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?

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    $\begingroup$ 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. $\endgroup$ May 26, 2016 at 13:44

1 Answer 1


Spontaneous emission and stimulated emission are cases related to measurement, but they are not identical to measurement. We might ask how the measurement of states and the preparation of states are related to measurement, or the decoherence of states. Both are necessary aspects of quantum physics. In order to do experiments we need to prepare identical states, where the laser does the job of duplicating photon states very well.

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.

A measurement is a form of decoherence, such as other events can cause decoherence. In a strict sense something like spontaneous emission is not a measurement however. In the Bohr sense a measurement means the reduction of states and the recording of this in a classical fashion. The mechanisms of preparing quantum states and measuring them requires a classical input, and the spontaneous and stimulated emission of photons have this feature. However, if nobody actually records the state of a photon then while these classical "driving mechanisms" are at play, there is not recorded classical piece of information. That in a strict sense defines a measurement.

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    $\begingroup$ The system doesn't really care whether a physicist has been paying attention, or not, so the to nature there is no difference between the two scenarios. It may be a crappy measurement from the standpoint of error calculus, but it's still a measurement. $\endgroup$
    – CuriousOne
    May 26, 2016 at 16:59

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