# Is it possible to decompose into eigenstates of Dirac Hamiltonian?

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?

• The same as the decomposition of $L^2(\mathbb{R}^3,\mathbb{R})$ into eigenstates of the free Hamiltonian $H\propto p^2$. – ACuriousMind Feb 20 '16 at 20:33
• @AccidentalFourierTransform: I believe OP's problem is that the solutions are plane waves which are not square-integrable. – ACuriousMind Feb 20 '16 at 20:34
• @ACuriousMind I am unfamiliar with this decomposition, which I would also say does not exist. I know that sometimes we extend the Hilbert space with distrubtions that are not in it to be able to write certain formulas easier, but a decomposition $\mathcal H = \mathcal H_+ \oplus \mathcal H_-$ is something I have never done when the eigenstates are not part of the Hilbert space. – s.harp Feb 20 '16 at 20:39
• I'll write an answer. – ACuriousMind Feb 20 '16 at 20:39
• $\mathcal{H}_{\pm}$ are the eigenspaces corresponding to eigenvalues $\pm 1$ of the operator $\hat{H} = H/\sqrt{\textbf{p}^2+m^2}$. I'm not sure whether this operator is well-behaved mathematically. – higgsss Feb 20 '16 at 21:28

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.
• Generally, $\gamma^5$ doesn't commute with the Dirac Hamiltonian $H$. Even in the massless limit, in which $[\gamma^5,H]=0$, each eigenspace of $\gamma^5$ contains both positive- and negative-energy solutions of $H$. That is, there are positive- and negative-energy solutions with either handedness. – higgsss Feb 20 '16 at 21:21
• Looking back I no longer think that this is the best way to go about it. While I am not too familiar with Rigged Hilbert spaces, I think they are the wrong approach. The spectral theorem for unbounded self-adjoint operators gives you a thing called a resolution of identity, that is some projections $E_\lambda$, $\lambda\in\Bbb R$ so that the operator is equal to $\int_{\Bbb R} \lambda dE_\lambda$ on its domain (modulo details about what kind of convergence we are talking about etc). The split of the Hilbert space into positive modes will the subspace associated to... – s.harp Apr 5 '17 at 20:53
• ... $\Bbb1-E_0=\int_0^\infty dE_\lambda$ and the one for the negative modes will be the space corresponding to $E_0=\int_{-\infty}^0dE_\lambda$. In my opinion this is the most natural way to understand a statement like "decompose the Hilbert space into the positive and negative eigenstates of a self-adjoint (not necessarily bounded) operator". – s.harp Apr 5 '17 at 20:54