# Does spontaneous emission count as a measurement?

It is my understanding that in the density matrix formalism for open quantum systems the environment-induced measurements/collapses/projections are accounted for by the Lindblad superoperator (from What is the physical meaning of the Lindblad operator? in PSE).

The Lindblad operator is notably used to describe spontaneous emission.

Does this mean that, in the context of the so-called "measurement problem", a spontaneous emission is to be considered as a measurement?

Would that be also true for alpha decay and other similar quantum tunnelling processes?

• The interaction of an atom with the field is not a measurement, unless... you make a measurement. If you define spontaneous emission as a process where the atom ends up in its ground state with an extra photon in the field, then you are already assuming that you have made a measurement. If you do not observe the system in any way, then it does not make sense to talk about spontaneous emission, there is just a unitarily evolving entangled state of the atom and the field. A problem arises here only if you discuss "events" independently of measurements: in QM the only events are measurements. – Mark Mitchison May 26 '16 at 13:44

The Lindblad operator describes the attentuation of a superposition of states so the density matrix is $\rho_{ij} \simeq e^{-\Gamma_{ik}}\rho_{kj}$ for some states so the density matrix is reduced in a type of collapse mechanism.
Let us look at the Einstein coefficients for photon absorption $B_{12}$ and induced emission $B_{21}$. These contribute the absorption coefficient $k = \frac{E}{4\pi} (n_1B_{12} - n_2B_{21})$ for photons of energy $E = \hbar\omega$. This is a sort of negative absorption, so that $$\dot n_1|_{-abs} = B_{21}n_2\rho(E),$$ for $\rho(E)$ the energy density. This is coupled to the absorption rate $$\dot n_1|_{+abs} = -B_{21}n_1\rho(E),$$ and the spontaneous emission rate for $n_i$ $\dot n_i = (-1)^{i+1}A_{21}n_i$. If I assume the rates of spontaneous emission are constant then the population $n_1$ grows linearly with time. This means there number of these states is growing.
The reduction of a wave function is an aspect of spontaneous emission, and we can think of this as at least related to a measurement. With the Einstein coefficients, there is the spontaneous emission coefficients $A_{21}$, which must be nonzero. This is required in order the population inversion of states occurs. If you had $A_{21} = 0$ and were generating these $n_1$ states this would be a sort of quantum cloning. We are all aware of the no-cloning theorem. We then can think of this as a sort of thermodynamics process; in order to generate more of these $n_1$ states we must have the input of energy and by the decoherence of states in spontaneous emission entropy. So the spontaneous emission process shares features common to wave function reduction, whether that be a real or dynamic collapse or a phenomena seen in MWI. I am not getting into interpretations here.