Butch

Schrodinger equation and a half wave?

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9 hours ago, swansont said:

I don't know what this means. In the steady-state solution, these are standing waves. the energy is a constant.  

"the energies of the particle and the wave are out of phase" makes no sense.

Ok, an electron in an orbital of an atom as a particle would have higher average kinetic energy as the orbital decreased. Despite uncertainy the electron still has a particle nature, kinetic energy and mass.

I suppose perhaps I am stuck between a particle and a hard wave...

Probably just need to read some more, you have aided me very much.

Thank you.

Edited by Butch

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Hmm... At least one of my thoughts was correct, Schrodinger's next step was to incorporate the particle energy into the equation.

Two standing waves can be 180 degrees out of phase, the resultant would be the difference in energies...

Let's check our phase at this point, the math we have discussed, resolves the wave function. What I will be moving on to next is inclusion of the kinetic energy of the electron, potential energy and angular momentum?

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11 hours ago, Butch said:

Hmm... At least one of my thoughts was correct, Schrodinger's next step was to incorporate the particle energy into the equation.

Two standing waves can be 180 degrees out of phase, the resultant would be the difference in energies...

Let's check our phase at this point, the math we have discussed, resolves the wave function.

Solve the Schrödinger's equation and show this, then.

11 hours ago, Butch said:

What I will be moving on to next is inclusion of the kinetic energy of the electron, potential energy and angular momentum?

The KE and PE are already included, and the angular momentum comes from the solution.

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5 hours ago, swansont said:

Solve the Schrödinger's equation and show this, then.

I am not saying that it is there, just that standing waves can be out of phase. 

Just to address your understanding of my statement, not to propose any real system... If the electron followed a path through the center of the atom and out the other side to the altitude of the Bohr radius and then followed the opposite path back through center and to the Bohr radius opposite, it's kinetic energy would be greatest as it passed through the nucleus and least as it reached the Bohr radius.

Regardless the actual path of the electron it would have it's greatest k the closer it was to the nucleus.

This would be 180 degrees out of phase with the wave function.

Perhaps I still do not correctly comprehend wave function?

6 hours ago, swansont said:

The KE and PE are already included, and the angular momentum comes from the solution.

These are included in the full equation... I am not there yet.

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4 minutes ago, Butch said:

Perhaps I still do not correctly comprehend wave function?

Indeed.

Nor will you until you can abandon the miniature solar system view of this subject.

 

FYI there were three different approaches to the subject.

There were subsequently shown to end in the the same model although they came from different premises.

 

1) Quantisation.

The proposition that energy is quantised, and that the smooth classical laws should be represented by a series of permissible energy levels and forbidden zones.
There was little or no justification for this except that it fits the observed (empircal) evidence.
This results in the ladder type diagrams but it is only a small part of the story and nowadays considered a result that pops out of the equations offered in the second and third approaches.

2) Schroedinger.

This is a complicated differential equation of motion (with many many solutions) from which the necessary quantisation emerges directly from solutions and the boundary conditions.
That is why I started towards that, but you wished to plough on in your own furrow.

3) Heisenberg Matrix Mechanics

The uncertainly principle emerges directly from this approach.
Further (advanced) work can convert between this method and Schroedinger.
It also leads directly to the tensor statement linking the energy tensor to the hamiltonian and to the linear algebra required for special topics such as entanglement.

 

Method (1) does not need or introduce the non locality required in QM.

Method (2) does this by means of assigning a meaning of probability density to the wave function.

Method (3) does this directly.
 

 

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1 hour ago, Butch said:

I am not saying that it is there, just that standing waves can be out of phase. 

And how is that relevant?

1 hour ago, Butch said:

Just to address your understanding of my statement, not to propose any real system... If the electron followed a path through the center of the atom and out the other side to the altitude of the Bohr radius and then followed the opposite path back through center and to the Bohr radius opposite, it's kinetic energy would be greatest as it passed through the nucleus and least as it reached the Bohr radius.

There is no path. Electrons in atoms do not follow trajectories.

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On 2/16/2018 at 12:34 PM, studiot said:

Indeed.

Nor will you until you can abandon the miniature solar system view of this subject.

 

FYI there were three different approaches to the subject.

There were subsequently shown to end in the the same model although they came from different premises.

 

1) Quantisation.

The proposition that energy is quantised, and that the smooth classical laws should be represented by a series of permissible energy levels and forbidden zones.
There was little or no justification for this except that it fits the observed (empircal) evidence.
This results in the ladder type diagrams but it is only a small part of the story and nowadays considered a result that pops out of the equations offered in the second and third approaches.

2) Schroedinger.

This is a complicated differential equation of motion (with many many solutions) from which the necessary quantisation emerges directly from solutions and the boundary conditions.
That is why I started towards that, but you wished to plough on in your own furrow.

3) Heisenberg Matrix Mechanics

The uncertainly principle emerges directly from this approach.
Further (advanced) work can convert between this method and Schroedinger.
It also leads directly to the tensor statement linking the energy tensor to the hamiltonian and to the linear algebra required for special topics such as entanglement.

 

Method (1) does not need or introduce the non locality required in QM.

Method (2) does this by means of assigning a meaning of probability density to the wave function.

Method (3) does this directly.
 

 

I think the reason I got lost was I did not realize that x represented a point in 3d space... Can you confirm this?

 

On 2/16/2018 at 1:28 PM, swansont said:

And how is that relevant?

There is no path. Electrons in atoms do not follow trajectories.

It may not be relevant... However I think it is the phase of potential energy v kinetic energy.

As far as path, my understanding is that the time independent equation eliminates velocity as a consideration and with it we deal only with potential energy and position. Can we not also modify the equation so that we can consider kinetic energy and velocity(as an oscillation) and discount position?

 

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4 minutes ago, Butch said:

It may not be relevant... However I think it is the phase of potential energy v kinetic energy.

Repeating this will not make it make sense.

4 minutes ago, Butch said:

As far as path, my understanding is that the time independent equation eliminates velocity as a consideration and with it we deal only with potential energy and position. Can we not also modify the equation so that we can consider kinetic energy and velocity(as an oscillation) and discount position?

Energy is not a function of position in the QM solution.

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34 minutes ago, swansont said:

Repeating this will not make it make sense.

Energy is not a function of position in the QM solution.

Then I am unclear on amplitude, I have a background in electronics and Ohms law is etched into my mind, sorry. 

34 minutes ago, swansont said:

Repeating this will not make it make sense.

Energy is not a function of position in the QM solution.

Does not Coulomb's law and position indicate potential energy? 

Ok... I get that velocity would indicate linear transition between energy levels.

There is a lot of science I need to incorporate here.

Edited by Butch

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1 hour ago, Butch said:

Then I am unclear on amplitude, I have a background in electronics and Ohms law is etched into my mind, sorry. 

The solution to the Schrödinger equation yields eigenvalues and eigenstates. You get one number per state. e.g. The ground state of Hydrogen is 13.6 eV below the unbound state.

Quote

Does not Coulomb's law and position indicate potential energy? 

Yes. You need to solve the Schrödinger equation, not Coulomb's law. That's an input, not the output. The position information is lost.

These solutions (and the derivation) exist online. You should familiarize yourself with them.

Quote

Ok... I get that velocity would indicate linear transition between energy levels.

There is a lot of science I need to incorporate here.

Velocity has lots of implications that are contrary to what we observe.

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3 minutes ago, swansont said:

The solution to the Schrödinger equation yields eigenvalues and eigenstates. You get one number per state. e.g. The ground state of Hydrogen is 13.6 eV below the unbound state.

Yes. You need to solve the Schrödinger equation, not Coulomb's law. That's an input, not the output. The position information is lost.

These solutions (and the derivation) exist online. You should familiarize yourself with them.

Velocity has lots of implications that are contrary to what we observe.

Again, thank you for your patience.

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1 hour ago, Butch said:

I think the reason I got lost was I did not realize that x represented a point in 3d space... Can you confirm this?

 

Well really x represents only 1D space.

It is easier to work in a single space dimension to start with.

But you also need time to set up the equation.

going back to an earlier post of mine in this thread, which I think you didn't properly catch,

You start with a solution

Let the solution be a function of space and time y(x,t)

Do you understand this notation?

Now to find a time independent solution you separate the variables by

 

On 11/02/2018 at 11:46 PM, studiot said:

You look for a solution of the form y(x,t) = g(x)f(t) so that you have two separate functions (one of x and one of t) multiplied together.

 

Do you understand this ?

 

This allows you to reduce the solution to y = A f(x), where A is a constant.

 

But you are not yet done because x can take on any value between plus and minus infinity it is encompasses all 1D space.

What you want is for your x to refer not to a single point but the region (of 1D space) where the wave acts.

If you like to picture it this way, the whole of the wave must be concentrated between two values of x,

say between x = -0.5 x10-11 metres and x = +0.5 x10-11 metres

which is the diameter of a hydrogen atom.

So for the Schroedinger equation the value of the solution y (I am trying to avoid greek letters) between x and (x+dx) give the electron density between these values.

So if we add up (sum) all the small dx from x = -0.5 x10-11 to x = +0.5 x10-11 this must add up to 1 stating the obvious fact that a bound electron must be within the atom.

You can also regards it as the probability of finding an electron at a particular x.

But then you loose the time independence because you must specify how long you are prepared to wait observing that particular point!

 

dammed if you do and dammed if you don't.

 

:)

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The hydrogen atom, using radial coordinates, is limited to non-negative values of r. 

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1 hour ago, studiot said:

Indeed time does not have negative values.

r is spatial. Hydrogen solution typically uses spherical coordinates. "r" rather than "x"

And if we're looking at the time-independent solution, we need not worry about the time solution separation of variables.

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10 hours ago, swansont said:

r is spatial. Hydrogen solution typically uses spherical coordinates. "r" rather than "x"

And if we're looking at the time-independent solution, we need not worry about the time solution separation of variables.

I think we are talking at cross purposes here.

I put it down to this poxy website software that insists on presenting everything to me so tiny I get it mixed up.

Looking closely in the clear light of day I see that last night i mistook your small r for a small t.

Sorry for that.

 

Don't forget I am starting with the one dimensional case so 'radial' has no meaning.

I also indicated that I took some trouble to post a derivation of the Schroedinger equation starting with the classical wave equation but can't find it again.

This derivation does indeed use an 'inspired guess' solution by separation of the variables.

Edited by studiot

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3 hours ago, studiot said:

 Don't forget I am starting with the one dimensional case so 'radial' has no meaning.

Which is fine for a square well, but for the hydrogen atom probably only confuses things.

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15 hours ago, studiot said:

 

Well really x represents only 1D space.

It is easier to work in a single space dimension to start with.

But you also need time to set up the equation.

going back to an earlier post of mine in this thread, which I think you didn't properly catch,

You start with a solution

Let the solution be a function of space and time y(x,t)

Do you understand this notation?

Now to find a time independent solution you separate the variables by

 

 

Do you understand this ?

 

This allows you to reduce the solution to y = A f(x), where A is a constant.

 

But you are not yet done because x can take on any value between plus and minus infinity it is encompasses all 1D space.

What you want is for your x to refer not to a single point but the region (of 1D space) where the wave acts.

If you like to picture it this way, the whole of the wave must be concentrated between two values of x,

say between x = -0.5 x10-11 metres and x = +0.5 x10-11 metres

which is the diameter of a hydrogen atom.

So for the Schroedinger equation the value of the solution y (I am trying to avoid greek letters) between x and (x+dx) give the electron density between these values.

So if we add up (sum) all the small dx from x = -0.5 x10-11 to x = +0.5 x10-11 this must add up to 1 stating the obvious fact that a bound electron must be within the atom.

You can also regards it as the probability of finding an electron at a particular x.

But then you loose the time independence because you must specify how long you are prepared to wait observing that particular point!

 

dammed if you do and dammed if you don't.

 

:)

Yes, this all makes good sense, thank you! It is difficult to discern all this from texts.

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1 hour ago, Butch said:

Yes, this all makes good sense, thank you! It is difficult to discern all this from texts.

 

Yeah you have to read a lot of texts to find the stuff between the lines.

 

So order a large drum of midnight oil.

 

:)

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