I Justification of a trick in solving PDEs arising in Physics

[fluidistic]

The answer is you can't do such a thing. These are linearly independent, and orthogonal, functions.

The spherically symmetric solution of the time-independent Schrodinger equation, ignoring the radial component, is the $\ell=0$ and $m=0$ spherical harmonic. There is no other linear combination of other spherical harmonics which are equivalent to $\ell=0$,$m=0$. If there were, then it would defeat the whole point of being a set of basis functions.

That's exactly what I realized, while offline! Thank you very much for pointing this out.
Indeed, you cannot use the incomplete basis to create any function. In particular, you cannot recreate the element you removed from the complete basis. It doesn't matter that you still have an infinite amount of basis elements, it is incomplete.

 

[fluidistic]

The answer is you can't do such a thing. These are linearly independent, and orthogonal, functions.

The spherically symmetric solution of the time-independent Schrodinger equation, ignoring the radial component, is the $\ell=0$ and $m=0$ spherical harmonic. There is no other linear combination of other spherical harmonics which are equivalent to $\ell=0$,$m=0$. If there were, then it would defeat the whole point of being a set of basis functions.

So this just strenghten my point that if any spherical harmonic is involved, then there is no way to get rid of the angle dependence. PeroK's point is that with infinitely many terms involving spherical harmonics, you can remove the angle dependency.

 

[PeroK]

The answer is you can't do such a thing. These are linearly independent, and orthogonal, functions.

The spherically symmetric solution of the time-independent Schrodinger equation, ignoring the radial component, is the $\ell=0$ and $m=0$ spherical harmonic. There is no other linear combination of other spherical harmonics which are equivalent to $\ell=0$,$m=0$. If there were, then it would defeat the whole point of being a set of basis functions.

What about $l = 1$ and an equal superposition of $m_x = 1, m_y = 1,m_z = 1$? That must be spherically symmetric as well.

In any case, if you can only have a single spherically symmetric function from the spherical harmonics, then they are clearly not a basis of eigenfunctions. See here:

https://en.wikipedia.org/wiki/Spherical_harmonics#Spherical_harmonics_expansion

 

[Haborix]

What about $l = 1$ and an equal superposition of $m_x = 1, m_y = 1,m_z = 1$? That must be spherically symmetric as well.

In any case, if you can only have a single spherically symmetric function from the spherical harmonics, then they are clearly not a basis of eigenfunctions. See here:

https://en.wikipedia.org/wiki/Spherical_harmonics#Spherical_harmonics_expansion

First, there is only one $m$, as in you need one l and one m to specify a spherical harmonic. Second, by spherically symmetric I mean doesn't depend on the angular variables. You can search images to see that $Y_1^1$ is not spherically symmetric, nor is any combination of spherical harmonics with the same $\ell$ but different values of $m$ (its on the wikipedia link you gave). I do not know what you mean by your final statement.

 

[PeroK]

First, there is only one $m$,

So if, just for the hell of it, we measured the orbital angular momentum about the x-axis (instead of the z-axis)?

That's not allowed?

 

[Haborix]

I don't see the relevance of any of the talk in this thread about measuring things. But you are free to calculate the expectation value of any component of angular momentum you like, and you are also free to take the spherical harmonics as being simultanous eigenfunctions of $\hat{L}^2$ and anyone of the $\hat{L}_i$.

 

[PeroK]

I don't see the relevance of any of the talk in this thread about measuring things. But you are free to calculate the expectation value of any component of angular momentum you like, and you are also free to take the spherical harmonics as being simultanous eigenfunctions of $\hat{L}^2$ and anyone of the $\hat{L}_i$.

Which means you have three states corresponding to $L_i = 1$ for $i = 1, 2, 3$. And that means you can take a superposition of those states. And an equal superposition of these states must be spherically symmetric.

But, more fundamentally, the spherical harmonics (if expressed in the z-basis, say) can still be used to represent ANY function (whether spherically symmetric or not).

If you say they cannot, then they are not a basis. Note that the $l = 0, m = 0$ constant function is not the only spherically symmetric function.

 

[Haborix]

Which means you have three states corresponding to $L_i = 1$ for $i = 1, 2, 3$. And that means you can take a superposition of those states. And an equal superposition of these states must be spherically symmetric.

You would be adding things together in different bases if you did that, which would be strange and unhelpful, and they would not be spherically symmetric. They would all have $\ell=1$ (or any $\ell>0$)

But, more fundamentally, the spherical harmonics (if expressed in the z-basis, say) can still be used to represent ANY function (whether spherically symmetric or not).

True.

If you say they cannot, then they are not a basis. Note that the constant function is not the only spherically symmetric function.

What other function is spherically symmetric besides the constant function?

 

[PeterDonis]

What other function is spherically symmetric besides the constant function?

Any function of radius $r$ only.

 

[Haborix]

Any function of radius $r$ only.

Yes, that's the point, a constant function for the angular functions.

 

[PeroK]

What other function is spherically symmetric besides the constant function?

Yes, of course, the penny's dropped. All the variations are in the radial function. Apologies.

 

[PeterDonis]

you say that if the problem is spherically symmetric, then so must be its solution?

No. It is not the case that every solution of an equation must have the same symmetry as the equation.

What is the case is that the entire set of solutions, taken together, must have the same symmetry as the equation. So, for example, if a particular equation is time symmetric, although individual solutions might not be time symmetric, solutions that are not time symmetric will occur in pairs, with each being the time reverse of the other.

Similarly, a solution of a spherically symmetric equation that is not spherically symmetric will occur as part of a family of solutions that are complements of each other in some way. For example, the $p$ orbital solutions for the electron in a hydrogen atom are not spherically symmetric; they have axial symmetry about some particular axis. The entire family of $p$ orbital solutions (for a given energy level) consists of one such solution for each possible axis (i.e., each possible direction in space), so all of them taken together have spherical symmetry (because every possible axis is present in the family of solutions and no axis is preferred over any other) even though each individual solution does not.

 

[fluidistic]

No. It is not the case that every solution of an equation must have the same symmetry as the equation.

What is the case is that the entire set of solutions, taken together, must have the same symmetry as the equation. So, for example, if a particular equation is time symmetric, although individual solutions might not be time symmetric, solutions that are not time symmetric will occur in pairs, with each being the time reverse of the other.

Similarly, a solution of a spherically symmetric equation that is not spherically symmetric will occur as part of a family of solutions that are complements of each other in some way. For example, the $p$ orbital solutions for the electron in a hydrogen atom are not spherically symmetric; they have axial symmetry about some particular axis. The entire family of $p$ orbital solutions (for a given energy level) consists of one such solution for each possible axis (i.e., each possible direction in space), so all of them taken together have spherical symmetry (because every possible axis is present in the family of solutions and no axis is preferred over any other) even though each individual solution does not.

I am not sure why you are quoting me as if I had affirmed that a solution to an equation must possesses the same symmetries, I haven't affirmed that, and I think it is pretty clear in my first post, where I mention that this doesn't hold for Schrodinger's equation. I was replying to PeroK, from whom I had understood that statement, I just wanted to have his confirmation to make sure I had understood him fine.

Otherwise, I think you give the answer to my question. (I am very glad, thank you very much!)
Why is it though, that the set of solutions, taken together, must have the same symmetry as the equation? Is it from a theorem in group theory? Where does this come from?

 

[PeterDonis]

Why is it though, that the set of solutions, taken together, must have the same symmetry as the equation?

Because the equation is equivalent, mathematically, to the entire set of its solutions.

 

[bob012345]

No. It is not the case that every solution of an equation must have the same symmetry as the equation.

What is the case is that the entire set of solutions, taken together, must have the same symmetry as the equation. So, for example, if a particular equation is time symmetric, although individual solutions might not be time symmetric, solutions that are not time symmetric will occur in pairs, with each being the time reverse of the other.

Similarly, a solution of a spherically symmetric equation that is not spherically symmetric will occur as part of a family of solutions that are complements of each other in some way. For example, the $p$ orbital solutions for the electron in a hydrogen atom are not spherically symmetric; they have axial symmetry about some particular axis. The entire family of $p$ orbital solutions (for a given energy level) consists of one such solution for each possible axis (i.e., each possible direction in space), so all of them taken together have spherical symmetry (because every possible axis is present in the family of solutions and no axis is preferred over any other) even though each individual solution does not.

I'm having trouble seeing what you mean for $p$ orbitals. I'm not seeing how this is spherically symmetric.

Are you assuming an infinite number of possible axis orientations in space thus smearing these functions out?

 

[PeterDonis]

Are you assuming an infinite number of possible axis orientations in space thus smearing these functions out?

Of course. There are an infinite number of possible $p$ orbitals, one for each possible orientation of the axis. The three usually shown in textbooks are just the three most convenient ones for illustration.

 

[bob012345]

Of course. There are an infinite number of possible $p$ orbitals, one for each possible orientation of the axis. The three usually shown in textbooks are just the three most convenient ones for illustration.

Thanks. Is that smearing identical to or in addition to the time dependent solutions of the Schrödinger equation?
$$\Psi(\mathbf{r,\theta,\phi},t) = \psi(\mathbf{r,\theta,\phi}) e^{-iEt/\hbar}$$

 

[PeterDonis]

Is that smearing

What smearing? There is no smearing in what you quoted from me. I simply made a statement about how many $p$ orbital states there are.

 

[PeterDonis]

identical to or in addition to the time dependent solutions of the Schrödinger equation?

I have no idea what this means.

 

[bob012345]

What smearing? There is no smearing in what you quoted from me. I simply made a statement about how many $p$ orbital states there are.

I'm just trying to understand your statements that there are an infinite number of possible $p$ orbitals and all of them taken together have spherical symmetry. To me that implies a kind of smearing over many possible states to be spherically symmetric. Then I asked if that is the effect of the time dependence that comes from the Schrödinger equation or something else.

 

[PeterDonis]

I'm just trying to understand your statements that there are an infinite number of possible orbitals and all of them taken together have spherical symmetry.

The first part should be obvious: there is one $p$ orbital for each possible axis direction, and there are an infinite number of possible axis directions.

The second part just means that you can take the $p$ orbital for any chosen axis direction and transform it by a spherically symmetric rotation to the $p$ orbital for any other chosen axis direction; in other words, the set of $p$ orbitals (more precisely, the set of $p$ orbitals for a given energy level) as a whole has spherically symmetric rotations as a symmetry group. (In more technical language, the set of $p$ orbitals for a given energy level is symmetric under SO(3) transformations.) Another way of putting this is that no particular axis direction is picked out as different from any other as far as $p$ orbitals are concerned; the "shape" of the $p$ orbital is the same regardless of the axis direction in which it is oriented.

To me that implies a kind of smearing over many possible states to be spherically symmetric.

No, it doesn't. See above.

 

[bob012345]

0

The first part should be obvious: there is one $p$ orbital for each possible axis direction, and there are an infinite number of possible axis directions.

The second part just means that you can take the $p$ orbital for any chosen axis direction and transform it by a spherically symmetric rotation to the $p$ orbital for any other chosen axis direction; in other words, the set of $p$ orbitals (more precisely, the set of $p$ orbitals for a given energy level) as a whole has spherically symmetric rotations as a symmetry group. (In more technical language, the set of $p$ orbitals for a given energy level is symmetric under SO(3) transformations.) Another way of putting this is that no particular axis direction is picked out as different from any other as far as $p$ orbitals are concerned; the "shape" of the $p$ orbital is the same regardless of the axis direction in which it is oriented.
No, it doesn't. See above.

The first part was obvious. I get everything your say above but I don't see how the it relates to the questions in the OP. That was discussing spherical symmetric solutions and PDE's, not spherically symmetric rotations as a symmetry group. It seems like you are trying to argue something that isn't spherically symmetric ($p$ orbitals), really is because your can consider SO(3) transformations. Is that a fair statement?

 

[PeterDonis]

That was discussing spherical symmetric solutions and PDE's, not spherically symmetric rotations as a symmetry group.

If a PDE has a symmetry, which is what the thread discussion is about, then its solutions must form a group under the transformations of that symmetry. The two statements are mathematically equivalent. This is another way of saying what I said in post #44.

It seems like you are trying to argue something that isn't spherically symmetric ($p$ orbitals), really is because your can consider SO(3) transformations. Is that a fair statement?

No. I'm not saying any individual $p$ orbital is spherically symmetric. I'm saying that the set of all $p$ orbitals forms a group under SO(3), which is mathematically equivalent to saying that the PDE that the $p$ orbitals are all solutions of has spherical symmetry. An obvious shorthand way of saying that is saying that the set of $p$ orbitals has spherical symmetry, even though no individual $p$ orbital does. But if you don't like the shorthand way of saying it, just say it the other way. The meaning is the same.

 

[PeterDonis]

If a PDE has a symmetry

Perhaps it will help to clarify the meaning of this statement as well. The original statement in the OP was that the PDE is "independent of some variable" (or variables). However, it was pointed out that the Schrodinger Equation in spherical coordinates, assuming a potential that only depends on $r$, does not have this property; the angular variables still appear in the Laplacian. Even so, the Schrodinger Equation for this case still has spherical symmetry, in the sense that the equation is invariant under SO(3) transformations of the coordinates; those transformations leave functions of $r$ only invariant, and they also leave the Laplacian invariant, so the Schrodinger Equation for this case does have spherical symmetry, even though the angular coordinates still appear in the Laplacian.

 

[hutchphd]

Would the invocation of Unsold's Theorem shed any light here?$$4\pi \sum_{m=-l}^l Y_{lm}^*Y_{lm}=2l+1$$ No angular dependence.
I have otherwise lost the thread of this thread.

 

[bob012345]

Perhaps it will help to clarify the meaning of this statement as well. The original statement in the OP was that the PDE is "independent of some variable" (or variables). However, it was pointed out that the Schrodinger Equation in spherical coordinates, assuming a potential that only depends on $r$, does not have this property; the angular variables still appear in the Laplacian. Even so, the Schrodinger Equation for this case still has spherical symmetry, in the sense that the equation is invariant under SO(3) transformations of the coordinates; those transformations leave functions of $r$ only invariant, and they also leave the Laplacian invariant, so the Schrodinger Equation for this case does have spherical symmetry, even though the angular coordinates still appear in the Laplacian.

Ok, Thanks. I think we are on the same wavelength now. Do you know of a counter example where the PDE does not have spherical symmetry and would the classical orbital equations be an example?

 

[PeterDonis]

Do you know of a counter example where the PDE does not have spherical symmetry

The Schrodinger Equation with a potential that is not a function of $r$ only (for example, inside a parallel plate capacitor) would be an example.

would the classical orbital equations be an example?

Orbital equations with what potential/force?

 

[bob012345]

The Schrodinger Equation with a potential that is not a function of $r$ only (for example, inside a parallel plate capacitor) would be an example.
Orbital equations with what potential/force?

Thanks. I meant the Newtonian gravitational potential $U(r)= -\frac{GMm}{r}$.

 

[PeterDonis]

I meant the Newtonian gravitational potential $U(r)= -\frac{GMm}{r}$.

The Newtonian gravitational equations are spherically symmetric, so the set of solutions will be as well.

 

[Demystifier]

when the equation itself ... independent of some variable

What does it mean that the equation itself is independent of some variable? If the equation contains a partial derivative with respect to some variable, then I would say that the equation "depends" on the variable. In that case, the general solution depends on this variable, but a particular solution (the one that describes a concrete physical situation) may or may not depend on it.