Can't the Negative Probabilities of Klein-Gordon Equation be Avoided? - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2019-06-16T13:41:51Z https://physics.stackexchange.com/feeds/question/340023 http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://physics.stackexchange.com/q/340023 9 Can't the Negative Probabilities of Klein-Gordon Equation be Avoided? Feynmans Out for Grumpy Cat https://physics.stackexchange.com/users/20427 2017-06-18T12:09:53Z 2017-06-18T13:28:49Z <p>I came across <a href="https://arxiv.org/abs/quant-ph/0608140" rel="noreferrer">these notes</a> of Dyson on Relativistic Quantum Mechanics. There on p. 3, he mentions that the issue with the Klein-Gordon equation is that the only way to relate $\psi$ with a probability density (that has a continuity equation) is to define $$\rho=\dfrac{\iota}{2m}\bigg(\psi^*\dfrac{\partial \psi}{\partial t}-\psi \dfrac{\partial \psi^*}{\partial t}\bigg)$$ with the continuity equation $$\nabla \cdot \vec{j}+\dfrac{\partial \rho}{\partial t}=0$$ where $$\vec{j}=\dfrac{1}{2m\iota}\big(\psi^*\nabla\psi-\psi\nabla\psi^*\big).$$ He says that the issue with such a probability density is that since the Klein-Gordon equation is a second-order equation, both $\psi$ and $\dfrac{\partial \psi}{\partial t}$ constitute the initial condition and thus, are arbitrary - leading to the unavoidable negative probability densities. </p> <p>But can't the requirement of a positive probability density be thought of as a restriction on the initial conditions themselves? Like in Special Relativity, velocity of a particle is a part of the initial condition but the theory restricts what kind of initial conditions one can have. Similarly, can't we restrict the form of the initial condition to ensure the non-negative nature of the probability density?</p> https://physics.stackexchange.com/questions/340023/-/340032#340032 7 Answer by J.G. for Can't the Negative Probabilities of Klein-Gordon Equation be Avoided? J.G. https://physics.stackexchange.com/users/96494 2017-06-18T12:58:36Z 2017-06-18T12:58:36Z <p>I'll work in the $+---$ convention so $\mathbf{k}\cdot\mathbf{x}-k_0t=-k\cdot x$. You should find that $\psi=\exp{\mp ikx}$ obtains $j^\mu=\pm m^{-1}k^\mu$. Complex conjugation of $\psi$ still gives a solution for the KGE (the TDSE doesn't work like this) and changes the signs of energy and probability. The problem with trying to magic away half the solutions is it breaks invariance under time reversal. This is related to the fact that $E^2=p^2+m^2$ is invariant under $E\to -E$.</p> <p>The general solution can be written as a Fourier transform. If you generalise a linear combination of planar-wave solutions with constant coefficients so that, after quantization, these coefficients are operator-valued, you encounter another difficulty: $\hat{\psi}$ won't be Hermitian. By contrast, if you keep all solutions the integrand has two terms: one with annihilation operators, the other with creation operators. If either term is deleted, $\psi$ will annihilate the vacuum bra but not the vacuum ket or vice versa, depending on the term retained.</p>