Does wave-function collapse cause the entropy of the atom (ie. the sub-atomic particle system that makes up the atom) to increase?
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Disclaimer. I'm not sure it even makes sense to talk about changes in entropy of systems that undergo wavefunction collapse unless one also includes the measuring apparatus as part of the system. Having said this, here are my two cents that I hope are informative: The (von-Neumann) entropy of a quantum system prepared in a state (density operator) is defined as $$ S(\rho) = -k_B\mathrm {tr}(\rho\ln\rho) $$ Where $k_B$ is Boltzmann's constant. In particular, If a quantum system is described by a pure state (the notion of state as an element of a Hilbert space that you learn when you start out in QM), then its entropy is zero. As a result, if you prepare an atom such that its quantum state is pure, then its entropy will not depend on which pure state it is prepared in. For example, whether it's prepared in an energy eigenstate, or a linear combination of energy eigenstates, its entropy after being prepared in any pure state will be zero. In this sense, perhaps it can be said that the entropy of an atom that collapses from one pure state to another via projective measuement does not change. In order for the entropy of a quantum system consisting of an atom to be greater than zero, you would have to prepare the atom in a statistical mixture of pure states. For example, you could envision coupling a large sample of atoms to a heat bath. |
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