I Why does an electron orbital have a preferred z axis?

[Happiness]

Why is there a preferred z axis even though the potential energy function is perfectly spherical? Shouldn't the electron be around the nucleus in a spherically symmetrical way?

 

[Vanadium 50]

Before we can explain why something is true, we need to know if it is true. Why do you believe there is a preferred z-axis?

 

[Happiness]

These are some electron orbitals.

Except for the first two, the rest are all not spherically symmetrical, hence a preferred z axis. This is unexpected since the potential energy function is perfectly spherically symmetrical.

For an isolated hydrogen atom, which direction (in the physical world) will turn out to be the preferred z axis?

 

[dextercioby]

The z axis is just a very human made convention. It has to do with humans and the fact that the "z" label is the one going up, if you allow me the metaphor. If the "right handed orthogonal system" had the x label going up, and z to the right, then angular momentum would have Lx wherever Lz is now used. Mathematics requires only one preferred direction (rotation axis in classical mechanics). By convention this the "up" one, which, by another convention, is called z.

 

[Happiness]

From the shapes of the orbitals, this mathematical choice of a z axis produces a physical effect: the orbitals is now symmetrical about this particular axis. But this is absurd.

Which physical direction would this z axis be? Are you saying it's random? That is, in an ensemble of isolated hydrogen atoms, some has their z axis pointing up, some right, some slanted to the left by some angle, etc.

And why is the spherical symmetry destroyed in the physical world?

 

[dextercioby]

Mathematics requires a preferred direction. It stems from the symmetries of flat (Galilean/Minkowski) space-time. Yes, mathematics properly applied to physical laws/principles produces physical effects/consequences. Think of Newton's laws. The third law plus a definition plus the second law plus mathematics imply total momentum conservation.

 

[Cthugha]

What is so strange about this? The three numbers in the name of the orbitals are the quantum numbers. The first one is the principal quantum number, the second one corresponds to orbital angular momentum and the third one tells about the projection of the orbital angular momentum along some preferred axis.

As you will notice, all the orbitals with 0 orbital angular momentum are spherically symmetric, while the other ones are not. This is quite easy to see when considering the classical analogue: of a rotating sphere. Can you create rotational motion without introducing one preferred axis (or several of them) as the rotation axis? What do you get if you consider a "spherically symmetric" rotation axis by just considering every possible orientation of the rotation axis and summing over all of these orientations with equal weight?

 

[Happiness]

But the choice of the z axis is arbitrary, so it cannot have a physical effect. Other arbitrary choices such as reference frame, coordinate system, basis vectors, units of measurements, whether I start counting time now or 5 minutes ago, all these do not have physical effects.

 

[Happiness]

This is quite easy to see when considering the classical analogue: of a rotating sphere. Can you create rotational motion without introducing one preferred axis (or several of them) as the rotation axis?

But in this classical problem, the rotation axis is given by the problem. It is part of the input data.

Whereas for these hydrogen wave functions, no preferred axis is given by the problem. The only input data are Schrodinger equation and the spherically symmetrical potential energy function (Coulomb's law). The z axis was arbitrarily chosen by the problem solver (when he chose the spherical coordinate system with a z axis). So we would expect the solutions to be independent of this z axis.

 

[Vanadium 50]

Since this is I-level, I am assuming you know some QM. What feature do all the spherically symmetric wavefunctions have in common?

Follow-up: which wavefunctions have the same energy?

 

[PeterDonis]

the choice of the z axis is arbitrary, so it cannot have a physical effect

That's right, so which direction in space the z axis points cannot have a physical effect. But that does not mean there is no such thing as rotation or that rotation does not have an axis. If you have a ball spinning, not interacting with anything else, it's physically the same regardless of which direction in space its spin axis is pointing. But that does not mean it's not spinning and does not have a spin axis at all.

Similarly, if I have an electron in, say, one of the 2p orbitals of a hydrogen atom, not interacting with anything else, it's physically the same no matter which way in space the axis of the 2p orbital is pointing. But that does not mean there isn't any 2p orbital or that it doesn't have nonzero orbital angular momentum or that it doesn't have an axis.

 

[Happiness]

What feature do all the spherically symmetric wavefunctions have in common?

They are all spherically symmetric. In other words, they do not have a preferred z axis.

Follow-up: which wavefunctions have the same energy?

Wave functions that have the same n, provided you accept that these wave functions are the correct solutions. But since these wave functions are not spherically symmetrical, it makes more sense to reject them as the correct solutions to a spherically symmetrical problem.

 

[PeterDonis]

since these wave functions are not spherically symmetrical, it makes more sense to reject them as the correct solutions to a spherically symmetrical problem

This is wrong. A problem with a given symmetry does not have to have all solutions satisfy that symmetry individually. It just has to have the complete set of solutions, taken as a whole, satisfy that symmetry.

For example, you showed pictures of electron wave functions that are not spherically symmetrical. Take one of them, say again the 2p orbital wave function for the hydrogen atom. How many such wave functions are there? Answer: an infinite number, one for each possible direction in space that the 2p orbital can point. No one of these solutions is spherically symmetric, but the set of all of them is.

Once again, if your logic were correct, it would show that there could be no such thing as rotation, since rotation breaks spherical symmetry by picking out an axis. That's obviously absurd. The correct logic is that, since there is such a thing as rotation, rotation about any axis, pointing in any direction in space, must be treated the same, so that the set of all possible rotations is spherically symmetric.

 

[Cthugha]

Wave functions that have the same n, provided you accept that these wave functions are the correct solutions. But since these wave functions are not spherically symmetrical, it makes more sense to reject them as the correct solutions to a spherically symmetrical problem.

You are misunderstanding something. Nobody said that a SINGLE of these orbitals would yield the complete solution if the problem is spherically symmetric. Have a look at the several orbitals for n=2 and angular momentum of 1. What would you get if you superpose them?

 

[Happiness]

You are misunderstanding something. Nobody said that a SINGLE of these orbitals would yield the complete solution if the problem is spherically symmetric. Have a look at the several orbitals for n=2 and angular momentum of 1. What would you get if you superpose them?

I think I get what you are saying. So for an isolated hydrogen atom with n=2, angular momentum=1, its wave function should be the superposition of all such wave functions with their z axis pointing in all directions, equally weighed? And that solution would be spherically symmetrical. Right?

 

[ZapperZ]

Why is there a preferred z axis even though the potential energy function is perfectly spherical? Shouldn't the electron be around the nucleus in a spherically symmetrical way?

This is a mathematical construct IF the atom is placed, say, in a magnetic field. We then call that the z-direction. So this is where the anisotropic symmetry is broken.

Zz.

 

[Vanadium 50]

What feature do all the spherically symmetric wavefunctions have in common?

They are all spherically symmetric.

Walked into that one.

 

[Cthugha]

I think I get what you are saying. So for an isolated hydrogen atom with n=2, angular momentum=1, its wave function should be the superposition of all such wave functions with their z axis pointing in all directions, equally weighed? And that solution would be spherically symmetrical. Right?

Well, roughly. As angular momentum is quantized, the story is a bit more difficult. In a nutshell, whatever z-axis you choose as the reference and measurement axis, you will have three possible values for the orbital angular momentum component along this axis that you might measure: +1, 0 and -1. The three corresponding orbitals always look the same and the superposition of these three orbitals will be spherically symmetric again for any choice of your z-axis.

 

[Happiness]

Well, roughly. As angular momentum is quantized, the story is a bit more difficult. In a nutshell, whatever z-axis you choose as the reference and measurement axis, you will have three possible values for the orbital angular momentum component along this axis that you might measure: +1, 0 and -1. The three corresponding orbitals always look the same and the superposition of these three orbitals will be spherically symmetric again for any choice of your z-axis.

$p_x$ 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?

 

[Vanadium 50]

And the superposition of these 3 orbitals alone, exactly as they are shown in the diagram, doesn't give spherical symmetry right?

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.

 

[Happiness]

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.

I still get an angular dependence: the result is proportional to $(\sqrt{2}\cos\theta+2\sin\theta\cos\phi)^2$.

 

[strangerep]

@Happiness: let's take a step back...

What do think is meant by the phrase "A (particular) class of physical system is spherically symmetric" ?

 

[Happiness]

@Happiness: let's take a step back...

What do think is meant by the phrase "A (particular) class of physical system is spherically symmetric" ?

It means there exists a point from which everything is the same in all directions.

 

[strangerep]

It means there exists a point from which everything is the same in all directions.

That's only valid for zero angular momentum. Here's the answer I was looking for:

A class of physical system is said to spherically symmetric iff the set of solutions to its governing equations of motion (EoM) is invariant under 3D rotations, i.e., under SO(3) transformations. I.e., any solution will map into another under an arbitrary rotation. In the case of a linear system, any solution will map (in general) to a linear combination of other solutions under the rotation. E.g., a solution which is an eigensolution of the $L_z$ operator is a linear combination of eigensolutions of (say) the $L_x$ operator (for fixed eigenvalue of $L^2$).

IOW, the crucial concept is invariance of the entire set of solutions of the relevant EoM's.

 

[Vanadium 50]

I still get an angular dependence

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.

 

[Happiness]

That's only valid for zero angular momentum. Here's the answer I was looking for:

A class of physical system is said to spherically symmetric iff the set of solutions to its governing equations of motion (EoM) is invariant under 3D rotations, i.e., under SO(3) transformations. I.e., any solution will map into another under an arbitrary rotation. In the case of a linear system, any solution will map (in general) to a linear combination of other solutions under the rotation. E.g., a solution which is an eigensolution of the $L_z$ operator is a linear combination of eigensolutions of (say) the $L_x$ operator (for fixed eigenvalue of $L^2$).

IOW, the crucial concept is invariance of the entire set of solutions of the relevant EoM's.

I don't understand you.

 

[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.

Were you saying $(\psi_{210}+\psi_{211}+\psi_{21-1})^2$ is a function of $r$ only? $\psi_{210}+\psi_{211}+\psi_{21-1}$ is real; there isn't any imaginary part to consider.

 

[PeterDonis]

I don't understand you.

Consider a simpler example from classical physics: a spinning ball.

The space in which the ball is spinning is still spherically symmetric. So are the laws of physics that govern the spinning ball. But a single particular spinning ball is not. It has a definite rotation axis that picks out a particular direction. How is this possible? (Note that, as I said before, with the definition of "spherically symmetric" that you are implicitly using in your head, it should be impossible to ever have anything rotate at all.)

The answer to the spinning ball question I just asked that corresponds to what @strangerep told you is that there is an infinite set of solutions describing spinning balls with axes pointing in all possible directions (assuming we have fixed all the properties of the ball like its mass, radius, moment of inertia, etc.), and you can map any single solution to any other single solution by a rotational transformation (what he called an SO(3) transformation). So the entire set of solutions has to have SO(3) symmetry. And that is how the underlying spherical symmetry of space and the laws of physics manifests itself for spinning balls. A single spinning ball does not have to be spherically symmetric by itself; that doesn't violate the spherical symmetry of space and the laws.

 

[Happiness]

Consider a simpler example from classical physics: a spinning ball.

The space in which the ball is spinning is still spherically symmetric. So are the laws of physics that govern the spinning ball. But a single particular spinning ball is not. It has a definite rotation axis that picks out a particular direction. How is this possible? (Note that, as I said before, with the definition of "spherically symmetric" that you are implicitly using in your head, it should be impossible to ever have anything rotate at all.)

The answer to the spinning ball question I just asked that corresponds to what @strangerep told you is that there is an infinite set of solutions describing spinning balls with axes pointing in all possible directions (assuming we have fixed all the properties of the ball like its mass, radius, moment of inertia, etc.), and you can map any single solution to any other single solution by a rotational transformation (what he called an SO(3) transformation). So the entire set of solutions has to have SO(3) symmetry. And that is how the underlying spherical symmetry of space and the laws of physics manifests itself for spinning balls. A single spinning ball does not have to be spherically symmetric by itself; that doesn't violate the spherical symmetry of space and the laws.

If a single particular spinning ball is not spherically symmetrical, then there is no issue with my definition of "spherically symmetrical".

From strangerep's reply, I thought he was saying a single particular spinning ball is still spherically symmetrical since the spinning ball no longer has zero angular momentum and he said my definition only works in cases with zero angular momentum.

My concern is with the spherical symmetry of the problem, not with the the spherical symmetry of space and the laws. If the problem and its input data do not pick out a certain direction to be special, then the solution shouldn't too. What I am getting from reading all these replies is that the solution is still spherically symmetrical even though the eigenfunctions are not spherically symmetrical, because the solution is a superposition of the eigenfunctions such that the solution still turns out to be spherically symmetrical.

To make an analogy with the classical spinning ball. If the problem does not specify how the ball is spun, then the solution is a superposition of all the states that are spinning along an axis in all directions.

 

[PeterDonis]

From strangerep's reply, I thought he was saying a single particular spinning ball is still spherically symmetrical since the spinning ball no longer has zero angular momentum and he said my definition only works in cases with zero angular momentum.

No, that's not what he was saying. He was not talking about a single spinning ball at all; he explicitly said a class of physical systems in his post.

My concern is with the spherical symmetry of the problem

Yes, and the problem is not the same as a single solution taken from an infinite set of possible solutions. The former can be spherically symmetric even if the latter is not. You already accept this in the case of the spinning ball, since you agree a single spinning ball is not spherically symmetric: the general "problem" of which a particular spinning ball is a single solution is spherically symmetric (roughly, it is the "problem" of how things rotate in ordinary 3-space, and that problem is spherically symmetric since it doesn't pick out any particular axis of rotation as special). So what's the problem with it for quantum systems?

To make an analogy with the classical spinning ball. If the problem does not specify how the ball is spun, then the solution is a superposition of all the states that are spinning along an axis in all directions.

You're conflating "a solution" with "the set of solutions". You're also missing a key distinction between the classical case and the quantum case.

In the classical case, a ball spinning around a single axis is a solution, for any orientation of the axis. But a superposition of all those solutions is not a solution. The classical equations of motion don't work like that. There is no such thing as an actual physical state of a ball with this property.

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.

In short, in the classical case, there is no single solution for a spinning ball that is spherically symmetric. Only the set of all solutions is.

In the quantum case, if we talk about, say, a spin-1/2 particle like an electron, there are states which are states of definite spin about a single axis, for all possible orientations of the axis. These are eigenstates of the corresponding spin operators, and correspond to the classical spinning ball states described above. (Note that these are not the same as the electron orbitals that you showed pictures of; those are states of electrons bound in a hydrogen atom, and we're not talking about that case here, just the simpler case of a single isolated electron where we're only looking at its spin.)

However, in the quantum case, since the Schrodinger Equation is linear, any superposition of solutions is also a solution. That means there is an actual, physical state that an electron could be in which is literally a superposition of spin eigenstates about all possible axes. (I don't know that anyone has ever actually prepared a real electron in such a state, but by the laws of QM it must exist.) And this state will in fact be spherically symmetric.

So in the quantum case, we actually can find a physical state that solves our "problem" and is spherically symmetric. 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.

What I am getting from reading all these replies is that the solution is still spherically symmetrical even though the eigenfunctions are not spherically symmetrical

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.

 

[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 Schrodinger equation. An eigenfunction such as $\psi_{211}$ is a solution to the Schrodinger 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 Schrodinger equation.

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

Yes, the eigenstates satisfy the Schrodinger 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 Schrodinger equation

The eigenstates of the Hamiltonian satisfy the time-independent Schrodinger equation. But all states satisfy the time-dependent Schrodinger 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".