tag:blogger.com,1999:blog-8148573551417578681.post1393428749048004358..comments2019-03-11T10:35:25.796-07:00Comments on Dark Buzz: Why Cosmologists hate CopenhagenRogerhttp://www.blogger.com/profile/03474078324293158376noreply@blogger.comBlogger1125tag:blogger.com,1999:blog-8148573551417578681.post-30807006934257125132019-03-11T10:35:25.796-07:002019-03-11T10:35:25.796-07:00In quantum mechanics in practice, we prepare many ...In quantum mechanics in practice, we prepare many states and measure them in many ways, however in quantum mechanics as a global metaphysics we more think of there being a single state measured in many ways, in which case in solving the equation A_i = Tr[\hat M_i\hat\rho], with the A_i being all summary statistics of all the experimental raw data we have, the \hat M_i being an operator that represents the measurement that results in that summary statistic, and \hat\rho being the density matrix that represents the state, there is at least one basis in which \hat\rho is a diagonal matrix. If there is only one state then all measurement operators effectively commute because off–diagonal entries in such a basis make no contribution to the trace, so the one–state metaphysics of quantum mechanics is equivalent to the one–state metaphysics of classical (statistical) physics.<br />I make this argument (with LaTeX, so easier to read!) as a small part of my arXiv:1901.00526. I'd like to know if this argument appears elsewhere.Peterhttps://www.blogger.com/profile/08654675777726560464noreply@blogger.com