Dirac equations (5)

In the Dirac representation (the standard representation) $$\begin{align} \alpha^{i } = \begin{pmatrix} 0 & \sigma^{i } \\ \sigma^{i} & 0 \end{pmatrix} \; , \quad \beta = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \; , \quad \pmb{\gamma } = \begin{pmatrix} 0 & \pmb{\sigma} \\ - \pmb{\sigma } & 0 \end{pmatrix} \end{align}$$ and the boost a frame in the ${\bf n}$ direction, $$\begin{align} S( \Lambda ) = \cosh{ \frac{\varphi}{2} } I - \pmb{\alpha } \cdot {\bf n} \sinh{ \frac{ \varphi }{2} } \end{align}$$ Then making use of $\cosh{ \varphi } = \gamma = \frac{\epsilon }{m}$ and $\sinh{\phi} = \beta \gamma$, $$\begin{align} \cosh{ \frac{\varphi }{2} } & = \sqrt{ \frac{ \cosh{\varphi} + 1 }{2} } = \sqrt{ \frac{ \epsilon + m }{2m} } \\ \sinh{ \frac{\varphi }{2} } & = \sqrt{ \frac{ \cosh{\varphi} - 1 }{2} } = \sqrt{ \frac{ \epsilon - m }{2m} } \end{align}$$ The boost transformation is $$\begin{align} S(\Lambda ) = \begin{pmatrix} \cosh{\varphi} & - \pmb{ \sigma } \cdot {\bf n} \sinh{\varphi } \\ - \pmb{ \sigma } \cdot {\bf n} \sinh{\varphi } & \cosh{\varphi } \end{pmatrix} = \begin{pmatrix} \sqrt{ \frac{\epsilon + m}{2m} } & - \pmb{\sigma } \cdot {\bf n} \sqrt{ \frac{\epsilon - m}{2m} } \\ - \pmb{\sigma } \cdot {\bf n} \sqrt{ \frac{\epsilon - m}{2m} } & \sqrt{ \frac{\epsilon + m}{2m} } \end{pmatrix} \end{align}$$ Use $(-1) \times {\bf n}'$ for boosting from the rest frame of a particle into the frame of an observer. Using $\psi (x') = S( \Lambda) \psi (x)$, obtain the plane wave solutions. I have still a question: the Dirac field $\phi$ has the dimension $[ \psi ] = L^{ 1/2 }$ and we can see from the commutation relation that $[b_{\alpha } ] = [d_{\alpha }] = L^{1/2}$. Therefore, can we confirm that $[u] = [v] = L^{0}$? (If this is true, it is consistent.) $$\begin{align} \psi (x) & = \int \frac{ d {\bf k} }{ (2 \pi)^{3} } \frac{m }{ k_{0 } } \sum_{ \alpha = 1,2 } \left[ b_{\alpha } (k ) u^{ (\alpha ) } (k ) e^{- i k \cdot x} + d^{\dagger }_{\alpha } (k ) v^{ ( \alpha ) } (k) e^{ik \cdot x } \right] \\ \overline{ \psi } (x) & = \int \frac{ d { \bf k } }{ ( 2 \pi )^{3} } \frac{m }{ k_{0 } } \sum_{ \alpha = 1, 2 } \left[ b^{\dagger }_{ \alpha } (k ) \overline{u }^{ (\alpha ) } (k ) e^{ i k \cdot x} + d^{ }_{\alpha } (k ) \overline{v }^{ ( \alpha ) } (k) e^{- ik \cdot x } \right] \end{align}$$ Anticommutation relations are $$\begin{align} \lbrace b_{\alpha }(q) , b^{ \dagger }_{\beta } (q' ) \rbrace & = (2 \pi)^{3} \frac{ k_{0} }{ m} \delta ( {\bf q} - {\bf q}' ) \delta_{\alpha \beta } \\ \lbrace d_{\alpha }(q) , d^{ \dagger }_{\beta } (q' ) \rbrace & = (2 \pi)^{3} \frac{ k_{0} }{ m} \delta ( {\bf q} - {\bf q}' ) \delta_{\alpha \beta } \end{align}$$ and all other anticommutators give zero. $$\begin{align} \lbrace \psi_{\xi } (t, {\bf x}) , \psi^{\dagger }_{\eta } ( t , {\bf y} ) \rbrace = \delta_{ \xi \eta } \delta ( {\bf x} - {\bf y } ) \end{align}$$ $$\begin{align} \lbrace \psi_{a} (x) , \overline{\psi}_{b } (x') \rbrace = (i \gamma^{\mu } \partial_{ \mu } + m )_{ab} i \Delta (x- x') \end{align}$$