Let's consider the solutions $\psi$ of the Klein-Gordon equation: $$\bigg{(}\frac{\partial^2}{\partial t^2}-\Delta + m^{2}\bigg{)}\psi(x) = 0$$ and define: $$\rho = \frac{i}{2m}\bigg{(}\psi^{*}\frac{\partial \psi}{\partial t}-\frac{\partial \psi^{*}}{\partial t}\psi\bigg{)} \quad \mbox{and} \quad {\bf{j}} = -\frac{i}{2m}(\psi^{*}\nabla\psi -\nabla \psi^{*}\psi)$$ Then, with a little algebra one can show that the following continuity equation is satisfied: $$\frac{\partial \rho}{\partial t} + \nabla \cdot {\bf{j}} = 0$$

Question: I've heard that because the above continuity equation has both positive and negative solutions, the solutions of the Klein-Gordon equation do not have a probabilistic interpretation. Why is that? More precisely, what is the connection between probabilistic interpretations and continuity equations? Does it have something to do with Noether's Theorem?

  • $\begingroup$ Check "canonical quantization" and "Dirac Sea", those keywords should lead you somewhere... $\endgroup$
    – ohneVal
    Commented Mar 22, 2021 at 14:17

2 Answers 2


It simply does not make sense to speak of negative probability. Under the general axiomatic framework of probability, called the Kolmogorov axioms of probability, the first axiom is non-negativity.

What would it mean to say that there is a $-20\%$ probability of finding a particle in the region $x\in[x_1,x_2]$ upon the measurement of its position? Exactly, nothing!


The description of the issue by the OP is a bit fuzzy so I would clarify here the description that I have in mind of the issue.

The definition of a probability density associated with the wavefunctions satisfying the KG equations and admitting a continuity equation can be found to be $$\rho=\dfrac{i}{2m}\bigg(\psi^*\dfrac{\partial \psi}{\partial t}-\psi \dfrac{\partial \psi^*}{\partial t}\bigg)$$ with the continuity equation $$\nabla \cdot \vec{j}+\dfrac{\partial \rho}{\partial t}=0$$ where $$\vec{j}=\dfrac{1}{2mi}\big(\psi^*\nabla\psi-\psi\nabla\psi^*\big)$$

Since the KG equation is a second-order equation, one can freely choose the $\frac{\partial\psi}{\partial t}$ as a part of the initial condition -- and thus, there is nothing to ensure that the supposed probability density $\rho$ is non-negative. Moreover, even if you start out with carefully chosen initial conditions that lead to a non-negative initial probability density, the KG equations can evolve the system to such a state that the probability densities become negative. See, this old question of mine.

  • $\begingroup$ Thanks for your answer! I guess the intention of my question is to clarify the connection between $\rho$ and $\psi$. I mean, the statemente is "because $\rho$ is not necessarily positive, then $\psi$ has no probabilistic interpretation". But why one implies the other? $\endgroup$
    – MathMath
    Commented Mar 22, 2021 at 14:30
  • $\begingroup$ @MathMath Good point, no, the probability density is not $\vert\psi\vert^2$ here. It is the $i/2m (...)$ expression that I wrote down. The reason as to why that is the probability density and not $\vert\psi\vert^2$ is because one cannot find a probability density current satisfying the continuity equation if one were to take $\vert\psi\vert^2$ as the probability density. $\endgroup$
    – user87745
    Commented Mar 22, 2021 at 14:33

Noether's theorem tells us that there is a continuity equation corresponding to any symmetry the system has. For the Schrodinger equation, the global phase symmetry corresponds to a continuity equation involving a quantity which can be interpreted as a probability density current. If you try to do the same thing for the Klein-Gordon equation you find that this interpretation cannot be right because it would imply negative probabilities, which are absurd by definition of what probabilities are.


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