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Symmetries as the definition of particle charges Modern realistic particle physics theories are constructed from the requirement that there is such symmetry group which defines the quantities which are conserved in all processes which are described by theory (free propagation, interactions). This symmetry group is given as the direct product of subgroups of ...


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An observable in quantum mechanics is a measurable quantity in an experiment or observation. A postulate for building the mathematical theory of quantum mechanics is that 2.With every physical observable q there is associated an operator Q, which when operating upon the wavefunction associated with a definite value of that observable will yield that ...


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Random variables satisfy the Kolmogorov axioms for probability; quantum observables do not. In particular, any four-tuple of binary random variables (with any joint distribution) satisfies Bell's Inequality, while there are four-tuples of quantum observables that don't.


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An observable in quantum mechanics is an operator (say $\widehat{\mathcal{O}}$) on the Hilbert space (Say $\mathcal{H}$) of physical states, such that eigenkets in (say $\widehat{\mathcal{O}}$) in $\mathcal{H}$ span $\mathcal{H}$. The eigenvalues of $\widehat{\mathcal{O}}$ are then the observable values of some classical variable $\mathcal{O}$, even though ...


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What you say is quite reasonable. At the risk of being slightly more pedantic, I would say that physical observables are only those random variables that are Hermitean. Any operator (Hermitean or not) is a random variable -- in quantum mechanics these might be various properties of a particular state like spin, energy, etc. In quantum field theory, the ...


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This subtlety is related to the fact that the momentum operator $\hat{P}$ (unlike the Hamiltonian $\hat{H}=\frac{\hat{P}^2}{2m}$) has no eigenfunctions compatible with the Dirichlet boundary conditions, and $\hat{P}$ is not a self-adjoint operator. This is essentially Example 4 in F. Gieres, Mathematical surprises and Dirac’s formalism in quantum mechanics, ...


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I would say the answer is rather simple. If you do various experiments to measure the position (or momentum) of a particle, then QM only gives you consistent and accurate predictions when the mathematical symbols $x$ and $p$ are associated with the measured position or momentum. However, position and momentum are not the only canonically conjugate operators ...


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The answer may be deeper than you expect. The operator $K$ that the OP defined --- here it will be called $-i\nabla_x$ --- is the generator of space translations. And the operator $X$ --- here called $x$ --- is the generator of momentum translations. If you exponentiate them, to form the corresponding one-parameter unitary groups of translation, you get ...



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