Proving the Gamma Matrix Identity in QFT: Is There a Mistake in My Attempt?

[Catria]

Homework Statement

Prove that \gamma^{a}\gamma^{b}\gamma^{c}\gamma^{d}\gamma^{e}\gamma_{a} = 2\left(\gamma^{e}\gamma^{b}\gamma^{c}\gamma^{d}+\gamma^{d}\gamma^{c} \gamma^{b}\gamma^{e}\right)

Each of the \gamma^{i}s are as used in the Dirac equation.

Homework Equations

\gamma^{a}\gamma^{b}\gamma^{c}\gamma^{d}\gamma_{a} = -2\gamma^{d}\gamma^{c}\gamma^{b}

\gamma^{a}\gamma^{b} + \gamma^{b}\gamma^{a} = 2g^{ab}

The Attempt at a Solution

\gamma^{a}\gamma^{b}\gamma^{c}\gamma^{d}\gamma^{e}\gamma_{a} = 2g^{ab}\gamma^{c}\gamma^{d}\gamma^{e}\gamma_{a} - \gamma^{b}\gamma^{a}\gamma^{c}\gamma^{d}\gamma^{e}\gamma_{a}

= 2\gamma^{c}\gamma^{d}\gamma^{e}\gamma^{b} + 2\gamma^{b}\gamma^{e}\gamma^{d}\gamma^{c}

Perhaps I mixed up something or there is a typo...

 

[TSny]

I think what you wrote so far is correct. But, as you can see, it is not yet very close to what you want.

Instead, start over and try commuting $\gamma^{e}$ with $\gamma_{a}$. To do this, you will need to use the identity obtained by lowering $a$ in the identity \gamma^{a}\gamma^{b} + \gamma^{b}\gamma^{a} = 2g^{ab}

 

[Catria]

Thank you.