Transforming a QFT identity (quantisation of the Dirac field)

Question

When (falsely) quantizing the Dirac-field, Peskin/Schroeder (Introduction to Quantum Field Theory) get with $$\psi(\vec{x})=\int\dfrac{d^3p}{(2\pi)^3}\dfrac{1}{\sqrt{2E_{\vec{p}}}}e^{i\vec{p}\cdot\vec{x}}\sum\limits_{s=1,2}\left(a_{\vec{p}}^su^s(\vec{p})+b_{-\vec{p}}^sv^s(-\vec{p})\right)$$ and $$ \left[a_{\vec{p}}^r,~a_{\vec{q}}^{s\dagger}\right] = \left[b_{\vec{p}}^r,~b_{\vec{q}}^{s\dagger}\right] = (2\pi)^3\delta^{(3)}(\vec{p}-\vec{q})\delta^{rs}$$ the relationship $$\left[\psi(\vec{x}),~\psi^{\dagger}(\vec{y})\right] = \int\dfrac{d^3p~d^3q}{(2\pi)^6}\dfrac{1}{\sqrt{2E_{\vec{p}}2E_{\vec{q}}}}e^{i(\vec{p}\cdot\vec{x}-\vec{q}\cdot\vec{y})}\times\sum\limits_{r,s}\left(\left[a_{\vec{p}}^r,~a_{\vec{q}}^{s\dagger}\right]u^r(\vec{p})\bar{u}^s(\vec{q})+\left[b_{-\vec{p}}^r,~b_{-\vec{q}}^{s\dagger}\right]v^r(-\vec{p})\bar{v}^s(-\vec{q})\right)\gamma^0$$

This is all easy to see but I am wondering where the $\gamma^0$ at the end comes from?

Answer 1

$$\psi(\vec{x})=\int\dfrac{d^3p}{(2\pi)^3}\dfrac{1}{\sqrt{2E_{\vec{p}}}}e^{i\vec{p}\cdot\vec{x}}\sum\limits_{s=1,2}\left(a_{\vec{p}}^su^s(\vec{p})+b_{-\vec{p}}^sv^s(-\vec{p})\right)$$ so $$\psi^{\dagger}(\vec{y})=\int\dfrac{d^3q}{(2\pi)^3}\dfrac{1}{\sqrt{2E_{\vec{q}}}}e^{-i\vec{q}\cdot\vec{y}}\sum\limits_{s=1,2}\left(a_{\vec{q}}^{\dagger s}u^{\dagger s}(\vec{q})+b_{-\vec{q}}^{\dagger s}v^{\dagger s}(-\vec{q})\right)$$

This gives the relationship $$\left[\psi(\vec{x}),~\psi^{\dagger}(\vec{y})\right] = \int\dfrac{d^3p~d^3q}{(2\pi)^6}\dfrac{1}{\sqrt{2E_{\vec{p}}2E_{\vec{q}}}}e^{i(\vec{p}\cdot\vec{x}-\vec{q}\cdot\vec{y})}\times\sum\limits_{r,s}\left(\left[a_{\vec{p}}^r,~a_{\vec{q}}^{s\dagger}\right]u^r(\vec{p})u^{\dagger s}(\vec{q})+\left[b_{-\vec{p}}^r,~b_{-\vec{q}}^{s\dagger}\right]v^r(-\vec{p})v^{\dagger s}(-\vec{q})\right)$$

Now, $$\gamma^0 = \begin{pmatrix} I & 0 \\ 0 & -I \end{pmatrix}$$ and so we can define the Dirac adjoint of $u$, $\bar{u} = u^{\dagger}\gamma^0$ which also gives $u^{\dagger} = \bar{u}\gamma^0$ as $\left(\gamma^0\right)^{-1} = \gamma^0$ and we obtain the required result

Answer 2

Because $\bar u \gamma^0 = u^\dagger$