Solution of Klein-Gordon equation admit no probability interpretation

Question

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?

Answer 1

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!

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Edit

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.

Answer 2

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.