Hermitian conjugate of spinors

Question

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?

Answer 1

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".

Answer 2

$(\xi^{\alpha}\sigma^{\mu}_{\alpha\dot\beta}\bar\psi^{\dot\beta})^\dagger=(\xi^{\alpha}\sigma^{\mu}_{\alpha\dot\beta}\bar\psi^{\dot\beta})^*=(\sigma^{\mu}_{\alpha\dot\beta}\bar\psi^{\dot\beta})^*(\xi^\alpha)^*=(\bar{\psi}^{\dot\beta})^*(\sigma^{\mu}_{\alpha\dot\beta})^*\bar\xi^{\dot\alpha}=\psi^{\beta}\sigma^{\mu}_{\beta\dot\alpha}\bar\xi^{\dot\alpha}$