# Deriving energy eigenstates of free Weyl field

I'm wondering if it's possible to derive energy eigenstates for a fermion field without guessing the anti-commutation relations from the start. I'm taking the Hamiltonian for a Weyl field $$\psi$$ to be $$H=\sum_k \sum_s H_{k,s}, \qquad \sum_s H_{k,s} = \psi_k^\dagger (\vec k \cdot \vec\sigma) \psi_k$$ For $$\vec k$$ and spin aligned with the $$z$$ axis, this becomes $$H_{k,+} = k (\psi_{k,1r}^2 + \psi_{k,1i}^2)$$ The conjugate momentum to $$\psi_k$$ is $$\pi_k = i \psi_k^\dagger$$, so the conjugate momenta for the components of $$\psi_{k,1}$$ would be $$\pi_{k,1r} + i \pi_{k,1i} = \psi_{k,1i} + i \psi_{k,1r}$$ I assume we need to pick either the real or imaginary component to take as our canonical coordinate, with the other becoming its conjugate momentum. Let's take $$\psi_{k,1r}$$. $$\hat H_{k,+} = 2 k \left( \frac12 \hat \pi_{1r}^2 + \frac12 \hat \psi_{1r}^2 \right)$$ The expression in parentheses looks like the Hamiltonian for a harmonic oscillator, which maybe is okay if there's some criterion that excludes all excited states above the first (leaving two states---occupied and unoccupied).

Is the derivation right so far? Or did we need to know ahead of time that $$\psi_{k,1r}$$ and $$\psi_{k,1i}$$ aren't like ordinary real numbers---or that $$\hat\pi_{k,1r} \not\equiv -i \, \partial/\partial\psi_{k,1r}$$?

Now I'm thinking that rather than picking one of $$\{\psi_{k,1r},\psi_{k,1i}\}$$ as the canonical coordinate, we should recognize that there's an ambiguity and leave it intact. For any $$0 \leq \alpha \leq 1$$, the Hamiltonian can be decomposed as: $$\hat H_{k,+} = k \left[ \alpha \left( - \frac{\partial^2}{\partial \psi_{k,1r}^2} + \hat\psi_{k,1r}^2 \right) + (1-\alpha) \left( - \frac{\partial^2}{\partial \psi_{k,1i}^2} + \hat\psi_{k,1i}^2 \right) \right]$$ The eigenstates are then $$\Psi_n(\psi_{k,1r},\psi_{k,1i}) = \exp(-\tfrac12 (\psi_{k,1r}^2 + \psi_{k,1i}^2)) H_n(\psi_{k,1r}) H_n(\psi_{k,1i})$$ where $$H_n$$ are Hermite polynomials---they are products of harmonic oscillator eigenstates with the same energy level. The states also need to respect the global phase symmetry $$\psi \to e^{i\phi} \psi$$ $$\psi_{k,1r} \to \psi_{k,1r} \cos\phi - \psi_{k,1i} \sin\phi \qquad \psi_{k,1r} \to \psi_{k,1i} \cos\phi + \psi_{k,1r} \sin\phi$$ This means that after the transformation, the wave function needs to still be an eigenstate of the Hamiltonian (for any choice of $$\alpha$$. The ground state $$\Psi_0$$ does respect the symmetry, since $$H_0(x) = 1$$ and $$(\psi_{k,1r}^2 + \psi_{k,1i}^2)$$ is invariant.
The first excited state $$\Psi_1$$ contains an overall factor $$\psi_{k,1r} \psi_{k,1i}$$, so if we later make a gauge choice $$\psi_{k,1i} \equiv 0$$, that state would be zero everywhere.
The key is to use the phases of the spinor components as the canonical coordinates, as that provides a clean division into coordinates and momenta (versus above where both the real and imaginary parts are integrated over in the action). The Lagrangian for a Fourier mode is $$\mathcal L(k) = i \psi_k^\dagger \sigma^\mu k_\mu \psi_k$$ Let's decompose the spinor as magnitude-phase: $$\psi_k = \begin{pmatrix} R_1 e^{-i\phi_1} \\ R_2 e^{-i\phi_2} \end{pmatrix}$$ In the action we will have an integral over $$i\psi^\dagger \dot\psi \,\mathrm dt = i \psi^\dagger \mathrm d \psi$$. Keeping the real part, for the top component we have $$\mathrm{Re}[i (R_1 e^{i\phi_1})(\mathrm d R_1 \, e^{-i\phi_1} - i R_1 \mathrm d\phi_1 e^{-i\phi_1})] = R_1^2 \, \mathrm d \phi_1$$ So the action looks like $$S = \int R_1^2 \, \mathrm d \phi_1 + \int R_2^2 \, \mathrm d \phi_2 - \int H \, \mathrm d t$$ For a $$k$$ whose spatial part is oriented along the $$+z$$ axis, the Hamiltonian is $$H_k = k (R_1^2 - R_2^2)$$ The magnitude-squared of each spinor component is the conjugate momentum to the phase angle $$\phi_1$$ or $$\phi_2$$. $$\hat{(R_1^2)} = -i\hbar \frac{\partial}{\partial\phi_1} \qquad \hat{(R_1)^2} = -i \hbar \frac{\partial}{\partial\phi_2}$$ $$\hat H = -i\hbar k \left( \frac{\partial}{\partial\phi_1} - \frac{\partial}{\partial\phi_2} \right)$$ We require that $$\Psi(2\pi) = \Psi(0)$$, which limits valid eigenfunctions of $$\hat H_k$$ to $$\Psi_{m,n}(\phi_1,\phi_2) = \exp \left( \frac{i}{\hbar} (m \phi_1 + n \phi_2) \right)$$ for $$m,n \in \mathbb Z$$. The magnitude-squared cannot be negative. Its eigenvalues are $$r_1^2 = m \qquad r_2^2 = n$$ so we are constrained to those solutions with $$m \geq 0, \, n \geq 0$$, with energy eigenvalues $$E_{m,n} = \hbar k(m-n)$$ At this point, it seems that we can have any number of particles with positive or negative energy contributions of $$\hbar k$$. If $$m$$ and $$n$$ are unbounded, then the Hamiltonian is not bounded from below. Only integer values are allowed for the $$R^2$$'s, meaning that (if we take the magnitude to be nonnegative) $$R_1,R_2 \in \{0,1,\sqrt{2},\sqrt{3},\cdots\}$$ For the $$0,1$$ values, we have that $$R = R^2$$, which actually lets us translate the $$\hat H_k$$ to a second order Hamiltonian. Hamiltonians like to be quadratic in the momenta. $$\hat H = \hbar k \left( -\frac{\partial^2}{\partial\phi_1^2} + \frac{\partial^2}{\partial\phi_2^2} \right)$$ For these special values, it is also possible to represent other parameterizations of the spinor with finitely many terms. For example, the real part of the first component of $$\psi$$: $$\psi_{1,R} = \cos(\phi) \, \sqrt{R^2}$$ The square root function does not have a Taylor series at $$0$$ (already a red flag?), so let's take it at $$1$$. $$\psi_{1,R} = \cos(\phi) \left[ 1 + \frac{R^2-1}{2} - \frac{(R^2-1)^2}{8} + \cdots \right]$$ $$\hat\psi_{1,R} = \cos(\phi) \left[ 1 + \frac{1}{2}\left( -i\hbar\frac{\partial}{\partial\phi_1} - 1 \right) - \frac{1}{8}\left( -\hbar^2 \frac{\partial^2}{\partial\phi_1^2} + 2 i\hbar \frac{\partial}{\partial\phi_1} + 1 \right) + \cdots \right]$$ If we allow $$\lvert{m}\rvert,\lvert{n}\rvert > 1$$, then we need to take infinitely many derivatives just to pull out the components of the spinor. For $$R_1^2 \in \{0,1\}$$, this is simply $$\hat\psi_{1,R} = -i\hbar\cos(\phi)\, \frac{\partial}{\partial\phi_1}$$ I don't know if this is sufficient reason to exclude all $$m,n > 1$$, but I find it convincing enough. With just two states $$m = 0$$ and $$m=1$$ for each wave number and spin orientation, the creation/annihilation operators can be expressed as $$2\times2$$ matrices $$\hat b^\dagger = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \qquad \hat b = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$$ which exhibit the anti-commutation relations $$bb^\dagger + b^\dagger b = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \quad bb = b^\dagger b^\dagger = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$$