The Feynman lectures on physics are special for many reasons.
Feynman delineates the big picture rather clearly; is able to be very concise about various points that make many other authors excessively talkative; isn't afraid to address questions that some people incorrectly consider a domain of philosophers even though they have become a domain of physics many years ago; isn't afraid to clarify many widespread misconceptions and explain why e.g. Einstein was wrong about quantum mechanics; offers some original cute research which may have had pedagogical motivations but that is more important than that e.g. the derivation of general relativity from a consistent completion of the coupling of the stressenergy tensor to a new spin2 field (it was a really beautiful, pragmatic, modern, and stringtheorylike approach to general relativity); and for many other reasons.
Twostate systems are among the topics in quantum mechanics that Feynman dedicated much more attention than most other textbooks of quantum mechanics which is one reason why Feynman's students were much more likely to understand the foundations of quantum mechanics properly.
On this blog, twostate systems have played a central role in many older articles such as the introduction to quantum computation (a qubit is a very important twostate system); twofermion doublewell system (a ramification of the Brian Cox telekinetic diamond insanity), and several others.
There are hundreds of important examples of twostate systems. The underlying maths governing their evolution is isomorphic in all these cases but we still tend to think about the individual examples "differently" because we have different intuitions about the different system. The electron's spin is one example; the ammonia molecule is another. The mathematical isomorphism between the two systems is a fact; nevertheless, people tend to incorrectly assume that the mathematics of the ammonia molecule we will focus on is much more "classical" than the example of the electron's spin. Well, it's not. All states in the Universe obey the same laws of quantum mechanics.
Looking at Feynman's chapter on MASERs
The third, final volume of the Feynman lectures on physics is mostly dedicated to quantum mechanics. There are many chapters on twostate systems. Chapter 9 of Volume III is about MASERs – cousins of LASERs that use a particular transition between two states of the ammonia molecule.
Ammonia, \(NH_3\), is a stinky gas excreted in urine. It escapes from the liquid which is why you could smell it on most toilets, especially in the third world and the Soviet Bloc during socialism. ;) I apologize to the French and others but Germans have the cleanest toilets. But we want to discuss a closely related issue, namely the states of the ammonia molecule. :)
The model above shows us that the molecule looks like a pyramid with a triangle of hydrogen atoms at the base and a nitrogen atom at the top of the pyramid (above the center of the triangle). All the nuclei may move relatively to each other, in principle. However, the relative motion that would change the distances between the four atoms looks like a collection of harmonic oscillators with pretty high frequencies i.e. energies. That's why if we only look at lowlying energy states (at most a tiny fraction of an electronvolt above the minimum possible energy), we are restricted to the ground states of the harmonic oscillators, to the lowestlying state in which the distances between the four nuclei are fixed at values that minimize the energy with this accuracy. The vibrations changing the distances are thus forbidden. The wave functions for the electrons are completely determined by the lowenergy constraints.
How many states of the ammonia molecule are there?
Well, even if the lengths of the six edges of the pyramid are determined, we may do something with the pyramid: we may rotate it. If you define the vector \(\vec d\) pointing from the nitrogen atom to the center of the hydrogen triangle, \(\vec d\) may be any direction on the twosphere \(S^2\). I've chosen the direction of the arrow in the opposite way than what you may find natural because \(\vec d\) is really proportional to the dipole moment. Note that the hydrogens are "somewhat more positive" (much like they tend to lose the electrons elsewhere: compare the \(H^+\) or \(H_3 O^+\) and \((OH)^\) ions in water) while the nitrogen is more "negative", like in \(N^ H_3^+\), so the dipole moment has to start from the negative nitrogen.
For each direction on the sphere, there is a distinct quantum state orthogonal to others. Also, the orientation of the hydrogen triangle within its plane may change (rotate) and is given by an arbitrary angle defined modulo 120 degrees (because the rotation by 120 degrees maps the triangle to itself). The lowlying states will pick a superposition of these states (triangles rotated by \(\gamma\)) that minimize the energy, pretty much the "symmetric combination" of all directions.
So the vector \(\vec d\) is the only "light degree of freedom" we are allowed to vary without raising the energy too much.
In classical physics, you could take \(\vec d\) to point along any axis, e.g. the negative \(z\) semiaxis ("nitrogen is above the triangle"). If the nitrogen had the same properties as the classical plastic model, the value of \(\vec d\) would be conserved. We could just forget about all the states with different directions of \(\vec d\); we could consistently demand that \(\vec d\) has a preferred direction.
Let's try to impose as similar a condition as we can in quantum mechanics, too. Quantum mechanics allows \(\vec d\) to change arbitrarily (the dipole moment isn't conserved) so the true energy eigenstates would be some spherical harmonics that depend on \(\vec d\). However, the transitions to different values of \(\vec d\) are unusually slow and we may forget about them.
However, there is one transition that is actually fast enough: the transition from \(+\vec d\) to \(\vec d\). We can't neglect it at all. Why does exist? You may explain it in many ways but one of them is the "quantum tunneling". It's just possible for the nitrogen atom to be pushed through the triangle and appear on the opposite side of the pyramid. The ammonia molecule is exactly the kind of a microscopic object where things like quantum tunneling are inevitable.
So there's a significant probability that an ammonia molecule starting with a direction of \(+\vec d\) ends up with the opposite direction, \[
\vec d\to \vec d.
\] Note that the position of the center of mass is conserved. You can't "ban" this process which is fast enough (the frequency is high enough for the purposes we consider but low enough to be compatible with our lowenergy constraints). That's why you can't study the state of the ammonia molecule with a given value of \(\vec d\) only. You must study the states with \(\vec d\) and \(\vec d\) at the same moment.
The states "nitrogen above the triangle" and "nitrogen below the triangle" will be called \(\ket 1\) and \(\ket 2\), respectively. The picture above should really have the same orientation of the hydrogen triangle in both states, if you wanted it to be really natural, but it's a detail because I said that the relevant states are averaged over the orientation of the triangle, anyway.
At any rate, it is legitimate to require that the dipole moment \(\vec d\) lies in a particular line – the transitions to nonparallel values of the dipole moments are so slow that they can be neglected. However, the flipping of the sign of \(\vec d\) is something that simply cannot be neglected. It's possible, allowed, inevitable, and it's a vital reason why the MASERs work at all.
You see that this is already a big deviation from classical physics in which there's no tunneling effect. In classical physics, \(\vec d\) seems to be conserved. Some people could argue that the ammonia molecule is already pretty large so classical physics should more or less apply. They would be wrong. Classical physics never applies exactly. Any physical system in this Universe, regardless of the size, obeys the laws of quantum mechanics. Classical physics may sometimes be a good enough approximation but it's never precise and it's always wrong at least for some questions when we talk about small molecules.
We have restricted our attention to a twostate system. The different shapes of the molecule (different lengths of edges) were banned because the change of the geometry (or excited states of the electrons) would raise the energy too much; different directions of the dipole moment than \(\vec d\) and \(\vec d\) where \(\vec d\) is the (measured) initial value may be ignored because the transition to nonparallel values of \(\vec d\) is too slow.
We know that the state vector will belong to a twodimensional Hilbert space. The relevant Hamiltonian is\[
\hat H = \pmatrix{ E_0 & A\\ A& E_0 }.
\] The offdiagonal matrix elements \(A\) are due to the quantum tunneling. In classical physics, we would have to have \(A=0\) but that's definitely not the case in quantum physics. By redefining the phase of \(\ket 1\) and \(\ket 2\) (only the relative phase matters), we could change the phase of \(A\) and choose a basis in which \(A\) is real and positive. Note that in general, \(\hat H\) would be Hermitian so the upperright matrix element would be the complex conjugate of the lowerleft matrix element.
The two diagonal entries are the "best approximations for the energy" of the pyramids in which we neglect the tunneling. These matrix elements are equal to each other due to the rotational symmetry of the laws of physics. After all, \(\ket 2\) may be obtained from \(\ket 1\) by a rotation and because rotations are symmetries, they don't change the expectation value of the energy. It has to be the same for both states, \(E_0\).
However, \(E_0\) isn't an eigenvalue of the energy. Instead, to find the eigenvalues of the energy, we have to diagonalize \(\hat H\). We find out that the eigenstates are\[
\frac{\ket 1 \ket 2}{\sqrt{2}}, \quad
\frac{\ket 1+ \ket 2}{\sqrt{2}}.
\] I divided the sum and difference by \(\sqrt{2}\) to make the states normalized but I didn't really have to do that. Again, note that the phases (and, if you choose this convention, general normalization factors) of the eigenstates are undetermined. In the column notation, the eigenstates are\[
\pmatrix{ +\frac{1}{\sqrt 2} \\ \frac{1}{\sqrt{2}} },\quad
\pmatrix{ +\frac{1}{\sqrt 2} \\ +\frac{1}{\sqrt{2}} }.
\] If you multiply the matrix \(\hat H\) by these vectors, you will get multiples of themselves. The coefficients in the multiples are the energy eigenvalues:\[
E_I = E_0  A, \quad E_{II} = E_0 + A.
\] Now you see why I chose the first sign to be minus – I wanted the first eigenvalue to be the lower one. In other words, I wanted the first eigenstate to be the genuine ground state. The Roman numerals were picked to label the eigenvalues from the lowest one to the highest one.
So if you really cool down the molecule, it won't sit in the shape of a particular pyramid. It just can't sit there because there exists quantum tunneling, even at vanishing temperature \(T=0\,{\rm K}\). It is an inevitable process of quantum mechanics. Instead, the molecule will ultimately emit photons and drop to the lowestenergy state which means the ground state, the energy eigenstate with the lowest energy, and it has the same probability amplitude to be in the "pyramid up" and "pyramid down" states.
Because many people make a mistake, let me emphasize one more thing. The energy eigenstates \[
\ket{I,II} = \frac{\ket 1 \mp \ket 2}{\sqrt 2}
\] are not just some dull "statistical mixtures" that have a 50% probability to be in the state \(\ket 1\) and 50% probability to be in the state \(\ket 2\) (pyramid up or down). Instead, the relative phase between the states \(\ket 1\) and \(\ket 2\) is absolutely crucial for the physical properties of the linear superposition state.
In particular, if the relative phase is anything else than \(+1\) or \(1\), the superposition fails to be an energy eigenstate. It is only an energy eigenstate if the relative phase is real and when it's real, it's damn important whether the relative sign is \(+1\) or \(1\). The negative relative sign gives us the lowerenergy state – the quantum tunneling has the effect of maximally lowering the energy from the "intermediate" level \(E_0\) down to \(E_0A\), while the positive relative sign does exactly the opposite: it raises the energy to \(E_0+A\).
Almost all the antiquantum zealots would try to talk about a preferred basis and because they're obsessed with position eigenstates, they would probably allow \(\ket 1\) and \(\ket 2\) to be the only states that may be "truly" realized. But this is of course completely wrong because even if you start with \(\ket 1\), you can't have \(\ket 1\) forever. Because of the quantum tunneling, the initial state \(\ket 1\) inevitably evolves into general linear superpositions of \(\ket 1\) and \(\ket 2\): it oscillates between \(\ket 1\) and \(\ket 2\).
As emphasized repeatedly on this blog, there's no way to ban general linear superpositions. The states in a given basis inevitably evolve into general complex combinations of such basis vectors. And all the complex coefficients, including the relative phases – and especially the relative phases – are absolutely critical. Only a basis of energy eigenstates is completely "sustainable": each energy eigenstate only evolves into a multiple of itself (well, it evolves into itself with a different phase).
And it is not true that the relevant energy eigenstates are always "equal mixtures" of \(\ket 1\) and \(\ket 2\). For example, in the case of our molecule, things change e.g. when we add the electric field.
Adding electric fields: MASER
The surrounding electric field \(\vec E\) doesn't change anything about the existence of the tunneling. However, it will add the interaction energies between the electric dipole and the electric field so that the Hamiltonian is\[
\hat H = \pmatrix{ E_0+d E & A\\ A& E_0  dE }.
\] The expectation value of energy of \(\ket 1\) i.e. the upperleft matrix element (associated with the pyramid pointing up, the dipole is down) was increased by the product \(\vec d\) and \(\vec E\) because the two vectors are pointing in the opposite direction; the expectation value of energy of \(\ket 2\) was lowered for the same reason.
Again, this matrix (the Hamiltonian) may be diagonalized. In this case of a real matrix (it was made real by our having redefined the phases of the basis vectors), the coordinates of the eigenvectors are still real. But it is no longer the case that the absolute values of both coordinates are equal. They're rather general. The energy eigenvalues are\[
E_I = E_0  \sqrt{A^2+ d^2 E^2},\quad
E_{II} = E_0 + \sqrt{A^2+ d^2 E^2}.
\] Note that those formulae reduce to \(E_0\mp A\) for \(E=0\). Also, they reduce to \(E_0\mp dE\) for \(dE\gg A\). Note that if you drew graphs of \(E_0\mp dE\) as a function of the electric field \(E\), you would get two straight lines that intersect each other. However, when you draw the graphs of the exact results \(E_I,E_{II}\) written in the displayed equation above, the two curves never cross. One of them never drops below \(E_0+A\) and the other one never jumps above \(E_0A\) so they obviously cannot cross. In fact, the right upper arm of the curve (which is clearly a hyperbola) "bends" and continuously connects to the left upper arm of the curve; the same thing holds for the lower curve. See Avoided crossing (= "repulsion of eigenvalues" or "level repulsion") at Wikipedia.
A few final steps are needed to explain why such molecules are able to emit and absorb some radiation whose frequency is \(f\) where\[
hf = E_{II}  E_{I} \sim 2A + \frac{d^2 E^2}{A}
\] where the final form of the expression is a Taylor expansion of the square root that is OK for all realistic (small enough) values of the electric field \(E\). No, you won't be really able to produce \(dE\gg A\) in the lab: these would be too strong electric fields.
For example, start with the absorption. Place the ammonia molecule to a variable field \(\vec E\) which has the right frequency \(f\sim 2A/h\). In the \(\vec E=0\) energy eigenstate basis which is composed of the vectors \(\ket 1\mp \ket 2\), the Hamiltonian was diagonal (that's what the energy eigenstates mean). When you add the electric field \(\vec E\) going like \(\cos (2\pi f t)\), it will contribute some offdiagonal elements in this basis and when the frequency is right, the resonance will be able to "accumulate" the probability amplitude of \(\ket{II}\) even if the initial state is \(\ket{I}\). If you choose a wrong frequency, the amplitude of the state to be in \(\ket{II}\) will receive contributions with different phases each cycle and they will cancel out after a while.
So the ammonia molecule is able to increase its energy by capturing some energy from electromagnetic waves at the right frequencies. That's absorption. There's also stimulated emission, the timereversed process to the absorption. If the molecule is already mostly at the higher level \(\ket{II}\), the electric field oscillating at the right frequency will encourage the molecule to drop to \(\ket{I}\) and deposit the energy difference to the electromagnetic waves.
In the text above, the electric field \(\vec E\) was assumed to be large enough and treated as a classical background that only affects the Hamiltonian of the molecule via "classical parameters". However, if the electric field is weak enough, it becomes important that \(\vec E\) is a quantum observable as well. The energy carried by the electromagnetic waves of frequency \(f\) is no longer continuous – it is quantized i.e. forced to be a multiple of the photon's energy \(E_\gamma = hf\).
When you calculate the transitions properly, you will find out that there's actually a nonzero probability amplitude for the ammonia molecule in the excited state \(\ket{II}\) to emit a photon even if there's no oscillating electric field around the molecule to start with. This is the spontaneous emission. A proper look at the "time reversal argument" is enough to see why the spontaneous emission has to exist. Even one last photon may be absorbed (absorption is always "stimulated") and we end up with zero photons; the timereversed process therefore has to start with zero photons and end with one photon and its "invariant" probability amplitude has to be the same i.e. nonzero. The ability of systems to emit even if the initial number of photons is zero is called "spontaneous emission"; the total "stimulated plus spontaneous emission" has the probability proportional to \(N_\gamma+1\) – it's the squared matrix element of a harmonic oscillator's position matrix element.
However, for MASERs, the stimulated emission is more important because the intensity of the electromagnetic waves is high – "stimulated" is what the letter "S" in "MASER" (or "LASER") stands for. I recommend you e.g. Feynman's treatment in the Volume III of his lectures for sketched calculations of the formulae for the transition rates, why there is a resonance, what is the width of the curve, and so on.
My main goal was more specific here: to convince you that the superpositions of "classically intuitive" states are absolutely inevitable and natural. This wisdom holds for all systems in quantum mechanics, including manylevel and manybody systems and including infinitedimensional Hilbert spaces whose bases are labeled by positions or any other continuous or discrete observables. Physical systems may be found in any superpositions and if you need to identify a basis that is a bit more "sustainable" than other bases, it's a/the basis of the energy eigenstates, not e.g. position eigenstates. What these slightly preferred energy eigenstates are depends on the Hamiltonian. For example, in our case, the form of the eigenstates depended on the surrounding electric field. So there can never be any "a priori preferred basis". There is never a preferred basis but if some basis is more wellbehaved than others, it's because it's closer to a basis of energy eigenstates, and such a basis always depends on the Hamiltonian as well as the environment: it can only be determined "a posteriori". In particular, the wave functions for the ground states are more important than all others and that's the wave functions into which cool enough systems want to sit down. These ground state wave functions describe what the degrees of freedom are doing – and it doesn't matter that you may find these wave functions "complicated" or "different from those you would like to prescribe".
The ammonia molecule is just another system that invalidates any nonquantum or "realist" (antiquantum zealots prefer to use the term "realist" for themselves over the much more accurate but less flattering term "classical and optimized for cranks eternally and dogmatically stuck in the concepts of the 17th century physics") replacement for the proper rules of quantum mechanics.
For example, coherence of the ammonia molecule is totally essential and testable so if the Universe decided to "split" during any of the processes discussed above, as Everett liked to imagine, it would immediately have tangible consequences that disagree with the experiment.
Also, if there were any pilot waves envisioned by de Broglie or Bohm, one couldn't explain "where" the photon is created during the spontaneous emission. Note that during absorption or emission of electromagnetic waves, the number of photons isn't conserved, in contradiction with an elementary property of the flawed pilot wave paradigm. The photon emitted by an excited ammonia molecule is "everywhere". You can't fix this problem of the pilot wave theory by forcing the system to remember the "truly real classical field configuration" instead of the position of particles because that would be similarly incompatible with the particlelike properties of the electromagnetic field. The electromagnetic field – and all other fields in the world – exhibits both particlelike and wavelike behavior and which of them is more relevant depends on the situation, on the relevant terms in the Hamiltonian, on the frequency of the waves and the occupation numbers etc. If you want to declare one of the behaviors (particlelike or wavelike) to be "more real" than the other, you're guaranteed to end up with a fundamentally wrong theory. If you keep on doing such things for years, then you become an irreversible crackpot.
Theories with GRW collapses would also predict unacceptable effects whose existence may be safely ruled out experimentally. And I am not even talking about theories that would love to completely "ban" the complex superpositions because the authors of these theories must be misunderstanding everything, including the content of Schrödinger's equation that makes the evolution into general complex combinations inevitable.
Of course, the main point of all such texts is always the same: quantum mechanics has been demonstrated to be the right framework to describe the world for more than 85 years and everyone who is hoping that a completely, qualitatively different description will replace quantum mechanics – e.g. the incredible idiots that clearly gained a majority in similar threads on the Physics Stack Exhange (holy crap, the users such as "Ron Maimon" and "QuestionsAnswers" are just so unbelievably stupid!) – is a crackpot.
And that's the memo.
How NASA’s James Webb Space Telescope Will Answer Astronomy’s Biggest Questions (Synopsis)

“The [James Webb] telescope is basically designed to answer the big
questions in astronomy, the questions Hubble can’t answer.” Amber Straughn
Have you ev...
3 hours ago
snail feedback (12) :
Thanks for the ongoing QM lessonsdespite one graduate course in it, QM was a hole in my math/physics trek which I am slowly remedying after many years.
I went to the Physics Stack link to read the t'Hooft (if it is indeed him) and comments. Holy crap, indeed.
Questions and Answers tells you basically to f$%k off,
and calls what you say "pathetic lies and mischaracterisations"...Maybe it is a good thing that I havent been following it for ages.
I see nothing wrong with t'Hooft (again, him or a troll?)
framing models and testing them and tinkering with them. t'Hooft deserves a lot of slack because of his accomplishments. Maimon and QuestionsandAnswers are a different matter. What all three are doing is simply
looking for a deterministic substructure to QM, and denying locality rather than reality. For QandA to say you are denying objective reality is a distortion. Obviously there is an objective reality. That is not the same as saying that reality we experience is not crystallized until a "measurement", ie any interaction, occurs. We live in the reality of events that have and are interacting resulting in the decoherence of probability wave amplitudes.
IMO a problem that many people (me included) have is that describing QM using words rather than math leads to confusion about determinism and other things, particularly foundational issues.
QandA's responses to you are amazingly condescending.
Thanks, Gordon, for the comment which is still entirely offtopic and I will refrain from responding to any physics in your comment because it's only a path to trouble.
Off topic, sorry, but there is an interesting new paper by Connes:
http://arxiv.org/abs/1208.1030v1
Any comments ? I am having hard times to understand it. Is it a threat to string theory ?
Dear Raisonator,
the paper makes no sense. Connes (and collaborator) has been trying to claim that the Standard Model may be reformulated as a compactification on a noncommutative manifold which is able to predict relationships between some coupling constants in the Standard Model. It wasn't really ever true  the values of the coupling constants that may look "more natural" in his variables are still not physically privileged over other values so it isn't possible for a quantum field theory to become predictive in this way.
At any rate, Connes cared about this "new simplicity in the noncommutative variables"  a formalism that wasn't really worked out beyond the classical level (he uses ordinary tools of QFT for the loops only)  so he had predicted the Higgs mass of 170 GeV. It just happened that 170 GeV was the very first value of the Higgs mass that got excluded by the Tevatron:
http://motls.blogspot.com/2008/08/tevatronfalsifiesconnesmodelof.html?m=1
So he was shooting into the wrongest place of the real axis. One must be pretty lucky for that. ;) Now, when we know that the Higgs is near 126 GeV, he's trying to retroactively correct the wrong prediction by adding random components that would make it bad science even if they were used in an otherwise healthy context  but when they're used within the framework that makes invalid claims about the possibility to predict things that cannot be predicted, it's even worse.
Best wishes
Lubos
sorry , OT
zdravim Lubosi,
I have an old problem of understanding, perhaps trivial.
Probably you know the solution immediately.
The stimulated emission, one photon comes in, interacts somehow with the atomic field (how is not important), and if everything fits, two identical photons come out. OK !
On the other hand, quantum
cryptography (one photon communication) protects themselves with
statements like : due to the noncloning theorem , one photon can not
be.... => quantum crypto is secure.
But what is with Laser/Maser, Radio ?
All these devices produce identical photons from in principle one
photon.
Or is it so, that thouse photons are in
reality not perfectly identical, but identical enough for Laser etc.?
Thanks and ciao
Ludek
Dear Luďku, an extremely good question.
Dear Lubos,
stupid question maybe, Would anything change if we used the good old H2O molecule as the basis for our discussion? After all it has a dipole moment as well.
Dear Mikael, H2O molecule isn't a 2state system in any sense, so you can't construct a MASER out of it and you can't use in discussions on 2state systems.
I may be puzzled by your question but this whole article, every single topic in this blog entry, is about 2state systems. They're systems for which you can actually count the relevant microstates satisfying certain conditions that may evolve into each other  and you get two. I got 2 because the states essentially come from the pyramidup and pyramiddown states of the ammonia.
But H2O isn't a pyramid. Its shape can't ever be organized "just in two ways". There isn't any group of atoms (the triangle) that would define a plane and allow another atom to be above or below it. For H2O, the counting just doesn't give 2. It gives either 1 or infinity. Your suggestion that this discussion boils down to a nonzero dipole moment indicates that you haven't even started to understand this topic yet. Pretty much every system in the world has a nonzero dipole moment. That doesn't mean that every system is a 2state system.
Dear Lubos,
I can see that H2O has an infinite number of states. But if you allow for rotations NH3 has them as well. Only if you keep the H atoms in place it becomes a two state system. So my question is why can I neglect the rotations in one case but not in the other?
There's still a maser transition in water, around 22 GHz, similar to the 24 GHz ammonia maser frequency, but the rotational modes for H2O have almost the same frequency so the decoupling of the problem isn't legitimate.
For ammonia, the triangle has a larger number of Hatoms and may be approximated by a circle. Due to the identical character of the atoms, forcing the 120degree periodicity the rotational modes of NH3 correspond to higher frequencies that may be considered "parameterically higher". For H2O, this decoupling of the scale is not possible as both energies are of the same order.
Thanks, Lubos. I think it is the kind of answer I should have guessed from the geometrical initution. But it is always reassuring to get a true expert answer.
Dear Lubos,
after spending more time on this topic I think your answer is just not right, at least partially.
For the rotational modes the only thing what should matter is the quantized angelar momentum
and therefore the moment of inertia of the molecule for the rotation axis.
The bigger the moment of inertia the lower the transition frequencies of course.
Instead the key point for the two states of the Ammonia molecule should be fact
that if you push the N throught the triangle of the three H you create a state which is different
to the original state not just by a rotation. Instead you must additional exchange to
H atoms to come to the original situation.
Regarding the maser frequency of 22 GHz for water are you sure about its origin?
Post a Comment