# What do the density function and the current vector in the continuity eq. form of the Klein-Gordin eq. mean?

In explaining Dirac's attempt to arrive at a relativistic Schrodinger equation, it is claimed that Dirac has started from

$$\left(-\frac{1}{c^2}\frac{\partial^2}{\partial t^2} + \nabla^2\right)\phi = \frac{m^2c^2}{\hbar^2}\phi$$ to arrive at $$\nabla\cdot J + \frac{\partial\rho}{\partial t} = 0$$ where $$\rho = \frac{i\hbar}{2m}(\psi^*\partial_t\psi - \psi\partial_t\psi^*)$$ $$J^k = \frac{i\hbar}{2m}(\psi^*\partial^k\psi - \psi\partial^k\psi^*)$$

Is there a mathematical/physical motivation for turning the first equation into a continuity equation or Dirac only blindly played with equations? What are the physical interpretations for $\rho$ and $J^{k}$ in the continuity equation? Do they have any physical significance?

• In quantum mechanics, there is a continuity equation as well. And those means a current of probability, and a density of probability. Basically meaning: "Probability is conserved". Perhaps its the same here... Oct 14, 2015 at 15:59

Nice questions, you ask all the right things.

Is there a mathematical/physical motivation for turning the first equation into a continuity equation or Dirac only blindly played with equations?

Yes, there is a very good reason why Dirac looked at the continuity equation. At the time, the relativistic extension of the Schroedinger equation was an open and pressing problem. According to the Copenhagen Interpretation, the immediate recipe for extension was to start with the relativistic equation for a particle's energy, $$E^2 = {\bf p}^2c^2 + (mc^2)^2$$ and replace the energy and 3-momentum by their operatorial counterparts, as quantum observables applied to a wave function $\psi$. In analogy to the non-relativistic case, energy would be replaced by the $\hat{E} = i\hbar \frac{\partial}{\partial t}$ operator and momentum by $\hat{p} = -i\hbar\nabla$. This yields the Klein-Gordon equation, $$\left( \nabla^2 - \frac{1}{c^2}\frac{\partial^2}{\partial t^2} \right)\psi = \left( \frac{mc}{\hbar}\right)^2\psi$$
If this were to be the correct equation, by analogy with the non-relativistic case it would have to produce the time-evolution of a probability density obeying a conservation equation. But if we try to use $\psi^*\psi$ as probability density, we see that it would have to satisfy a very strange higher order equation and there wouldn't be any obvious proper probability flux. On the other hand, it is not hard to notice that we can easily obtain from the Klein-Gordon equation the divergence of the probability flux as known from the non-relativistic case. Just write $$-i\psi^*\left( \nabla^2 - \frac{1}{c^2}\frac{\partial^2}{\partial t^2} \right)\psi + i\left[\left( \nabla^2 - \frac{1}{c^2}\frac{\partial^2}{\partial t^2} \right)\psi^*\right]\psi = -i\left( \frac{mc}{\hbar}\right)^2\psi + i\left( \frac{mc}{\hbar}\right)^2\psi = 0$$ and rearrange it into $$\nabla \cdot \left[ - \frac{i\hbar}{2m} \left( \psi^* ( \nabla \psi) - ( \nabla \psi^* ) \psi \right) \right] + \frac{\partial}{\partial t} \left[\frac{i\hbar}{2m}\left(\psi^*\frac{\partial \psi}{\partial t} - \frac{\partial \psi^*}{\partial t}\psi\right) \right] = 0$$ or $$\nabla \cdot {\bf j} + \frac{\partial}{\partial t}\rho = 0$$ for $${\bf j} = - \frac{i\hbar}{2m} \left( \psi^* ( \nabla \psi) - ( \nabla \psi^* ) \psi \right) \;\; \text{and} \;\; \rho = \frac{i\hbar}{2mc^2}\left(\psi^*\frac{\partial \psi}{\partial t} - \frac{\partial \psi^*}{\partial t}\psi\right)$$ Now we have something that looks like a probability flux, ${\bf j}$, but nothing that looks like a probability density, since obviously we cannot say that $\rho \ge 0$. Instead, $\rho$ can take both positive and negative values. So now we are left with the question: "What is actually the meaning of this continuity equation, and what is the meaning of the relativistic $\psi$, assuming the Klein-Gordon equation itself is physically meaningful?" This is why Dirac "played" with the continuity equation: to summarize the inconsistencies of the Klein-Gordon equation as a relativistic extension of the Schroedinger equation.

What are the physical interpretations for $\rho$ and $J_k$ in the continuity equation? Do they have any physical significance?

Yes, after all it turns out that $\rho$ and $J_k$ do have physical significance and the Klein-Gordon equation does provide a valid description of relativistic spin-zero particles. After Dirac developed his theory for relativistic spin-$\frac{1}{2}$ particles with mass, it was realized that, just as the Dirac equation has both positive and negative energy solutions, and the corresponding spinors must be interpreted separately, the Klein-Gordon too has positive and negative solutions that might just make sense if they were interpreted separately. Lo' and behold, the idea works out beautifully:

The energy of plane wave solutions to the Klein-Gordon, $$\psi = A e^{\frac{i}{\hbar}({\bf p}\cdot {\bf x}\; - \;Et)}$$ satisfies the relativistic dispersion $E = \pm \sqrt{{\bf p}^2 c^2 + (mc^2)^2} = \pm |E|$. Then the corresponding plane waves can be written as $$\psi^{(\pm)}_{\bf p} = A_{\pm} e^{\frac{i}{\hbar}({\bf p}\cdot {\bf x}\; \mp \;|E|t)}$$
and any solution of the Klein-Gordon equation can be regarded as a superposition of "positive energy" and "negative energy" solutions: $$\psi = \sum_{\bf p}{c^{(+)}_{\bf p}\psi^{(+)}_{\bf p}} + \sum_{\bf p}{c^{(-)}_{\bf p}\psi^{(-)}_{\bf p}} = \psi^{(+)} + \psi^{(-)}$$ Let's take a look at what the quantity $\rho$ means for plane wave solutions of only positive or negative E: $$\rho_{\bf p}^{(\pm)} = \frac{i\hbar}{2mc^2}\left( \left(\psi_{\bf p}^{(\pm)}\right)^* \frac{\partial \psi_{\bf p}^{(\pm)}}{\partial t} - \frac{\partial \left(\psi_{\bf p}^{(\pm)}\right)^*}{\partial t}\psi_{\bf p}^{(\pm)}\right) = \frac{i\hbar}{2m}\left( \left(\psi_{\bf p}^{(\pm)}\right)^* \left( \mp\frac{i}{\hbar}|E|\right)\psi_{\bf p}^{(\pm)} - \left( \mp\frac{i}{\hbar}|E|\right)^*\left(\psi_{\bf p}^{(\pm)}\right)^* \psi_{\bf p}^{(\pm)} \right) = \pm \frac{|E|}{2mc^2}\left(\psi_{\bf p}^{(\pm)}\right)^* \psi_{\bf p}^{(\pm)}$$ In other words, $\rho$ is positive for positive E plane waves, and negative for negative E waves. But how to interpret this? Again inspired by Dirac's theory, the positive E solutions are interpreted as positive charge solutions, while negative E solutions are assigned negative charge. As for the Dirac spinors, we now have, for instance, positive charge particles and negative charge anti-particles (yeah, it's backwards, but only a matter of convention). Then $\rho$ can be interpreted as the corresponding charge density, provided we redefine it to adjust for a "charge" factor. That is, for particles/antiparticles with elementary charge $\pm e$, we redefine $$\rho = \frac{ie\hbar}{2mc^2}\left(\psi^*\frac{\partial \psi}{\partial t} - \frac{\partial \psi^*}{\partial t}\psi\right)$$ Similarly, we need to add a charge factor to the current ${\bf j}$, which now becomes a charge current: $${\bf j} = - \frac{ie\hbar}{2m} \left( \psi^* ( \nabla \psi) - ( \nabla \psi^* ) \psi \right)$$ Accordingly, a general solution of the Klein-Gordon equation must be interpreted as a superposition of positive and negative charge components.

So the short answer to your question is: $\rho$ is a charge density for a spin-zero field and ${\bf j}$ is the associated current density. The wave function $\psi$ is a "charge amplitude" that can be separated into positive and negative charge components.

The formalism behind this interpretation is a little bit more involved then presented here. For instance, it can be shown from the stress-energy tensor for a Klein-Gordon field that both the positive charge and the negative charge components carry positive energy after all (for plane waves this is just |E|). You can also check that the Klein-Gordon field $\psi$ can describe neutral particles too, provided $\psi = \psi^*$, in which case both the charge density and the current density vanish identically. For charged fields, $\psi$ is complex, and $\psi$ and $\psi^*$ represent opposite charges. One example commonly cited as described by a Klein-Gordon field is the pion triplet $(\pi^+, \pi^0, \pi^-)$.

If you are interested, Walter Greiner's volume on "Relativistic Quantum Mechanics" has an excellent in-depth chapter on the Klein-Gordon equation with a lot of examples and finer points discussed.