Axis of quantization is such a lame choice of words. Quantization is a physical phenomenon, exes are something we choose like notations. Quantization does not depend on our choice. This is close to my favorite dumb project "Quantization of the reference frame".   

I can offer comments, but no sources.

AS: can this single-body molecular shaped object rotate in the simulation box (which in fact means that the axis of quantization can rotate)? or the axis of quantization should always be in an upright position?

To my understanding (and I'm sorry, I can't really back this up with sources) electron spins tend to line up either parallel or anti-parallel with external magnetic fields; in particular, the highest magnetic moment in the atom would tend to align, and other magnetic moments would tend to either be parallel or antiparallel.  As for why it happens, I don't know.  

AS: Why can/can't the quantization axis rotate in the space?

If I understand the question, by "quantization axis" you mean the orientation of the magnetic moment of the molecule.  I would think it would have to do with alignment with some external magnetic field.  However, if there is no external magnetic field, it might still tend to line up with the magnetic moment of other atoms in the area.

I don't know if that was any help, but I wonder if I could get you to help me with a related question.  I have been asking another question about diatomic hydrogen here:

https://www.researchgate.net/post/Where_can_I_find_the_emission_and_absorption_spectra_of_diatomic_hydrogen_gas/1

I had thought that there was only 1 stable form of hydrogen, but I see in this wikipedia article "The ratio between the ortho and para forms is about 3:1 at standard temperature and pressure – a reflection of the ratio of spin degeneracies. However, if chemical equilibrium between the two forms is established, the para form dominates at low temperatures (approx. 99.8% at 20 K).[3]"

Could you help me to determine what are the energies related to the "ortho" and "para" states of diatomic hydrogen, and whether an exchange from the higher state to the other could spontaneously release a photon?  Or are these energy exchanges only occurring through convection?

I see in https://en.wikipedia.org/wiki/Spin_isomers_of_hydrogen that the equation for the energies is

E_J = {J(J+1) hbar^2}/{2I} 

My guess is that "I" is the moment of inertia of the hydrogen molecule, and I should plug in J=0 and J=1, for the ortho and para states.  

Axis of quantization is such a lame choice of words. Quantization is a physical phenomenon, exes are something we choose like notations. Quantization does not depend on our choice. This is close to my favorite dumb project "Quantization of the reference frame".   

This question is a great question in the theory of observation that involves what quantum mechanics (QM) is all about. QM is not really about the systems being observed but what observers can say about those systems. Two entirely different things.

To observe electron spin, observers need apparatus. So that is where your focus should be. First decide what you want to observe or simulate. That should tell you on the orientation of  your apparatus and then  that apparatus can be described by parameters. These are classical numbers in general. In your case, the parameters involved are the angles defining the orientation of the apparatus (relative to say the laboratory) detecting the spin (to use crude imagery). The observer CAN  change those parameters. Rotating apparatus happens all the time (the earth is rotating, going round the sun, orbiting the centre of the galaxy, and so on.)

If you are using a box to simulate spin, you will have a problem, because boxes do not have rotational symmetry per se. But a box can be rotated, relative to some fixed frame of observation. You should be able to simulate that.

I would take some care to distinguish between the abstract mathematical formalism used to describe a wave function with angular momentum, and the specific apparatus that is being simulated. The apparatus always defines its own preferred axis of quantization.

I see in https://en.wikipedia.org/wiki/Spin_isomers_of_hydrogen that the equation for the energies is

E_J = {J(J+1) hbar^2}/{2I} 

 I calculated the Delta E for transitions from 1-0, 2-1, and 2-0, and the expected wavelengths of resulting photons, using the formula and got wavelengths of 41, 13 and 20 micrometers, respectively.  

Are these transitions regularly observed?

Thank you all very much for the answers. I really appreciate it! They are very helpful. I need to think about your answers and I might be back with some more questions in the future.

Johnathan,

to answer your question, these transitions that are calculated from the rigid-rotor model are just an approximation of what is actually observed and there is always some deviation from the actual transitions. In general, they depend on the system studied. Here I add that I have seen in some papers that people use this rigid-rotor approximation to make sure that the spectra observed in their experiments is related to the rotation transitions (not other transitions like the vibrational ones), which in fact shows how helpful these approximations are specially in analyzing the experimental data. To calculate the accurate transitions, you usually need to include some other factors into account. Please see the non_linear rigid rotor section here:  https://en.wikipedia.org/wiki/Rigid_rotor.

Also, in the systems where other forces are present and are strong enough to affect the free rotation of the H2 molecule, incorporating the perturbation into the Schrodinger equation is essential, the method that has been very well formulated in quantum mechanics.

Thank you all again.

Independent of the location and the orientation in the universe, the physics of elementary particle is similar. This is the so-called cosmological principle. 

https://en.wikiversity.org/wiki/Hilbert_Book_Model_Project

I seem to recall in the derivation of formulae for the Schrodinger wave-function, there was an assumption made that the function should be expressible in terms of Psi(r,θ,ϕ)=Θ(θ)R(r)Φ(ϕ), that is; the variables were separable.  It always struck me as a bit of a leap, but one that seemed to have been experimentally confirmed.

 However, the confirmation seems to come, largely, from putting molecules in a strong electric field, such as a discharge tube.  Could it be that the discharge tube aligns and polarizes the atoms in the θ=0 direction, setting up an opportunity for precession that wouldn’t otherwise have happened?  George Jaroszkiewicz said "The apparatus always defines its own preferred axis of quantization."

Igor Goliney seems sure that the theta and phi axes are just arbitrary choices, saying "axes are something we choose like notations. Quantization does not depend on our choice".  

But if my recollection is correct, changes in the quantum number related to the azimuthal, and changes in the quantum number related to the polar coordinates resulted in different energies.  

Jonathan> But if my recollection is correct, changes in the quantum number related to the azimuthal, and changes in the quantum number related to the polar coordinates resulted in different energies.

Perhaps you should refresh your memory. In the absence of symmetry breaking fields, energies do not depend on the axis of quantisation. For a fixed energy there are usually a number of degenerates states. States naturally defined with respect to one axis can be expressed as linear superposition of states naturally defined with respect to another. They are related by finite dimensional matrices, which form (irreducible) representations of the rotation group.

There is a certain confusion concerning use of the  term ' quantization axis'  in the answers above, apparently due to the fact one has to use space-fixed and body-fixed quantization numbers for polyatomic molecules. For diatomic molecules we only need the quantization  in space.  A choice of the quantization axis in space is arbitrary. One can introduce body-fixed quantum numbers (similarly to polyatomic molecules) to account for rotation-electronic interactions but this a separate question.  

KO: Perhaps you should refresh your memory.

Ha, indeed.  It's been over ten years.

KO: In the absence of symmetry breaking fields, energies do not depend on the axis of quantisation.

I don't think I disagree.  

But I'm looking at Griffith's "Introduction to Quantum Mechanics" Equation 4.50 it says something about the allowed energies (this is for monatomic hydrogen) 

𝐸𝑛𝑙=ℏbar^2/(2𝑚𝑎^2 ) 𝛽𝑛𝑙^2 

Here, 𝑙, m, and n are all quantum numbers. 𝛽𝑛𝑙 is the nth root of the 𝑙th bessel function.

The energies do depend on those quantum numbers, 𝑙, m, n.  

And if a photon is absorbed or emitted, wouldn't its axis of electrical polarization be related somehow to the θ=0 axis?  You wouldn't need an external electric field, but the polarization of the emitted or absorbed photon is a defacto electric field.

In the formula m is the hydrogen mass, not the azimuth quantum number.  The energy of the molecule is indeed independent of the azimuth quantum number until there an external field breaking the symmetry. The polarization of the emitted or absorbed photon is a defacto electric field as already pointed to by Jonathan.

Jonathan> (this is for monatomic hydrogen)

𝐸𝑛𝑙 = ℏ^2/(2𝑚𝑎^2 ) 𝛽𝑛𝑙^2 

Here, 𝑙, m, and n are all quantum numbers. 𝛽𝑛𝑙 is the nth root of the 𝑙th Bessel function.

I don't have Griffith's book book available, but this is absolutely, definitely, certainly not  the spectrum of the hydrogen atom. Not even close! It seems to be the spectrum of a scalar particle inside a hard-wall spherical cavity. And, as already pointed out, m is the mass of that particle, not a quantum number (as is also indicated by the notation, with no mention of m in the energy specification).

Jonathan> And if a photon is emitted, wouldn't its axis of electrical polarization be related somehow to the θ=0 axis?

For a mathematical description, a coordinate system may be needed. However if the hydrogen atoms (we are always predicting the statistical distribution of many independent observations) are unpolarised, one averages over all possible "quantisation axes" in the theoretical description. It is also possible to consider the same process with (completely or partly) polarised atoms; this leads to more detailed theoretical predictions.

It's always ``allowed''. Coordinate choices aren't physical. 

KO:  I don't have Griffith's book book available, but this is absolutely, definitely, certainly not  the spectrum of the hydrogen atom. Not even close! It seems to be the spectrum of a scalar particle inside a hard-wall spherical cavity.

That's correct.  I was looking at a formula from Section 4.1.3 under the header "Example:  Consider the infinite spherical well, V(r) = 0 if ra."  I had thought that the results would be similar enough they would be close.

I should have been looking in section 4.2 where Griffiths introduces the electric potential between a proton and electron

V(r)=-e^2/(4 pi \epsilon_0 r)

And from that derives the Bohr formula for the energy of hydrogen atoms.   There's a note, which I'm paraphrasing here:

"E(n=2) = -3.4 eV is the first excited state--or rather, states, since we can have (l,m,n) = {(0,0,2),(1,-1,2),(1,0,2),(1,1,2)} so there are actually four different states that share this energy, 

For n>2, the degeneracy is Sum(from l=0 to n-1) of (2l+1) = n^2

This seems interesting, though.  For the hard-wall spherical cavity, the energy seems to be a function of two quantum numbers, both l, and n.  However, with the inverse square potential of an actual proton, the energy can be expressed as a function only of n.  

KO: And, as already pointed out, m is the mass of that particle, not a quantum number (as is also indicated by the notation, with no mention of m in the energy specification).

You're right.  I see now, a footnote "Beware: The letter m is now doing double-duty, as mass and as the so-called magnetic quantum number.  There is no graceful way to avoid this since both are standard.  Some authors now switch to M or mu for mass, but I hate to change notation in midstream, and I don't think confusion will arise as long as you are aware of the problem."

Confusion arose because I was unaware of the problem.

KO: (we are always predicting the statistical distribution of many independent observations) are unpolarised, one averages over all possible "quantisation axes" in the theoretical description.

Again I don't disagree.  But that does not prevent someone from talking about the polarization of an individual photon, or the orientation of an individual hydrogen atom.  

Jonathan> However, with the inverse square potential of an actual proton, the energy can be expressed as a function only of n.

Yes, that is the beauty of the Coulomb potential (and the Newton's gravity potential). There is an extra symmetry which increases the degeneracy of each energy level, and leads to a very simple spectrum and the accompanying Balmer formula. I believe this symmetry has been very helpful for the development of first Newton's mechanics, and Quantum Mechanics a few hundred years later. The corresponding signature in planetary motion is the prediction of closed elliptic orbits (in the simplest approximation).

On closer look, the hydrogen spectrum is not exactly n2-fold degenerate. The degeneracies are lifted by relativistic effects, in a way which depends on the existence of electron spin (leaving an interesting trace of supersymmetry), due to the existence of vacuum fluctuations (Lamb shift), and finally by a small amount due to the proton spin. The last hyperfine splitting causes the famous 21-cm hydrogen line. 

Jonathan> But that does not prevent someone from talking about the polarization of an individual photon, or the orientation of an individual hydrogen atom.

Yes, certainly. And one can make a (probabilistic) prediction for each single event, based on (f.i.) an assumed polarisation of the hydrogen atom. But that is of little help if one does not know that polarisation, and in particular  if it varies from atom to atom. Then the best one can do is to average over all possibilities.

Whatever the rotational state, there always exists one generator of the rotation group that is diagonal, that's all. This generates rotations about an axis that's called the ``axis of quantization''. 

Depending on the physical system this can be globally defined, or not. That's where the details enter. This doesn't mean that the physical object can't rotate, however. And it doesn't mean that the diagonal generator can't be globally defined. 

Nor does it mean that it's not possible to relate two different choices. Therefore the whole point of studying the putative effects of such a rotation of the axis of quantization  seems vacuous. No effects of such a rotation are physical. 

JD

http://www.fisica.net/quantica/Griffiths%20-%20Introduction%20to%20quantum%20mechanics.pdf

NOTHING on hydrogen atom or molecule?

There are thousands of books on spectroscopy and quantum chemistry dealing with the molecule

many are for free and searchable via google as I did for you gifiith

HtB: There are thousands of books on spectroscopy and quantum chemistry dealing with the molecule  many are for free and searchable via google as I did for you grifiith

That's news to me.  Got any links to good ones?

KO: The degeneracies are lifted by relativistic effects, in a way which depends on the existence of electron spin (leaving an interesting trace of supersymmetry), due to the existence of vacuum fluctuations (Lamb shift), and finally by a small amount due to the proton spin. The last hyperfine splitting causes the famous 21-cm hydrogen line. 

I'm not sure what is meant by "vacuum fluctuations" here, but to my knowledge, the electron spin represents a fourth quantum number. which marks whether the magnetic dipole moment of the electron spin parallel or anti parallel to the dipole moment of the whole atom.  

If the two spins are anti-parallel, it represents a higher energy state, and there is a slight emission of energy when the two magnets align.

JD

I found YOUR Griffith

Try first to find books with QM approaches for H2 that is most suitable for you with your knowledge

http://web.pdx.edu/~pmoeck/books/Tipler_Llewellyn.pdf

section 9.2, perhaps.

JD

google for Pauling (yes the double nobel price winner) & Wilson

INTRODUCTION QUANTUM MECHANICS

With Applications to Chemistry

dating from 1935

Hello everyone, thank you all again for your time and your great comments. I have put careful thought into them one by one and here is what I have come up with so far. I would like to share it with you guys and ask to correct it if I'm wrong.

From what I can gather, the quantization axis is an arbitrary axis which is just used to visualize the difference between rotationally quantized H2 molecules. In fact, these molecular axis spatial orbitals are something like electron spatial orbitals (s, p, d, f orbitals). I use the p orbital as an example. Well, it has three sub-orbitals px, py, pz. Let’s say that we rotate the px orbital by theta= 90 and phi= 90 degrees, such that it goes to an orientation similar to py, but the orbital is still px. There should not be any orientational states in between these two orientations, otherwise, we would have orbitals other than the three px, py, pz orbitals. In fact, this orientational change (rotation) cannot be explained in any ways unless one uses the fact that the rotated px orbital has indeed jumped straight from a px orbital to a py one. We can examine this by somehow calculating the energy of a by-90°-90°-rotated px orbital. If the calculated energy is similar to py’s energy, then the rotated px orbital is in fact a py orbital.