I Justification of a trick in solving PDEs arising in Physics

[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.

 

[Demystifier]

The trick won't work for the Schrodinger's equation in 3D spatial regions where it works for many different PDE's.

Try by yourself! Take the free (that is $V=0$) 3D Schrodinger equation in cylindrical/spherical coordinates and assume that $\psi$ depends only on $r$. What do you get?

 

[PeroK]

Perhaps to take a fresh look at this. If we consider the gravitational field of the Sun, then it must be spherically symmetric; but, if we add a planet, then the orbit of the planet is not spherically symmetric - it doesn't even make much sense to talk about the Earth having a spherically symmetric orbit around the Sun. Likewise, the electron itself breaks the spherical symmetry of the system with just a proton and its Coulomb potential.

 

[Demystifier]

This is in contrast with all the other PDE's I have quoted so far. With them, if you use the trick, you will find a complete basis from which you can write the full solution.

No, that's wrong! If you do the trick, you will get an ordinary 2nd order differential equation. It has only two independent solutions, so you cannot get a complete basis. For a complete basis, you must consider solutions that depend on all coordinates, not just one.

 

[fluidistic]

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

Yes. I agree. That is why the trick does not work for the Schrodinger's equation in that case. Because if you use the trick and assume that the solutions do not depend on the angles, you will only find the ground state, and this eigenfunction does not form a complete basis from which you can write any solution to the problem.
This is in contrast with all the other PDE's I have quoted so far. With them, if you use the trick, you will find a complete basis from which you can write the full solution. The major difference is that the eigenfunctions won't depend on the angle(s) for those equations, while they do, in general, for the Schrodinger's equation.

I am still waiting to see an example where this fails for the heat equation, you pointed out "Likewise, in the heat case there are solutions which do depend on theta and phi even when the container has a spherical shape.". I claim that this is not true, unless of course you put an angle dependence in the inititial/and/or boundary conditions like in the Wikipedia article you linked.

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

We agree on this. You therefore cannot solve the problem of finding "the solution" to the equation. In that case you get lucky and get 1 eigenfunction, but nothing more. This is different than in the case of all other PDE's I have quoted so far.

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.

You get the full set of eigenfunctions that can construct "the solution". You aren't missing out any solution. So the trick works for these PDE's. Again, this doesn't work for the Schrodinger's equation.

 

[Demystifier]

You get the full set of eigenfunctions that can construct "the solution". You aren't missing out any solution.

Come on, how can solutions that depend only on 1 coordinate be a complete basis for functions that depend on 3 coordinates?

 

[fluidistic]

No, that's wrong! If you do the trick, you will get an ordinary 2nd order differential equation. It has only two independent solutions, so you cannot get a complete basis. For a complete basis, you must consider solutions that depend on all coordinates, not just one.

I am not convinced. If we look at the document I linked, p. 644 where we use trick to get rid of the angle dependence. The full solution, written as an infinite sum of eigenfunctions, is a function that does not depend on any angle, and where all, absolutely all of the eigenfunctions also lack this angle dependence.
The same applies for many of the electrostatics problem we find in undergraduate Physics.

That is what is contrasted with the case of the Schrodinger's equation.

 

[fluidistic]

Come on, how can solutions that depend only on 1 coordinate be a complete basis for functions that depend on 3 coordinates?

Because the assumptions are that:
1) The geometry possesses a symmetry (say azimuthal symmetry)
2) The boundary conditions possesses the same symmetry.
3) The initial conditions (if we consider the time dependence problem) also possesses the symmetry.

In that particular case, any solution will also have the symmetry (for the heat equation and many other PDE's). This does not hold for the Schrodinger's equation, as Peter Donis pointed out, it is the set of all solutions to the PDE that does possesses the symmetry. He answered in post #42, and thereby resolved the question I raised in my OP.

 

[Demystifier]

I am not convinced. If we look at the document I linked, p. 644 where we use trick to get rid of the angle dependence. The full solution, written as an infinite sum of eigenfunctions, is a function that does not depend on any angle, and where all, absolutely all of the eigenfunctions also lack this angle dependence.
The same applies for many of the electrostatics problem we find in undergraduate Physics.

That is what is contrasted with the case of the Schrodinger's equation.

Ah, now I finally see what confuses you. Yes, they get an infinite number of eigenfunctions (not just two) because they consider dependence on 2 variables, not just 1. The first variable is $r$ and the second variable is the time $t$.

But the same is true for the Schrodinger equation if time dependence is included. For instance, consider Schrodinger equation for free particle in 3 dimensions. Assume that $\Psi=\Psi(x,t)$ does not depend on $y,z$. Looking for the solution in the form
$$\Psi(x,t)=\psi(x)e^{-iEt}$$
the Schrodinger equation (in units $\hbar=1$) reduces to
$$-\frac{1}{2m}\frac{\partial^2\psi(x)}{\partial x^2}=E\psi(x)$$
It has two independent solutions
$$\psi(x)=e^{\pm ikx}$$
where
$$k=\sqrt{2mE}$$
But $E$ is arbitrary non-negative number so each $E$ corresponds to another pair of solutions. The arbitrariness of $E$ is equivalent to the arbitrariness of $k$, so the full $y,z$-independent solution is
$$\Psi(x,t)=\int_{-\infty}^{\infty} dk \,c(k) e^{-iE(k)t} e^{ikx}$$
where
$$E(k)=\frac{k^2}{2m}$$
and $c(k)$ is arbitrary. The solution above involves the complete set $e^{ikx}$ for all possible $k$.

I hope it helps.

 

[fluidistic]

Ah, now I finally see what confuses you. Yes, they get an infinite number of eigenfunctions (not just two) because they consider dependence on 2 variables, not just 1. The first variable is $r$ and the second variable is the time $t$.

I am not sure about the confusion part. :) Hope this will get resolved.
I do not care about the number of eigenfunctions. I am interested to check whether they depend on all possible variables, or whether the symmetry of the problem is included in those eigenfunctions. For this particular example of p.644 of the document I linked, it is question of the heat equation, with angular symmetry. In that case, all eigenfunctions (which happen to be infinite in numbers, but this doesn't matter at all) are independent of $\theta$. If we were solving the Schrodinger's equation on that same geometry with the same initial condition, I am not sure whether the eigenfunction would be independent of $\theta$. It happens that for the H-atom, this isn't the case. And that is very particular to that equation, comparing with many PDE's. I guess it is not the only one having this particularity though. What matters, as Peter Donis pointed out, is that the set of all possible solutions must possesses the same symmetry than the equation + boundary conditions + initial conditions.

But the same is true for the Schrodinger equation if time dependence is included. For instance, consider Schrodinger equation for free particle in 3 dimensions. Assume that $\Psi=\Psi(x,t)$ does not depend on $y,z$. Looking for the solution in the form
$$\Psi(x,t)=\psi(x)e^{-iEt}$$
the Schrodinger equation (in units $\hbar=1$) reduces to
$$-\frac{1}{2m}\frac{\partial^2\psi(x)}{\partial x^2}=E\psi(x)$$
It has two independent solutions
$$\psi(x)=e^{\pm ikx}$$
where
$$k=\sqrt{2mE}$$
But $E$ is arbitrary non-negative number so each $E$ corresponds to another pair of solutions. The arbitrariness of $E$ is equivalent to the arbitrariness of $k$, so the full $y,z$-independent solution is
$$\Psi(x,t)=\int_{-\infty}^{\infty} dk \,c(k) e^{-iE(k)t} e^{ikx}$$
where
$$E(k)=\frac{k^2}{2m}$$
and $c(k)$ is arbitrary. The solution above involves the complete set $e^{ikx}$ for all possible $k$.

I hope it helps.

I am not sure what you are trying to point out with the Schrodinger's equation example. I can mentally follow you. Although I would have to think deeper. For example, I think you can find "eigenfunctions" or part of a basis but it will lack the y and z dependence that a full solution may have. Don't quote me on that part though please :) I am in a rush.
But I don't think that is the problem.

The "strange" thing about the Schrodinger's equation, is that, given a problem consisting of a PDE + boundary conditions + region (domain) + initial conditions (if any) that possesses a symmetry, for many PDE's any solution will have the same symmetry. For example, the one given in p.644 points out the general solution to the heat equation, and you can see that not only it does have the symmetry (lacking the angle's dependence), but also any eigenvalue possesses this symmetry. The same cannot be said for Schrodinger's equation in general, and this fails for the H-atom. It fails in such a way that one cannot use the trick to assume solutions of the form $\psi=\psi(r,t)$ or $\psi=\psi(r)$ even though the problem is "radial" in that there is no symmetry breaking. If you do so, you only get the ground state, not enough to write down the full solution. For the heat equation, this is different, you would find all eigenstates, and this allows you to write the full solution, as has been done in all the links I have provided in post #65. At this point I feel like repeating myself, so I will stop here, and unfortunately I don't think this will clear things up.
I may not explain myself well enough, since I think there is a misunderstanding between us (and PeroK also seems not to have understood me). If someone else would like to help me, feel welcome. :)

 

[Demystifier]

The same cannot be said for Schrodinger's equation in general, and this fails for the H-atom.

The Schrodinger equation, in general, has a potential $V(x,y,z)$, while other equations (heat etc) you consider don't have an analog of that. I think that's the whole source of difference. But if you consider free Schrodinger equation with $V(x,y,z)=0$, it should not differ from the heat equation. In fact, the heat equation and the free Schrodinger equation are the same equation up to a change of variables $t\to t'=it$. So if you found a solution of one of them, you automatically know the corresponding solution of the other.

 

[Demystifier]

It fails in such a way that one cannot use the trick to assume solutions of the form $\psi=\psi(r,t)$ or $\psi=\psi(r)$ even though the problem is "radial" in that there is no symmetry breaking. If you do so, you only get the ground state, not enough to write down the full solution. For the heat equation, this is different, you would find all eigenstates, and this allows you to write the full solution, as has been done in all the links I have provided in post #65.

Here is another possible source of confusion. If you consider $\psi(r,t)$ in a non-Schrodinger equation, you will find an infinite set of solutions. But it is misleading to say that this infinite set is complete. It is complete for functions that depend only on $r$, but it is not complete for functions that depend on all 3 variables $r,z,\theta$.

 

[fluidistic]

Here is another possible source of confusion. If you consider $\psi(r,t)$ in a non-Schrodinger equation, you will find an infinite set of solutions. But it is misleading to say that this infinite set is complete. It is complete for functions that depend only on $r$, but it is not complete for functions that depend on all 3 variables $r,z,\theta$.

I see what you mean, but I insist that the assumptions are that there is a lot of symmetry (at boundary condition level, initial condition and domain of the equation). Under these assumptions, for the common PDE we see in undergraduate physics, the general solution will have the same symmetry, as well as the eigenfunctions making it up. And so you do end up with the complete solution that depends on $r$, $z$ but not on $theta$ (it's in every link I posted in my previous post(s)). I am not saying that in general the general solution cannot depend on $\theta$. I am saying that under those assumptions of symmetry, the general solution as well as its eigenfunctions all possesses that symmetry. And that this doesn't hold for the Schrodinger's equation.

 

[fluidistic]

The Schrodinger equation, in general, has a potential $V(x,y,z)$, while other equations (heat etc) you consider don't have an analog of that. I think that's the whole source of difference. But if you consider free Schrodinger equation with $V(x,y,z)=0$, it should not differ from the heat equation. In fact, the heat equation and the free Schrodinger equation are the same equation up to a change of variables $t\to t'=it$. So if you found a solution of one of them, you automatically know the corresponding solution of the other.

Yeah please don't go that way :) I will get confused very early, as I have read that the Schrodinger's equation is a wave equation, not a diffusion one(!). This will lead to a tangent to this thread, I suppose.

 

[PeterDonis]

if the geometry in which the equation is defined, as well as its boundary conditions, are independent of some variable

The geometry of space in classical (pre-relativity) physics is independent of all variables.

The correct criterion is the one I gave in an earlier post: that the equation is invariant under a particular group of transformations of the coordinates--for spherical symmetry, the group is SO(3). And, as I pointed out, the Schrodinger Equation with a potential that depends only on $r$ has this property.

 

[fluidistic]

The geometry of space in classical (pre-relativity) physics is independent of all variables.

The correct criterion is the one I gave in an earlier post: that the equation is invariant under a particular group of transformations of the coordinates--for spherical symmetry, the group is SO(3). And, as I pointed out, the Schrodinger Equation with a potential that depends only on $r$ has this property.

Ok.
And as you pointed out, the set of all possible solutions must possesses this symmetry property. It does not change the fact that for Schrodinger's equation, regarding the H-atom in vacuum problem, each eigenfunction (except the ground state) depend on the angle(s). This differs from eigenfunctions of the heat equation, in a symmetric problem where the domain is a sphere. Similarly for Laplace equation we see in electrostatics and many more PDE's. Schrodinger's equation was "weird" at a first glance because even though the set of all eigenfunctions for the H-atom problem possesses the rotational symmetry, each individual eigenfunction does not.

 

[PeterDonis]

the assumptions are that there is a lot of symmetry (at boundary condition level, initial condition and domain of the equation)

You are mixing up several different kinds of symmetry. That's not a good idea; you need to keep them separate.

The first kind of symmetry is invariance of the equation itself under some group of transformations. As I have already pointed out in previous post, if an equation has this kind of symmetry, the full set of its solutions will also have the symmetry (in the sense that they will form a group under the symmetry transformations); but not all individual solutions, by themselves, will have the symmetry. This is true for any equation, not just the Schrodinger Equation; it's true for the heat equation, for example, if you consider the full set of all possible solutions.

Imposing a symmetry on boundary conditions or initial conditions changes things because it restricts the set of solutions that you are considering. For example, the set of solutions of the heat equation if a spherically symmetric boundary condition is imposed is not the same as the set of all solutions of the heat equation, with no constraint imposed on boundary conditions.

In your original comparison between the heat equation and the Schrodinger Equation, you were imposing spherically symmetric boundary conditions on the heat equation, but not on the Schrodinger Equation; for the Schrodinger Equation, you were only assuming spherical symmetry for the equation itself (i.e., invariance under SO(3)). So of course you're going to see individual solutions of the Schrodinger Equation that are not spherically symmetric, even though you're not seeing such solutions of the heat equation; you restricted the available solutions to the heat equation to rule out the ones that weren't spherically symmetric, by imposing boundary conditions.

You do this again in your latest post; see below.

for Schrodinger's equation, regarding the H-atom in vacuum problem, each eigenfunction (except the ground state) depend on the angle(s).

Wrong. There are an infinite number of spherically symmetric solutions for the H-atom--all of the $s$ orbitals. The ground state is just the lowest energy $s$ orbital (1s); it's not the only one.

This differs from eigenfunctions of the heat equation, in a symmetric problem where the domain is a sphere.

"The domain is a sphere", if I'm understanding correctly what you mean by that, is a boundary condition. You're not imposing any corresponding boundary condition on the Schrodinger Equation, so of course, as above, you're going to find individual solutions of the Schrodinger Equation that aren't spherically symmetric, whereas you won't for the heat equation because you've explicitly eliminated them by imposing a boundary condition.

If you eliminate that boundary condition from the heat equation, you will find solutions that aren't spherically symmetric, even if you maintain the spherical symmetry (invariance under SO(3)) of the equation itself. And similarly for the other PDEs you mention.

 

[PeterDonis]

I have read that the Schrodinger's equation is a wave equation, not a diffusion one(!).

As a matter of mathematics, a "wave equation" is one that relates the second time derivative to second space derivatives. The Schrodinger Equation, like the heat equation (and diffusion equations generally) relates the first time derivative to second space derivatives.

It's unfortunate that the QM view of solutions of the Schrodinger Equation as being "wave functions" has muddied the waters in this respect.

 

[fluidistic]

Now that's a big lesson (to me). Thank you immensily for this input. Oops, yes, I didn't realize any s-orbital also lacked angles dependence.
But more importantly, I did not realize that I was not imposing spherical symmetry in the H-atom.

In your original comparison between the heat equation and the Schrodinger Equation, you were imposing spherically symmetric boundary conditions on the heat equation, but not on the Schrodinger Equation; for the Schrodinger Equation, you were only assuming spherical symmetry for the equation itself (i.e., invariance under SO(3)). So of course you're going to see individual solutions of the Schrodinger Equation that are not spherically symmetric, even though you're not seeing such solutions of the heat equation; you restricted the available solutions to the heat equation to rule out the ones that weren't spherically symmetric, by imposing boundary conditions.

(Emphasis mine). That's something I totally overlooked. I still do not see it. For the H-atom, all I do is ask for any solution (eigenfunction or not) to vanish at infinity. I did not realize this wasn't a spherically symmetric boundary condition. So now everything makes sense to me, I think.

And yes, I am aware that even in a spherical region, if the boundary conditions are not symmetric, the solution and eigenfunction will/might depend on the angles.

So, all along, the Schrodinger's equation was no different than any other PDE. It was tricky to me to realize that the H-atom problem was not spherically symmetric (more precisely, its boundary conditions).

And so the overall lesson is that the "trick" to assume solutions of a form where some variable is missing due to some symmetry is justified, provided one makes sure that the symmetry is really there. In the H-atom, it isn't there, unless we impose some spherical boundary conditions. But asking psi to vanish at infinity isn't such a spherically symmetric boundary condition. (I still do not see it entirely, but I buy your argument).

 

[bob012345]

As a matter of mathematics, a "wave equation" is one that relates the second time derivative to second space derivatives. The Schrodinger Equation, like the heat equation (and diffusion equations generally) relates the first time derivative to second space derivatives.

It's unfortunate that the QM view of solutions of the Schrodinger Equation as being "wave functions" has muddied the waters in this respect.

The Schrödinger equation has been referred to as a diffusion equation with an imaginary diffusion coefficient. But what is diffusing? The probability density?

Baym, G. Lectures on Quantum Mechanics, New York: W. A. Benjamin, 1969. Pages 50-53.
https://en.wikipedia.org/wiki/Schrödinger_equation

 

[PeterDonis]

The Schrödinger equation has been referred to as a diffusion equation with an imaginary diffusion coefficient. But what is diffusing? The probability density?

No, the wave function itself, since that's what the equation is governing. The wave function is a probability amplitude, so you could say probability amplitude is diffusing.

 

[PeterDonis]

I still do not see it entirely

A function that vanishes at infinity in three dimensions does not have to vanish at the same "rate" (dependence on $r$) in different directions as $r \rightarrow \infty$.

 

[fluidistic]

Thanks for all, PeterDonis. I think this is much clearer to me now. I would love to study some group theory, as it is used in Physics in a lot of areas, and my knowledge is lacking there.

 

[PeterDonis]

Thanks for all, PeterDonis. I think this is much clearer to me now.

You're welcome! Glad I could help.

 

[vanhees71]

Isn't this a bit too complicated? For one-particle wavefunctions, a (pure) state can only be spherically symmetric around the origin if it depends only on $r=|\vec{x}|$ which automatically makes to an eigenstate of $\hat{\vec{L}}^2$ to the eigenvalue $0$, i.e., $\ell=0$.

Any superposition of wave functions with $\ell=1$ (dipole or $p$ waves) is thus not spherically symmetric.