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  Hermitian conjugate of spinors

+ 1 like - 0 dislike

In any textbook, hermitian conjugate of spinor is defined like $ \psi_{\alpha}^{+}=\bar\psi_{\dot{\alpha}} $ and $(\psi^{\alpha})^{+}=\bar{\psi}^{\dot\alpha}$. We have Pauli matrices $\sigma^{\mu}_{\alpha\dot\beta}$ and $(\bar\sigma^{\mu})^{\dot\alpha\beta}$ they are hermitian matrices i.e. $(\sigma^{\mu}_{\alpha\dot\beta})^{+}=\sigma^{\mu}_{\alpha\dot\beta}$ and $((\bar\sigma^{\mu})^{\dot\alpha\beta})^{+}=(\bar\sigma^{\mu})^{\dot\alpha\beta}$.

Now consider the known expression $$(\xi\sigma^{\mu}\bar\psi)^{+}=\psi\sigma^{\mu}\bar\xi$$

that is $(\xi^{\alpha}\sigma^{\mu}_{\alpha\dot\beta}\bar\psi^{\dot\beta})^{+}=(\bar\psi^{\dot\beta})^{+}(\sigma^{\mu}_{\alpha\dot\beta})^{+}(\xi^{\alpha})^{+}=\psi^{\beta}\sigma^{\mu}_{\alpha\dot\beta}\bar\xi^{\dot\alpha}=???$

there is no more possible to sum over indices at all. Where is the mistake?

And one more question: By definition hermitian conjugate of two spinors$(\psi\xi)^{+}=\xi^{+}\psi^{+}$. But how is defined $(P^{\mu}Q_{\alpha})^{+}$ where $P^{\mu}$ is momentum operator and $Q_{\alpha}$ is spinor. On which space ${"+"}$ acts?

This post imported from StackExchange Physics at 2015-02-11 11:52 (UTC), posted by SE-user Paramore
asked Nov 11, 2014 in Theoretical Physics by Paramore (5 points) [ no revision ]

1 Answer

+ 0 like - 0 dislike

The matrices $\sigma^{\mu}$, being hermitian, satisfy:$(\sigma^{\mu}_{\alpha\dot{\beta}})^{*}$=$( \sigma^{\mu *})_{\dot{\alpha}\beta}$=$(\sigma^{\mu\dagger})_{\beta\dot{\alpha}}$=$(\sigma^{\mu})_{\beta\dot{\alpha}}$. So $(\xi\sigma^{\mu}\bar\psi)^{*}$=$(\xi^{\alpha}\sigma^{\mu}$$_{\alpha\dot{\beta}}$$\bar{\psi}$$^{\dot{\beta}})$$^{*}$=$\dots$, the rest follows. See the Binetruy's book "Supersymmetry".

This post imported from StackExchange Physics at 2015-02-11 11:52 (UTC), posted by SE-user ioannis
answered Dec 1, 2014 by ioannis (10 points) [ no revision ]

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