Dirac equation, $\alpha_i$, $\beta$ hermitian The argument I've seen is the one given here: http://epx.phys.tohoku.ac.jp/~yhitoshi/particleweb/ptest-3.pdf under (3.10):
$$H=\vec{\alpha}\cdot(-i\vec{\nabla})+\beta m$$
$H$ is hermitian, $-i\vec{\nabla}$ is hermitian, so $\vec{\alpha},\beta$ are hermitian.
This is not convincing, because a hermitian operator being a sum of two operators does not imply that the two are also hermitian, for example in the harmonic oscillator, the position and momentum operators can be written as a sum or difference of the annihilation/creation operators, which are non-hermitian.
So is the presented argument wrong? Why are the $\alpha_i, \beta$ really hermitian?
 A: I think its important to say that the Dirac equation cannot be derived using standard QM: it can be motivated, but ultimalely the definitive argument for/against its correction is a matter of experimental confirmation.
This means that the arguments that the link gives you in favour of the equation are just formal: they might seem very logical and unavoidable, but if the resulting equation doesnt work, its wrong.
I think its important to remember this, as we novice physicists tend to forget/ignore the infinite number of proposed equation, which felt very logical at the time, but soon ruled out by experiments. Dirac equation remained because it works. $\boldsymbol \alpha,\beta$ are hermitian because the resulting equation works.
That being said, we can give some arguments as to why $\boldsymbol \alpha,\beta$ cannot be non-hermitian: if we want a Hamiltonian like
$$
H=\boldsymbol\alpha\cdot\boldsymbol p+m\beta
$$
to be hermitian, then we are forced to take both $\boldsymbol \alpha$ and $\beta$ hermitian. One way of proving this is to see that $H$ must be hermitian regardless of the value of $m$. In particular, $H$ must be hermitian for massless particles, so
$$
H_\text{massless}=\boldsymbol\alpha\cdot\boldsymbol p
$$
must be hermitian. But as $\boldsymbol p$ is hermitian, the three matrices $\boldsymbol \alpha$ must be hermitian. As $\boldsymbol \alpha$ is independent of $m$, $\boldsymbol\alpha$ must also be hermitian in the massive case:
$$
H=\boldsymbol\alpha\cdot\boldsymbol p+m\beta
$$
Finally, as we already proved, the first term in the expression above is hermitian, so if $H$ is to be hermitian, the so must the term $m\beta$. This proves that $\beta$ must be hermitian.
A: The Hilbert space in which the Dirac equation acts is a product of the infinite-dimensional space corresponding to position (or momentum) and a finite-dimensional spinor space, in which the matrices $\alpha_{j}$ and $\beta$ act.  We know the momentum operator is Hermitian, so we may look at a subspace for fixed $\vec{p}$; in order for $H$ to be Hermitian, the matrix $\vec{\alpha}\cdot\vec{p}+\beta m$ must be Hermitian in the finite-dimensional spinor space for every value of $\vec{p}$. This means that an arbitrary linear combination of the $\alpha_{j}$ and $\beta$ matrices is Hermitian, which is only possible if all four matrices are Hermitian themselves.
