Why does an electron orbital have a preferred z axis?

[Happiness]

It could be the case that you have a ball which is spinning, but you don't know what axis it's spinning around. In that case, you could construct a probability distribution that described that state of knowledge. But that would still not be the same as attributing an actual physical state to the ball that was a superposition of spinning around all possible axes. As above, no such state exists.

But such a state exists in our world, right? Since our world obeys quantum mechanics and not classical physics. Provided that we don't make a measurement to cause the collapse of the wave function, then the ball is in a superposition of states.

 

[Happiness]

Then you made a mistake soemwhere, most likely in ignoring the imaginary parts.

Given that the three states are orthogonal, the cross-terms will end up cancelling each other, so you will get

(\sin \theta \sin \phi)^2 + (\sin \theta \cos \phi)^2 + \cos^2 \theta = 1

This is a consequence of Unsöld's theorem.

Wrong. If you get the properly normalized wavefunctions (those you posted are not), add them and square them they will only be a function of r.

What Unsöld's theorem says is $|\psi_{210}|^2+|\psi_{211}|^2+|\psi_{21-1}|^2$ is a function of $r$ only. But this is not a superposition of states. What I was saying was a superposition of $p_x$, $p_y$ and $p_z$ does not give spherical symmetry. So Unsöld's theorem would not be relevant in this case.

 

[Happiness]

But not all states will have that property, even for a spherically symmetric problem, and states that don't have that property are certainly physically relevant. So to fully see the spherical symmetry of the problem, you still have to look at the set of all possible solutions.

Not quite. There is no such thing as the solution. There are an infinite number of possible solutions. Some are spherically symmetric and some are not. The eigenfunctions themselves are solutions, and are in fact the ones that get physically realized whenever you actually measure the spin. But the ones that are not are always part of some set of solutions that, taken as a whole, is.

"The solution is still spherically symmetrical", the solution here means the solution to the problem, which was set up to be spherically symmetrical. (The set up was an isolated hydrogen atom, without any applied magnetic field or any such history of application.) I wasn't referring to the solution to the Schrödinger equation. An eigenfunction such as $\psi_{211}$ is a solution to the Schrödinger equation but is not a solution to the problem. Only a superposition (of eigenfunctions) that is spherically symmetrical can be a solution to the problem.

 

[Happiness]

Consider a nitrogen atom, which has a partially filled p subshell, that has never been exposed to a magnetic field or any influence that could cause it to distinguish a particular direction from another. Consider those 3 electrons in its p subshell. Are they all in the same state? Are they all in the same state that is spherically symmetrical, formed by the superposition of eigenstates corresponding to p subshell?

Or is one of them in $p_x$, one in $p_y$ and one in $p_z$? By Unsöld's theorem, which says $|\psi_{p_x}|^2+|\psi_{p_y}|^2+|\psi_{p_z}|^2$ is a function of $r$ only, this case still gives a spherically symmetrical distribution of the 3 electrons when considered as a whole, but not so if each electron is considered separately at a time. But if this is the case for the nitrogen atom, then which direction in the physical world is the z axis?

 

[Cthugha]

View attachment 247757
$p_x$ and $p_y$ orbitals have their $m$ negative of each other, implying their $L_z$ are pointing in opposite directions. But I don't see how they are opposite from the diagram. How do you see it?

And the superposition of these 3 orbitals alone, exactly as they are shown in the diagram, doesn't give spherical symmetry right? Because it would give a higher probability along the 3 axes of the coordinate system. To produce spherical symmetry, we have to continue adding more triplets of orbitals ($p_x$, $p_y$ and $p_z$) whose axes point in all other directions. Right?

No, very obviously adding the squared modulus of the three orbitals you mention above removes any angular dependence.

Just for the record: Where did you get this picture from? It is completely wrong. The px, py and pz orbitals do NOT correspond to m=-1,0 and +1. The following image from wikipedia gives the orbitals corresponding to the different m states:

Note that only the p-orbital for m=0 looks like the ones you posted. So why is that the case? The orbitals are in principle complex functions. The value of the real part is given by the shape, while the colors indicate the imaginary part. As you can see, the m=0 orbital is already real, while the ones for m=-1 and +1 are not. In fact, they have a 2 pi rotation of the phase around the axis. As phase gradients are roughly representing the local wavevector, this is indicative of the orbital angular momentum.
However, in many situations you want real orbitals or ones with a directionality that matches the symmetry of your problem. This is how px, py and pz-orbitals come into play. If you build the two possible superpositions of the m=-1 and m=+1 modes, the corresponding superposition orbitals will be real. And they will look like px and py. These orbitals are great, e.g. for chemistry, where you may have a lattice and bonds predominantly form along the lattice directions. However, they do not represent the orbital angular momentum projections.

 

[vanhees71]

I have not read the entire thread, but my advice is to make the issue clear from a completely "classical point of view" first.

The spherical harmonics also occur in classical field theory (electrodynamics), though often it's not treated in the introductory electrodynamics lecture. How does one come intuitively to investigate these important class of functions?

Take electrostatics as the most simple example. The characteristic equation is Poisson's equation for the electrostatic potential
$$\Delta V(\vec{x})=-\rho(\vec{x}),$$
where $\rho(\vec{x})$ is a given charge distribution, and you look for solutions for $V$.

One ansatz for the solution is a separation of coordinates in some given coordinate system. Now if you have something spherical symmetric, it's a good guess to use spherical coordinates for this separation ansatz, i.e., you write
$$V(\vec{x}) \equiv V(r,\vartheta,\varphi)=R(r) \Theta(\vartheta) \Phi(\varphi),$$
with functions $R$, $\Theta$, and $\Phi$ depending just on one of the coordinates.

Now of course Euclidean space is isotropic, but the spherical coordinates introduce a preferred direction, namely the polar axis of the coordinate system. Usually you start from a Cartesian coordinate system and taking the 3-axis as this polar axis. That's how you break the spherical symmetry by choice of a coordinate system, and because the preferred axis is the 3-axis that's a preferred direction. It's of course not in any way preferred in Euclidean space but it's just our choice to prefer this direction to be able to define the spherical coordinates. These are even singular along this polar axis. This are of course also only coordinate singularities since there's nothing dramatically happening along this axis in Euclidean space which is perfectly homogeneous and isotropic.

 

[Happiness]

Just for the record: Where did you get this picture from? It is completely wrong. The px, py and pz orbitals do NOT correspond to m=-1,0 and +1.

The diagram is from the University of Pennsylvania: https://www.sas.upenn.edu/~milester/courses/chem101/AQMChem101/AQMPages/AQMIIIc.html

No, very obviously adding the squared modulus of the three orbitals you mention above removes any angular dependence.

But since adding square moduli is not a superposition, would you then agree that a superposition of $p_x$, $p_y$ and $p_z$ will not produce spherical symmetry, unless you also add in those triplets ($p_x$, $p_y$ and $p_z$) whose longitudinal axes point in other directions?

 

[Cthugha]

The diagram is from the University of Pennsylvania: https://www.sas.upenn.edu/~milester/courses/chem101/AQMChem101/AQMPages/AQMIIIc.html

Then I would use different resources for learning physics.

But since adding square moduli is not a superposition, would you then agree that a superposition of $p_x$, $p_y$ and $p_z$ will not produce spherical symmetry, unless you also add in those triplets ($p_x$, $p_y$ and $p_z$) whose longitudinal axes point in other directions?

The orbitals are orthogonal. If you find that the squared modulus of the sum and the sum of the individual squared moduli differs, your math is erroneous.

 

[PeterDonis]

such a state exists in our world, right?

Such a state of a quantum particle exists. Whether such a state of a spinning ball exists is not so clear. A spinning ball is a macroscopic object, and whether exact linear superposition of states works for macroscopic objects the way it does for quantum particles depends on which interpretation of QM you favor.

Provided that we don't make a measurement to cause the collapse of the wave function, then the ball is in a superposition of states.

A macroscopic object like a ball, once you take decoherence into account, is continualy measuring and collapsing itself, if you adopt a collapse interpretation of QM.

 

[PeterDonis]

the solution here means the solution to the problem, which was set up to be spherically symmetrical.

I wasn't referring to the solution to the Schrödinger equation.

This doesn't make sense; in order to get a solution to the problem, you have to solve the Schrödinger equation. You can't have one without the other.

 

[Happiness]

The orbitals are orthogonal. If you find that the squared modulus of the sum and the sum of the individual squared moduli differs, your math is erroneous.

The cross terms are sure to vanish only when integrating over the volume of the whole sphere, right? In other words, $\int_{all\,V}|\psi_1|^2+|\psi_2|^2dV=\int_{all\,V}|\psi_1+\psi_2|^2dV$, for a pair of orthogonal wave functions. But if I integrate over say a quadrant of the sphere, then $\int_{quadrant}|\psi_1|^2+|\psi_2|^2dV\neq\int_{quadrant}|\psi_1+\psi_2|^2dV$, right?

 

[Vanadium 50]

If you integrate over half a sphere you won't get anything spherically symmetric, since you're only looking at half a sphere.

 

[Happiness]

If you integrate over half a sphere you won't get anything spherically symmetric, since you're only looking at half a sphere.

If you integrate over the whole sphere, you won't get know if there is spherical symmetry or not, since you're adding up the contribution of the whole sphere.
Take 2 light bulbs. Bulb A shines uniformly in all direction and so produces a spherical symmetry. Bulb B shines more brightly to the right side and so does not produce a spherical symmetry. If you integrate the light intensity over the whole spherical surface area (for a certain radius), then both A and B will always give you the same answer no matter how you rotate the bulbs. But if you integrate the light intensity over a part of the spherical area, A gives the same answer but B does not in general, when the bulbs are rotated. This shows A has spherical symmetry and B does not.

 

[Happiness]

This doesn't make sense; in order to get a solution to the problem, you have to solve the Schrödinger equation. You can't have one without the other.

Yes, the eigenstates satisfy the Schrödinger equation. But I don't see them as the solution to the problem. I see them as the basis vectors that I could use to represent the state that is a solution to the problem. I could have very well used other basis vectors to represent the same state that is a solution to the problem.
But previously I mistook those eigenstate basis vectors to be themselves individually a solution to the problem. This was the reason for my confusion.

 

[Cthugha]

I lost track of what you are trying to achieve. Do you try to find a single wave function that has spherical symmetry itself for a state with finite angular momentum? That is obviously impossible. The angular momentum operator acting on some state is \hat{L}\psi=(r\times\hat{p})\psi=r\times\frac{\hbar}{i}\nabla\psi.
So you have the cross product of r with the gradient of the wave function. Obviously for a wave function that is spherically symmetric, the gradient must necessarily point in the radial direction, while any gradient along the sphere is 0. Accordingly any wave function that is itself spherically symmetric must necessarily correspond to an angular momentum of 0.

 

[PeterDonis]

the eigenstates satisfy the Schrödinger equation

The eigenstates of the Hamiltonian satisfy the time-independent Schrödinger equation. But all states satisfy the time-dependent Schrödinger equation. And if your problem involves the possibility of states that aren't eigenstates of the Hamiltonian, then the latter is the correct equation for your problem.

 

[Happiness]

I lost track of what you are trying to achieve. Do you try to find a single wave function that has spherical symmetry itself for a state with finite angular momentum? That is obviously impossible. The angular momentum operator acting on some state is \hat{L}\psi=(r\times\hat{p})\psi=r\times\frac{\hbar}{i}\nabla\psi.
So you have the cross product of r with the gradient of the wave function. Obviously for a wave function that is spherically symmetric, the gradient must necessarily point in the radial direction, while any gradient along the sphere is 0. Accordingly any wave function that is itself spherically symmetric must necessarily correspond to an angular momentum of 0.

This only applies to orbital angular momentum, right? It doesn't apply to spin angular momentum. In other words, it is possible for an object to have spherical symmetry and non-zero spin angular momentum, right? Again, when I say an object has spherical symmetry, I mean the object experiences the same condition/influence from all directions. The world it sees (and has been seeing) is the same in all directions.

 

[PeterDonis]

when I say an object has spherical symmetry, I mean the object experiences the same condition/influence from all directions. The world it sees (and has been seeing) is the same in all directions.

This is vague. What does it mean in precise math?

 

[PeterDonis]

when I say an object has spherical symmetry

You've talked about a lot of things other than "objects" in this thread. For example, in the OP you said the potential was spherically symmetric. But the potential is not an "object".

I think you haven't fully thought through what exactly you are asking.

 

[Happiness]

This is vague. What does it mean in precise math?

It means the potential energy function is spherically symmetrical. It depends only on $r$, the distance from the "object", or the centre of the "object". Object here means the hydrogen atom, or some other element, or an electron, or some other elementary "particle".

 

[PeterDonis]

It means the potential energy function is spherically symmetrical.

Ok, but the potential energy is not an "object".

Object here means the hydrogen atom

Which is not the same as the potential energy. Nobody disputes that the potential energy is spherically symmetrical (a function of $r$ only in spherical coordinates) for the cases under discussion, and that's the only interpretation of "spherically symmetric" that you've given a precise mathematical definition for. Yet you keep talking about "objects", and evidently have in mind some other definition of "spherically symmetric" for objects, which can't possibly be "the potential energy function is spherically symmetrical" since the potential energy function is not an object. So what are you talking about?

 

[Happiness]

Ok, but the potential energy is not an "object".
Which is not the same as the potential energy. Nobody disputes that the potential energy is spherically symmetrical (a function of $r$ only in spherical coordinates) for the cases under discussion, and that's the only interpretation of "spherically symmetric" that you've given a precise mathematical definition for. Yet you keep talking about "objects", and evidently have in mind some other definition of "spherically symmetric" for objects, which can't possibly be "the potential energy function is spherically symmetrical" since the potential energy function is not an object. So what are you talking about?

An electron has spherical symmetry when it experiences a potential energy function that is spherically symmetrical.

 

[PeterDonis]

An electron has spherical symmetry when it experiences a potential energy function that is spherically symmetrical.

If this is your definition, it's quite perverse (not to mention different from any definition used in the literature), since we have already seen that there are plenty of solutions to the Schrödinger equation with a spherically symmetric potential that are not spherically symmetric (the wave functions are not just functions of $r$). I see no good reason to call those solutions "spherically symmetric" simply because the potential is (and neither, as I mentioned, does any of the literature).

 

[PeterDonis]

The OP question has been more than sufficiently addressed. Thread closed.