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Did Dirac derive the correct equation for the wrong reasons? This is a question about the historical discovery of the Dirac equation and how it was deduced. Looking back at that discovery with our current knowledge.

Dirac wanted to find an equation of the electron which had only a first derivative of time. The motivation seems to be that it would serve as a wavefunction is the quantum equation $i\hbar\dot{\psi} =\hat{H}\psi$.

However, it turned out that the Dirac equation really was the equation of a quantum field not a wave function. So in that respect there was no real reason for it to only have first derivatives. It could have been like a scalar field with second derivatives. (e.g. Klein Gordon.)

On the other hand, Dirac's reasoning led him to the correct equation. So is there some deeper meaning why the fermion quantum field equation must look much like a wave function equation? If the Dirac equation hadn't been discovered, what would be a "correct" way to deduce it?

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closed as primarily opinion-based by Aaron Stevens, AccidentalFourierTransform, Jon Custer, Noiralef, G. Smith Jul 19 at 2:25

Many good questions generate some degree of opinion based on expert experience, but answers to this question will tend to be almost entirely based on opinions, rather than facts, references, or specific expertise. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Well, one possible response is that the wavefunction of the one-particle sector of a free quantum field obeys the same differential equation that the field as a whole does. I'm not sure if that counts as the truly deeper reason though. $\endgroup$ – knzhou Jul 16 at 18:33
  • $\begingroup$ RE: the title on this question, the mod consensus was that the edited title was definitely better, but if the OP insists it's not worth taking control away from the OP just to make the title less clickbaity. $\endgroup$ – Chris Jul 26 at 19:33
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  1. The Dirac equation is both the Schrödinger equation of a fermionic wave function and the equation of motion of a quantum field. It is questionable to claim that it is "really" the equation for a field and not for a wavefunction; it is both.

  2. Every equation of motion for a free quantum field is also the free evolution equation for its one-particle wavefunctions, see e.g. this answer by Andrew, where the argument is made for a scalar field but goes through unaltered for any other field with a mode expansion. It is therefore unavoidable that the Dirac equation appears as both the equation of motion of a wavefunction and of a field.

  3. There is no "correct" way to "deduce" new physical theories as they are not deduced from existing postulates, but are extensions of the existing theory, just as there e.g. is no objectively correct way to deduce the form of Hamiltonians for specific system. They are justified post-hoc by their predictions matching experiment, but the heuristics used to arrive at them cannot be right or wrong, just useful or not.

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Actually, the Dirac equation, which is a system of four first-order equations for the four components of the Dirac spinor, is generally equivalent to one fourth-order equation for just one component. Please see below the derivation from my article (Journ. Math. Phys. 52, 082303 (2011)). As Dirac looked for a first order equation, it does look like he derived the correct equation for a wrong reason.

Let us consider the Dirac equation in electromagnetic field: $$\require{\cancel} \begin{equation}\label{eq:pr25} \nonumber (i\cancel{\partial}-\cancel{A})\psi=\psi, \end{equation}$$ where, e.g., $\cancel{A}=A_\mu\gamma^\mu$ (the Feynman slash notation). For the sake of simplicity, a system of units is used where $\hbar=c=m=1$, and the electric charge $e$ is included in $A_\mu$ ($eA_\mu \rightarrow A_\mu$). In the chiral representation of $\gamma$-matrices $$\begin{equation}\label{eq:d1} \nonumber \gamma^0=\left( \begin{array}{cc} 0 & -I\\ -I & 0 \end{array} \right),\gamma^i=\left( \begin{array}{cc} 0 & \sigma^i \\ -\sigma^i & 0 \end{array} \right), \end{equation}$$ where index $i$ runs from 1 to 3, and $\sigma^i$ are the Pauli matrices.

If $\psi$ has components $$\begin{equation}\label{eq:d2} \nonumber \psi=\left( \begin{array}{c} \psi_1\\ \psi_2\\ \psi_3\\ \psi_4\end{array}\right), \end{equation}$$ the Dirac equation can be written in components as follows: $$\begin{eqnarray}\label{eq:d3} \nonumber (A^0+A^3)\psi_3+(A^1-i A^2)\psi_4 +i(\psi_{3,3}-i\psi_{4,2}+\psi_{4,1}-\psi_{3,0})=\psi_1, \end{eqnarray}$$ $$\begin{eqnarray}\label{eq:d4} \nonumber (A^1+i A^2)\psi_3+(A^0-A^3)\psi_4 -i(\psi_{4,3}-i\psi_{3,2}-\psi_{3,1}+\psi_{4,0})=\psi_2, \end{eqnarray}$$ $$\begin{eqnarray}\label{eq:d5} \nonumber (A^0-A^3)\psi_1-(A^1-i A^2)\psi_2 -i(\psi_{1,3}-i\psi_{2,2}+\psi_{2,1}+\psi_{1,0})=\psi_3, \end{eqnarray}$$ $$\begin{eqnarray}\label{eq:d6} \nonumber -(A^1+i A^2)\psi_1+(A^0+A^3)\psi_2 +i\psi_{2,3}+\psi_{1,2}-i(\psi_{1,1}+\psi_{2,0})=\psi_4. \end{eqnarray}$$

Obviously, the latter two equations can be used to express components $\psi_3,\psi_4$ via $\psi_1,\psi_2$ and eliminate them from the first two equations (a well-known result).

The resulting equations for $\psi_1$ and $\psi_2$ are as follows: $$\begin{eqnarray}\label{eq:d7} \nonumber -\psi_{1,\mu}^{,\mu}+\psi_2(-i A^1_{,3}-A^2_{,3}+A^0_{,2}+A^3_{,2}+ i(A^0_{,1}+A^3_{,1}+A^1_{,0})+A^2_{,0})+\\ \nonumber +\psi_1(-1+A^{\mu} A_{\mu}-i A^{\mu}_{,\mu}+i A^0_{,3}-A^1_{,2}+A^2_{,1}+i A^3_{,0})-2 i A^{\mu}\psi_{1,\mu}=0, \end{eqnarray}$$ $$\begin{eqnarray}\label{eq:d8} \nonumber -\psi_{2,\mu}^{,\mu}+i\psi_1( A^1_{,3}+i A^2_{,3}+i A^0_{,2}-i A^3_{,2}+ A^0_{,1}-A^3_{,1}+A^1_{,0}+i A^2_{,0})+\\ \nonumber +\psi_2(-1+A^{\mu} A_{\mu}-i( A^{\mu}_{,\mu}+A^0_{,3}+i A^1_{,2}-i A^2_{,1}+ A^3_{,0}))-2 i A^{\mu}\psi_{2,\mu}=0. \end{eqnarray}$$

As the first equation contains $\psi_2$, but not its derivatives, it can be used to express $\psi_2$ via $\psi_1$, eliminate it from the second equation and obtain an equation of the fourth order for $\psi_1$.

$$\begin{eqnarray}\label{eq:d8n1} \nonumber \left(\left(\Box'-i F^3\right)\left(i F^1+F^2\right)^{-1}\left(\Box'+i F^3\right)-i F^1+F^2\right)\psi_1=0, \end{eqnarray}$$ where $F^i=E^i+i H^i$; $E^i$ and $H^i$ are electric and magnetic fields: $$\begin{eqnarray}\label{eq:d8n9} \nonumber F^{\mu\nu}=A^{\nu,\mu}-A^{\mu,\nu}=\left( \begin{array}{cccc} 0 & -E^1 & -E^2 & -E^3\\ E^1 & 0 & -H^3 & H^2\\ E^2 & H^3 & 0 & -H^1\\ E^3 &-H^2 & H^1 & 0 \end{array} \right), \end{eqnarray}$$ and the modified d'Alembertian $\Box'$ is defined as follows: $$\begin{eqnarray}\label{eq:d8n2} \nonumber \Box'=\partial^\mu\partial_\mu+2 i A^\mu\partial_\mu+\imath A^\mu_{,\mu}-A^\mu A_\mu+1. \end{eqnarray}$$

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