USeptim said:
What means "fluctuations on the vacuum due to electric field been quantized"? The reason that excited states may fall into others with lower energy is explained by Fermi's golden rule and it can be explained with the existence of a "virtual photon" with energy \hbarw_{ab} (a is the current state and b the possible new state) whose hamiltonian would cause a perturbation that gives Fermi's transition rate.
With this approach it's only needed to add a new phenomena to classical QM.
Well, if you add a photon as a perturbation, then you're not dealing with a hydrogen atom in a
vacuum. You're dealing with a hydrogen atom with a photon flying by. I was referring to an absolutely empty vacuum, with no electric field besides that of the proton and electron.
You are right that a photon can cause "stimulated emission," but even in the absence of any photon, an excited hydrogen atom can decay to its ground state in a process called "spontaneous emission."
Let me quote Shankar's
Principles of Quantum Mechanics, Second Edition, Chapter 18 "Time Dependent Perturbation Theory." This comes after a discussion of stimulated emission and Fermi's Golden Rule. From pp.506-507:
The general formalism, illustrated by the preceding example, may be applied to a host of other phenomena involving the interaction of atoms with radiation. The results are always in splendid agreement with experiment as long the [sic] electromagnetic field is of macroscopic strength. The breakdown of the above formalism for weak fields is most dramatically illustrated by the following example. Consider a hydrogen atom in free space (the extreme case of weak field) in the state |2,l,m>. What is the rate of decay to the ground state? Our formalism gives an unambiguous answer of zero, for free space corresponds to A=0 (in the Coulomb gauge), so that H1=0 and the atom should be in the stationary state |2,l,m> forever. But it is found experimentally that the atom decays at a rate R≃109 second-1, or has a mean lifetime τ≃10-9 second. In fact, all excited atoms are found to decay spontaneously in free space to their ground states. This phenomenon cannot be explained within our formalism.
So are we to conclude that our description of free space (which should be the simplest thing to describe) is inadequate? Yes! The description of free space by A=Ȧ=0 is classical; it is like saying that the ground state of the oscillator is given by x=p=0. Now, we know that if the oscillator is treated quantum mechanically, only the average quantities <0|X|0> and <0|P|0> vanish in the ground state, and that there are nonzero fluctuations (ΔX)2=<0|X2|0> and (ΔP)2=<0|P2|0> about these mean values. In the same way, if the electromagnetic field is treated quantum mechanically, it will be found that free space (which is the ground state of the field) is described by <A>=<Ȧ>=0 (where A and Ȧ are operators) with nonvanishing fluctuations (ΔA)2, (ΔȦ)2. The free space is dormant only in the average sense; there are always quantum fluctuations of the fields about these mean values. It is these fluctuations that trigger spontaneous decay.
A very eloquent passage I couldn't hope to improve on (except fixing his typo). He's using mostly the standard notation:
H1 is the perturbation Hamiltonian,
A is the magnetic vector potential, and |
n,
l,
m> are the eigenstates of the hydrogen atom.
"Quantizing the field" is just terminology which means: treat
A and
Ȧ as
operators with nonzero commutators, in contrast to the classical versions, which are real valued and have identically zero commutators. It's exactly the same concept as treating
X and
P as operators with nonzero commutators.