\{\partial_{\mu}L_{\nu},L_{\rho}\}(\partial_{\sigma}H)H^{\dagger})$ component-wise and I consider a part of the expression. And now I don't know how one can differentiate this expression with respect to $\partial_{\mu}$ and to say the mathematica that $L_{\mu}$, $\pi^{j}$ is functions.

Thank you for your kind replies .

additions: Let

\[Pi]\[Tau][x_]=(Subscript[\[Pi], 0]    Sqrt[2] Subscript[\[Pi], -]
Sqrt[2] Subscript[\[Pi], +] -Subscript[\[Pi], 0]

)/.{Subscript[\[Pi], 0]->Subscript[\[Pi], 0][x], Subscript[\[Pi], +]->Subscript[\[Pi], +][x], Subscript[\[Pi], -]->Subscript[\[Pi], -][x]}

and

\[Xi][x_]=(1    0
0   1

)+I/(2 f) \[Pi]\[Tau][x] -1/(8 f^2) \[Pi]\[Tau][x] . \[Pi]\[Tau][x]

then [Xi]'[x] give: $\left( \begin{array}{cc} \frac{i \left(\pi _0' x\right)}{2 f}-\frac{2 \left(\pi _+ x\right) \left(\pi _-' x\right)+2 \left(\pi _- x\right) \left(\pi _+' x\right)+2 \left(\pi _0 x\right) \left(\pi _0' x\right)}{8 f^2} & \frac{i \left(\pi _-' x\right)}{\sqrt{2} f} \\ \frac{i \left(\pi _+' x\right)}{\sqrt{2} f} & -\frac{i \left(\pi _0' x\right)}{2 f}-\frac{2 \left(\pi _+ x\right) \left(\pi _-' x\right)+2 \left(\pi _- x\right) \left(\pi _+' x\right)+2 \left(\pi _0 x\right) \left(\pi _0' x\right)}{8 f^2} \\ \end{array} \right)$

But $'$ is not four-dimensional derivative. And I don't know how one may arrive at the problem with more general consideration.

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    $\begingroup$ What Mathematica code have you tried? Please post it so the answer can build on the code you already have. $\endgroup$
    – Jens
    Feb 2 '17 at 18:05
  • $\begingroup$ @Jens I add the code. $\endgroup$
    – illuminato
    Feb 2 '17 at 21:30

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