Can a wave function that represents a particle of spin that is not 1/2 be a solution of the Dirac equation?

Question

During a QFT course, we were deriving the Dirac equation using the relativistic quantum mechanics' approach. Dirac was well aware of the Klein-Gordon $$\frac{1}{c^2}\frac{\partial^2}{\partial t^2}\psi-\nabla^2\psi +\frac{m^2c^2}{\hbar^2}\psi=0$$ which can be derivated from the energy-momentum relation of a free particle \begin{equation}E^2=m^2c^4+p^2c^2\end{equation} and that is somewhat of a generalization of the Schrodinger equation to the theory of special relativity. However, as one learn in a Quantum Mechanics' course, the Klein-Gordon equation leads to many absurd results, which mostly come from the fact that it is not a first order differential equation . Thus, Dirac suggested that a wave function had to respect the equation $$(\alpha \cdot p +\beta mc^2)\psi=E\psi$$ which is linear in $\frac{\partial}{\partial t}$ and $\nabla$, where $p=-i\hbar \nabla$ and $E=i\hbar \frac{\partial}{\partial t}$, and where $\alpha$ and $\beta$ are unknown at this point. Since the equation has to respect the energy-momentum relation, the form of $\alpha$ and $\beta$ are restricted. To establish the form of $\alpha$ and $\beta$, we can multiply the left hand side by itself and the right hand side by itself, and then require that the energy-momentum relation holds. We thus optain that the following equations have to be respected: $$\alpha_i \alpha_j+\alpha_j \alpha_i=2\delta_{ij}$$ $$\alpha_i \beta +\beta a_i=0$$ $$\beta^2=1$$ The choice of $\alpha_i$ and $\beta$ is not unique. Moreover, the dimension of these objects is not unique either. In my course, we chose the following solution: $$\alpha_i=\begin{pmatrix} 0& \sigma_i\\ \sigma_i &0 \end{pmatrix}$$ $$\beta= \begin{pmatrix} 1 & 0\\ 0 & -1\end{pmatrix}$$ where $\sigma_i$ are the Pauli matrices. $\sigma_i$ and $\beta$ are thus four dimensional matrices. It turns out that with the Dirac equation and this particular choice, the wave function which is a solution of the equation represents a particle with spin 1/2. My question is: if we had chosen a different solution for $\alpha_i$ and $\beta$ with matrices of higher dimensions, would we then possibly have, as a solution of the Dirac equation, wave functions that representation particles of spins that are different than 1/2?

Answer 1

You can absolutely represent different kinds of particles if you choose different $\alpha$ or $\beta$ in your first-order wave equation. As a trivial example, consider using $8 \times 8$ block diagonal matrices with $\alpha$ (or $\beta$) in the top-left and bottom-right blocks. That represents a pair of spin $1/2$ particles, which could have spin $0$ or $1$. By applying an additional constraint, you could get rid of one of those components if desired.

A more serious example, which uses $16 \times 16$ matrices, is the Rarita-Schwinger equation for spin $3/2$ particles. In this case, you need to apply even more additional constraints to get just the spin $3/2$ component. You can go even further and write down similar wave equations for arbitrarily high spin (though at this point nobody would write them in terms of $\alpha$'s and $\beta$'s).

Answer 2

This is a comment.

The Dirac equation as defined, has wavefunctions that can model spin 1/2 particles.

Any changes define a different differential equation and it has to be checked what the solutions are good for. For example:

In theoretical physics, the Rarita–Schwinger equation is the relativistic field equation of spin-3/2 fermions. It is similar to the Dirac equation for spin-1/2 fermions. This equation was first introduced by William Rarita and Julian Schwinger in 1941.

Answer 3

You can write massive spin-0 and spin-1 wave equations using the Dirac equation with the standard Dirac matrices, but with the fields occupying different subspaces of the Clifford algebra than the Dirac field does.

The trick is essentially the same as that used to turn the second-order harmonic oscillator equation $\phi'' = -k^2\phi$ into the first-order equation $iψ' = kψ$, where $ψ = \phi + i\phi'/k$.

The harmonic oscillator equation is the 0+1 dimensional massive Klein-Gordon equation. Complexifying the field doesn't work in 1+1 or more dimensions because there isn't enough room for the gradient, but Cliffordizing it works in any number of dimensions.

If the scalar field is real, and you pick the spacetime signature that gives $(γ^0)^2 = -1$, then you can use the Clifford algebra over the (mathematical) field of the (physicists') field, either $\mathbb R$ or $\mathbb C$. In the case of a real field in 0+1 dimensions, the Clifford algebra is isomorphic to $\mathbb C$ and you get the same complex-number solution as above.

If you pick $(γ^0)^2 = +1$ then you must complexify the algebra even if the field is real. I wish Dirac hadn't chosen this option, because it obscures the fact that the Clifford algebra essentially functions as a generalization of the complex numbers, not an additional structure on top of them. But I'll stick with Dirac's algebra from here on. If you want to use the other signature then just take $i\to 1$ below.

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The massive Klein-Gordon equation can be written as a pair of first-order equations, $$\begin{eqnarray} \partial_μ \phi &=& A_μ \\ \partial^μ\! A_μ &=& -m^2\phi \end{eqnarray}$$ and if you define $ψ = \phi + i\rlap{\,/}A/m$, then these can be combined into the single first-order equation $i \rlap/\partial ψ = mψ$.

You can pull a similar trick with the massive spin-1 (Proca) equation $$\partial^μ (\partial_μ B_ν - \partial_ν B_μ) = -m^2 B_ν$$ because $\rlap/\partial$ applied to a vector results in its divergence as a scalar (here zero) plus its curl as a bivector, and $\rlap/\partial^2 = \partial^2$. You can write the Proca equation as a pair of first-order equations, $$\begin{eqnarray} \partial_μ B_ν - \partial_ν B_μ &=& F_{μν} \\ \partial^μ F_{μν} &=& -m^2 B_ν \end{eqnarray}$$ and if you define $ψ = \rlap{\,/}B + i\rlap{\,/}F/m$ (where $\rlap{\,/}F = γ^μγ^νF_{μν}$), then these can be combined into the single equation $i \rlap/\partial ψ = mψ$.

You can also write the massless Klein-Gordon equation and Maxwell's equations with or without a source as Dirac-like equations, but it's somewhat more awkward and you don't get exactly the Dirac equation.

All of these fields transform in the way any Clifford-valued field should: by conjugation by the same Lorentz transformation matrices that are used to transform the Dirac field.

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For completeness, in the spin-½ case, you can promote $ψ$ to a Clifford-valued field by right-multiplying it by a nonzero row vector, such as $(1\;0\;0\;0)$. Regardless of your choice, $ψ$ can then be seen as belonging to a peculiar left ideal of the Clifford algebra, whose meaning is quite a bit more obscure than the scalar+vector and vector+bivector subspaces of the spin-0 and spin-1 cases.

I've heard it suggested that the spin-½ field can be interpreted as a rotation from a "canonical spinor" to a physical spinor. The canonical spinor has (by convention) an orientation defined relative to your current coordinates, while the physical spinor has a fixed physical orientation, and this explains why the Dirac field transforms on only one side.

At any rate, Dirac fields don't have to transform in that odd one-sided way. If you promote them to Clifford fields, which also entails writing $\mathrm{tr}(\barφψ)$ instead of $\barφψ$ in the Lagrangian, then transforming them on both sides like ordinary Clifford fields doesn't change the physics as long as you're consistent about it. However, you then have a nontrivial constraint on the valid values of the fields, which I don't know how to express.

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I don't know the significance, if any, of any of this.