Justification of a trick in solving PDEs arising in Physics

[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.

 

[fluidistic]

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.

Oops! I meant that if the geometry in which the equation is defined, as well as its boundary conditions, are independent of some variable, say $\theta$, then we assume that particular solutions (for example the eigenfunctions) are also independent of that variable. That trick simplifies greatly the problem (for example when using separation of variables, you already start the problem with 1 less variable to deal with).
This trick works for the heat equation and others too. But it doesn't work for the Schrodinger's equation.

So I would not say "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." because this doesn't hold for the heat equation. As I wrote earlier in this thread, take the case of a cylinder in which we solve the heat equation. The temperature of its boundaries is considered to be fixed (Dirichlet boundary condition). The geometry and the boundary conditions as well, are invariant under any $\theta$ rotation around the symmetry axis. I.e. you rotate the cylinder and the physics of the problem remains unchanged. In that case neither the particular solutions (eigenfunctions) nor the general solution depend on $\theta$ even though the Laplacian in cylindrical coordinate contains this angle's dependence. You can just equate to 0 the Laplacian term involving $\theta$ right away, that's the trick I'm talking about (employed all over the place in mathematical methods for physicists). And this trick doesn't work for Schrodinger's equation.
Scratch this, I misunderstood what you meant by the general solution I guess. I am not sure.

 

[Demystifier]

Oops! I meant that if the geometry in which the equation is defined, as well as its boundary conditions, are independent of some variable, say $\theta$, then we assume that particular solutions (for example the eigenfunctions) are also independent of that variable. That trick simplifies greatly the problem (for example when using separation of variables, you already start the problem with 1 less variable to deal with).
This trick works for the heat equation and others too. But it doesn't work for the Schrodinger's equation.

There is no mathematical difference between the heat and the Schrodinger equation in that regard. The difference is only in "physics", namely in what kind of solution are you usually looking for. Loosely speaking, in the heat case you usually look for the solution, while in the Schrodinger case you usually look for all solutions (compatible with given geometry). But "usually" is the keyword, because in some situations it can be reversed. In the Schrodinger case you can look for the ground state, which for the hydrogen atom does not depend on $\theta$ and $\varphi$. Likewise, in the heat case there are solutions which do depend on $\theta$ and $\varphi$ even when the container has a spherical shape.

 

[fluidistic]

There is no mathematical difference between the heat and the Schrodinger equation in that regard. The difference is only in "physics", namely in what kind of solution are you usually looking for. Loosely speaking, in the heat case you usually look for the solution, while in the Schrodinger case you usually look for all solutions (compatible with given geometry). But "usually" is the keyword, because in some situations it can be reversed. In the Schrodinger case you can look for the ground state, which for the hydrogen atom does not depend on $\theta$ and $\varphi$. Likewise, in the heat case there are solutions which do depend on $\theta$ and $\varphi$ even when the container has a spherical shape.

For the heat equation, when we solve it with separation of variables, we first find the eigenfunctions, not "the solution" that satisfies the given boundary condition. Those eigenfunctions are found using the trick to assume that they do not depend on the angle(s) in the first place. The solution to the full problem (equation + boundary conditions) will not depend on the angles either.
If we consider the time independent heat equation (so there is no initial conditions), if the geometry where the equation is defined and if the boundary conditions do not depend on angles, I think that any solution to the equation (including the eigenfunctions) will not depend on the angles. Maybe you have in mind a problem where initially (so you consider the time-dependent equation?) the temperature profile possesses some angular dependence? If not, then I would like to see a mathematical example of a given function that satisfies the heat equation in a symmetric region where no angle dependence is to be found in neither initial condition nor boundary conditions.

 

[Demystifier]

Those eigenfunctions are found using the trick to assume that they do not depend on the angle(s) in the first place.

I think it's simply not true. See e.g.
https://en.wikiversity.org/wiki/Hea..._3-D_Heat_Equation_in_Cylindrical_Coordinates
where the solutions depend on all coordinates. Can you make a reference/link to a source where we can see equations that you have in mind?

 

[fluidistic]

I think it's simply not true. See e.g.
https://en.wikiversity.org/wiki/Hea..._3-D_Heat_Equation_in_Cylindrical_Coordinates
where the solutions depend on all coordinates. Can you make a reference/link to a source where we can see equations that you have in mind?

Sure. Page 644 of https://www.et.byu.edu/~vps/ME505/IEM/08 03.pdf. Neither the eigenfunctions nor "the solution" to the problem depend on the angle variable.

In your wikipedia example, there is an explicit dependence of $u$ with $\theta$ as initial condition. I am not sure about the boundary conditions, it looks like they might also depend on $\theta$, but it doesn't matter at this point, since the initial conditions already depend on $\theta$. No wonder the solution depends on $\theta$. But that is not the case I am considering.

Edit: I have cited the heat equation, but I have also stated others had this trick too. If I remember well, we often used this trick with Laplace and Poisson equations regarding electrostatics problems. Even Griffiths did this in his famous textbook (as far as I remember he only dealt with problem with azimuthal symmetry. Jackson went further.). Another (less known) equation where the trick works: $\nabla ^2 n+\lambda n =\frac{1}{\kappa } \frac{\partial n }{\partial t}$ where the region of interest (i.e. the domain of the equation) is a sphere and $n$ is a concentration such that $n=0$ on the surface of the sphere. This particular problem is symmetric (when the initial conditions are angle independent, i.e. depend only on $r$), the solution shouldn't depend on any of the 2 angles in spherical coordinates. It turns out that the eigenfunctions of that problem, as well as the general solution, do not depend on the zenithal and azimuthal angles.
In that aspect, Schrodinger's equation is different.

Edit2: Another source, with Laplace equation and the symmetry trick used: https://dslavsk.sites.luc.edu/courses/phys301/classnotes/laplacesequation.pdf. You can't do that with Schrodinger's equation.

Edit3: Another equation where the trick is used: https://pages.physics.wisc.edu/~craigm/toroid/toroid/node4.html.

 

[Demystifier]

Page 644 of https://www.et.byu.edu/~vps/ME505/IEM/08 03.pdf.

In the second line it says that it refers to the case of cylinder with angular symmetry.
But the angular symmetry is not a necessity, it's just a special case. Page 650 considers different cases.

 

[fluidistic]

In the second line it says that it refers to the case of cylinder with angular symmetry.
But the angular symmetry is not a necessity, it's just a special case. Page 650 considers different cases.

I think you are missing the point. Page 650 only emphasizes my point. In general $u$ is a function of all variables.
The 1st case considered on that page is that there is a symmetry along the z-axis. Any translation along that axis leaves the problem unchanged. The result is that $u$ is a function of all variables but $z$. In other words, the trick works. You can start the problem by assuming that $u$ does not depend on $z$.
The 2nd case considered on that page has a finite height, but possesses angular symmetry. Again, the trick works and it turns out that you can assume that the solution(s) don't depend on $\theta$, and similarly for the 3rd case on that page.

My point is that this reasoning does not work for the Schrodinger's equation.

Edit: The situation is different with the Schrodinger's equation, in that, even if everything possesses spherical symmetry, the eigenfunctions, individually, do not have this symmetry. It is only the "set" of these eigenfunction that does, as PeterDonis wrote. The trick does not work. You cannot assume that the solution(s) to the equation are independent of the azimuthal angle even though absolutely everything from the initial conditions to the geometry of the spatial region with its boundary conditions do have that angular non dependence.

 

[Demystifier]

The situation is different with the Schrodinger's equation

I claim that it isn't. For instance, in my recent paper https://arxiv.org/abs/2003.14049 we study the Ginzurg-Landau equation, which is a generalization of the Schrodinger equation. Just put $b=\beta=0$ in our equations and you will get the Schrodinger. In the text before Eq. (9) we assume that solution depends only on $x$ and we find the solution. Likewise, in the text before Eq. (11) we assume that the solution depends only on $r$ and we find the solution. So the trick works even for the Schrodinger equation.

 

[fluidistic]

I claim that it isn't. For instance, in my recent paper https://arxiv.org/abs/2003.14049 we study the Ginzurg-Landau equation, which is a generalization of the Schrodinger equation. Just put $b=\beta=0$ in our equations and you will get the Schrodinger. In the text before Eq. (9) we assume that solution depends only on $x$ and we find the solution. Likewise, in the text before Eq. (11) we assume that the solution depends only on $r$ and we find the solution. So the trick works even for the Schrodinger equation.

I think you haven't read the whole thread. It has been pointed out that for the H-atom, a problem with spherical symmetry (i.e. both azimuthal and zenithal symmetry), the eigenfunctions do depend on the angles. In other words, you cannot assume that the solution $\psi$ does not depend on $\theta$ and $\phi$ as you would with the heat equation, Laplace and so on.
So, while it may be possible that you found a case where the trick can be applied for the Schrodinger's equation, the trick doesn't work in general for that equation, while it seems to work for many others. Also, at first glance it seems that in your case, you are dealing with a 2D problem. And symmetries of $\psi$ in different dimensions have different properties, so that may be why you obtained a psi independent of the angles. But that's irrelevant at this point, the fact will still stay as it is. The trick won't work for the Schrodinger's equation in 3D spatial regions where it works for many different PDE's.

Edit: See the 15th misconception. It isn't exactly what we're dealing with, but it is very similar. https://www.physics.umd.edu/courses/Phys401/bedaque07/misconnzz.pdf It discusses the difference between 1D and 3D symmetry and the probability density (so not $\psi$ but almost).

 

[Demystifier]

It has been pointed out that for the H-atom, a problem with spherical symmetry (i.e. both azimuthal and zenithal symmetry), the eigenfunctions do depend on the angles.

The ground state eigenfunction does not depend on the angles. Other eigenfunctions do.

In other words, you cannot assume that the solution $\psi$ does not depend on $\theta$ and $\phi$

Yes you can. If you do that, you get a special solution called the ground state.

as you would with the heat equation, Laplace and so on.

When you do that for heat equation, Laplace and so on, you also get a special solution. It's just not called the "ground state" anymore.