# Does the collapse of the wave function increase entropy of the atomic system itself?

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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It does. W.f. collapse means that you've made a measurement so the environment (e.g an emitted photon) has info about the internal state. Tracing out the environment leaves the atom in an entropyful mixed state –  Slaviks May 10 '13 at 4:33
And if you do postselection instead of a trace (ie detect and measure the photon) you can prepare a pure state with $S=0$ –  Slaviks May 10 '13 at 4:39

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.