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I've just learned that the Dirac Hamiltonian in the Dirac Equation is given by $H = c\alpha \cdot \overrightarrow{p} + \beta mc^2$ and $\alpha$ is a 4 X 4 represented as $\begin{bmatrix} 0 & \sigma \\ \sigma & 0 \end{bmatrix}$ where the $\sigma$s are Pauli Spin matrices. We know that $\overrightarrow{p} = -i\hslash \nabla$ is a 3 vector/operator. I dont understand this product between a 4X4 matrix and a 3 vector.

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You missed the vector arrow on $\vec{\alpha}$. So $\vec{\alpha}$ is a "vector" having 3 components ($\alpha_x,\alpha_y,\alpha_z$}. And each of these 3 is a $4\times 4$ matrix.

Or more explicitly: $$\alpha_x=\begin{pmatrix} 0 & \sigma_x \\ \sigma_x & 0 \end{pmatrix}$$ $$\alpha_y=\begin{pmatrix} 0 & \sigma_y \\ \sigma_y & 0 \end{pmatrix}$$ $$\alpha_z=\begin{pmatrix} 0 & \sigma_z \\ \sigma_z & 0 \end{pmatrix}$$

The Dirac Hamiltonian $$H=c\vec{\alpha}\cdot\vec{p}+\beta mc^2$$ is short-hand notation for $$H=c(\alpha_x p_x+\alpha_y p_y+\alpha_z p_z)+\beta mc^2$$

So finally the Hamiltonian is a $4\times 4$ matrix where each of its components is a differential operator. You can apply this thing to a $4$-component spinor field \begin{pmatrix} \psi_1(\vec{r},t) \\ \psi_2(\vec{r},t) \\ \psi_3(\vec{r},t) \\ \psi_4(\vec{r},t) \end{pmatrix} to obtain another $4$-component spinor field.

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  • $\begingroup$ Yeah, I actually didn't use the vector sign because the idea of a vector having matrices as it's components was bothering me ;p. I get it that it's not a three vector. So do the derivatives in the momentum operators just act individually in all the components of the spinor? Also, could you help me with resources that are more mathematically complete since the notations are really putting me off?(Idk if that makes sense) $\endgroup$ Aug 11, 2021 at 15:15
  • $\begingroup$ @NabaneetSharma Yes, the spatial derivatives act equally on the individual spinor components. $\endgroup$ Aug 11, 2021 at 15:35
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Shortly $\alpha$ is a vector of $4\times 4$ matrices, it is better denoted by $\vec{\alpha}$ perhaps, then the structure is more evident:

$$\vec{\alpha} \cdot \vec{p} = \alpha_x p_x + \alpha_y p_y + \alpha_z p_z$$

where $$\alpha_{i} = \begin{pmatrix} 0 & \sigma_{i} \\ \sigma_{i} & 0 \end{pmatrix}$$ for $i\in \{x,y,z\}$

At the end of the day you have a matrix of size $4\times 4$ acting on spinors as the other answer describes.

The Hamiltonian given by the OP is probably not the most enlightening version since it does not display the Lorentz invariance of the Dirac equation.

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