# Why do we need matrices in the Dirac equation?

Consider the following equation:

\begin{equation} \nabla^2 - \frac{1}{c^2}\frac{\partial^2}{\partial t^2} = \left(A \partial_x + B \partial_y + C \partial_z + \frac{i}{c}D \partial_t\right)\left(A \partial_x + B \partial_y + C \partial_z + \frac{i}{c}D \partial_t\right) \end{equation}

I am confused why Dirac thought matrices were needed in place of $A,B,C,D$. I understand why, in order to get all the cross-terms such as $\partial_x\partial_y$ to vanish, we must assume $AB + BA = 0\;\ldots$ and why $A^2 = B^2 = \ldots = 1$. But Wikipedia then says

Dirac, who had just then been intensely involved with working out the foundations of Heisenberg's matrix mechanics, immediately understood that these conditions could be met if $A, B, C$ and $D$ are matrices, with the implication that the wave function has multiple components.

Can someone explain why the above equations and the conditions suggest matrices should be used?

I suppose one sees that the result must involve objects that do not commute under multiplication in order to satisfy the anticommutation property, so that might be a clue. But aren't there other non-commutative things it could have been?

• Maybe the fact the Pauli equation existed would be another clue. If an object had two have two components, then one might suspect in order to be able to multiple a multi-component wavefunction, A,B,CD must be matrices. – Stan Shunpike Feb 4 '15 at 22:26
• Dirac was of a generation that studied things like projective geometry and matrices extensively. His and the previous generation (his teachers) "explored" matrices for symmetries and useful properties. See Klein's books on elementary mathematics from an advanced standpoint for examples - the geometry one is very interesting. One benefit is that Dirac easily saw situations where a matrix will fit a problem. – C. Towne Springer Feb 4 '15 at 23:04
• I suspect the Wiki author means that because Dirac had been recently working with matrices, which have commutation and anticommutation rules. Thus he saw a connection to matrices. As I understand it, matrices were very uncommon in physics until Heisenberg came along. – Ryan Unger Feb 4 '15 at 23:10

In his "Recollections of an Exciting Era" (History of Twentieth Century Physics: Proceedings of the International School of Physics "Enrico Fermi". Course LVII.-New York; London: Academic Press, 1977. P. 109-146), Dirac wrote that, to derive what is now known as the Dirac equation, he had played with mathematical formulas, and he needed the Pauli matrices $\sigma_1,\sigma_2,\sigma_3$ to describe the spin of electron. He noticed that a square of $p_1\sigma_1+p_2\sigma_2+p_3\sigma_3$ (where $p_1,p_2,p_3$ are the three components of momentum) yields $p_1^2+p_2^2+p_3^2$ and was excited by this mathematical result. However, he could not extend this result to a sum of 4 squares (required for special relativity) until he switched from the Pauli matrices to 4x4 matrices. Therefore, I would say, Dirac started with matrices due to the previous success of the Pauli matrices.

As for your question "Do we need matrices in the Dirac equation?", my recent result might be relevant: the Dirac equation is generally equivalent to one fourth order equation for one unknown (see my article http://akhmeteli.org/wp-content/uploads/2011/08/JMAPAQ528082303_1.pdf (JOURNAL OF MATHEMATICAL PHYSICS 52, 082303 (2011)). So, technically, you can do without matrices.

Regular numbers could never fulfill $$AB + BA = 0, \quad AA = 1 = BB.$$ The only way to fulfill the first eq. is to have either $A = 0$ or $B = 0$ but this violates the second equation.

Matrices on the other hand can fulfill such equations and since Dirac knew about matrices he did not discard his idea after finding something that's impossible with regular numbers.

My guess is that Dirac realised that if he wanted to define the square root of the d'Alembert operator he needed to change the algebra of coefficients from a commutative one to a non-commutative one. The work of Clifford on Clifford algebras was more or less contemporary, so Dirac probably knew nothing about them. Nonetheless he postulated that the coefficient should have been from the image of the representation of a Clifford algebra (again, he probably didn't think in these terms, but we now know that Dirac matrices are just a representation, hence matrices with complex entries in general, of a certain Clifford algebra).

• +1, although Clifford was a fair bit before Dirac, but I think you're right that Dirac wouldn't have thought in these terms. – Selene Routley Feb 6 '15 at 8:16
• I think I read this somewhere, perhaps Penrose's Road to reality, but I can't recall exactly – Phoenix87 Feb 6 '15 at 8:30
• What you say is highly plausible: we kind of get used to having so much knowledge at our fingertips that we forget how long it took knowledge to diffuse across field boundaries. I read somewhere that Heisenberg was creating his own noncommutative algebras from scratch and first principles until Born, his adviser, told him he was dealing with matrices. Also, my impression of Penrose is that he would be fairly a reliable historian. – Selene Routley Feb 6 '15 at 8:41