Is it possible to decompose into eigenstates of Dirac Hamiltonian?

Question

If we have the Hilbert space $\mathcal H = L^2(\mathbb R^3, \mathbb C^4)$ and a Hamiltonian:

$$H=\gamma^i p_i + m \gamma^0$$

where $\gamma^i$ are matrices and $\{\gamma^i,\gamma^j\}=\delta^{ij}$. A statement I found in a book is that the Hilbert space decomposes into the orthogonal sum

$$\mathcal H = \mathcal H_+ \oplus \mathcal H_-$$

Where $\mathcal H_\pm$ is given by the positive/negative (including zero) eigenspaces of the Dirac Hamiltonian.

As far as I can tell, the solutions of the Dirac equation are not in $L^2$. Is this correct?

What is meant then with this decomposition?

Answer 1

1. It is generally the case that the "eigenstates" $\lvert p\rangle$ of free Hamiltonians (and free states in general, also called scattering states) are not inside the Hilbert space, but only inside a larger distributional space. They "span" the actual Hilbert space in the sense that all states in it are obtained as "wavepackets" $$ \int f(p)\lvert p\rangle \mathrm{d}p $$ for some square-integrable function $f$. For more on this, see Are all scattering states un-normalizable?, Why are eigenfunctions which correspond to discrete/continuous eigenvalue spectra guaranteed to be normalizable/non-normalizable?, Rigged Hilbert space and QM and links therein.

2. The decomposition of the Dirac spinor space into the positive/negative eigenspaces is then gotten as the spaces where the function $f$ has support only on the respective part of the modes.