- #1

welcomeblack

- 13

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Hi all, I've been playing around with spin 1/2 Lagrangians, and found the very interesting

Fierz identities. In particular for the S x S product,

[itex]

(\bar{\chi}\psi)(\bar{\psi}\chi)=\frac{1}{4}(\bar{\chi} \chi)(\bar{\psi} \psi)+\frac{1}{4}(\bar{\chi}\gamma^{\mu}\chi)(\bar{\psi}\gamma_{\mu} \psi)-\frac{1}{4}(\bar{\chi}\sigma^{\mu\nu}\chi)(\bar{\psi}\sigma_{\mu\nu} \psi)-\frac{1}{4}(\bar{\chi}\gamma^{\mu}\gamma^{5}\chi)(\bar{\psi}\gamma_{\mu}\gamma_{5} \psi)+\frac{1}{4}(\bar{\chi}\gamma^{5}\chi)(\bar{\psi}\gamma_{5} \psi)

[/itex]

I assume this is valid for any spinors chi and psi. If I then set chi equal to psi,

[itex]

(\bar{\psi}\psi)(\bar{\psi}\psi)=\frac{1}{4}(\bar{\psi} \psi)(\bar{\psi} \psi)+\frac{1}{4}(\bar{\psi}\gamma^{\mu}\psi)(\bar{\psi}\gamma_{\mu} \psi)-\frac{1}{4}(\bar{\psi}\sigma^{\mu\nu}\psi)(\bar{\psi}\sigma_{\mu\nu} \psi)-\frac{1}{4}(\bar{\psi}\gamma^{\mu}\gamma^{5}\psi)(\bar{\psi}\gamma_{\mu}\gamma_{5} \psi)+\frac{1}{4}(\bar{\psi}\gamma^{5}\psi)(\bar{\psi}\gamma_{5} \psi)

[/itex]

Since adjoint*spinor is a scalar, I can divide by psibar*psi and rearrange to get

[itex]

\bar{\psi}\psi=\frac{1}{\bar{\psi}\psi}[\frac{1}{3}(\bar{\psi}\gamma^{\mu}\psi)(\bar{\psi}\gamma_{\mu} \psi)-\frac{1}{3}(\bar{\psi}\sigma^{\mu\nu}\psi)(\bar{\psi}\sigma_{\mu\nu} \psi)-\frac{1}{3}(\bar{\psi}\gamma^{\mu}\gamma^{5}\psi)(\bar{\psi}\gamma_{\mu}\gamma_{5} \psi)+\frac{1}{3}(\bar{\psi}\gamma^{5}\psi)(\bar{\psi}\gamma_{5} \psi)]

[/itex]

Plugging this in to the Dirac Lagrangian,

[itex]

\mathcal{L}=i\bar{\psi}\gamma^{\mu}\partial_{\mu}\psi - m\bar{\psi}{\psi} \\

\mathcal{L}=i\bar{\psi}\gamma^{\mu}\partial_{\mu}\psi - \frac{m}{\bar{\psi}\psi}[\frac{1}{3}(\bar{\psi}\gamma^{\mu}\psi)(\bar{\psi}\gamma_{\mu} \psi)-\frac{1}{3}(\bar{\psi}\sigma^{\mu\nu}\psi)(\bar{\psi}\sigma_{\mu\nu} \psi)-\frac{1}{3}(\bar{\psi}\gamma^{\mu}\gamma^{5}\psi)(\bar{\psi}\gamma_{\mu}\gamma_{5} \psi)+\frac{1}{3}(\bar{\psi}\gamma^{5}\psi)(\bar{\psi}\gamma_{5} \psi)]

[/itex]

Is everything I've done mathematically allowed? How about physically allowed? If the two representations of psibar*psi are equivalent, shouldn't they give back the same equations of motion?

Fierz identities. In particular for the S x S product,

[itex]

(\bar{\chi}\psi)(\bar{\psi}\chi)=\frac{1}{4}(\bar{\chi} \chi)(\bar{\psi} \psi)+\frac{1}{4}(\bar{\chi}\gamma^{\mu}\chi)(\bar{\psi}\gamma_{\mu} \psi)-\frac{1}{4}(\bar{\chi}\sigma^{\mu\nu}\chi)(\bar{\psi}\sigma_{\mu\nu} \psi)-\frac{1}{4}(\bar{\chi}\gamma^{\mu}\gamma^{5}\chi)(\bar{\psi}\gamma_{\mu}\gamma_{5} \psi)+\frac{1}{4}(\bar{\chi}\gamma^{5}\chi)(\bar{\psi}\gamma_{5} \psi)

[/itex]

I assume this is valid for any spinors chi and psi. If I then set chi equal to psi,

[itex]

(\bar{\psi}\psi)(\bar{\psi}\psi)=\frac{1}{4}(\bar{\psi} \psi)(\bar{\psi} \psi)+\frac{1}{4}(\bar{\psi}\gamma^{\mu}\psi)(\bar{\psi}\gamma_{\mu} \psi)-\frac{1}{4}(\bar{\psi}\sigma^{\mu\nu}\psi)(\bar{\psi}\sigma_{\mu\nu} \psi)-\frac{1}{4}(\bar{\psi}\gamma^{\mu}\gamma^{5}\psi)(\bar{\psi}\gamma_{\mu}\gamma_{5} \psi)+\frac{1}{4}(\bar{\psi}\gamma^{5}\psi)(\bar{\psi}\gamma_{5} \psi)

[/itex]

Since adjoint*spinor is a scalar, I can divide by psibar*psi and rearrange to get

[itex]

\bar{\psi}\psi=\frac{1}{\bar{\psi}\psi}[\frac{1}{3}(\bar{\psi}\gamma^{\mu}\psi)(\bar{\psi}\gamma_{\mu} \psi)-\frac{1}{3}(\bar{\psi}\sigma^{\mu\nu}\psi)(\bar{\psi}\sigma_{\mu\nu} \psi)-\frac{1}{3}(\bar{\psi}\gamma^{\mu}\gamma^{5}\psi)(\bar{\psi}\gamma_{\mu}\gamma_{5} \psi)+\frac{1}{3}(\bar{\psi}\gamma^{5}\psi)(\bar{\psi}\gamma_{5} \psi)]

[/itex]

Plugging this in to the Dirac Lagrangian,

[itex]

\mathcal{L}=i\bar{\psi}\gamma^{\mu}\partial_{\mu}\psi - m\bar{\psi}{\psi} \\

\mathcal{L}=i\bar{\psi}\gamma^{\mu}\partial_{\mu}\psi - \frac{m}{\bar{\psi}\psi}[\frac{1}{3}(\bar{\psi}\gamma^{\mu}\psi)(\bar{\psi}\gamma_{\mu} \psi)-\frac{1}{3}(\bar{\psi}\sigma^{\mu\nu}\psi)(\bar{\psi}\sigma_{\mu\nu} \psi)-\frac{1}{3}(\bar{\psi}\gamma^{\mu}\gamma^{5}\psi)(\bar{\psi}\gamma_{\mu}\gamma_{5} \psi)+\frac{1}{3}(\bar{\psi}\gamma^{5}\psi)(\bar{\psi}\gamma_{5} \psi)]

[/itex]

Is everything I've done mathematically allowed? How about physically allowed? If the two representations of psibar*psi are equivalent, shouldn't they give back the same equations of motion?

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