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The usual explanation of spontaneous radiation is that the energy eigenstates are perturbed by QED interaction, so that the eigenstates obtained from single-particle QM are no longer eigenstates of the full Hamiltonian, and in turn different (original) eigenstates can mix after some time-evolution. However I'm not quite convinced, why in the first place the states must be in single-particle Hamiltonian eigenstates instead of full Hamiltonian?

EDIT:I need to rephrase it a bit since the answers reflect that my statement has been misleading. I'm not questioning the fundamental principle of superposition, I'm more concerned about the phenomenological fact that spontaneous emissions always happen experimentally after you excite an atom(if I'm not mistaken). But isn't it possible to excite the atom to an QED eigenstate such that it stays there forever?

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I think that is possible, such an eigenstate would be something like a mixture of (excited atom/no photon) + (unexcited atom/photon). However then you have the problem of keeping the photon portion from interacting with the rest of the universe and exciting other atoms. If you have a single atom in a high-Q cavity you can create this situation, but it will inevitably decay through cavity losses. –  user2963 Sep 4 '12 at 21:25
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4 Answers 4

First of all, concerning the question

Why the atom is not physically in the full-Hamiltonian eigenstates which we can't calculate exactly,

akhmeteli correctly writes that this question and all analogous questions and proposals saying "a physical system must always be in some preferred basis vectors" is a fundamental misunderstanding of quantum mechanics. A basic universal postulate of quantum mechanics is the superposition principle: whenever a system is allowed to be in certain states, the same system is always allowed to be in an arbitrary superposition (complex linear combination) of these states! So it's manifestly wrong to say that an atom is "always in an energy eigenstate" or "always in one of the basis vectors of another basis you randomly picked".

It may be hard to prepare a large enough physical system, like Schrödinger's cat, in a general superposition and it is even harder for many people to imagine that it's possible at all. However, in principle, it's always possible; it's what quantum mechanics guarantees. It just doesn't allow atoms or other physical systems to be "constrained" to "physically be" in a previously chosen basis.

Now, concerning the emission of radiation, first we must ask: What does it mean that there was an emission? It means that the energy carried by the radiation is different (higher) at the end than what it was at the beginning. It's possible to reformulate this proposition (which may be right or wrong after a particular time interval and in a particular physical system) in terms of photons: the emission of radiation occurs when the number of photons increases. These equivalences are completely general; they encode what the very phrase "emission of radiation" means.

We should ask: Why does a quantum mechanical theory allow the number of photons to be increased? It's because the Hamiltonian has the form

$$ H = \sum_{i} \hbar\omega_i \cdot N_i + \sum_i \hbar a_i \cdot d_i + \sum_i a^\dagger_i\cdot d_i + \text{non-photon-related} $$

The first term which is a sum over one-photon states just counts the total energy stored in the photons: each photon contributes $\hbar\omega$ to the total energy. This first term preserves the number of photons: an $X$-photon state is evolving to an $X$-photon state. Note that $N_i=a^\dagger_i a_i$ where $a_i,a^\dagger_i$ are properly normalized annihilation and creation operators.

The two following terms containing the letter $d$ change the number of the photons and they're needed for the emission and absorption processes because, as argued previously, emission or absorption is, by definition, the same thing as the change in the number of photons. So these terms contain either the annihilation operator or the creation operator without its Hermitian conjugate partner. I chose the letter $d_i$ for the dependence of these terms on the quantum number of the atoms – this $d_i$ acts on the atoms' degrees of freedom etc. and in some approximations, the most important piece is expressed by the atoms' electric dipoles etc.

It's important to emphasize once again that I am not saying that the number of photons $N_i$ must always be sharply defined in any state allowed for the physical system. It's not true at all; all superpositions are possible. We're just trying to calculate a particular meaningful answer – the probability of emission – and we find out that the probability of emission depends on the initial state, especially on the number of photons in various states. So we're considering these initial states with a certain number of photons because they're a part of the formulation of the question. One may also calculate the probability of emission for a general initial state that is a superposition of states of different values of $N$ but in that case, one must be a bit more careful in defining what the term "emission" really means.

The emission and absorption processes change one particular number of photons $N_i$ for one value of the index $i$. The Hamiltonian above shows that even if $N_i=0$ in the initial state, there is a nonzero probability that we will get a $N_i=1$ final state: so the number of photons may spontaneously jump even if it is zero to start with.

Albert Einstein was the first one who figured out the most general argument why it's so: the absorption has to be "stimulated" because one must absorb photons that actually exist to start with. No photons means nothing to absorb. But the probability of absorption is a function $f(N)$ of the number of photons in the initial state $N$ which therefore gets changed to $N-1$: one photon is subtracted.

Now, the microscopic probability (classically) or probability amplitude (quantum mechanically) must be the same for the time-reversed (more precisely: CPT-conjugated) process. The time-reversed process to absorption is emission. But the initial state gets mapped to the final state and vice versa.

Because the probability of absorption goes like $C\cdot N$ where $N$ is the number of photons in the initial state and because the number of photons in the final state is $N-1$ and because this $N-1$ becomes the number of photons in the initial state of the time-reversed process, we see that for $N-1$ photons in the initial state, the probability of emission is proportional to $N$. In other words, for $n=N-1$ photons in the initial state, the probability of emission goes like $n+1=N$. In the form $n+1$, the term $n$ corresponds to the stimulated emission and the term $1$ corresponds to the spontaneous emission.

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I guess my original post is somewhat misleading, please read the EDIT. –  Jia Yiyang Sep 4 '12 at 11:58
    
Hi @Jia, concerning your added question, no, it is not possible to excite an atom in such a way that it would become stationary. For a state to be stationary, it has to be an energy eigenstate. But because the energy contains terms proportional to $a$ and $a^\dagger$ that change the number of photons, energy eigenstates either have to be annihilated by this whole photon-non-conserving part or they have to be superpositions of states with different values of $N$ - and if you look at the coefficient, you will see that the average number of photons in the state would have to diverge. –  Luboš Motl Sep 5 '12 at 16:05
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I think nobody has answered why the probability of spontaneous emission is high and the atom quickly decays to its ground state. The reason is that there is only one state with the atom excited and no photons, while there are a huge number of states with the atom in its ground state and a photon emitted. For every angle of emission there is one state (or two states if one takes into account the helicity) and the photon may be emitted in all directions. In statical mechanical terms, the entropy of the full system increases when the atom emits a photon.

Sometimes one listens that the atom spontaneously decays because it wants to decrease its energy, but the argument is false because one has to considerer the full system atom plus electromagnetic field and in this case the energy is conserved as it must be in an isolated system (and the momentum is of course conserved as well, the atom recoils). The good argument is the increasing of entropy instead of the decreasing of energy.

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Ok, this is right, but in this case it isn't "entropy" which is increasing, but "quantum spreadoutness", the final superposed radiation field is still in a pure state. –  Ron Maimon Sep 6 '12 at 1:55
    
@RonMaimon Thank you for the correction. I just realize the term "entropy" is at least confusing because there are not statistical mixed states as you have noticed. What I meant is simply that the number of accessible states for the system is greater if the atom decays. So that the number the microstates is bigger. Is not correct to use the term "entropy"? I do not know the term (and what you mean by) "quantum spreadoutness". –  drake Sep 6 '12 at 2:09
    
There is no term. It becomes entropy once you measure outgoing photon direction, so entropy is right in some sense (hence +1), but it's not entropy until you make a measurement, it's just a quibble. –  Ron Maimon Sep 6 '12 at 2:14
    
This is a more convincing argument, so +1. But this only makes sense from a statistical point of view. If we do an experiment on one atom, we should adhere to quantum mechanical principles, that is, we excite an atom, then we measure energy(actually I'm not even sure what we are measuring in this case), and see if emission happens. In this scenario the number of microstates is not in the picture, the only issue is whether we have made a QED eigenstate. Or are you suggesting people have never actually excited an atom to a QED eigenstate because there are too many other possible states? –  Jia Yiyang Sep 7 '12 at 12:49
    
@JiaYiyang Let me give a very crude and heuristic argument. If one has an only dice (atom) with a lot of faces (final states which include emission in any direction and no emission) and throws it, the probability of getting a "one" (the atom remains in its excited state without emitting) is very low. Moreover, one is throwing the dice continuously, so maybe one is very lucky and the first time gets a "one" (the atom remains in its excited state during a very short period of time) but the next time one throws the dice one is going to get any other number (emission in some direction). –  drake Sep 7 '12 at 17:47
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The states don't have to be single-particle Hamiltonian eigenstates, but the single-particle Hamiltonian is still a very good and simple approximation, so it is widely used.

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The question is not about how to do the calculation, it's about why the atom is not physically in the full-Hamiltonian eigenstates which we can't calculate exactly. –  Jia Yiyang Sep 4 '12 at 3:10
    
@Jia Yiyang: It is my understanding that in general a system can be in a superposition of Hamiltonian eigenstates (in that case, if the eigenvalues do not equal each other, the system does not have a definite energy), unless you reject the principle of superposition. –  akhmeteli Sep 4 '12 at 3:46
    
Ok I guess my original post is somewhat misleading, please read the EDIT. –  Jia Yiyang Sep 4 '12 at 11:57
    
I don't have a clear idea of how the QED Hamiltonian eigenstates look in this case, but to get such an eigenstate you have to control a lot of phonon modes, and that does not seem feasible. It may be feasible for an atom in a cavity, where very few modes are relevant, and I vaguely remember that something like that was indeed done, but that is a different story. –  akhmeteli Sep 4 '12 at 13:30
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I believe the question is addressing the fact that for a hydrogen atom in the 2p state, for example, there is no oscillating dipole moment; and therefore it ought to be stable. This presents us with the problem of why it spontaneously emits radiation. It's not hard to trace the time evolution of the oscillating dipole moment once the process begins, and we have a superposition of the 1s and 2p states. Then the atom is an antenna, and it oscillates. (Note that the OP does not seem to have a problem with radiation taking place when the atom is in a mixed state.) But for the pure state, how does the oscillation get started?

The question is difficult to answer if we insist on thinking about a single atom; however, we can get a different point of view if we look at an ensemble of 100 hydrogen atoms. Let us excite one of those atoms to a 2p state. Then we can plausibly argue that it has no reason to radiate, and that the system ought to be stable.

Now let us consider an alternate system where all 100 atoms are in the mixed state, each of them excited to 1% of the 2p state? In this case all the atoms would be radiating classically via their oscillating dipole moments, and there would be no mystery.

In the first case, the system is stable, but in the second case, it radiates. But what is the difference between the two systems. In particular, there any experimental way in which we can distinguish the two cases?

In fact, I think I am correct in suggesting that the density matrix description of the two systems is identical; that in fact they are the same system; and depending only on how we choose to describe it, the mechanism of the radiation is apparent in one case and obscure in the other case.

By the way, I am not resorting to the ensemble example because I want to take advantage of collisions to bump the atoms out of their pure states and obscure the equilibrium. I think my argument is just as true for a single atom, but I don't know how to frame it in a convincing way for that case.

I have a more detailed analysis of these cases on my blog, where I calculate the amount of radiation expected semi-classically from Case B and equate it to the amount of radiation in Case A according to the law of spontaneous emission in a series of posts starting here.

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