Long lifetimes of the Rydberg atoms seems counterintuitive to me.  Intuitive explanation: the higher electron was excited, the more levels below became available for spontaneous decay. So the first thought is the probability of decay should increase.  But the experiments say that highly excited Rydberg atoms are extremely stable (lifetimes ~ seconds). Moreover, the lifetime tends to grow with the increase of quantum number n.  Is there any mistake in my intuitive explanation?  Edited: I do not agree that my question duplicates the one some users pointed at. My question is about possible ways to deal with inconsistency between intuitive explanation and experimental facts.
121359
How to explain the long lifetime of  with Fermi's golden rule? Wikipedia says it is partly due to tiny wavefunction overlap with inner orbitals, but what about the outer ones?
16822
In section 134 of Vol. 3 (Quantum Mechanics), Landau and Lifshitz make the energy complex in order to describe a particle that can decay:  $$ E = E_0 - \frac{1}{2}i \Gamma. $$  The propagator $U(t) = \exp(-i H t)$ then makes the wavefunction die exponentially with time.  But also, $H$ is non-Hermitian.  My question: Do we have to modify the basic postulates of quantum mechanics (as described by Shankar, say, or the earlier sections of Landau &amp; Lifshitz) to describe unstable particles?
43336
Let's say we have a emitter, emitting light that has frequency f, less than the threshold frequency of a metal.  If you leave light shining onto that metal, for long enough, does the energy of the individual photons accumulate, on the electrons, so eventually they will ionize, or does this not happen? What am I missing?
24695
the spectrum of the Gamma and Alpha decays are both discrete, i.e. the $\alpha$-particles and the $\gamma$-rays take on only discrete values when emitted from a decaying nucleus.  Why is it then, that the $\beta^{\pm}$ can take on continuous values?  The main thing that distinguishes the beta decay from the other two is, that it is a three body problem, i.e. the nucleus does not only decay into an electron/positron, but also into a electron-neutrino/antineutrino. I don't see yet how this immediately implies that the spectrum of the electrons is continuous though.  The way I understand the two body decays is, that the initial nucleus spontaneously decays into the two bodies, i.e. a smaller nucleus and a gamma or alpha particle. As the energy levels in both the nuclei are quantized, only certain values for the energy of the photon and the Helium core are allowed, sine Energy and momentum need to stay conserved.  Does the third particle change things in the way that there basically are two things (i.e. the electron and the neutrino in the beta decay) that are not restricted by an inner energy level hierarchy in the way the nuclei are, thus allowing the energy given by the nuclei to be split arbitrarily (and continuously) between the electron and the neutrino?  If this is the wrong explanation, please correct me.
22936
So, I was reading Atom: Journey Across The Subatomic Cosmos by Isaac Asimov in order to better understand quantum mechanics when I came across this sentence:      The electron couldn't spiral into the proton because it couldn't take up an orbit with a length less than a single wave.    I was wondering why that is? Why can't an electron not have non-whole number wavelengths in atomic energy levels?
6880
One of the problems with Bohr's theory to describe the hydrogen atom, was that the electron orbiting around the nucleus has an acceleration. Therefore it radiates and loses energy, until it would collapse with the nucleus.  Now Schrödinger describes the electron as a wave function. His theory is able to describe all atoms (in contrast to Bohr's model), but how is the radiation problem solved? I understand that the wave has not an exact position in the time anymore. But the electron still "moves", so it has an acceleration anyway (because of vibrations or so).   Why is in this theory the electron not radiating anymore? And if it is, why doesn't the atom collapse?
6431
I was reading some popular science book and there was this sentence. It says that energy has to be emitted in discrete portions called quanta. Otherwise the whole energy in the universe would be converted into  high frequency waves.  I'm not a physicist, so this conclusion seems to me like a huge leap.  First, we assume that energy is emitted in a continuous way (not in quanta).  And how do we get to the statement "the whole energy in the universe would be converted into  high frequency waves"?
11211
I came across a discussion of atomic Rydberg states, which appear to be defined as having their outer electron in a highly excited state. In the same text, it mentioned that these excited electrons could be modeled rather well by the Bohr model.  I don't see why this should be the case. The Bohr model treats the electron as a circularly orbiting body around the nucleus, whereas the electron cloud model seems to produce orbits which are highly anisotropic. In my understanding, the Bohr model also proposes electron orbits with fixed radii.  I don't see how this can be deduced from taking the limit of large $n$. How is Bohr's model related to electron cloud models via the correspondence principle?  On a further side note, I am curious why the different orbital energy splitting as a result of atomic effects (Fine structure, Hyperfine etc) which causes non-degenerate orbitals wrt $\ell$ is ignored for high $n$, where orbital energies are only dependent on the principal quantum number.  References:   Randall G. Hulet and Daniel Kleppner, Rydberg Atoms in "Circular" States, Phys. Rev. Lett. 51 (1983) 1430  R.J. Brecha, G. Raithel, C. Wagner, H. Walther,  Circular Rydberg states with very large n,
53800
I was watching the Cosmos documentary where Neil deGrasse Tyson explained how certain energy photons get absorbed by an atom, which causes the electrons of that atom to climb into a higher energy state.  He then says that an atom produces a photon when those electrons drop to a lower energy state, but that we don't know why this happens. I understood that as 'we don't know what triggers this to happen'.  Is that true? And if so, are there any feasible theories that explain this phenomenon?
114896
I was thinking about the many interpretations of quantum physics, and one thing that never made sense to me was the many world's interpretation. Basically at any given moment for which something exists in a superposition, isn't this interpretation basically saying that there would have to exist a new universe for every possible discrete outcome that could result once the measurement is made?   So let's say an electron's location is a superposition of two distinct locations. Doesn't that mean there would at that moment have to be two distinct universes for each distinct location that is superposed? If so, then clearly the amount of universes being spawned every moment must be astronomical.   There could be trillions upon trillions of electrons(for instance) having superposed states, and then every moment in time (I suppose every planck time?), it seems there would have to be yet another set of universes spawning for every superposition.  How could this be? Surely there must be some kind of thermodynamic cost to replicate an entire universe. What would fuel or supply the additional mass and energy for every universe at every moment for every superposed state?  To my lay person's mind, this seems to be a very big weakness of this interpretation, but I needed to ask people with the physics background required to answer this question. Could there be some kind of loophole that could allow alternate universes to spawn so rapidly? Otherwise, doesn't this present a bit of a problem for taking this interpretation seriously?
21899
The question itself is pretty self explanatory. I asked this to my chemistry teacher when he was doing the photoelectric effect while teaching atomic structure, and he just shrugged it off.  One possible mechanism that I thought up, is an atom with a unpaired electron (say hydrogen) to absorb one photon, and jump to a higher energy level, and then just immediately be hit by another photon with a different energy which raises it to an even higher level. Then finally the electron falls back into its proper non exited state releasing not two but only one photon which has the energy of both the initial photons, ideally combining photons.  Is this mechanism correct? Does it occur in nature? Do photons have other ways in which they merge? or do they not merge at all?
25883
I could understand that the definition of a second wouldn't have an uncertainty when related to the transition of the Cs atom, so it doesn't have an error because it's an absolute reference and we measure other stuff using the physical definition of a second, like atomic clocks do.  But why doesn't the speed of light have uncertainty? Isn't the speed of light something that's measured physically?  Check out that .
34288
The higher the number of the shell (n), the higher is the energy level of the electron. However, why was it necessary to have negative values. So for example, when $n=1$, the energy could be $5 eV$ and for $n=2$, $6 eV$... having positive values could also have supported the idea that as $n$ increases, energy of electron increases. What is the point of having negative numbers, does it somehow aid calculations?
28486
This press release by NASA:      When the universe was still very young, only a few kinds of atoms existed. Scientists believe that around 100,000 years after the big bang, helium and hydrogen combined to make a molecule called helium hydride for the first time. Helium hydride should be present in some parts of the modern universe, but it has never been detected in space — until now.   Naively I would expect that the first molecule to form would be the H$_2$. A claim that HeH$^+$ was before that should mean that it has larger binding energy that allowed the molecules to withstand more radiation. However, if I check Wikipedia for dissociation energies of these molecules, I see $436\,\rm kJ/mol$ for H$_2$ and $360 \,\rm kJ/mol$ for HeH$^+$.  Additionally, helium hydride is charged and should interact with lower energy photons that will destabilize the bond due to sheer quantity.   And finally, the temperature mentioned in the article is $4000\rm\, K \sim 0.34 \, eV$ — which is quite below the dissociation energy of molecular Hydrogen ($4.52 \,\rm eV$). Why was the formation of molecules delayed so much? Is this one of those situations where baryon-to-photon ratio is important and photons overwhelm the baryons?  Could someone please explain this situation?
52282
It is said in many textbooks that alpha decay involves emitting alpha particles, which are very stable. Indeed, the binding energy () is higher than for $Z$-neighboring stable isotopes. But the binding energy is lower than, for example, ${}^9\mathrm{Be}$ (). My question is why aren't other nuclear compounds ejected from heavy nuclei, e.g. ${}^9\mathrm{Be}$?  The   $$e^{-\frac{4\pi}{\hbar}\frac{Ze^2}{4\pi\epsilon_0}\frac{1}{v_\alpha}}$$ decreases exponentially in $Z$, so it explains intuitively why lower-$Z$ particles would tunnel more often. Specifically, it would explain why we would see ${}^9\mathrm{Be}$ emission ${e^{-2}}\simeq 0.14$ times as often compared to ${}^4\mathrm{He}$ emission. Also, the particles need to form in the nucleus prior emissions; but with a similar binding energy per nucleon (~7.08 MeV for ${}^4\mathrm{He}$ vs. 6.47 MeV for ${}^9\mathrm{Be}$) and higher total binding energy for the ${}^9\mathrm{Be}$ nucleus, I would expect that its formation in the same order of prevalence as the alpha particle (according to Ohanian, between 0.1 and 1 alpha particles are in existence at any moment in time).  Can anyone explain this? A reference to an article/textbook would be preferable.  EDIT Same goes for ${}^{16}\mathrm{O}$ which is also a double magic isotope, as 'anna v' pointed out. For it, the Gamow factor is smaller by $e^{-4}\simeq 0.02$, and emission should still be viable.
119915
I have almost no background in physics and I had a question related to Schrodinger's Equation. I think, it is not really research level so feel free to close it, but I would request you to kindly suggest some existing literature which can help me develop a better understanding for the same.   While reading up about it from an introductory text on Quantum Physics, I wondered for a little while how does one derive this equation. Very soon, the author () added the detail that he just took it on faith when he was himself a student. He goes on to say Schrodinger equation rightly predicts the behavior of atomic transitions etc and people believe that it is a fundamental law of nature for quantum systems.  Then he raises the question himself which I wanted to know the answer to. Namely     What made Schrodinger write such an equation which became a fundamental equation, not to be derived from more fundamental equations?   He then adds that it will be an interesting topic for students of history of science which does not answer my question. Could you please try answering this question (or is this question really useless) ?  Thanks for your time
53679
It's an old-new question (I found only one similar question with unsatisfactory (for me) answer: )  It's strange for me how all books simply pass by such an important question, and mentioning strange and mathematically unsupported reasons such as:   orbits are stationary (while as I know this is just idealization, there is no stationary orbits in reality even for Hydrogen)  electrons are actually not localized due to uncertainty principle, thus they have no acceleration (while obviously in a non-spherically symmetric orbits a kind of "charge acceleration distribution" always exist) vacuum fluctuations play a major role (according to QED).   I'm not interested in how Bohr or Schroedinger explained it, I want to see a rigorous proof with QM, QED or maybe even the standard model as whole. I would like to see how this question was closed.