# Why Klein-Gordon Equation violate postulate of quantum mechanics?

In QFT for gifted amateur, authors stated the probability current may be negative, a negative probability violates the Copenhagen interpretation. However, in the book QFT by Srednicki mentioned the second order derivative in time causing the norm of state vectors is not in general time independent.

Can anyone explain why even time-dependent hamiltonian for Schrodinger equation norm is constant, also illustrate how Klein-Gordon Equation results in norm of state vectors generally time dependent?

The GK "norm" $$\langle \phi|\phi\rangle = \int d^3 x (\phi^*\partial_t \phi-(\partial_t \phi^*)\phi)$$ is time independent for solutions of KG equation, but may be negative. So whatever it is, it is not a probability.

It is not difficult to show that the Schroedinger norm $$\langle \psi|\psi\rangle = \int d^3x |\psi|^2$$ is time independent for $$i\partial_t \psi(x,t) = \left(- \frac 12 \nabla^2+ V(x,t)\right)\psi(x,t)$$ even for $$V$$ depending on $$t$$, but more importantly $$|\psi|^2$$ is positive, so $$\langle \psi|\psi\rangle$$ can have a probability interpretation. However if $$\phi$$ satifies $$KG$$ rather than Shroedinger there is no reason for $$\int |\phi|^2d^3x$$ to be time independent.

It's easiest to answer this in terms of conserved densities and currents. A general local conservation law has the form $$\frac{\partial \rho}{\partial t}=-\nabla\cdot\mathbf{J}$$ For the Schrodinger equation, multiplying through by the complex conjugate, we have $$\psi^\star i\hbar\partial_t\psi=-\frac{\hbar^2}{2m}\psi^\star\nabla^2\psi+V(x)\psi^\star\psi$$ Taking the complex conjugate of the above expression and then subtracting it from the above, we have \begin{align}i\hbar (\psi^\star\partial_t\psi+\psi\partial_t\psi^\star)&=-\frac{\hbar^2}{2m}(\psi^\star\nabla^2\psi+\psi\nabla^2\psi^\star)\\ \partial_t(\psi^\star\psi)&=\nabla\cdot\left(\frac{i\hbar}{2m}(\psi^\star\nabla\psi+\psi\nabla\psi^\star)\right) \end{align} Thus, we have a conserved quantity, $$\rho=\phi^\star\phi$$. This can be interpreted as the probability density of finding the particle; it is positive definite. If we do this for Klein-Gordon, though, we have \begin{align}(\phi^\star\partial^2_t\phi-\phi\partial^2_t\phi^\star)&=c^2(\psi^\star\nabla^2\psi-\psi\nabla^2\psi^\star)\\ \partial_t(\phi^\star\partial_t\phi-\phi\partial_t\phi^\star)&=\nabla\cdot(c^2(\psi^\star\nabla\psi-\psi\nabla\psi^\star)) \end{align} Thus, the conserved quantity here is $$\rho=\phi^\star\partial_t\phi-\phi\partial_t\phi^\star$$; this is not always positive definite, and thus can't possibly be interpreted as probability.

Note that this lines up with Mr. Stone's answer. For Schrodinger, we have \begin{align} \frac{d}{dt}\langle\psi|\psi\rangle &= \int_\mathcal{V} \psi^\star\psi\\ &=\int_\mathcal{V}dV\partial_t\rho\\ &=\int_\mathcal{V}dV\, \nabla\cdot\left(\frac{i\hbar}{2m}(\psi^\star\nabla\psi+\psi\nabla\psi^\star)\right)\\ &=\int_\mathcal{S}d\mathbf{a}\cdot\left(\frac{i\hbar}{2m}(\psi^\star\nabla\psi+\psi\nabla\psi^\star)\right)\\ &= 0 \end{align} because $$\mathcal{V}$$ is over all space. Thus, the Schrodinger norm is time-independent. For Klein-Gordon, it's the exact same business \begin{align} \frac{d}{dt}\langle\psi|\psi\rangle &= \int_\mathcal{V} (\phi^\star\partial_t\phi-\phi\partial_t\phi^\star)\\ &=\int_\mathcal{V}dV\partial_t\rho\\ &=\int_\mathcal{V}dV\, \nabla\cdot\left(c^2(\psi^\star\nabla\psi+\psi\nabla\psi^\star)\right)\\ &=\int_\mathcal{S}d\mathbf{a}\cdot\left(c^2(\psi^\star\nabla\psi+\psi\nabla\psi^\star)\right)\\ &= 0 \end{align}

• Are you missing a minus in the first equation?
– Cryo
Apr 15, 2021 at 17:12
• Oh, yes, thank you. Apr 15, 2021 at 18:48