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Happiness
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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?
Cthugha said: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?
Happiness said:the choice of the z axis is arbitrary, so it cannot have a physical effect
They are all spherically symmetric. In other words, they do not have a preferred z axis.Vanadium 50 said:What feature do all the spherically symmetric wavefunctions have in common?
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.Vanadium 50 said:Follow-up: which wavefunctions have the same energy?
Happiness said:since these wave functions are not spherically symmetrical, it makes more sense to reject them as the correct solutions to a spherically symmetrical problem
Happiness said: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.
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?Cthugha said: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 said: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 said:What feature do all the spherically symmetric wavefunctions have in common?
Happiness said:They are all spherically symmetric.
Happiness said: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?
Cthugha said: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 said:And the superposition of these 3 orbitals alone, exactly as they are shown in the diagram, doesn't give spherical symmetry right?
I still get an angular dependence: the result is proportional to ##(\sqrt{2}\cos\theta+2\sin\theta\cos\phi)^2##.Vanadium 50 said: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.
It means there exists a point from which everything is the same in all directions.strangerep said:@Happiness: let's take a step back...
What do think is meant by the phrase "A (particular) class of physical system is spherically symmetric" ?
That's only valid for zero angular momentum. Here's the answer I was looking for:Happiness said:It means there exists a point from which everything is the same in all directions.
Happiness said:I still get an angular dependence
I don't understand you.strangerep said: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 said: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
[tex](\sin \theta \sin \phi)^2 + (\sin \theta \cos \phi)^2 + \cos^2 \theta = 1[/tex]
This is a consequence of Unsöld's theorem.
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.Vanadium 50 said: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 said:I don't understand you.
If a single particular spinning ball is not spherically symmetrical, then there is no issue with my definition of "spherically symmetrical".PeterDonis said: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 said: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.
Happiness said:My concern is with the spherical symmetry of the problem
Happiness said: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.
Happiness said: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
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.PeterDonis said: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.
Vanadium 50 said: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
[tex](\sin \theta \sin \phi)^2 + (\sin \theta \cos \phi)^2 + \cos^2 \theta = 1[/tex]
This is a consequence of Unsöld's theorem.
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.Vanadium 50 said: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.
"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.PeterDonis said: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.
Happiness said: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?