This is actually a very interesting question.  Detectors that have preferences for certain frequencies I'm not sure fall in the usual category of measurement devices described by LCSA operators (Cohen-Tannoudji).  Nevertheless you can then ask is the inefficiency in the process of collapse or in the process of transmission of the signal.  I have a framework for such a theory but it might be equivalent to that of imperfect measurement devices.  http://academic.reed.edu/physics/faculty/wheeler/documents/Quantum%20Mechanics/Miscellaneous%20Essays/Generalized%20Quantum%20Measurement/Imperfect%20Quantum%20Measurements.pdf

So my vote is that you will see more results in the higher density region and there will be more uncollapsed amplitude in the other.  

Experimentally this is studied in ion traps, e.g. http://www.nature.com/nature/journal/v412/n6842/full/412052a0.html

I don't know if i understand this question, but if i do, isn't the answer basically contained in Fermi's "golden rule"?

I think Bulnes is referring to a set of measurement devices of finite volume in disjoint regions with different responsive properties.  Is this right? 

This is likely off thread, but it gives me a chance to vent about the comment i hear so often that nature likes to be in the lowest energy state. No it doesn't! Energy is conserved. For example, when an excited atom spontaneously emits a photon, it transitions to a lower energy state, but the EM field has gained the energy of the emitted photon. The total energy is unchanged. Why should the EM field be favored? It's more of a case of the second law of thermodynamics. There are a lot more photon states than atomic states, so the probabilities are in favor of a photon being emitted. Nature is not lazy, just messy.

You can define the energy as a projection over the full set of superpositions =\sum (as I'm sure you know) but experimentally this is not possible.  One projects out well defined values.  This naturally leads to the question of conservation of energy, etc in such measurements.  This is not resolved but I am pretty sure it is not so except over many such events.  Many worlds gives a net conservation over the bifurcating spaces.  

Experimentally, the issue of the conservation of energy (and momentum)  in quantum mechanics was the content of the experiment known since as the Compton effect (in response to a paper by Bohr, Kramers and Slater). Theoretically, it's expressed by the the fact that time translation invariance is a global symmetry, also,  in quantum mechanics and this has consequences for the correlation functions, i.e. transition amplitudes and corresponding probabilities, that can be calculated in any convenient way (whether canonical, where one constructs the corresponding unitary operator or path integral, where one shows that it implies identities between correlation functions), i.e. also for configurations that do not solve the classical equations of motion.  This is, now, standard textbook material and can be obtained as a consequence of Noether's theorem applied to quantum mechanics as a quantum field theory in one spacetime dimension, for example. It is, of course, a subgroup of the Galilean transformations, that are the global symmetries of non-relativistic quantum mechanics, that, in turn, are the appropriate limits of the Lorentz and Poincaré transformations, that are the global symmetries of the Standard Model and whose validity is routinely checked as background. 

Everyone knows about the Compton effect but when it is applied to near plane waves (energy eigenstates) it is not really relevant to the question.  The problem is easier to explain in terms of position measurement and center of mass conservation.  In this case, collapse generates a system with a shifted center-of-mass unless there is some hidden back reaction on the measurement device.  

I understand the conservation of energy (at least in Minkowski space) as follows. An atom can be in a combination of several energy eigenstates (a consequence, i believe, of the uncertainty principle), but when a photon, say, is emitted, it occurs over some time delta t, during which the energy is uncertain by an amount delta E (again, the uncertainty principle). The emitted photon also has an energy uncertainty delta E until it is measured. The measurement requires delta t to be zero (within experimental error), therefore the energy of the photon is determined (within experimental error). The second law of thermodynamics says that you can create order so long as you create a greater disorder elsewhere. The randomly emitted photon is disorder. The atom in a specific eigenstate is order (but only after the photon is detected!). The detection of the photon has to create disorder in the detecting device. Energy in the quantum world is not conserved during a quantum transition where the uncertainty principle obtains but only after it has occurred. I think this is consistent with quantum theory.

Dear Francis,

I guess I must disagree with you at least on one point: the expectation value of energy is always rigorously constant, whether a state is in an eigenstate of the Hamiltonian or otherwise. So the statement that ``when a photon, say, is emitted, it occurs over some time delta t, during which the energy is uncertain by an amount delta E'' is misleading as it stands. This statement, however, is only true for the full interacting Hamiltonian. If you were to ask by how much the energy contained in the field and that in the atom may vary in time during the emission process, I do believe your claim to be correct. But the ``time-energy'' uncertainty must be used with rather greater care than the usual one.

Similarly ``Energy in the quantum world is not conserved during a quantum transition where the uncertainty principle obtains'' is misleading. In fact, the idea that energy might only be conserved ``in the average'' was the idea central to the paper by Bohr, Kramers and Slater, which turned out to be incorrect.

On the other hand, I very much like your take on the second law. Sorry for nitpicking.

@Leyvraz, one question.  If you do a single experiment, do you measure the expectation value every time?  If not, how is energy conserved?

``Do you measure the expectation value every time?'' Clearly not, since you (usually) measure an eigenvalue of the observable you measure, whereas the expectation value is in general no eigenvalue.

To clarify my meaning, I was not speaking of measurement. When I spoke of energy, I meant that Hamiltonian which defines the time evolution. In a measurement, you have a complex interchange of energy between the apparatus and the system, which cannot in general be made arbitrarily small. The statement that a measurement which lasts less than a certain time cannot measure energy to arbitrary accuracy may well be true. But to state that in an emission process, which is not a measurement, energy can fluctuate via the time-energy uncertainty principle, is misleading: total energy, involving both field, atom and interaction, is exactly conserved. And even the statement I made above (concerning energy measurements made in a short time) is probably incorrect in general: for details, see Aharonov and Bohm, Phys. Rev. 122, 1649 (1961)

I think the conservation in BKS was for scattering events not in measurements in general (typically treated as non unitary events).  I gave an example above of another conserved quantity from classical physics that seems manifestly not conserved in measurements.  

The energy-time uncertainty relations have certainly suffered the greatest abuse over time (and deservedly so).  

I think i misunderstood the question or else the question evolved. Am i abusing poor old Heisenberg by offering this opinion (or question)? Wouldn't an atom in a state that is a combination of eigenstates have an uncertainty of energy such that the lifetime of the state would be finite? You might then measure an emitted photon and get a definite energy, which would be different from the expectation value. But how would we even know what the expectation value is in the first place? I would think you would have to have a large enough population of similarly prepared atoms, measure the emitted photon energies, and then figure out the expectation value. Does this make any sense?

@Francis, I think it mostly makes sense to me.  The problem is that QM is so formulaic when it comes to measurement that we have these sorts of ambiguities.  The question of why an isolated 2p H atoms state decays is an old one.  Often QED is evoked to get an answer.  If there is a slight superposition to start with then one has an oscillating dipole and the semiclassical derivation gives reasonable results for decay times although these derivations are rather delicate.  The exponential behavior we derive is known to not be valid for long times (Sakurai appx B).  What I never got was if we actually only end up with a superposition of some states is the 1s with radiation and some left in the 2p with no radiation.  The observation then lets us only see one of these states as the result.  I never liked evoking Heisenberg to get these results.  It looks no better a dimensional analysis justification.  

I think you only know about the expectation value by theory and your preparation or by examining many identical systems like you say.  

Im at first tempted to answer the question by stating that humans have

preferences, not particles.

But suppose that on one side you have a metal, and an electron comes

comes along and tries to enter a gap region of a semiconductor, clearly it will not

propagate much further. But suppose it enters the conduction band of a semiconductor, then it keeps on going.

@Francis: ``Wouldn't an atom in a state that is a combination of eigenstates have an uncertainty of energy such that the lifetime of the state would be finite? ''

I am afraid that is not correct in general. You are thinking of an excited atom, in which case there are other issues, to which I shall returnin the next paragraph. But if you take, say, a particle in an infinite square well in a combination of the first two eigenstates, you can solve the time-dependent Schroedinger equation and see that the solution is an undamped oscillation in the space generated by the two eigenstates. There is no finite lifetime, and that is always the case when the system has purely discrete spectrum.

On the other hand, if you wish to explain spontaneous emission, or in other words, why a H in the 2p state decays to 1s and a photon, you need  to consider its coupling to the quantized electromagnetic field (semiclassical will not really work, since you can only have a truly semiclassical approximation if the photon number is large, which is not well satisfied if you emit only one photon). You do not need the truly difficult parts of QED, though, because you can stick with a limited description of the interaction, namely the dipole approximation. Within this approximation, the eigenstate 2p of the hydrogen Hamiltonian is not an eigenstate of the full interacting Hamiltonian any more. On the other hand, there exists a continuum of states having the same energy as the initial state, and this, indeed, is a recipe for decay. The reason for this is, in fact, related to the original question: the presence of a (very) large number of states into which the initial state can decay makes it impossible to get back to the original state, so that decay indeed occurs.

Professor Leyvraz, you make some good points and also made me think. The situation of the infinite square well is, to my mind, not realistically quantum mechanical. It is, qualitatively, about the same as a string fastened at both ends that vibrates in several modes at once. My understanding of the uncertainty principle is that there always has to be an interaction. Certainly a system isolated from the environment and in its lowest entropy state will last indefinitely. I sort of assumed (but didn't mention) an interaction with the external virtual EM field when i wrote about the atom (undoubtedly excited) in a combination of states. Thank you for clarifying things for me.

Well, you can think semiclassically, that a potential barrier higher than

the original electron energy will repell it, but it stays in a region where the

potential is lower than its energy.

Unless there is an inelastic event,  normal

Quantum theory assumes that the electron has some a priori definite energy.

As to the number of available energy states, I think I already answered. A gap

has almost no states, but a band has many.

If your question is different, please explain.