Justification of a trick in solving PDEs arising in Physics

In summary, the initial condition doesn't matter for the solution of a PDE if the boundary conditions are not dependent on any angle.
  • #1
fluidistic
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In solving some PDEs such as the heat/diffusion equation or the wave equation, when the equation itself, as well as its associated boundary conditions, are independent of some variable (for example the azimuthal angle), we often use the trick to assume that the solution (and eigenfunctions) are independent of that variable too. We use no justification whatsoever and since we find "the" solution, it doesn't matter which way we obtained it, it is the only solution. That is perfectly fine to me.

However why does such a trick fail for example with Schrödinger equation? There, for example for the H-atom problem where the potential is central, there is no angle dependence anywhere. However the eigenfunctions involve spherical harmonics, and so the solution (the wavefunction) is not angle independent.

My question is thus, how can we justify or get to know beforehand whether this trick will work, without having to go through the whole process of checking whether the trick actually works?
 
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  • #2
In the Schrodinger equation, there is angular dependence in the Laplacian coming from the momentum operators. I don't see how you distinguish this from the other PDEs you list. In general, when written in coordinates with an angular variable, they will also have derivatives with respect to those angular variables. It may be that the boundary/initial conditions make it so that the only allowed angular dependence is a constant function. If you plug this constant angular dependence into the full separable PDE, then the angular parts seem to disappear. But that's a relic of boundary/initial conditions identifying the solution, not that the PDE didn't a priori have some angular dependence.

I would say you just get a feel for this when you setup the problem. For a separable PDE you can quickly see that some of the conditions will force you to pick ##\ell=0## and ##m=0## (thinking in spherical harmonics).
 
  • #3
Haborix said:
In the Schrodinger equation, there is angular dependence in the Laplacian coming from the momentum operators. I don't see how you distinguish this from the other PDEs you list. In general, when written in coordinates with an angular variable, they will also have derivatives with respect to those angular variables. It may be that the boundary/initial conditions make it so that the only allowed angular dependence is a constant function. If you plug this constant angular dependence into the full separable PDE, then the angular parts seem to disappear. But that's a relic of boundary/initial conditions identifying the solution, not that the PDE didn't a priori have some angular dependence.

I would say you just get a feel for this when you setup the problem. For a separable PDE you can quickly see that some of the conditions will force you to pick ##\ell=0## and ##m=0## (thinking in spherical harmonics).
Right, maybe I didn't explain myself well enough. Of course the Laplacian in spherical coordinate contains derivative with respect to angles, and this Laplacian operator appears in Schrödinger's equation as well as the heat equation. However, it turns out that when the potential is radial (does not depend on any angle) in the Schrödinger's equation as is the case for the H-atom, one would be tempted to assume a solution that does not depend on any angle as well. But we would miss any excited state of the atom (or we would only get the ground state of the system, missing out the set of eigenfunctions).
However for the heat equation and others, if the boundary conditions and the geometry in which we solve the equation do not depend on any angle, then the solution (and eigenfunctions) will not depend on any angle. That's clearly different than in the case of the Schrödinger's equation. So, even though the Laplacian still appears, the derivative with respect to the angles can be assumed to vanish (and it turns out that they do vanish if one goes through the great pain to check this out), saving a lot of time and efforts (i.e. using a trick).

So my question still stands (to me at least).
 
  • #4
fluidistic said:
We use no justification whatsoever and since we find
the justification is called "existence and uniqueness theorems
fluidistic said:
However why does such a trick fail for example with Schrödinger equation?
The corresponding IVP for the Schrödinger equation has a unique solution as well
 
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  • #5
wrobel said:
The corresponding IVP for the Schrödinger equation has a unique solution as well
Sure, but I do not understand why you mention this. Also, we can focus on the time independent equations without changing anything to the discussion.
 
  • #6
wrobel said:
If you really think that evolution and non evolution equations are the subject of the same discussion , my remarks are useless indeed:)
Yes, I haven't changed my mind on my take (yet at least).
Take a cylinder in which its boundaries are kept at a constant temperature ##T_b## (Dirichlet b.c. to keep things simple). It doesn't matter an iota if initially the temperature of the interior of the cylinder was, or not, at ##T_b##. The thing is that the temperature profile will not depend on the azimuthal angle ##\theta## (here using cylindrical coordinates). In other words, it doesn't matter at all whether you are solving ##\nabla \cdot (\kappa \nabla T) - C_p\partial_t T=0## or ##\nabla \cdot (\kappa \nabla T)=0##. The fact that ##T## will not depend on ##\theta## is unaltered.
The same reasoning applies for Schrödinger's equation, with the (huge) difference that the solution ##\psi## to that equation, will depend on angles, even though neither the geometry nor the boundary conditions do. It doesn't matter an iota whether you consider the time dependent or time-independent equation, it will not change the above stated fact.

I probably missed something... But I do not see what.
 
  • #7
I think you slightly overstate your temperature example. There is no reason that the initial temperature profile in the cylinder couldn't depend on ##\theta##. Now comes in your second PDE. If the initial configuration evolves to the steady state solution (solving the second PDE), then the angle dependence disappears across time. I wonder if the issue isn't something to do with the eigenvalue being real or complex in the eigenvalue problem. The time evolution for the angular modes being an exponential decay in the Temperature problem, but being oscillatory in the Schrodinger equation. Anyway, I haven't had an extended block of time to sit down and iron this out, but I thought I'd share a few more thoughts. I'm interested to know the "solution" to the puzzle if you happen to find one that satisfies you.
 
  • #8
Haborix said:
I think you slightly overstate your temperature example. There is no reason that the initial temperature profile in the cylinder couldn't depend on ##\theta##. Now comes in your second PDE. If the initial configuration evolves to the steady state solution (solving the second PDE), then the angle dependence disappears across time. I wonder if the issue isn't something to do with the eigenvalue being real or complex in the eigenvalue problem. The time evolution for the angular modes being an exponential decay in the Temperature problem, but being oscillatory in the Schrodinger equation. Anyway, I haven't had an extended block of time to sit down and iron this out, but I thought I'd share a few more thoughts. I'm interested to know the "solution" to the puzzle if you happen to find one that satisfies you.
Sure, you can introduce an angle depency through the initial conditions, but this is not something I want to do. I assume that the given problem (with its initial conditions if, for some reason you insist in considering the time dependent problem) lacks any dependency on angles. The strange thing with Schrödinger equation compared to the other I mentioned, is that the solution will depend on the angles even though nothing else does (including your initial conditions).
 
  • #9
Are you sure what you say about the Schrodinger equation is true, assuming no angular dependence in the initial value problem?
 
  • #10
Haborix said:
Are you sure what you say about the Schrodinger equation is true, assuming no angular dependence in the initial value problem?
Yes, and you don't even need to bring the time dependence at all, it still works. (For simplicity sake, would we please focus on the time independent equations instead? The time dependence brings nothing of importance to the discussion...)
Look at the eigenfunctions https://en.wikipedia.org/wiki/Hydrogen_atom, the angular dependence, as I wrote in the very first post in this thread, are the spherical harmonics. The potential is central (depends only on the distance between the electron and the proton, no angle dependence!), yet the solutions to the equation involve angles. You don't get this feature with the heat equation, etc.
 
  • #11
fluidistic said:
However, it turns out that when the potential is radial (does not depend on any angle) in the Schrödinger's equation as is the case for the H-atom, one would be tempted to assume a solution that does not depend on any angle as well.
You are misinterpreting the solutions to the SDE. What you find, via the Spherical Harmonics, is a complete set of basis solutions that are related to AM (angular momentum) eigenfunctions about the chosen z-axis.

Any particular solution of the SDE wil be a linear combination of these spherical harmonics and, if the imposed conditions are spherically symmetric, then so will the overall linear combination of eigenfunctions.

Now, we also need to look more closely at why we have this particular basis of eigenfunctions. If we measure AM about the chosen z-axis, then that measurement breaks the spherical symmetry. The solution then comprises the relevant non-spherically symmetric eigenfunctions.

In other words, the SDE was solved to produce non-spherically symmetric solutions relating to non-spherically symmetric measurements of AM.
 
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  • #12
I second @PeroK's first two paragraphs. I'm not sure of the relevance of the second two paragraphs, though.
 
  • #13
PeroK said:
You are misinterpreting the solutions to the SDE. What you find, via the Spherical Harmonics, is a complete set of basis solutions that are related to AM (angular momentum) eigenfunctions about the chosen z-axis.

Any particular solution of the SDE wil be a linear combination of these spherical harmonics and, if the imposed conditions are spherically symmetric, then so will the overall linear combination of eigenfunctions.

Now, we also need to look more closely at why we have this particular basis of eigenfunctions. If we measure AM about the chosen z-axis, then that measurement breaks the spherical symmetry. The solution then comprises the relevant non-spherically symmetric eigenfunctions.

In other words, the SDE was solved to produce non-spherically symmetric solutions relating to non-spherically symmetric measurements of AM.
I see, good point.
However, I am not yet convinced. Can you really remove the angular dependency with a linear combination of these eigenfunctions? I don't think so.

Are you also implying that, if I assume that the wavefunction must vanish at infinity (spherical symmetry "boundary condition") then, say, the 2nd excited state cannot be a solution, because it lacks the spherical symmetry? I don't really buy that, if that is what you mean.
 
  • #14
fluidistic said:
I see, good point.
However, I am not yet convinced. Can you really remove the angular dependency with a linear combination of these eigenfunctions? I don't think so.
They form a complete basis for all square-integrable 3D functions. You can represent any solution with them.
 
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  • #15
fluidistic said:
Are you also implying that, if I assume that the wavefunction must vanish at infinity (spherical symmetry "boundary condition") then, say, the 2nd excited state cannot be a solution, because it lacks the spherical symmetry? I don't really buy that, if that is what you mean.
Vanishing at infinity is a property of the radial part of the wave-function. That holds for all solutions for the H atom owing to the need for square integrability.
 
  • #16
PeroK said:
They form a complete basis for all square-integrable 3D functions. You can represent any solution with them.
Yes, but the thing is, what if all the solutions have an angle(s) dependence?

Since the spherical harmonics, which involve angles, are involved in the basis of all ##\Psi##s that satisfy Schrödinger's equation, they are linearly independent. How do you cancel, say a ##\cos(3\theta )## by using a weighted sum of sines and cosines of ##n\theta## where ##n\neq 3##?
 
  • #17
fluidistic said:
Yes, but the thing is, what if all the solutions have an angle(s) dependence?

Since the spherical harmonics, which involve angles, are involved in the basis of all ##\Psi##s that satisfy Schrödinger's equation, they are linearly independent. How do you cancel, say a ##\cos(3\theta )## by using a weighted sum of sines and cosines of ##n\theta## where ##n\neq 3##?
With an infinite sum. In the same way that you can construct any function from sines and cosines in Fourier analysis. E.g. the eigenfunctions in the infinite square well.

Similarly, the spherical harmonics form a basis and any wavefunction can be described as an infinite sum of them. Otherwise, the whole theory falls apart.
 
  • #18
PeroK said:
With an infinite sum. In the same way that you can construct any function from sines and cosines in Fourier analysis. E.g. the eigenfunctions in the infinite square well.

Similarly, the spherical harmonics form a basis and any wavefunction can be described as an infinite sum of them. Otherwise, the whole theory falls apart.
Thank you, that is very convincing.
Now, I still have to understand, if the problem given is the H-atom in vacuum, there is spherical symmetry. There is no unique solution, right? In that particular case, any eigenfunction satisfies the equation (and can be normalized). But you say that if the problem is spherically symmetric, then so must be its solution? Wouldn't it discard any eigenfunction to be a solution? In other words, are you saying that the 1st (or 2nd, etc.) excited state cannot be a solution to the equation because it lacks the spherical symmetry that the problem possesses?
 
  • #19
fluidistic said:
But you say that if the problem is spherically symmetric, then so must be its solution? Wouldn't it discard any eigenfunction to be a solution? In other words, are you saying that the 1st (or 2nd, etc.) excited state cannot be a solution to the equation because it lacks the spherical symmetry that the problem possesses?
The eigenstates of the ##L_z## operator are not spherically symmetric. You will always need a linear combination of those eigenstates in a spherically symmetric state.

Those eigenstates may only arise (as the state of the system) if you know (by measurement) the angular momentum about the z-axis - and that breaks the spherical symmetry.
 
  • #20
PeroK said:
The eigenstates of the ##L_z## operator are not spherically symmetric. You will always need a linear combination of those eigenstates in a spherically symmetric state.

Those eigenstates may only arise (as the state of the system) if you know (by measurement) the angular momentum about the z-axis - and that breaks the spherical symmetry.
I see what you mean. You are essentially involving quantum mechanics in some way. If I understand well, if I do not perform any measurement on the H-atom in vacuum, its state is a linear combination of the eigenstates, but this linear combination has to possesses a spherically symmetry because there's nothing that broke that symmetry (as far as my knowledge of the system goes).
However if my cousin performed an angular momentum measurement when I was blinking my eyes, he would break some spatial symmetry and his wavefunction knowledge would be different than mine. In his view, psi lacks any spherical symmetry. We would be both "right", since (according to some/most? interpretations) psi is not universal but observer-dependent.
I am still not quite convinced we must involve quantum mechanics interpretations to deal with the problem, which I tend to think can be casted as purely mathematical.
 
  • #21
fluidistic said:
I am still not quite convinced we must involve quantum mechanics interpretations to deal with the problem, which I tend to think can be casted as purely mathematical.
The explanation can be made purely mathematical, independent of any physics:

Any square-integrable wavefunction (whether symmetric or not) can be expressed as a (possibly infinite) linear combination of the basis eigenfunctions.

Note that this particular eigenbasis corresponds to the eigenfunctions of the Hamiltonian, Total Angular Momentum (AM) and AM about the z-axis; i.e. the operators ##H, L^2## and ##L_z##.

Note that we could have chosen any other basis, but that this basis (in terms of energy and angular momentum is the most useful to study the physics of the Hydrogen atom).

End of mathematics.

***

The physics comes in when we look at experiments involving the Hydrogen atom and we find the these operators apply to the most important observables: usually the emission spectrum. Note that the ##L_z## observable, corresponding to the ##m## quantum number is important in terms of the spectrum in an external magnetic field. That's what would break the spherical symmetry (as it is equivalent to measuring the angular momentum about a certain axis.)

fluidistic said:
If I understand well, if I do not perform any measurement on the H-atom in vacuum, its state is a linear combination of the eigenstates, but this linear combination has to possesses a spherically symmetry because there's nothing that broke that symmetry (as far as my knowledge of the system goes).
However if my cousin performed an angular momentum measurement when I was blinking my eyes, he would break some spatial symmetry and his wavefunction knowledge would be different than mine. In his view, psi lacks any spherical symmetry. We would be both "right", since (according to some/most? interpretations) psi is not universal but observer-dependent.

This misses the point. If you have a system in given a state and someone else does not know what you've done, then there's no mystery: you can predict what the system will do and they cannot! If there is an external magnetic field, then the atom will behave appropriately, whether or not you know about the field.

The actual point is much simpler. You can do things to a hydrogen atom that make its state not spherically symmetric.
 
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  • #22
I appreciate very much your input PeroK. I can understand everything in your last post, but I have some question regarding
PeroK said:
This misses the point. If you have a system in given a state and someone else does not know what you've done, then there's no mystery: you can predict what the system will do and they cannot! If there is an external magnetic field, then the atom will behave appropriately, whether or not you know about the field.
In the case where I have done things to the H atom in the past without my cousin knowing about it, and we are both observing that the H atom is now left unperturbed in vacuum, my wavefunction will differ from my cousin's (mine will be an eigenstate with broken symmetry, i.e. angle dependence if I measured ##L_z## as you point out), while his won't. His wavefunction would be a superposition of all eigenstates, with the coefficients either undetermined (as I thought they would be if no more information is provided to him) or determined as to make his psi spherically symmetric since for him the problem is angular independent. But in the end we should be both right in that if we were to perform an experiment, we should obtain a result that makes sense to both of us. I may have more predictive power than him, but that is fine.
Regarding your comment about a magnetic field, sure, if the observer is not aware of the system itself (H atom + magnetic field), but only part of it, his description of the outcome or behavior of the system will or might be erroneous. But that's different from the case where the perturbation has finished and two observers, each one with different information about the past of the system, are currently observing the same system.

Is my reasoning correct?
 
  • #23
fluidistic said:
His wavefunction would be a superposition of all eigenstates, with the coefficients either undetermined
His wavefunction will simply be wrong. This seems to me an irrelevant question.

The relevance of these issues is really to explain the fine and hyperfine structure and the Zeeman effect. For example, the emission spectrum of hydrogen splits for higher energy levels because of the small differences in the different orbital angular momentum states.

It's not about what information you may or may not have. It's to explain the variations in the observed emission spectrum.
 
  • #24
PeroK said:
His wavefunction will simply be wrong. This seems to me an irrelevant question.

The relevance of these issues is really to explain the fine and hyperfine structure and the Zeeman effect. For example, the emission spectrum of hydrogen splits for higher energy levels because of the small differences in the different orbital angular momentum states.

It's not about what information you may or may not have. It's to explain the variations in the observed emission spectrum.
You seem to assume that there is a correct wavefunction as if it is an objective function (which isn't a given, I think, and is matter of following QM interpretations). I do not see how/why his wavefunction would be wrong. After all, after a measurement is done, the outcome will not be inconsistent with his wavefunction (nor mine for that matter). For him, the state would have "collapsed" into an eigenstate (which was part of his wavefunction with an undetermined coefficient before the measurement). For me, the state would remain unchanged. Why would my wavefunction be correct and his, wrong?
About the Zeeman effect, sure, if all observers agree that the H atom is under an applied B field, then they could predict the line splitting.

Going back to the mathematical side of the question... at the end of the day, I still believe that an eigenstate lacking spherical symmetry is solution to the time independent (and dependent) Schrodinger's equation, even though the problem lacks the symmetry breakage. The case of the H atom in vacuum is an example, where the potential is central, and if psi is required to vanish at infinity as sole condition, then there is no symmetry breaking in the problem (or Hamiltonian + boundary conditions). We seems to differ in that, for you, if I understand well, in order to have a symmetry breaking in the solution, then in the past (or the present if a B field is applied without the observer knowing) there had been some symmetry breaking. In your case you involve quantum mechanics and interpretations of it, in my case I only stick to the mathematical aspect. Do you think I have understood you well?
 
  • #25
fluidistic said:
You seem to assume that there is a correct wavefunction as if it is an objective function (which isn't a given, I think, and is matter of following QM interpretations). I do not see how/why his wavefunction would be wrong. After all, after a measurement is done, the outcome will not be inconsistent with his wavefunction (nor mine for that matter). For him, the state would have "collapsed" into an eigenstate (which was part of his wavefunction with an undetermined coefficient before the measurement). For me, the state would remain unchanged. Why would my wavefunction be correct and his, wrong?
About the Zeeman effect, sure, if all observers agree that the H atom is under an applied B field, then they could predict the line splitting.

Going back to the mathematical side of the question... at the end of the day, I still believe that an eigenstate lacking spherical symmetry is solution to the time independent (and dependent) Schrodinger's equation, even though the problem lacks the symmetry breakage. The case of the H atom in vacuum is an example, where the potential is central, and if psi is required to vanish at infinity as sole condition, then there is no symmetry breaking in the problem (or Hamiltonian + boundary conditions). We seems to differ in that, for you, if I understand well, in order to have a symmetry breaking in the solution, then in the past (or the present if a B field is applied without the observer knowing) there had been some symmetry breaking. In your case you involve quantum mechanics and interpretations of it, in my case I only stick to the mathematical aspect. Do you think I have understood you well?
This all seems muddled to me. Mathematically, there is no issue: there is simply a basis of functions. And that's the end of the matter. If you don't want to talk about QM, then there is nothing more to say.

It's only when you look at experiments involving a hydrogen atom that you can talk about particular wavefunctions and particular solutions, spherically symmetric or otherwise.
 
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  • #26
PeroK said:
With an infinite sum. In the same way that you can construct any function from sines and cosines in Fourier analysis. E.g. the eigenfunctions in the infinite square well.

Similarly, the spherical harmonics form a basis and any wavefunction can be described as an infinite sum of them. Otherwise, the whole theory falls apart.
This isn't correct. There is no such infinite sum doing what fluidistic proposes.
 
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  • #27
Haborix said:
This isn't correct. There is no such infinite sum doing what fluidistic proposes.
What is he proposing?
 
  • #28
I took you to be responding to:
fluidistic said:
Yes, but the thing is, what if all the solutions have an angle(s) dependence?

Since the spherical harmonics, which involve angles, are involved in the basis of all ##\Psi##s that satisfy Schrödinger's equation, they are linearly independent. How do you cancel, say a ##\cos(3\theta )## by using a weighted sum of sines and cosines of ##n\theta## where ##n\neq 3##?
Specifically, the "How do you cancel,..." part.
 
  • #29
Haborix said:
I took you to be responding to:

Specifically, the "How do you cancel,..." part.
What is the answer?
 
  • #30
PeroK said:
What is the answer?
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.
 
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  • #31
Haborix said:
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.
 
  • #32
Haborix said:
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.
 
  • #33
Haborix said:
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
 
  • #34
PeroK said:
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.
 
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  • #35
Haborix said:
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?
 
<h2>1. What is the importance of justifying a trick in solving PDEs arising in Physics?</h2><p>Justifying a trick in solving PDEs is crucial because it ensures that the solution obtained is valid and accurate. This is especially important in physics, where the solutions are used to make predictions and understand complex phenomena.</p><h2>2. How do you determine if a trick is valid for solving a PDE in physics?</h2><p>To determine the validity of a trick in solving a PDE, one must carefully analyze the assumptions and limitations of the trick and compare it to the physical system being studied. This involves understanding the underlying physics and making sure the trick accurately captures the behavior of the system.</p><h2>3. Can a trick be used to solve any type of PDE in physics?</h2><p>No, not all tricks can be applied to every type of PDE encountered in physics. Some tricks may only be applicable to certain types of PDEs, while others may not be valid for any type of PDE. It is important to carefully consider the nature of the PDE and the assumptions of the trick before using it.</p><h2>4. Are there any risks associated with using a trick to solve a PDE in physics?</h2><p>Yes, there are potential risks associated with using a trick to solve a PDE in physics. If the trick is not valid for the given PDE, the solution obtained may be incorrect and lead to incorrect predictions. It is important to carefully validate and justify the use of any trick before applying it to a PDE.</p><h2>5. How can one ensure the reliability of a trick in solving PDEs arising in Physics?</h2><p>To ensure the reliability of a trick in solving PDEs, one must carefully validate and justify its use. This involves comparing the results obtained from the trick to known solutions or experimental data, as well as carefully considering the assumptions and limitations of the trick. It is also important to consult with experts in the field to verify the reliability of the trick.</p>

1. What is the importance of justifying a trick in solving PDEs arising in Physics?

Justifying a trick in solving PDEs is crucial because it ensures that the solution obtained is valid and accurate. This is especially important in physics, where the solutions are used to make predictions and understand complex phenomena.

2. How do you determine if a trick is valid for solving a PDE in physics?

To determine the validity of a trick in solving a PDE, one must carefully analyze the assumptions and limitations of the trick and compare it to the physical system being studied. This involves understanding the underlying physics and making sure the trick accurately captures the behavior of the system.

3. Can a trick be used to solve any type of PDE in physics?

No, not all tricks can be applied to every type of PDE encountered in physics. Some tricks may only be applicable to certain types of PDEs, while others may not be valid for any type of PDE. It is important to carefully consider the nature of the PDE and the assumptions of the trick before using it.

4. Are there any risks associated with using a trick to solve a PDE in physics?

Yes, there are potential risks associated with using a trick to solve a PDE in physics. If the trick is not valid for the given PDE, the solution obtained may be incorrect and lead to incorrect predictions. It is important to carefully validate and justify the use of any trick before applying it to a PDE.

5. How can one ensure the reliability of a trick in solving PDEs arising in Physics?

To ensure the reliability of a trick in solving PDEs, one must carefully validate and justify its use. This involves comparing the results obtained from the trick to known solutions or experimental data, as well as carefully considering the assumptions and limitations of the trick. It is also important to consult with experts in the field to verify the reliability of the trick.

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