Deriving energy eigenstates of free Weyl field

Question

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 $\hat\pi_{k,1r} \not\equiv -i \, \partial/\partial\psi_{k,1r}$?

EDIT: continuing the thread

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.

Answer 1

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 $$ bb^\dagger + b^\dagger b = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \quad bb = b^\dagger b^\dagger = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} $$