Preferred orientation of a nitrogen molecule in space?

[Happiness]

Consider an $N_2$ molecule. Chemists say that the triple bond is due to one $p_x - p_x$ overlap, one $p_y - p_y$ overlap and one $p_z - p_z$ overlap. The x-axis (the label is not important; I’m sure you know what I mean) is clear because it’s the longitudinal axis of the molecule. But how do you know what the directions of y and z axes in the physical world are?

Also, since any linear combination of $p_y$ and $p_z$ is also a solution to Schrodinger equation, how do you know the electrons are not in this kind of state? Why must they be purely in $p_y$ or $p_z$? (Suppose you believe the chemists are right.)

Are the electrons purely in $p_y$ and $p_z$? $p_y$ and $p_z$ are pointing in some definite direction in space, it’s just that those definite directions are unknown to us. Or are the electrons in a linear combination of $p_y$ and $p_z$ such that their orbital looks like a cylindrical shell surrounding the x axis? (Like a toilet paper roll with uniform probability amplitude all over the curved surface.)

 

[mfb]

If you measure the orbitals along some axis then you'll get a result based on that measurement, but if you don't then there is no preferred direction of the molecule.

 

[Happiness]

If you measure the orbitals along some axis then you'll get a result based on that measurement, but if you don't then there is no preferred direction of the molecule.

Which means in the N-N triple bond, two of the bonds (ie., those two $\pi$ bonds) “fuse” together into a uniform cylindrical-shell shaped orbital (ie., a superposition of $p_y$ and $p_z$ orbitals)? And it’s only upon a measurement that is in the y-z plane, that the cylindrical-shell shaped orbital “wave-functionally-collapses” into separate $p_y$ and $p_z$ orbitals?

 

[PeterDonis]

Chemists say

Can you give a specific reference for what chemists say? Chemists might not actually be saying quite what you think they are.

 

[Happiness]

Can you give a specific reference for what chemists say? Chemists might not actually be saying quite what you think they are.

"At the same time the $p_z$-orbitals approach and together they form a $p_z-p_z$ pi-bond. Likewise, the other pair of $p_y$-orbitals form a $p_y-p_y$ pi-bond." — https://en.wikipedia.org/wiki/Triple_bond

http://web.uni-plovdiv.bg/plamenpenchev/mag/books/inorgchem/Cotton-Wilkinson%20-%20Advanced%20Inorganic%20Chemistry_file1.pdf

"Both the $p_y$ and the $p_z$ orbitals on each carbon atom form pi bonds between each other. As with ethene, these side-to-side overlaps are above and below the plane of the molecule. The orientation of the two pi bonds is that they are perpendicular to one another (see figure below)." —https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Book:_Introductory_Chemistry_(CK-12)/09:_Covalent_Bonding/9.20:_Sigma_and_Pi_Bonds

https://web.ung.edu/media/Chemistry2/Chemistry-LR.pdf

 

[PeterDonis]

As with ethene, these side-to-side overlaps are above and below the plane of the molecule.

So there's an answer to your question in the OP:

how do you know what the directions of y and z axes in the physical world are?

You defined the $x$ axis as along the line between the two carbon atoms. The $y$ axis is then the direction perpendicular to that in the plane of the molecule, and the $z$ axis is the direction perpendicular to that plane.

For acetylene, which is linear and therefore does not pick out any particular direction perpendicular to $x$, the best way to visualize it is the "space-filling model" image in figure 9.20.6. Note that that image is symmetric about the $x$ axis, i.e., it picks out no preferred direction perpendicular to that axis. In other words, when sources talk about "y" and "z" directions for such cases, they're being somewhat sloppy.

 

[Happiness]

For acetylene, which is linear and therefore does not pick out any particular direction perpendicular to $x$, the best way to visualize it is the "space-filling model" image in figure 9.20.6. Note that that image is symmetric about the $x$ axis, i.e., it picks out no preferred direction perpendicular to that axis. In other words, when sources talk about "y" and "z" directions for such cases, they're being somewhat sloppy.

The space-filling model is not depicting the shape of the electron orbitals! It's only for showing the general shape of a molecule as though all its atoms are spherical in shape, and so it's not realistic, though it can be used as a good approximation to the realistic shape of a molecule. In other words, it always treats atoms as spheres even when their electron orbitals are not spherical, which is the case for $N_2$ with its $p$ orbitals. https://en.wikipedia.org/wiki/Space-filling_model

So with this understanding, you can see that the reference https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Book:_Introductory_Chemistry_(CK-12)/09:_Covalent_Bonding/9.20:_Sigma_and_Pi_Bonds
is not saying that the molecule is symmetrical about the x axis. It is saying there is a preferred direction for the sideway overlap of the $p$ orbitals, namely above and above the "plane" of the molecule, just like Figure 8.25(b).

So there's an answer to your question in the OP:

That sentence (ie., these side-to-side overlaps are above and below the plane of the molecule) was talking about ethene, which can't be compared to $N_2$. Ethyne—but not ethene—can be compared to $N_2$. In ethene, its 4 H atoms form a plane, destroying the cylindrical symmetry along the x axis, which exists in ethyne and $N_2$.

 

[PeterDonis]

The space-filling model is not depicting the shape of the electron orbitals!

True, but in the case under discussion it's a better visualization than the one you are trying to use. See below.

is not saying that the molecule is symmetrical about the x axis. It is saying there is a preferred direction for the sideway overlap of the $p$ orbitals, namely above and above the "plane" of the molecule

And as I said, when references seem to be saying that in such cases, they are being sloppy. You yourself can see why: because for a linear molecule, like N2 or C2H2, no particular direction perpendicular to the line of the molecule is picked out. So in the absence of any measurement perpendicular to the line of the molecule (and there isn't any such measurement for molecules in, say, nitrogen gas in the Earth's atmosphere or acetylene in a tank supplying a welding torch), the state of the molecule will be axially symmetric.

 

[Happiness]

So in the absence of any measurement perpendicular to the line of the molecule (and there isn't any such measurement for molecules in, say, nitrogen gas in the Earth's atmosphere or acetylene in a tank supplying a welding torch), the state of the molecule will be axially symmetric.

How about an acetylene molecule colliding at an angle with the wall of its container? Would it cause a collapse of the wave function to $p_y$ and $p_z$, since now the part of the molecule that collided with the wall is different experientially from the rest of the molecule, thereby breaking the cylindrical symmetry? And then after collision, the molecule moves off with the collapsed wave function. The directions of the $p_y$ and $p_z$ orbitals of the post-collision molecule are now definite, although this information is virtually inaccessible to a macroscopic observer, who may make a measurement on the molecule, but his measurement would change the wave function unless he so happens to make the measurement along the exact same direction of $p_y$ or $p_z$.

 

[PeterDonis]

How about an acetylene molecule colliding at an angle with the wall of its container?

How would you measure what happens?

Would it cause a collapse of the wave function to $p_y$ and $p_z$, since now the part of the molecule that collided with the wall is different experientially from the rest of the molecule, thereby breaking the cylindrical symmetry? And then after collision, the molecule moves off with the collapsed wave function

First, there is no such thing as "the part of the molecule that collided with the wall". The wall is also made of molecules; it's not an ideal plane.

Also, if the collision is elastic, as it would be expected to be, the molecule would return to the same internal state after the collision (all that would change is its center of mass momentum, which doesn't affect the orbitals). So no, I would not expect such a collision to collapse the wave function.

 

[Happiness]

Also, if the collision is elastic, as it would be expected to be, the molecule would return to the same internal state after the collision (all that would change is its center of mass momentum, which doesn't affect the orbitals). So no, I would not expect such a collision to collapse the wave function.

How about the potential energy function being asymmetrical? A molecule of the wall would develop partial (electric) charges at different parts of the molecule, as its electrons are not always evenly distributed. This causes asymmetrical electric potential energy function on the near end of the approaching acetylene molecule compared to its far end. Would this break of symmetry cause acetylene's wave function to collapse?

 

[PeterDonis]

How about the potential energy function being asymmetrical?

It's only asymmetrical during the collision, and if the collision is elastic, any changes during the collision are reversible so they wouldn't count as measurements and wouldn't collapse anything.

 

[Happiness]

It's only asymmetrical during the collision, and if the collision is elastic, any changes during the collision are reversible so they wouldn't count as measurements and wouldn't collapse anything.

But that only works if the uneven distribution of the electrons of the wall molecule is time-reversal symmetrical (the same in both forward time and backward time), right? The wall is constantly being influenced by other things through time in a time-irreversible manner. This would break the time-reversal symmetry of its electron distribution.

 

[PeterDonis]

that only works if the uneven distribution of the electrons of the wall molecule is time-reversal symmetrical (the same in both forward time and backward time), right?

No, all that's necessary is that the collisions are elastic. "Reversible" here doesn't mean "time symmetric", it just means "the molecule returns to the same internal state after the collision".

The wall is constantly being influenced by other things through time in a time-irreversible manner.

Yes, but the time scale over which this happens is much longer than the time scale of a single collision with a gas molecule. So it doesn't affect the fact that each individual collision is elastic.

 

[Happiness]

No, all that's necessary is that the collisions are elastic. "Reversible" here doesn't mean "time symmetric", it just means "the molecule returns to the same internal state after the collision".

Yes, but the time scale over which this happens is much longer than the time scale of a single collision with a gas molecule. So it doesn't affect the fact that each individual collision is elastic.

But if the wall experiences a time-irreversible change over a large time scale, then wouldn't this suggest that each individual collision must be slightly short of being elastic to accumulate such irreversibility over time? An irreversible change over an hour must be the result of small irreversible changes over microseconds, right?

 

[PeterDonis]

if the wall experiences a time-irreversible change over a large time scale, then wouldn't this suggest that each individual collision must be slightly short of being elastic to accumulate such irreversibility over time?

No, for two reasons.

First, the irreversible change does not have to come from the gas inside. There's a whole universe outside the wall. The gas can be in equilibrium with the container, meaning no time-irreversible change in their relationships, even if the container undergoes a time-irreversible change in its relationship with the environment.

Second, consider the kind of irreversible change we could be talking about. One example would be the gas container slowly cooling, let's say, because it starts out warmer than its environment. That means the molecules in the container (and therefore the molecules in the gas, since the gas should be in equilibrium with the container) are jiggling around a little faster than the molecules in the surrounding environment. So collisions between container molecules and environment molecules will, on net, transfer some kinetic energy from the container to the environment. That is an irreversible change, but every individual collision involved is still elastic. The reason kinetic energy gets transferred is that the average kinetic energy of the container molecules is higher, and it's straightfoward to show that under those conditions elastic collisions will on average transfer kinetic energy from the container to the environment. And similarly, as the container cools, it cools the gas by the same mechanism.

 

[Happiness]

True, but in the case under discussion it's a better visualization than the one you are trying to use. See below.

And as I said, when references seem to be saying that in such cases, they are being sloppy. You yourself can see why: because for a linear molecule, like N2 or C2H2, no particular direction perpendicular to the line of the molecule is picked out. So in the absence of any measurement perpendicular to the line of the molecule (and there isn't any such measurement for molecules in, say, nitrogen gas in the Earth's atmosphere or acetylene in a tank supplying a welding torch), the state of the molecule will be axially symmetric.

What if I introduce into a tank of acetylene, an acetylene molecule that already has $p_y$ and $p_z$ orbitals? Its wave function was collapsed by say a measurement previously done on it. Would collision with this newcomer molecule make the other molecules collapse their wave function into $p_y$ and $p_z$?

 

[PeterDonis]

What if I introduce into a tank of acetylene, an acetylene molecule that already has $p_y$ and $p_z$ orbitals?

How are you going to do that?

Its wave function was collapsed by say a measurement previously done on it.

What kind of measurement?

Also, even if such a measurement is made, how do you know the molecule will stay in that state? Not every state that results from a measurement will stay the same under time evolution.

Would collision with this newcomer molecule make the other molecules collapse their wave function into $p_y$ and $p_z$?

Why would it do that? The collisions are elastic so they don't affect the internal states of the molecules.

 

[Happiness]

Why would it do that? The collisions are elastic so they don't affect the internal states of the molecules.

Are the collisions brought about by the act of measurement that causes the wave function to collapse to $p_y$ and $p_z$, elastic?

If they are, then how do these elastic collisions cause a collapse of the wave function during measurement but not in this case? And in general, if an irreversible change consists of only elastic collisions, what is the origin of its irreversibility at the microscopic level?

If they aren’t, then what’s the reason for assuming that they aren’t but yet assuming those between the gas molecules themselves are?

 

[PeterDonis]

Are the collisions brought about by the act of measurement that causes the wave function to collapse to $p_y$ and $p_z$, elastic?

What collisions brought about by the act of measurement? You haven't even described what that act of measurement is, so how do you know it involves collisions?

 

[PeterDonis]

if an irreversible change consists of only elastic collisions, what is the origin of its irreversibility at the microscopic level?

There is no irreversibility at the microscopic level. The irreversibility is at the macroscopic level; in the example I gave, it's the flow of heat from the hotter object to the colder one. You can't see that heat flow in the individual collisions; you can only see it in the macroscopic average effect of the collisions over time. This is all standard statistical mechanics and thermodynamics.

 

[Happiness]

What collisions brought about by the act of measurement? You haven't even described what that act of measurement is, so how do you know it involves collisions?

The measurement of momentum of the electron along a certain direction, for example.

 

[PeterDonis]

The measurement of momentum of the electron along a certain direction, for example.

That wouldn't collapse the electron to a $p$ orbital state. It would collapse it to a linear momentum eigenstate (or, more realistically, a sharply peaked Gaussian in linear momentum space), which is not at all the same thing.

 

[Happiness]

That wouldn't collapse the electron to a $p$ orbital state. It would collapse it to a linear momentum eigenstate (or, more realistically, a sharply peaked Gaussian in linear momentum space), which is not at all the same thing.

But there must exist some measurement or collision or interaction that cause the wave function to collapse to $p_y$ and $p_z$, especially when chemists often talk about them and use them to explain many other things and observations.

And when one atom/molecule is in the collapsed state, it probably causes collapse in other atoms/molecules too, via its interaction with them.

 

[PeterDonis]

there must exist some measurement or collision or interaction that cause the wave function to collapse to $p_y$ and $p_z$,

There is such a measurement, yes, since those orbital states are eigenstates of observables. But first, it isn't the same observable (same measurement) for $p_y$ and $p_z$, because they're in different directions. And second, that observable is not the same as any interaction involved in elastic collisions.

when one atom/molecule is in the collapsed state, it probably causes collapse in other atoms/molecules too, via its interaction with them

Only if that interaction is the right kind to serve as a measurement of the same observable. Which it isn't.