Schrodinger equation and a half wave?

[Butch]

I have been studying Schrodinger's equation, and I have a question... Given the representation of a moving wave:

Eventually the quantum number corresponds to the number of wavelengths in the circuit of the electron in the orbital (discounting spin)? Is my understanding correct?

[studiot]

8 minutes ago, Butch said:

Is my understanding correct?

No.

[Butch]

16 minutes ago, studiot said:

No.

Okay, perhaps you could provide some assistance, what does the quantum number relate to, and is it demonstrated in the above equation (I know this is not the Schrodinger equation.)?

I am aware of constraints, however it seems that if the quantum number is related to wave function it must also be related to wavelength, where am I getting lost?

[studiot]

1 hour ago, Butch said:

Okay, perhaps you could provide some assistance, what does the quantum number relate to, and is it demonstrated in the above equation (I know this is not the Schrodinger equation.)?

I am aware of constraints, however it seems that if the quantum number is related to wave function it must also be related to wavelength, where am I getting lost?

What do you know about the sine function, because you are missing something from your posted equation ?

 

That something is usually given the synbol n.

[swansont]

The steady-state solution is going to be a standing wave, not a traveling wave.

[Butch]

22 hours ago, studiot said:

What do you know about the sine function, because you are missing something from your posted equation ?

 

That something is usually given the synbol n.

I am somewhat adept at the maths from the number line to basic calc... Quantum mechanics is new ground for me, I am trying to have a good understanding of the maths, not just rote rehearsel. From what I have studied this is where Schrodinger began, a time dependant traveling wave. The "n" is where my question arises... 2π/y represents a single cycle or an integral number of cycles. Should not "n" be represented by the same integral?

22 hours ago, swansont said:

The steady-state solution is going to be a standing wave, not a traveling wave.

Understood, but as I understand it the integral is a full wave.

Let me try getting to the point without being humbled traumatically... If n = .5 the result could be a low energy p orbital masked by the s orbital. This p orbital however would allow only a single electron. With my current understanding, I would apply inverse square for the energy of this orbital.

Edited February 11, 2018 by Butch
Elaboration

[swansont]

31 minutes ago, Butch said:

Understood, but as I understand it the integral is a full wave.

Let me try getting to the point without being humbled traumatically... If n = .5 the result could be a low energy p orbital masked by the s orbital. This p orbital however would allow only a single electron. With my current understanding, I would apply inverse square for the energy of this orbital.

There is no n = .5

The ground state of a square well potential is a half cycle. It's all about applying the boundary conditions to the integral.

[studiot]

OK so let's start at the beginning.

Both the simpler wave equation and the Schroedinger equation are (second order) differential equations connecting space and time via another variable that has a connection to each.

In the case of the wave equation, that's all it does.
Also its third 'connection variable' can be directly related to some physical property such as displacement (the y in your equation) or pressure or stress.

On the other hand the Schroedinger equation not only includes energy but also introduces a connection variable that is not physical property.

This variable is often referred to as a 'probability density' and has the units of the reciprocal of the square root of metres in one dimension.

I see that swansont has referred to your itnegral and boundary conditions.

 

A second order differential equation cannot be solved without boundary conditions as the two integrations needed introduces arbitrary constants or functions which are determined by the boundary conditions.

I am sure I derived Schroedinger from the wave equation here at SF not too long ago with full explanation of all this.
And I thought it was for you but perhaps it was for someone else.

The reason for my question about the sine wave is that.

1) It is periodic.

2) It is bounded.

A function like this one would not be acceptable as a solution.

 

Because although it repeats itself periodically, it also goes off to infinity and is discontinuous.

51 minutes ago, Butch said:

From what I have studied this is where Schrodinger began, a time dependant traveling wave.

All travelling waves are time dependent by definition.

A time independent wave is another name for a stationary or standing wave.

[Butch]

25 minutes ago, swansont said:

There is no n = .5

The ground state of a square well potential is a half cycle. It's all about applying the boundary conditions to the integral.

I believe you, but I don't see it... Are you saying that max amplitude is at either peak, akin to alternating electrical potential? Is that why I was asked about the sine function?

7 minutes ago, studiot said:

OK so let's start at the beginning.

Both the simpler wave equation and the Schroedinger equation are (second order) differential equations connecting space and time via another variable that has a connection to each.

In the case of the wave equation, that's all it does.
Also its third 'connection variable' can be directly related to some physical property such as displacement (the y in your equation) or pressure or stress.

On the other hand the Schroedinger equation not only includes energy but also introduces a connection variable that is not physical property.

This variable is often referred to as a 'probability density' and has the units of the reciprocal of the square root of metres in one dimension.

I see that swansont has referred to your itnegral and boundary conditions.

 

A second order differential equation cannot be solved without boundary conditions as the two integrations needed introduces arbitrary constants or functions which are determined by the boundary conditions.

I am sure I derived Schroedinger from the wave equation here at SF not too long ago with full explanation of all this.
And I thought it was for you but perhaps it was for someone else.

The reason for my question about the sine wave is that.

1) It is periodic.

2) It is bounded.

A function like this one would not be acceptable as a solution.

 

Because although it repeats itself periodically, it also goes off to infinity and is discontinuous.

All travelling waves are time dependent by definition.

A time independent wave is another name for a stationary or standing wave.

All understood, however if the polarity of L does not change wouldn't the amplitude (A) always be positive?

Ok... Not all understood, your plot is the tangent... Does this refer to the derivitave of the equation and containment must be applied because of infinity?

Edited February 11, 2018 by Butch
Correction

[studiot]

Where did L come from?

[Butch]

3 minutes ago, studiot said:

Where did L come from?

Intrinsic angular momentum... Sorry, I am trying to get my head around this.

By the way I quit smoking almost a month now, quite a feat while delving into quanta. Lol.

Edited February 11, 2018 by Butch

[studiot]

I shouldn't worry about angular momentum until you have got the basics sorted.

The n I referred to comes out of the wash as the principal quantum number.

 

When you solve the simple wave equation for a time independent solution you do this by a method known as 'separation of the variables'.

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.

Your original sine had x and t mixed up in the sine, which won't as you can't remove the time dependence.

[Butch]

16 minutes ago, studiot said:

I shouldn't worry about angular momentum until you have got the basics sorted.

The n I referred to comes out of the wash as the principal quantum number.

 

When you solve the simple wave equation for a time independent solution you do this by a method known as 'separation of the variables'.

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.

Your original sine had x and t mixed up in the sine, which won't as you can't remove the time dependence.

Thank you all very much.

[swansont]

11 hours ago, Butch said:

I believe you, but I don't see it... Are you saying that max amplitude is at either peak, akin to alternating electrical potential? Is that why I was asked about the sine function?

The peak is by definition the max amplitude, so I don't understand your question.

In an infinite square well, the wave function must go to zero at the boundary. In an orbital, the value at 0º has to match the value at 360º for the angle variable. (It also can't go to infinity). Those restrictions limit the solutions of the wave equation.

[Butch]

On 2/12/2018 at 6:16 AM, swansont said:

The peak is by definition the max amplitude, so I don't understand your question.

In an infinite square well, the wave function must go to zero at the boundary. In an orbital, the value at 0º has to match the value at 360º for the angle variable. (It also can't go to infinity). Those restrictions limit the solutions of the wave equation.

As concerns amplitude, are the peak amplitudes positive / negative or 0 / absolute? 

What is our reference?

If Amplitude is 0 @ theta = 90 the well would have a span of 1 wavelength, if however amplitude is 0 @ theta = 0 span would be 1/2 wavelength, if the latter is correct, what is our zero reference?

I apologise if I am being tedious, however in pursuing information on the web I have been witness to many erroneous assumptions and conflicting information.

[swansont]

27 minutes ago, Butch said:

As concerns amplitude, are the peak amplitudes positive / negative or 0 / absolute? 

For the wave function it can be either. You're going to square it at some point, for finding out anything.

27 minutes ago, Butch said:

What is our reference?

It depends on the system.

27 minutes ago, Butch said:

If Amplitude is 0 @ theta = 90 the well would have a span of 1 wavelength, if however amplitude is 0 @ theta = 0 span would be 1/2 wavelength, if the latter is correct, what is our zero reference?

Zeroes are often dictated by the boundary conditions. The square well mandates that the wave function be zero at the walls.

If the system is rotationally symmetric the zeroes won't matter. For an atom, you can say the solution at 0º is the same as at 360º, and they must have the same slope (i.e. it's a continuity condition). This precludes a half-wave solution.

 

 

[Butch]

16 minutes ago, swansont said:

If the system is rotationally symmetric the zeroes won't matter. For an atom, you can say the solution at 0º is the same as at 360º, and they must have the same slope (i.e. it's a continuity condition). This precludes a half-wave solution.

Great, I can pinpoint exactly where I am lost here... 

The amplitude is the same at 180 as it is at 0, the slope however is perpendicular... As long as we are at 0 at the wall why does slope matter?

[swansont]

Just now, Butch said:

Great, I can pinpoint exactly where I am lost here... 

The amplitude is the same at 180 as it is at 0, the slope however is perpendicular... As long as we are at 0 at the wall why does slope matter?

You are mixing the two situations. A square well does not have rotational symmetry, and thus does not have continuity conditions An atom is, and does.

[Butch]

2 minutes ago, swansont said:

You are mixing the two situations. A square well does not have rotational symmetry, and thus does not have continuity conditions An atom is, and does.

I am confused... How does the square well apply to the quantized energy states of the electron in an atom?

[swansont]

12 minutes ago, Butch said:

I am confused... How does the square well apply to the quantized energy states of the electron in an atom?

It's another example of a quantized system. You said you were trying to learn the concepts. A square well is a simple application of QM.

[Butch]

1 minute ago, swansont said:

It's another example of a quantized system. You said you were trying to learn the concepts. A square well is a simple application of QM.

I get that...(Thank you for your patience).

What I don't understand is this...

A wave function is described via a sinus function of the unit radius, a self reference would have amplitude that extended from 0 to A, and the principle quanta would be multiples of the wavelength?

In the atom would the reference be the nucleus, that is at 0 the electron would be at the center of the atom and at A it would be at the Bohr radius?

Or am I just a floating lump of protoplasm lost in space?

 

[swansont]

40 minutes ago, Butch said:

I get that...(Thank you for your patience).

What I don't understand is this...

A wave function is described via a sinus function of the unit radius, a self reference would have amplitude that extended from 0 to A, and the principle quanta would be multiples of the wavelength?

In the atom would the reference be the nucleus, that is at 0 the electron would be at the center of the atom and at A it would be at the Bohr radius?

Or am I just a floating lump of protoplasm lost in space?

 

Yes, the radial part of the wave function would go to zero at the center of the nucleus, and the most probable value is the Bohr radius. It depends only on r.

The two angular dimensions would be different functions (spherical harmonics, which depend only on theta and phi)

[Butch]

5 hours ago, swansont said:

Yes, the radial part of the wave function would go to zero at the center of the nucleus, and the most probable value is the Bohr radius. It depends only on r.

The two angular dimensions would be different functions (spherical harmonics, which depend only on theta and phi)

Phew! Thanks!

[Butch]

1 hour ago, Butch said:

Phew! Thanks!

I have the idea that the energies of the particle and the wave are out of phase, should I elaborate or has someone already been there?

1 hour ago, Butch said:

The two angular dimensions would be different functions (spherical harmonics, which depend only on theta and phi)

This is in my study materials, I will endeavor to gain a better understanding for later discussions.

Thank you.

[swansont]

10 hours ago, Butch said:

I have the idea that the energies of the particle and the wave are out of phase, should I elaborate or has someone already been there?

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.