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 Mar 1 comment How can (in Dirac's terminology) the product of two “real” linear operators be “not real”? And self-adjoint/Hermitian means the operator has real eigenvalues? Mar 1 comment How can (in Dirac's terminology) the product of two “real” linear operators be “not real”? I cannot find the word Hermitian on the page before. But maybe you are referring to this sentence: "A linear operator may equal its adjoint, and it is then called self-adjoint. It corresponds to a real dynamical variable, so it may be called alternatively a real linear operator." So for Dirac a real linear operator is a self-adjoint operator and an imaginary operator is then a non-self-adjoint operator? Mar 1 asked How can (in Dirac's terminology) the product of two “real” linear operators be “not real”? Feb 28 awarded Nice Question Feb 22 awarded Popular Question Jan 28 awarded Yearling Jan 12 revised Temperature below absolute zero? typo Jan 12 comment Less than absolute zero possible? A recent paper in Nature (nature.com/nphys/journal/v10/n1/full/nphys2815.html) claims, that negative temperature is a concept based on an inconsistent definition of entropy, see also my answer here: physics.stackexchange.com/a/93398/1648 Jan 12 answered Temperature below absolute zero? Jan 11 comment Is it possible to split baryons and extract useable energy out of it? So one cannot fire an electron onto a proton and by that converting it into a neutron (inverse beta decay/electron capture), waiting for the neutron to beta decay and hope that the electron that comes out has more energy than the electron that I used to begin with? But what if that happens in a radioactive nucleus or otherwise exited nucleus? Couldn't the electron that comes out have a bit more energy than the incoming one, by taking with it some energy from the nucleus? Jan 11 asked Is it possible to split baryons and extract useable energy out of it? Dec 16 awarded Notable Question Dec 14 awarded Popular Question Nov 3 comment Why are Only Real Things Measurable? I made further edits to my answers and removed Dirac's quote from two of them. Nov 3 revised About the complex nature of the wave function? deleted 680 characters in body Nov 3 revised Can one do the maths of physics without using $\sqrt{-1}$? deleted 517 characters in body Nov 3 comment QM without complex numbers I do not disagree with your answer. One can for sure formulate QM in a way that uses pairs of real numbers instead of complex ones. However as Dirac points out, one has to be aware that in QM measurements might not commute. If one would formulate QM with pairs of real numbers, it would be more difficult to distinguish between pairs of real numbers, where measurements commute and between pairs of real numbers, where measurements don't commute. Nov 3 comment QM without complex numbers I'm not sure if that is a good example, but think about the wave function described by Schrödingers equation. One could split Schrödingers equation into two coupled equations, one for the real and one for the imaginary part of the wave function. However one cannot measure the phase and the amplitude of the wave function simultaneously, because both measurements interfere with each other. To make this manifest, one uses a single equation with a complex wave function, and generates the observable real quantity by squaring the complex wave function. Nov 3 comment Why are Only Real Things Measurable? Dear Qmechanic, I believe otherwise and I'm willing to take the risk and let the crowd decide if my answers are relevant or not for the questions. I don't care about the reputation points. If I get any, I promise I will donate them to Dirac. Nov 3 comment Quantum version of the Galton Board Thank you! Your last reference also led me to this article in Wikipedia with a picture of the distribution: en.wikipedia.org/wiki/Quantum_walk . However, is it true that neither for the fermion nor for the boson case the resulting distribution can be expressed in closed form?