Another Dirac Equation

1. Basic Equation

We begin with a brief introduction to Dirac’s original paper [1,2]. Let $x^{\mu }$ and $p_{\mu }$ denote the four-dimensional coordinates of the whole system and their conjugate momenta, respectively. The new equation of motion is written as

$$\left({\alpha }^0\beta {\displaystyle \frac{\partial }{\partial x^0}}+{\alpha }^k{\displaystyle \frac{\partial }{\partial x^k}}+m\right)q\psi =0.$$

(1)

The quantities ${\alpha }^k(k=1,2,3)$ and $\beta$ are $4\times 4$ matrices, while q is a four-component quantity and $\psi$ is a scalar wave function. Unlike the usual Dirac equation, a four-component quantity q representing internal degrees of freedom appears instead of a four-component spinor. Its transpose is

$$q^t=(q_1,q_2,p_1,p_2),$$

and two pairs of canonically conjugate variables are introduced:

$$[q_i,p_j]=i{\delta }_{ij}(i,j=1,2).$$

At this stage, we do not specify the physical meaning of the internal variables. They may therefore be interpreted, for example, as representing Schwinger bosons, or some kind of “rotational” degree of freedom.

The parameter m is a mass parameter, and the coefficient matrices are defined by

$${\alpha }^1=\left(\begin{array}{cccc}0& 0& -1& 0\\ 0& 0& 0& 1\\ -1& 0& 0& 0\\ 0& 1& 0& 0\end{array}\right),{\alpha }^2=\left(\begin{array}{cccc}0& 0& 0& 1\\ 0& 0& 1& 0\\ 0& 1& 0& 0\\ 1& 0& 0& 0\end{array}\right),$$

$${\alpha }^3=\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& -1& 0\\ 0& 0& 0& -1\end{array}\right),\beta =\left(\begin{array}{cccc}0& 0& 1& 0\\ 0& 0& 0& 1\\ -1& 0& 0& 0\\ 0& -1& 0& 0\end{array}\right).$$

These four quantities mutually anticommute and satisfy

$${\left({\alpha }^1\right)}^2={\left({\alpha }^2\right)}^2={\left({\alpha }^3\right)}^2=-{\beta }^2=1.$$

Introducing the unit matrix ${\alpha }^0\equiv 1$, the equation of motion becomes

$$({\alpha }_{\mu }{\partial }^{\mu }+\beta m)q\psi =0.$$

(2)

Dirac derived the Klein–Gordon (KG) equation from the integrability condition of these four coupled equations and showed that negative-energy solutions are naturally excluded. However, the calculated particle mass spectrum decreased as the magnitude of the spin increased, and the approach did not lead to further development.

We therefore propose a new equation of motion from a different point of view. Following Dirac, we introduce the four-dimensional spin angular-momentum operator,

$$s_{\mu \nu }\equiv -{\displaystyle \frac{1}{8}}\left(q^t{\alpha }_{\mu }\beta {\alpha }_{\nu }q-q^t{\alpha }_{\nu }\beta {\alpha }_{\mu }q\right).$$

(3)

The right-hand side is an antisymmetric tensor, $s_{\mu \nu }=-s_{\nu \mu }$, representing the operator of the internal degrees of freedom of the particle. Multiplying Eq. (2) from the left by $q^t{\alpha }_{\mu }\beta$ gives

$$\left(q^t{\alpha }_{\mu }\beta {\alpha }_{\nu }q{\partial }^{\nu }-q^t{\alpha }_{\mu }qm\right)\psi =0.$$

(4)

This equation is a set of four coupled equations corresponding to the four values of the index $\mu$. Since it is obtained by multiplying Eq. (2) by $q^t{\alpha }_{\mu }\beta$, it may possess a wider class of solutions than Eq. (2). In fact, the purpose of this short paper is to investigate whether a more realistic mass spectrum can be obtained from this equation.

Before doing so, we introduce the following new spin vector appearing in the mass term:

$$s_{\mu 5}\equiv {\displaystyle \frac{1}{4}}\left(q^t{\alpha }_{\mu }q\right),$$

(5)

and rewrite the equation of motion in terms of the spin variables,

$$\left(i{\partial }_{\mu }-2s_{\mu \nu }{\partial }^{\nu }-2m{s}_{\mu 5}\right)\psi =0.$$

(6)

In what follows, $s_{\mu \nu }$ and $s_{\mu 5}$ will be referred to as spin variables or operators.

Following Dirac, we impose the integrability condition ${\partial }_{\mu }{\partial }_{\nu }-{\partial }_{\nu }{\partial }_{\mu }=0$ on Eq. (6). We then obtain

$$\left({\partial }_{\mu }{\partial }^{\mu }+i2m{\partial }^{\mu }{s}_{\mu 5}\right)\psi =0.$$

(7)

Requiring this equation to take the form of the KG equation, the relativistic mass operator is defined by

$$M^2\equiv i2m{\partial }^{\mu }{s}_{\mu 5}=2m{p}^{\mu }{s}_{\mu 5}.$$

(8)

In the rest frame $(p^k=0)$, this becomes $M^2=2m{p}^0{s}_{05}$, or equivalently $M=2m{s}_{05}$.

At the end of this section, we add two remarks. First, it should be necessary to verify the covariance of our basic equation. Under an infinitesimal Lorentz transformation,

$$x_{\mu }^{*}=x_{\mu }+{a_{\mu }}^{\nu }{x}_{\nu },{\partial }^{\mu *}={\partial }^{\mu }+{a^{\mu }}_{\nu }{\partial }^{\nu },$$

(9)

the spin operators must transform so that $s_{\mu \nu }$ is a second-rank tensor and $s_{\mu 5}$ is a vector:

$$\left\{\begin{array}{c}{s}_{\mu \nu }^{*}=s_{\mu \nu }+{a_{\mu }}^{\sigma }{s}_{\sigma \nu }+{a_{\nu }}^{\sigma }{s}_{\mu \sigma },\hfill \\ s_{\mu 5}^{*}=s_{\mu 5}+{a_{\mu }}^{\sigma }{s}_{\sigma 5}.\hfill \end{array}\right.$$

It then follows that the internal variables $q_a(a=1,2,3,4)$ should transform as

$$q^{*}=q-\beta Nq,N\equiv {\displaystyle \frac{1}{4}}{a}^{\mu \nu }{\alpha }_{\mu }\beta {\alpha }_{\nu }.$$

(10)

From these transformation laws, the covariance of our basic Eq. (6) can indeed be verified. The transformation law Eq. (10) can be rewritten as

$$q^{*}=(1-{\displaystyle \frac{i}{2}}{q}^{t}Nq)q(1+{\displaystyle \frac{i}{2}}{q}^{t}Nq)=(1+{\displaystyle \frac{i}{2}}{a}^{\mu \nu }{s}_{\mu \nu })q(1-{\displaystyle \frac{i}{2}}{a}^{\mu \nu }{s}_{\mu \nu })$$

(11)

This relation means that the four-dimensional angular-momentum operator is defined by

$$M_{\mu \nu }=x_{\mu }{\partial }_{\nu }-x_{\nu }{\partial }_{\mu }-i{s}_{\mu \nu }.$$

(12)

The second remark concerns the continuity equation of the system that is important with respect to the interpretation of the wave function. Following the usual procedure, Eq. (4) and its conjugate equation give

$$\left\{\begin{array}{c}\left({\partial }^{\mu }{\psi }^{†}\right)\left(q^t{\alpha }_{\mu }\beta {\alpha }_{\nu }q{\partial }^{\nu }\psi -m{q}^t{\alpha }_{\mu }q\psi \right)=0,\hfill \\ \left(-{\partial }^{\nu }{\psi }^{†}{q}^t{\alpha }_{\nu }\beta {\alpha }_{\mu }q-m{\psi }^{†}{q}^t{\alpha }_{\mu }q\right)\left({\partial }^{\mu }\psi \right)=0.\hfill \end{array}\right.$$

(13)

The minus sign in the first term of the second equation follows from ${\beta }^{†}=-\beta$. Adding these two equations, we obtain

$${\partial }^{\mu }\left({\psi }^{†}{q}^t{\alpha }_{\mu }q\psi \right)\equiv {\partial }^{\mu }{j}_{\mu }=0.$$

(14)

This continuity equation is the same as that derived by Dirac.

2. Algebra of Internal Operators

Our equation contains ten Hermitian internal operators, namely

$$\{s_k,s_{0k},s_{k5},s_{05}\}(k=1,2,3).$$

(15)

Here we have changed the notation for the spin angular momentum to the usual one,

$${ϵ}_{kij}{s}_{ij}\equiv 2s_k,$$

where ${ϵ}_{kij}$ is the totally antisymmetric tensor. These operators form the de Sitter group $SO(3,2)$. The commutation relations of the de Sitter group can be summarized as

$$[s_i,s_j]=i{ϵ}_{ijk}{s}_k,[s_{0i},s_{0j}]=-i{ϵ}_{ijk}{s}_k,[s_i,s_{0j}]=i{ϵ}_{ijk}{s}_{0k}.$$

(16)

These coincide with the commutation relations of the generators of the homogeneous Lorentz group. Similarly, for $s_{k5}$ we have

$$[s_{i5},s_{j5}]=-i{ϵ}_{ijk}{s}_k,[s_i,s_{j5}]=i{ϵ}_{ijk}{s}_{k5},[s_{0i},s_{j5}]=-i{\delta }_{ij}{s}_{05}.$$

(17)

From the first two commutation relations, the set $\{s_i,s_{i5}\}$ satisfies the same relations as $\{s_i,s_{0i}\}$ and represents another homogeneous Lorentz group. The last commutator would represent a canonical conjugate relation between $s_{0i}$ and $s_{i5}$ if $s_{05}$ were a constant. We shall discuss this point later. Next, let us find the commutation relations between $s_{05}$ and the other operators:

$$[s_i,s_{05}]=0,[s_{0i},s_{05}]=-i{s}_{i5},[s_{i5},s_{05}]=i{s}_{0i}.$$

(18)

The first relation shows that $s_{05}$ is a three-dimensional scalar. The second and third relations show that $s_{0i}$ and $s_{i5}$ are interchanged by taking their commutators with $s_{05}$. In other words, they represent canonical equations with $s_{05}$ as the internal Hamiltonian. For example, if $s_{i5}$ is regarded as a coordinate, $s_{0i}$ corresponds to its conjugate momentum. In fact, $s_{05}$ is written as

$$s_{05}={\displaystyle \frac{1}{4}}\left(q_1^2+q_2^2+p_1^2+p_2^2\right),$$

(19)

by the direct calculation. The $s_{05}$ represents the sum of the Hamiltonians of two harmonic oscillators. On the other hand, $(s_{0k},s_{k5},s_{05})$ satisfy the commutation relations of $SU(1,1)$ and play an important role in describing the internal motion. This will be discussed in detail in the next section.

We now have three Hermitian internal-operator vectors, $\{\mathbf{s},{\mathbf{s}}_0,{\mathbf{s}}_5\}$. Let us consider their orthogonality. Since each operator has three components $(k=1,2,3)$, direct calculation of their inner products gives zero:

$$\mathbf{s}·{\mathbf{s}}_0=\sum _k{s}_k{s}_{0k}=0,\mathbf{s}·{\mathbf{s}}_5=\sum _k{s}_k{s}_{k5}=0.$$

(20)

This result is independent of the order of the inner products. Thus the two internal operators ${\mathbf{s}}_0$ and ${\mathbf{s}}_5$ are orthogonal to the spin angular momentum $\mathbf{s}$. However, their mutual inner product is purely imaginary:

$${\mathbf{s}}_0·{\mathbf{s}}_5=\sum _k{s}_{0k}{s}_{k5}=-{\displaystyle \frac{3i}{2}}{s}_{05}=-{\mathbf{s}}_5·{\mathbf{s}}_0.$$

(21)

Although they are not orthogonal, their symmetrized inner product vanishes, ${\mathbf{s}}_0·{\mathbf{s}}_5+{\mathbf{s}}_5·{\mathbf{s}}_0=0$. Consequently, the three vectors $\{\mathbf{s},{\mathbf{s}}_0,{\mathbf{s}}_5\}$ may be regarded as forming a mutually quasi-orthogonal system.

Next, a direct calculation of the magnitude of the spin angular momentum gives an expression in terms of the sum of the Hamiltonians of two harmonic oscillators:

$${\mathbf{s}}^2=\sum _k{s}_k^2=s(s+1),s\equiv {\displaystyle \frac{1}{4}}\left(q_1^2+q_2^2+p_1^2+p_2^2\right)-{\displaystyle \frac{1}{2}}.$$

(22)

The magnitudes of the other two operators can also be written in terms of the square of the spin operator:

$${\mathbf{s}}_0^2={\mathbf{s}}_5^2={\mathbf{s}}^2+{\displaystyle \frac{3}{4}}=s(s+1)+{\displaystyle \frac{3}{4}}.$$

(23)

It follows from these results that the internal motion of this system is almost completely determined by the magnitude of the spin. In this sense the system is simple: internal degrees of freedom other than spin are frozen and do not appear explicitly.

3. Motion of the Particle

We now consider the motion of the particle. Let us first rewrite the $\mu =0$ component of the equation of motion:

$$i{\partial }_0\psi =\left(2s_{0k}{\partial }^k+2m{s}_{05}\right)\psi \equiv H\psi .$$

(24)

The quantity H appearing on the right-hand side corresponds to the Hamiltonian of the system. It should be noted, however, that the term containing the derivative on the right-hand side is non-Hermitian. To investigate its physical meaning, we consider an eigenstate of the total momentum:

$$\left(E_0+2i{s}_{0k}{p}^k-2m{s}_{05}\right)\phi \left(q\right)e^{-i{E}_{0}t-i{p}_k{x}^k}=0.$$

(25)

In the rest frame ($p^k=0$), this equation describes harmonic oscillators, and the energy eigenvalues are

$$E_0=2m{s}_{05}={\displaystyle \frac{m}{2}}\left(q_1^2+q_2^2+p_1^2+p_2^2\right)=m(n_1+n_2+1),$$

(26)

where $n_1,n_2=0,1,2,3,\dots$. We now introduce the generators of $SU(1,1)$ mentioned in the preceding section:

$$K_1={\mathbf{s}}_0·\widehat{\mathbf{p}},K_2={\mathbf{s}}_5·\widehat{\mathbf{p}},K_0=s_{05}.$$

(27)

where $\widehat{\mathbf{p}}$ is the unit vector in the direction of $\mathbf{p}$. Their commutation relations are

$$[K_1,K_2]=-i{K}_0,[K_2,K_0]=i{K}_1,[K_0,K_1]=i{K}_2.$$

(28)

These commutation relations closely resemble those of $SU\left(2\right)$, except for one sign. Consequently, this group is noncompact and its eigenvalues extend without bound. The Casimir operator is

$$\widehat{C}\equiv K_0^2-K_1^2-K_2^2=k(k-1)\widehat{I},k>0.$$

(29)

The quantity k is called the Bargmann index. For the present generators, it will be shown below that $k=1/2,1,3/2,2,5/2,\dots$. The states of the system are classified by the eigenvalues of $\widehat{C}$ and $K_0$ as $|k,\mu \rangle$:

$$K_0|k,\mu \rangle =\mu |k,\mu \rangle ,\mu =k+m,m=0,1,2,3,\dots .$$

(30)

Let us return to the eigenvalue problem of the Hamiltonian (3.1). To investigate the physical meaning of the previously neglected term containing the non-Hermitian momentum, we regard this term as a perturbation and calculate the second-order correction to the energy eigenvalue (the first-order correction vanishes). For the perturbation $H_1\equiv 2i{s}_{0k}{p}^k=2i{K}_{1}p$, the standard formula gives

$$E_0^{\left(2\right)}=\sum _{n\ne 0}{\displaystyle \frac{\langle 0|H_1|n\rangle \langle n|H_1|0\rangle }{E_0^{\left(0\right)}-E_n^{\left(0\right)}}}={\displaystyle \frac{p^2}{2m}}{\displaystyle \frac{1}{2k}}={\displaystyle \frac{p^2}{2E_0^{\left(0\right)}}}={\displaystyle \frac{p^2}{2M}}.$$

(31)

Thus we obtain the kinetic energy of classical mechanics. In ordinary perturbation theory, the second-order energy correction is generally negative; in the present case, however, the non-Hermiticity makes it positive and gives the reasonable energy correction.

To understand the physical meaning of the quantum number k, let us carry out a direct calculation of the Casimir operator $\widehat{C}$. Denoting the direction cosines of the unit vector $\widehat{\mathbf{p}}$ by $(e_1,e_2,e_3)$, we write

$$\widehat{C}=\sum _i{\widehat{C}}_i{e}_i^2-\sum _{i<j}\left(s_{0i}{s}_{0j}+s_{0j}{s}_{0i}+s_{i5}{s}_{j5}+s_{j5}{s}_{i5}\right)e_i{e}_j.$$

(32)

Here ${\widehat{C}}_i$ is the Casimir operator when $\widehat{\mathbf{p}}$ is directed along the ith axis. Its value is

$${\widehat{C}}_i=\left(s_i+{\displaystyle \frac{1}{2}}\right)\left(s_i-{\displaystyle \frac{1}{2}}\right),(i=1,2,3).$$

(33)

Substituting this into Eq. (3.9) and carrying out the lengthy calculation, we arrive at

$$\widehat{C}={(\mathbf{s}·\widehat{\mathbf{p}})}^2-{\displaystyle \frac{1}{4}}.$$

(34)

Since $h\equiv \mathbf{s}·\widehat{\mathbf{p}}$ is the helicity operator, comparison with Eq. (3.6) gives

$$k=h+{\displaystyle \frac{1}{2}}.$$

(35)

Thus the state vector $|k,\mu \rangle$ is an eigenstate of the magnitude of the spin and of the helicity through

$$h=k-{\displaystyle \frac{1}{2}},s=\mu -{\displaystyle \frac{1}{2}}.$$

Expressing these quantum numbers in terms of the harmonic-oscillator quantum numbers $(n_1,n_2)$, Eq. (2.8) gives $s=(n_1+n_2)/2$. For the helicity, the calculation is particularly simple when $\widehat{\mathbf{p}}=(0,1,0)$:

$$h=s_2={\displaystyle \frac{1}{4}}\left(q_1^2+p_1^2-q_2^2-p_2^2\right)={\displaystyle \frac{n_1-n_2}{2}}.$$

(36)

Thus the internal degrees of freedom are completely determined by the magnitude of the spin and the helicity.

Returning to the original equation, the KG equation can be written as

$$\left({\partial }_{\mu }{\partial }^{\mu }+M^2\right)\psi =\left(-E^2+p^2+2mE{K}_0-2mp{K}_2\right)\phi \left(q\right)e^{-iEt-i{p}_k{x}^k}=0.$$

(37)

To obtain the mass spectrum of the system, we perform the unitary transformation $e^{i\theta K_1}$. Then

$$\left[E^2-p^2-2mE(K_{0}cosh\theta -K_{2}sinh\theta )+2mp(K_{2}cosh\theta -K_{0}sinh\theta )\right]{\phi }^{\prime }=0,$$

where ${\phi }^{\prime }=e^{i\theta K_1}\phi$. Choosing $\theta$ so that the coefficient of $K_2$ vanishes, we obtain

$$cosh\theta ={\displaystyle \frac{E}{\sqrt{E^2-p^2}}},sinh\theta =-{\displaystyle \frac{p}{\sqrt{E^2-p^2}}}.$$

(38)

Substitution into the preceding equation gives

$$\left(E^2-p^2-2m\sqrt{E^2-p^2}{K}_0\right){\phi }^{\prime }=0.$$

Since $M=\sqrt{E^2-p^2}$ and $s=(n_1+n_2)/2$, the particle mass is finally obtained as

$$M=2m{K}_0=m(n_1+n_2+1)=m(2s+1).$$

(39)

The mass spectrum is equally spaced. If the spectrum is divided into the cases $n_1+n_2=\mathrm{even}$ and $n_1+n_2=\mathrm{odd}$, the two sectors are not mixed by the internal operators. This means that the states separate into a group with integer spin and a group with half-integer spin (with degeneracy). Since the wave function satisfies ${\phi }^{\prime }=e^{i\theta K_1}\phi$, it is described by an $SU(1,1)$ coherent state. In particular, in the rest frame $sinh\theta =0$ ($\theta =0$), and the wave function reduces to that of the harmonic oscillator.

4. Summary

We have discussed the equation proposed by Dirac in the final years of his life. This equation seems to be suitable for composite particles. The rotational motion is described in terms of the Schwinger boson. It would have been interesting if a Regge trajectory had emerged. But unfortunately the result is simply that the mass is proportional to the spin. One could also consider adopting a somewhat different point of view.

ChatGPT (OpenAI) was used to assist with the English translation, language editing, and LaTeX preparation of this manuscript. The scientific ideas, equations, analysis, and conclusions are the author’s own. The author reviewed and approved the final manuscript and takes full responsibility for its content.