A.14 The Klein-Gordon equation

The Schrö­din­ger equation for the quantum wave function is based on the nonrelativistic expression for the energy of a particle. This addendum looks at the simplest relativistic equation for wave functions, called the Klein-Gordon equation. The discussion will largely be restricted to a spinless particle in empty space, where there is no potential energy. However, the Klein-Gordon equation is the first step to more complex relativistic equations.

Recall first how the Schrö­din­ger equation arose. If there is no potential energy, classical physics says that the energy $E$ is just the kinetic energy $p^2$$\raisebox{.5pt}{$/$}$$2m$ of the particle. Here $p$ is the linear momentum of the particle and $m$ its mass. Quantum mechanics replaces the energy $E$ by the operator ${\rm i}\hbar\partial$$\raisebox{.5pt}{$/$}$$\partial{t}$ and the momentum ${\skew0\vec p}$ by $\hbar\nabla$$\raisebox{.5pt}{$/$}$${\rm i}$, where

\begin{displaymath}
\nabla = {\hat\imath}\frac{\partial }{\partial x} +
{\ha...
...\partial }{\partial y} + {\hat k}\frac{\partial }{\partial z}
\end{displaymath}

Then it applies the resulting operators on the wave function $\Psi$. That then produces the Schrö­din­ger equation

\begin{displaymath}
{\rm i}\hbar \frac{\partial\Psi}{\partial t} = -\frac{\hbar^2}{2m} \nabla^2 \Psi
\end{displaymath}

Solutions with a definite value $E$ for the energy take the form $\Psi$ $\vphantom0\raisebox{1.5pt}{$=$}$ $ce^{-{{\rm i}}Et/\hbar}\psi$. Substituting that into the Schrö­din­ger equation and rearranging produces the so-called Hamiltonian eigenvalue problem

\begin{displaymath}
-\frac{\hbar^2}{2m} \nabla^2 \psi = E \psi
\end{displaymath}

Here $\psi$ is called the energy eigenfunction.

According to classical relativity however, the energy $E$ of a particle in empty space is not just kinetic energy, but also rest mass energy $mc^2$, where $c$ is the speed of light. In particular, chapter 1.1.2 (1.2),

\begin{displaymath}
E = \sqrt{(mc^2)^2 + p^2c^2}
\end{displaymath}

The momentum can be identified with the same operator as before. But square roots of operators are very ugly. So the smart thing to do is to square both sides above. Making the same substitutions as for the Schrö­din­ger equation and cleaning up then gives the “Klein-Gordon equation”
\begin{displaymath}
\fbox{$\displaystyle
- \frac{1}{c^2} \frac{\partial^2\Ps...
...abla^2 \Psi = \left(\frac{mc^2}{\hbar c}\right)^2 \Psi
$} %
\end{displaymath} (A.43)

Solutions $ce^{-{{\rm i}}Et/\hbar}\psi$ with definite energy $E$ satisfy the time-independent Klein-Gordon equation or square Hamiltonian eigenvalue problem

\begin{displaymath}
- \hbar^2c^2 \nabla^2 \psi + (mc^2)^2 \psi = E^2 \psi
\end{displaymath}

This may be rewritten in a form so that both the Schrö­din­ger equation and the Klein-Gordon equation are covered:
\begin{displaymath}
\fbox{$\displaystyle
\mbox{empty space:}\quad
- \nabla...
...sqrt{E^2 - (mc^2)^2}}{\hbar c}
\end{array}
\right.
$} %
\end{displaymath} (A.44)

Here the constant $k$ is called the wave number. Note that the nonrelativistic energy does not include the rest mass energy. When that is taken into account, the Schrö­din­ger expression for $k$ above is the nonrelativistic approximation for the Klein-Gordon $k$ as it should be.

Further note that relativistic or not, the magnitude of linear momentum $p$ is given by the “de Broglie relation” $p$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\hbar}k$. That is because relativistic or not the momentum operator is $\hbar\nabla$$\raisebox{.5pt}{$/$}$${\rm i}$, so ${\widehat p}^2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vphantom0\raisebox{1.5pt}{$-$}$$\hbar^2\nabla^2$. Similarly, relativistic or not, the energy is associated the operator ${\rm i}\hbar\partial$$\raisebox{.5pt}{$/$}$$\partial{t}$. That means that the time-dependent factor in states of definite energy is $e^{-{{\rm i}}Et/\hbar}$. That allows the energy to be associated with an angular frequency $\omega$ by writing the exponential as $e^{-{\rm i}{\omega}t}$. The relationship between energy and frequency is then $E$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\hbar}\omega$. That is known as the “Planck-Einstein relation” when applied to photons. In short, relativistic or not,

\begin{displaymath}
\fbox{$\displaystyle
p = \hbar k \qquad E = \hbar\omega
$} %
\end{displaymath} (A.45)

The wave number $k$ is the quantum number of linear momentum, and the angular frequency $\omega$ is the one of energy. See addendum {A.19} for more on how these numbers arise physically from symmetries of nature.

It may be noted that Schrö­din­ger wrote down the Klein-Gordon equation first. But when he added the Coulomb potential, he was not able to get the energy levels of the hydrogen atom. To fix that problem, he wrote down the simpler nonrelativistic equation that bears his name. The problem in the relativistic case is that after you add the Coulomb potential to the energy, you can no longer square away the square root. Eventually, Dirac figured out how to get around that problem, chapter 12.12 and {D.82}. In brief, he assumed that the wave function for the electron is not a scalar, but a four-di­men­sion­al vector, (two spin states for the electron, plus two spin states for the associated antielectron, or positron.) Then he assumed that the square root takes a simple form for that vector.

Since this addendum assumes a particle in empty space, the problem with the Coulomb potential does not arise. But there are other issues. The good news is that according to the Klein-Gordon equation, effects do not propagate at speeds faster than the speed of light. That is known from the theory of partial differential equations. In classical physics, effects cannot propagate faster than the speed of light, so it is somewhat reassuring that the Klein-Gordon equation respects that.

Also, all inertial observers agree about the Klein-Gordon equation, regardless of the motion of the observer. That is because all inertial observers agree about the rest mass $m$ of a particle and the value of the speed of light $c$. So they all agree about the right hand side in the Klein-Gordon equation (A.43). And the left hand side in the Klein-Gordon equation is also the same for all inertial observers. You can crunch that out using the Lorentz transform as given in chapter 1.2.1 (1.6). (Those familiar with index notation as briefly described in chapter 1.2.5 recognize the entire left hand side as being simply $\partial^\mu\partial_\mu\Psi$. That is unchanged going from one observer to the next, because the upper index transforms under the Lorentz transform and the lower index under the inverse Lorentz transform. The operator $\partial^\mu\partial_\mu$ is called the “D’Alembertian,” much like $\nabla^2$ is called the Laplacian.)

But the bad news is that the Klein-Gordon equation does not necessarily preserve the integral of the square magnitude of the wave function. The Schrö­din­ger equation implies that, {D.32},

\begin{displaymath}
\int_{\rm all} \vert\Psi\vert^2{\,\rm d}^3{\skew0\vec r}= \mbox{constant, the same for all time}
\end{displaymath}

The wave function is then normalized so that the constant is 1. According to the Born statistical interpretation, chapter 3.1, the integral above represents the probability of finding the particle if you look at all possible positions. That must be 1 at whatever time you look; the particle must be somewhere. Because the Schrö­din­ger equation ensures that the integral above stays 1, it ensures that the particle cannot just disappear, and that no second particle can show up out of nowhere.

But the Klein-Gordon equation does not preserve the integral above. Therefore the number of particles is not necessarily preserved. That is not as bad as it looks, anyway, because in relativity the mass-energy equivalence allows particles to be created or destroyed, chapter 1.1.2. But certainly, the interpretation of the wave function is not a priori obvious. The integral that the Klein-Gordon equation does preserve is, {D.32},

\begin{displaymath}
\int_{\rm all}
\left\vert \frac{1}{c} \frac{\partial\Psi...
...si \right\vert^2
{\,\rm d}^3{\skew0\vec r}= \mbox{constant}
\end{displaymath}

It is maybe somewhat comforting that according to this expression, the integral of $\vert\Psi\vert^2$ must at least remain bounded. That does assume that the rest mass $m$ of the particle is not zero. Photons need not apply.

Another problem arises because even though the square energy $E^2$ is normally positive, the energy $E$ itself can still be both positive or negative. That is a problem, because then there is no lower limit to the energy, there is no ground state. The particle can then transition to states of lower and lower energy tending to minus infinity. That would release unbounded amounts of energy. (Since the kinetic energy can be arbitrarily large, the positive value of the energy can be arbitrarily large. That makes the negative value of the energy also arbitrarily large in magnitude.)

You might say, just ignore the negative energy possibility. But Dirac found that that does not work; you need both positive and negative energy states to explain such things as the hydrogen energy levels. The way Dirac solved the problem for electrons is to assume that all negative states are already filled with electrons. Unfortunately, that does not work for bosons, since any number of bosons can go into a state.

The modern view is to consider the negative energy solutions to represent antiparticles. In that view, antiparticles have positive energy, but move backwards in time. For example, Dirac’s negative energy states are not electrons with negative energy, but positrons with positive energy. Positrons are then electrons that move backward in time. To illustrate the idea, consider two hypothetical wave functions of the form

\begin{displaymath}
e^{-{\rm i}E t/\hbar} \psi_1 \quad \mbox{and} \quad e^{{\rm i}E t/\hbar} \psi_2
\end{displaymath}

where $E$ is the positive root for the energy. The first wave function is no problem; it is of the form of a wave function that you would get for a nonrelativistic particle of energy $E$. The second wave function is the problem. It is not considered to be a particle of negative energy $\vphantom0\raisebox{1.5pt}{$-$}$$E$. Instead it is considered an antiparticle of positive energy $E$ that moves backward in time. It is the reversal of the relevant direction of time that causes the sign change in the argument of the exponential.

You see why so much quantum physics is done using nonrelativistic equations.