Is it possible for me to use the Pauli matrices to show that they can give you the Klein-Gordon equation even though the KG equation isn't a matrix?

$${\partial \over \partial t}\Psi=\pm \vec \alpha\cdot \vec \nabla\Psi$$

Show that the choice of $$\alpha_i=\sigma_i$$ ($$\sigma_i$$ are the Pauli matrices) and squaring the operator equation above helps to recover the Klein-Gordon equation: $$\Box\Psi=0$$

My issue is that I don't know how to get a scalar value because all of my $$\alpha$$'s are matrices:

$$\begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix}, \begin{pmatrix} 0 & -i \\ i & 0 \\ \end{pmatrix}, \begin{pmatrix} 1 & 0 \\ 0 & -1 \\ \end{pmatrix}$$

When I try to take the dot product of the Pauli matrices all I end up with are more matrices:

$$\vec \alpha\cdot \vec \nabla\Psi= \begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix}{\partial\over \partial x}\Psi+ \begin{pmatrix} 0 & -i \\ i & 0 \\ \end{pmatrix}{\partial\over \partial y}\Psi+ \begin{pmatrix} 1 & 0 \\ 0 & -1 \\ \end{pmatrix}{\partial\over \partial z}\Psi$$ $$=\begin{pmatrix} {\partial \over \partial z} & {\partial \over \partial x} -i{\partial \over \partial y} \\ {\partial \over \partial x} +i{\partial \over \partial y} & -{\partial \over \partial z} \\ \end{pmatrix}$$

There is no way for me to turn this into a scalar, so I don't see how I can recover the Klein-Gordon equation. What am I doing wrong?

• Perhaps you can try $(\vec{\alpha} \cdot \vec{\nabla})^{2}$ and see if any interesting terms appear. – K_inverse Jan 24 '19 at 2:05
• The dot product reads like this $\vec{\alpha} \cdot \vec{\nabla} = \alpha_{x} \partial_{x} + \alpha_{y} \partial_{y} + \alpha_{z} \partial_{z}$, which is a matrix with the size dictated by $\alpha_{i}$ – K_inverse Jan 24 '19 at 2:19
• In this problem, yes – K_inverse Jan 24 '19 at 2:22
• Write $(\vec{\sigma}\cdot\vec{\nabla})^2$ as $\sigma_i \nabla_i \sigma_j \nabla_j$ and think about the anticommutation of Pauli matrices. – G. Smith Jan 24 '19 at 5:48
• What you're trying to get is a matrix equation in which each of the four components is either trivial or the KG equation. – G. Smith Jan 24 '19 at 5:50

I am not an authority in this field, but if I am not wrong, you need the Dirac matrices to obtain the Klein-Gordon equation by squaring a differential operator. If I remember correctly, the Klein-Gordon operator is the Laplace operator with respect to the Minkowski metric. Hence the differential operator whose square is Klein-Gordon is the Dirac operator, i.e. when you square the Dirac operator you get the Klein-Gordon operator. In the procedure of squaring, the matrices that work as coefficients of the Dirac operator are the Dirac matrices, while the Pauli matrices do not work. According to me, the mathematical reason for this fact is that for a set of four matrices to form the coefficients of the Dirac operator, they need to generate the so called Clifford algebra $$Cl(\mathbb{R}^{1,3})$$ of the Minkowski space $$\mathbb{R}^{1,3}$$. This is achieved via the Dirac matrices. The Pauli matrices generate a different type of Clifford algebra that is not isomorphic to $$Cl(\mathbb{R}^{1,3})$$. If you wish, see the Wikipedia article on classification of Clifford algebras. Observe that even the number of Pauli matrices is wrong: there are three non-identity generating matrices and you need four (three for space and one should be for time). You can use the Pauli matrices to form the classical version of the Dirac operator such that when you square it you obtain the usual, classical Laplace operator of $$\mathbb{R}^3$$.