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Let's see...that's a good revision exercise: 1) To begin with, one should define the quantities involved, namely the variance or its square root the standard deviation (As I see it, the framework is that the expectation value is the evaluation of a state $|\psi\rangle\in\mathcal{H}$ on an observable $A\in\mathcal{B}(\mathcal{H})$ self-adjoint, i.e. ... 2 There is no extra requirement for theses terms to vanish, everything works out if the calculation is done correctly. The most probable thing to forget is the product rule for the derivatives: \begin{align} ... 0 The Lagrangian is a Lorentz scalar. The terms you refer to show up like\phi(\partial_{\mu} \phi)$, these are Lorentz vectors and cannot show up in the Lagrangian. All vector indices are contracted as in$\phi(\partial_{\mu} \phi)(\partial^{\mu} \phi)$. 0 If you define your gamma matrices as the block matrices: $$\gamma^0:=\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}; \quad \gamma^i:=\begin{bmatrix} 0 & \sigma^i \\ -\sigma^i & 0 \end{bmatrix}$$ Then you can also define the block matrices ... 0 You don't define$\overrightarrow{\sigma}$. It looks like neither do Itzykson/Zuber, or, rather, it is defined there as a vector with Pauli matrices as components (if I am not mistaken). However, those matrices are 2x2, whereas all other matrices in your equation are 4x4. Therefore, such definition seems incompatible with your equation. I believe you should ... 0 Keep in mind, that in position space (where$\hat p$is proportional to$\partial_x$, you always have to think of the commutator as acting on a wave function. It is$\$ \langle x| \hat p A(\hat x) |\psi\rangle = \int dx^\prime \langle x|\hat p|x^\prime\rangle \langle x^\prime|A(\hat x) |\psi\rangle \\ = \int dx^\prime \delta(x-x^\prime) ...