# The Dirac Equation

The Schrödinger equation $$i\hbar\frac{\partial\psi}{\partial t}=\hat H\psi$$ is a non-relativistic approximation of what is supposed to be more realistic a relativistic equation. The first place one would look at to find a relativisitic generalization is the relativistic energy $$E=\sqrt{c^2p^2+m^2c^4}$$ Replacing $E$ and $p$ by operators $i\hbar\frac{\partial}{\partial t}$ and $-i\hbar\nabla$, respectively, we obtain the square-root Klein-Gordon equation $$-\hbar\frac{\partial\psi(t,x)}{\partial t}=\sqrt{-c^2\hbar^2\nabla^2+m^2c^4}\psi(t,x)$$ This equation is however not a desirable one. Due to the appearance of the radical in the right hand side, it is impossible to include external electromagnetic fields in a relativistically invariant way.

P.A.M. Dirac considered a linearization of the relativistic energy by writing $$\label{eq:linenergy}E=c\sum_{i=1}^3\alpha_ip_i+\beta mc^2=c\alpha\cdot p+\beta mc^2$$ where $\alpha=(\alpha_1,\alpha_2,\alpha_3)$ and $\beta$ have to be determined by comparing it with the relativistic energy.

Squaring \eqref{eq:linenergy}, we have \begin{aligned}E^2=&c^2[\alpha_1^2p_1^2+\alpha_2^2p_2^2+\alpha_3^3p_3^2+(\alpha_1\alpha_2+\alpha_2\alpha_1)p_1p_2+\\&(\alpha_2\alpha_3+\alpha_3\alpha_2)p_2p_3+(\alpha_3\alpha_1+\alpha_1\alpha_3)p_3p_1]+\\&mc^3[(\alpha_1\beta+\beta\alpha_1)p_1+(\alpha_2\beta+\beta\alpha_2)p_2+(\alpha_3\beta+\beta\alpha_3)p_3]+\\&\beta^2m^2c^4\end{aligned}\label{eq:linenergy2} \eqref{eq:linenergy2} must coincide with $c^2p^2+m^2c^4$. For that to happen we must require that \begin{align*}\alpha_1^2p_1^2+\alpha_2^2p_2^2+\alpha_3^2p_3^2&=p^2\\\alpha_i\alpha_j+\alpha_j\alpha_i&=0\ \mbox{for $i\ne j$}\\\alpha_i\beta+\beta\alpha_i&=0\\\beta^2m^2c^4&=m^2c^4\end{align*}If the $\alpha_i$’s and $\beta$ were numbers, we would have $\alpha_1=\alpha_2=\alpha_3=\beta=0$ which is not desirable. Since $\alpha_i$’s and $\beta$ are anticommuting, we may assume that they are $n\times n$ matrices. Now the $\alpha_i$’s and $\beta$, as $n\times n$ matrices, are required to satisfy \begin{aligned}\alpha_i\alpha_j+\alpha_j\alpha_i&=2\delta_{ij}{\bf 1},\ i,j=1,2,3\\\alpha_i\beta+\beta\alpha_i&=0,\ i=1,2,3\\\beta^2&={\bf 1}\end{aligned}\label{eq:linenergy3}where ${\bf 1}$ denotes the $n\times n$ identity matrix. In order for the Hamiltonian to be Hermitian, the $\alpha_i$’s and $\beta$ are required to be Hermitian. From \eqref{eq:linenergy3}, $$\mathrm{tr}\alpha_i=\mathrm{tr}\beta^2\alpha_i=\mathrm{tr}\beta(\beta\alpha_i)=-\mathrm{tr}\beta\alpha_i\beta=-\mathrm{tr}\alpha_i$$ Thus, $\mathrm{tr}\alpha_i=0$. Since $\alpha_i^2={\bf 1}$, $\alpha_i$ has eigenvalues $1,-1$. Together, we see that $n$ has to be an even number. The smallest $n$ is $n=2$, but this can’t be right as there are only three linearly independent anticommuting Hermitian matrices. For example, the Pauli matrices $$\sigma_1=\begin{pmatrix}0 & 1\\1 & 0\end{pmatrix},\ \sigma_2=\begin{pmatrix}0 & -i\\i & 0\end{pmatrix},\ \sigma_3=\begin{pmatrix}1 & 0\\0 & -1\end{pmatrix}$$ together with ${\bf 1}$ form a basis for the space of $2\times 2$ Hermitian matrices. For $n=4$, if we choose $$\label{eq:diracmat}\beta=\begin{pmatrix}{\bf 1} & {\bf 0}\\{\bf 0} & -{\bf 1}\end{pmatrix},\ \alpha_i=\begin{pmatrix}{\bf 0} & \sigma_i\\\sigma_i & {\bf 0}\end{pmatrix},\ i=1,2,3$$ then \eqref{eq:linenergy3} is satisfied.

Now replacing $E$ and $p$ by operators $i\hbar\frac{\partial}{\partial t}$ and $-i\hbar\nabla$, respectively, we obtain the Dirac equation $$i\hbar\frac{\partial\psi(t,x)}{\partial t}=H_o\psi(t,x)$$ where \begin{align*}H_0&=-i\hbar c\alpha\cdot\nabla+\beta mc^2\\&=\begin{pmatrix}mc^2{\bf 1} & -i\hbar c\sigma\cdot\nabla\\-i\hbar c\sigma\cdot\nabla & -mc^2{\bf 1}\end{pmatrix}\end{align*} Here, $\alpha=(\alpha_1,\alpha_2,\alpha_3)$ and $\sigma=(\sigma_1,\sigma_2,\sigma_3)$ are triplets of matrices. The Dirac equation acts of $\mathbb{C}^4$-valued wave functions $$\psi(t,x)=\begin{pmatrix}\psi_1(t,x)\\\psi_2(t,x)\\\psi_3(t,x)\\\psi_4(t,x)\end{pmatrix},\ \psi_i\in\mathbb{C},\ i=1,2,3,4$$

If $m=0$, then only three anticommuting $\alpha_i$ are needed, so it would be sufficient to use $2\times 2$ matrices. For example, one may choose $\alpha_i=\sigma_i$, $i=1,2,3$. Then we obtain the equation $$i\hbar\frac{\partial\psi(t,x)}{\partial t}=c\sigma\cdot\nabla\psi(t,x)$$ This equation is called the Weyl equation. The Weyl equation is thought to describe neutrinos. We will discuss more about this later.

If the space dimension is two, then we can also use Pauli matrices instead of Dirac matrices \eqref{eq:diracmat}. In this case, $H$ has the form $$H=-i\hbar c\left(\sigma_1\frac{\partial}{\partial x_1}+\sigma_2\frac{\partial}{\partial x_2}\right)+\sigma_3 mc^2$$

References:

[1] Walter Greiner, Relativistic Quantum Mechanics, 3rd Edition, Springer-Verlag, 2000

[2] Bernd Thaller, The Dirac Equation, Springer-Verlag, 1992

# Dirac Sea and Antiparticles

In here, I mentioned that one of the issues of the Klein-Gordon equation is that it admits solutions yielding negative energies.This issue continues to appear with the Dirac equation which is a relativistic generalization of the Schrödinger equation. (I will discuss the Dirac equation later in a different note.) The possibility that an electron can keep falling down to a higher negative energy level indefinitely seems unphysical and initially the suggestion had to face a huge backlash from physicists including Wolfgang Pauli. P.A.M. Dirac came up with a brilliant idea based on electron hole that each negative energy state is filled by an electron (remember that electrons are fermions so no more than one electron can occupy the same energy state due to Pauli’s exclusion principle). This idea is called Dirac sea of infinite electrons. Since all negative energy states are already occupied by electrons, an electron cannot fall down below zero energy level. This may not, however, definitely be true as David Hilbert has shown in his paradox of the Grand Hotel even after the negative energy states are all occupied by electrons it may accommodate additional electrons. Dirac also suggested that it may be possible that all negative energy levels are filled by electrons except for one. This would leave a hole with a negative energy. This hole was interpreted as a positron, the antiparticle on an electron. While it is a brilliant idea, Dirac sea also appears to be unphysical. Regardless, positron was discovered by Carl Anderson in 1932 and no one raised an issue about it afterwards. Move along, nothing to see here. Still it seems that many physicists are not very comfortable with the notion of Dirac sea and that they don’t believe that it is an actual physical reality. Dirac sea is nowadays introduced more for a pedagogical purpose rather than for the purpose of defining antiparticles. In modern quantum field theory, antiparticles are defined by wave functions traveling backward in time. If I remember correctly, this definition of antiparticles is due to John Archibald Wheeler. Note that those wave functions traveling backward in time do have negative energies.

Here is a thought. Physically an electron would have the minimum energy at rest and the rest energy is given by $E_0=mc^2$. This can be obtained by putting ${\bf p}\cdot{\bf p}=0$ in the relativistic energy-momentum relation. In fact, since ${\bf p}\cdot{\bf p}\geq 0$, we have $$E^2\geq m_0^2c^4$$ which implies that either $E\geq m_0c^2$ or $E\leq -m_0c^2$. So we can say that the energy of an electron cannot be negative and that the negative energy condition could be considered as merely a mathematical fluke. But if that were the case, what about antiparticles? That is a big question which seems to have no apparent answer within conventional quantum theory and for this reason physicists are still sticking to the negative energies.

I am currently working on an unconventional quantum theory and this might shed light on alternative possibility of antiparticles. Those who are curious can read about its brief idea and motivation here. In that quantum theory, antiparticles, while having positive energies, are described by wave functions that have negative probabilities. Of course the notion of negative probabilities sounds unphysical but it actually isn’t in this case. According to the theory, antiparticles do not live in our universe but in its twin parallel universe where the roles of time coordinate and a spatial coordinate are switched from those in our universe. While the wave function of an antiparicle is seen to have a negative probability in our universe, it actually has a positive probability in its own universe. I will write more details about it elsewhere in the very near future.

# The Klein-Gordon Equation

For the Schrödinger equation $$i\hbar\frac{\partial\psi}{\partial t}=\hat H\psi({\bf x},t),$$ the Hamiltonian $$\hat H=-\frac{\hbar^2}{2m_0}\nabla^2+V({\bf x})$$ corresponds to the nonrelativistic energy-momentum relation $$\hat E=\frac{\hat p^2}{2m_0}+V({\bf x})$$ where $$\hat E=i\hbar\frac{\partial}{\partial t},\ \hat p=-i\hbar\nabla$$ So, naturally considering the relativistic energy-momentum relation $$\label{eq:e-m}\frac{E^2}{c^2}-{\bf p}\cdot{\bf p}=m_0^2c^2$$ would be the starting point to obtain a relativistic generalization of the Schrödinger equation. Replacing $E$ and ${\bf p}\cdot{\bf p}$ in \eqref{eq:e-m} by operators $$\hat E=i\hbar\frac{\partial}{\partial t}\ \mbox{and}\ \hat p\cdot\hat p=-\hbar^2\nabla^2$$ acting on a wave function $\psi$, we obtain the Klein-Gordon equation for a free particle $$\label{eq:k-g}\left(\Box-\frac{m_0^2c^2}{\hbar^2}\right)\psi=0$$ where $$\Box=-\frac{1}{c^2}\frac{\partial^2}{\partial t^2}+\nabla^2$$

Free solutions of the Schrödinger equation with $V({\bf x})=0$ are of the form $$\psi=\exp\left[\frac{i}{\hbar}(-Et+{\bf p}\cdot{\bf x})\right]$$ They are also free solutions of the Klein-Gordon equation \eqref{eq:k-g} with the energy condition $$E=\pm c\sqrt{m_0^2c^2+p^2}$$ The solutions yielding negative energies appear to be unphysical and initially considered so by physicists, but later they were interpreted as antiparticles. Antiparticles are indeed seen in nature. In reality, antiparticles also have positive energies. Antiparticles as wave functions with negative energies is merely an interpretation of the mathematical representation of the energy condition. If antiparticles weren’t discovered, the negative energy condition would have been still thought to be unphysical.

Other than allowing solutions with negative energies, there was another issue with the Klein-Gordon equation noted by physicists. The conservation of four-current density $$j_\mu=\frac{i\hbar}{2m_0}(\psi^\ast\nabla_\mu\psi-\psi\nabla_\mu\psi^\ast),$$ where $\psi^\ast$ denotes the complex conjugate of $\psi$ and $\nabla_\mu=\left(-\frac{1}{c^2}\frac{\partial}{\partial t},\nabla\right)$, implies that the quantity $$\rho=\frac{i\hbar}{2m_0c^2}\left(\psi^\ast\frac{\partial\psi}{\partial t}-\psi\frac{\partial\psi^\ast}{\partial t}\right)$$ can be considered as a probability density. However, the problem is that $\rho$ can be negative. This is due to the appearance of first-order partial derivative $\frac{\partial\psi}{\partial t}$, which is the consequence of the Klein-Gordon equation being of second-order in time. Because of this, the Klein-Gordon equation was not regarded as a physically viable relativistic generalization of the Schrödinger equation and physicists were instead looking for a relativistic generalization of first-order in time like the Schrödinger equation. Such an equation was finally discovered by P. A. M. Dirac and is called the Dirac equation. On the other hand, the Klein-Gordon equation drew attention of physicists again after they realized that $\rho$ can be interpreted as charge density, and indeed charged pions $\pi^+$ and $\pi^-$ were discovered. Today, the Klein-Gordon equation is an important relativistic equation that describes charged spin-0 particles.

References:

[1] Walter Greiner, Relativistic Quantum Mechanics, 3rd Edition, Springer-Verlag, 2000