An Alternative Equation for the Wavefunction and its Eigenfunctions

[swansont]

58 minutes ago, John Henke said:

Okay, so you still haven't read the article you're criticizing.

And you still apparently haven’t read the rules you agreed to follow.

 

Quote

That's okay. I suggested it's tested with force. It has a well defined definition of gravity.

Gravity and QM being so tightly coupled...

Quote

That's why I said I might be able to find the value of n based on the value of gravity.

Why and how would gravity - something typically ignored in QM calculations - yield a QM parameter’s value? Why not use electrodynamics?

 

21 minutes ago, John Henke said:

And, Swansont, there is already an equation for a 1D particle in a well in the paper.

What section?

[Mordred]

For the OP this is just a primer to help understand the basics of QFT and how force is applied. Keep in mind this is a workup I did on this forum a few years back but it saves a lot of latex.

You should be able to get the gist of how QFT differentiates from QM.

One difference to recognize and I didn't cover is that the Schrodinger equation is first order while the Klien Gordon is second order. (This is a conflict of QFT to QM that requires a seperate fix). The repair comes into play when you factor in particle number density...

Quote

1) Field :A field is a collection of values assigned to geometric coordinates. Those values can be of any nature and does not count as a substance or medium.

2) As we are dealing with QM we need the simple quantum harmonic oscillator

3) Particle: A field excitation

 

Simple Harmonic Oscillator

[math]\hat{H}=\hbar w(\hat{a}^\dagger\hat{a}+\frac{1}{2})[/math]

the [math]\hat{a}^\dagger[/math] is the creation operator with [math]\hat{a}[/math] being the destruction operator. [math]\hat{H}[/math] is the Hamiltonian operator. The hat accent over each symbol identifies an operator. This formula is of key note as it is applicable to particle creation and annihilation. [math]\hbar[/math] is the Planck constant (also referred to as a quanta of action) more detail later.

 

Heisenberg Uncertainty principle

[math]\Delta\hat{x}\Delta\hat{p}\ge\frac{\hbar}{2}[/math]

 

[math]\hat{x}[/math] is the position operator, [math]\hat{p}[/math] is the momentum operator. Their is also uncertainty between energy and time given by

 

[math]\Delta E\Delta t\ge\frac{\hbar}{2}[/math] please note in the non relativistic regime time is a parameter not an operator.

 

Physical observable's are operators. in order to be a physical observable you require a minima of a quanta of action defined by

 

[math] E=\hbar w[/math]

 

Another key detail from QM is the commutation relations

 

[math][\hat{x}\hat{p}]=\hat{x}\hat{p}-\hat{p}\hat{x}=i\hbar[/math]

 

Now in QM we are taught that the symbols [math]\varphi,\psi[/math] are wave-functions however in QFT we use these symbols to denote fields. Fields can create and destroy particles. As such we effectively upgrade these fields to the status of operators. Which must satisfy the commutation relations

 

[math][\hat{x}\hat{p}]\rightarrow[\hat{\psi}(x,t),\hat{\pi}(y,t)]=i\hbar\delta(x-y)[/math]

[math]\hat{\pi}(y,t)[/math] is another type of field that plays the role of momentum

 

where x and y are two points in space. The above introduces the notion of causality. If two fields are spatially separated they cannot affect one another.

 

Now with fields promoted to operators one wiill wonder what happen to the normal operators of QM. In QM position [math]\hat{x}[/math] is an operator with time as a parameter. However in QFT we demote position to a parameter. Momentum remains an operator.

 

In QFT we often use lessons from classical mechanics to deal with fields in particular the Langrangian

 

[math]L=T-V[/math]

 

The Langrangian is important as it leaves the symmetries such as rotation invariant (same for all observers). The classical path taken by a particle is one that minimizes the action

 

[math]S=\int Ldt[/math]

 

the range of a force is dictated by the mass of the guage boson (force mediator)

[math]\Delta E=mc^2[/math] along with the uncertainty principle to determine how long the particle can exist

[math]\Delta t=\frac{\hbar}{\Delta E}=\frac{\hbar}{m_oc^2}[/math] please note we are using the rest mass (invariant mass) with c being the speed limit

 

[math] velocity=\frac{distance}{time}\Rightarrow\Delta{x}=c\Delta t=\frac{c\hbar}{mc^2}=\frac{\hbar}{mc^2}[/math]

 

from this relation one can see that if the invariant mass (rest mass) m=0 the range of the particle is infinite. Prime example gauge photons for the electromagnetic force.

 

Lets return to [math]L=T-V[/math] where T is the kinetic energy of the particle moving though a potential V using just one dimension x. In the Euler-Langrange we get the following

 

[math]\frac{d}{dt}\frac{\partial L}{\partial\dot{x}}-\frac{\partial L}{\partial x}=0[/math] the dot is differentiating time.

 

Consider a particle of mass m with kinetic energy [math]T=\frac{1}{2}m\dot{x}^2[/math] traveling in one dimension x through potential [math]V(x)[/math]

 

Step 1) Begin by writing down the Langrangian

 

[math]L=\frac{1}{2}m\dot{x}^2-V{x}[/math]

 

next is a derivative of L with respect to [math]\dot{x}[/math] we treat this as an independent variable for example [math]\frac{\partial}{\partial\dot{x}}(\dot{x})^2=2\dot{x}[/math] and [math]\frac{\partial}{\partial\dot{x}}V{x}=0[/math] applying this we get

 

step 2)

[math]\frac{\partial L}{\partial\dot{x}}=\frac{\partial}{\partial\dot{x}}[\frac{1}{2}m\dot{x}^2]=m\dot{x}[/math]

 

which is just mass times velocity. (momentum term)

 

step 3) derive the time derivative of this momentum term.

 

[math]\frac{d}{dt}\frac{\partial L}{\partial\dot{x}}=\frac{d}{dt}m\dot{x}=\dot{m}\dot{x}+m\ddot{x}=m\ddot{x}[/math] we have mass times acceleration

 

Step 4) Now differentiate L with respect to x

 

[math]\frac{\partial L}{\partial x}[\frac{1}{2}m\dot{x}^2]-V(x)=-\frac{\partial V}{\partial x}[/math]

 

Step 5) write the equation to describe the dynamical behavior of our system.

 

[math]\frac{d}{dt}(\frac{\partial L}{\partial\dot{x}}-\frac{\partial L}{\partial x}=0[/math][math]\Rightarrow\frac{d}{dt}[/math][math](\frac{\partial L}{\partial\dot{x}})[/math][math]=\frac{\partial L}{\partial x}\Rightarrow m\ddot{x}=-\frac{\partial V}{\partial x}[/math]

 

recall from classical physics [math]F=-\nabla V[/math] in 1 dimension this becomes [math]F=-\frac{\partial V}{\partial x}[/math] therefore [math]\frac{\partial L}{\partial x}=-\frac{\partial V}{\partial x}=F[/math] we have [math]m\ddot{x}-\frac{\partial V}{\partial x}=F[/math]

Now keep in mind this is just a beginning to understand QFT. However the advantage is your already employing the four momentum etc...

I still have to go through your papers. However one can employ the Langrangian to describe all particles and particle interaction and give the approximate number density etc including the Higgs field etc with the corresponding coupling constants while maintaining Lorentz invariance for different observers etc.

Just in case your not familiar with time derivatives (the overdots)

https://en.m.wikipedia.org/wiki/Time_derivative

 

This paper is rather advance but it will give you a good idea of how the Langrangian is used to describe the standard model.

https://www-d0.fnal.gov/results/publications_talks/thesis/nguyen/thesis.pdf

It should give you a good idea of what you are up against...

Edited April 28, 2020 by Mordred

[studiot]

Gsoh, Mordred, you've outdone yourself there   -    must be a +1

47 minutes ago, Mordred said:

Now in QM we are taught that the symbols φ,ψ are wave-functions however in QFT we use these symbols to denote fields. Fields can create and destroy particles. As such we effectively upgrade these fields to the status of operators. Which must satisfy the commutation relations

Interestingly the only place I have seen Fields (as opposed to simpler things) tested against the postulates of Relativity is in Wangsness ("Topics in theoretical Physics" )

[taeto]

It is a beautiful intro to QFT, very much appreciated.

Just please correct the spelling to Klein-Gordon (or Klein-Fock-Gordon for completeness). Klein is a very common name in German speaking countries, for some reason (klein = small).

[Mordred]

3 hours ago, taeto said:

It is a beautiful intro to QFT, very much appreciated.

Just please correct the spelling to Klein-Gordon (or Klein-Fock-Gordon for completeness). Klein is a very common name in German speaking countries, for some reason (klein = small).

I often tend to get that wrong lol I will strive to watch for it in the future good catch

7 hours ago, studiot said:

 

Interestingly the only place I have seen Fields (as opposed to simpler things) tested against the postulates of Relativity is in Wangsness ("Topics in theoretical Physics" )

Well in QFT all particles are field excitations. So how one defines a particle varies a bit in the the operators between QM ( position and momentum) to what is described above which is the first part above.

Scalar fields can also often use the symbol [math]\phi [/math] a good example is the Freidman equations of state for scalar fields in cosmology.

https://en.m.wikipedia.org/wiki/Equation_of_state_(cosmology)

If you look at the scalar field example you will note it also employs the potential and kinetic energy of a field. This is often employed to model the HIGGS field, the cosmological constant and inflation with variations on that equation in the link.

 If the OP likes I can offer to work up an example of a photon field to get a better feel of how the creation and annihilation operators will give rise to photon particle number density. (It's one of the simpler examples as bosons are symmetric) ie Pauli exclusion principle. As well as the photon being its own antiparticle which adds further simplifications.

Just for information on the [math]\psi[/math] this is commonly the Four component Dirac field where [math]\psi_L,\psi_R[/math] are two component left and handed spinor fields in particle physics.

Edited April 29, 2020 by Mordred

[Mordred]

Having looked over your article in more detail. I hate to point this out but if you studied nucleon Spectrography which led to the discovery of quarks. You would recognize you have energy levels not covered by 0, 1/2,1 particles see electric charge of the quark family and learn the isospin correspondence. This will correspond to the cross sections which is extremely important in particle physics and cannot be ignored. You will also discover a symbol conflict (one of many)

[math]\Gamma [/math] width of excited state/decay rate.

 

This statement in section 8 conflicts with GR.

Quote

All particles move along the t axis at the same rate, all of them being functions of the same thing. And so the only way to increase the amount of time through which a particle is moving is by increasing the amount of space it also moves through.

Your theory is not Lorentz invariant interaction rates will vary depending on observer under relativity. Common example muon decay rates.

Now even though no elementary particle has spins other than those mentioned you can have other spin values such as spin 3/2, spin 2, spin 5/2 etc. I would strongly suggest you look into spin in greater detail and what spin entails in  the intrinsic magnetic moment

Ie [math]\mu_s=g_e\frac {-e}{2m_e}S [/math] where g_e is the gyro magnetic factor of the electron. You will find that for example the gyro magnetic factor of the proton will be different. Yet both particles are spin 1/2. Then to confuse matters further the neutron will also have a different gyroscope magnetic moment.

Can you identify why this is the case? Secondly can your theory explain why this is the case ?

Edited April 29, 2020 by Mordred

[John Henke]

As to the letters currently in use, none of them equate to any others currently in use with the exceptions of C and h bar. I hope that clarifies things. I will work on using more subscripts (I ran out of letters).

As to increasing spin, an oversight on my part: you can just multiply in any arbitrary number of omega components to increase it.

As to Lorentz invariant--upon further reflection--yes, you're right, it's not. I’ll admit this is an aspect of the theory that has posed a considerable challenge as getting definitions of w overbar and k overbar to agree with energy, momentum, velocity, relativity, time etc—this is difficult. However, I believe it is doable and I’m going to continue working on it.

Now to a more general issue that should address most of the other concerns. It’s never been my goal to rewrite all of physics, which is absurd. My goal is to replace a part of them that I argue is problematic, but you guys have clarified that I need to fit the theory more explicitly into the existing framework. A simple solution seems to be, generally, to just argue that V overbar (my version of the wavefunction) or one of its eigenfunctions replaces every wavefunction or eigenfunction you would see in any equation in physics. I’m not sure, but I don’t think V overbar would still have w overbar and k overbar values that vary over t, but rather this work would be done by linear algebra. Thoughts? And similar is true of the V with an overbar and underbar which is a particle with boundary conditions and involves adding a matching complex conjugate of a lowercase gamma component to the existing one. At any rate, this replacement should work because the graphs of the two sets of equations, I believe, are identical. Then I’m not arguing anything has changed beyond the substitution. For instance, this substitution should have no impact on the energy states of hydrogen, isospin or a gyroscope magnetic moment at least not after n is calculated so that the units can be converted into SI. And I will continue to argue for an equation of a particle, but that’s more or less a separable theory that is not glued to the first.

The obvious question then is why this new wavefunction is preferable. First and foremost, it is far more streamlined. Now gamma components and omega components form the wavefunction, the x eigenfunction, the momentum and energy eigenfunctions (and perhaps even a particle). This is arguably more “natural” and philosophically viable.

Also, this model is less arbitrary. This gives a more specific description of spin and chirality, as it’s naturally occurring, this in contrast to the more arbitrary equations in use. Furthermore, measurements of time and length would be fundamental as opposed to the arbitrary units currently used.

And also gravity might be explained, possibly dark matter and energy as well, probably as curvatures in the x eigenfunction. These would serve as the preeminent tests of the theory—tests, mind you, that physics as a whole has not yet passed.

And Mordred that’s an impressive balance between brevity and clarity you gave in that overview. And thank you so much studiot and Mordred for taking the time to read the article and for your keen insights. They’ve inspired a lot of thought.

[studiot]

42 minutes ago, John Henke said:

As to the letters currently in use, none of them equate to any others currently in use with the exceptions of C and h bar. I hope that clarifies things. I will work on using more subscripts (I ran out of letters).

As a matter of interest, Professors Mott and Sneddon got there before you.

Swansont asked you to use your equations to derive the energy levels of the hydrogen atom.

Here is Dirac's solution.

In particular see the notation introduced in (99) in the middle extract.

[John Henke]

Oh, I'm well aware that's already been calculated.

What I'm essentially arguing is that you would just replace my wavefunction or its eigenfunctions into the appropriate locations, and then I too would have calculated the orbitals.

[swansont]

56 minutes ago, John Henke said:

 The obvious question then is why this new wavefunction is preferable. First and foremost, it is far more streamlined.

You've got perhaps a dozen variables (which you have not described in any detail). "Streamlined" is not the word that comes to mind.

56 minutes ago, John Henke said:

Also, this model is less arbitrary. This gives a more specific description of spin and chirality, as it’s naturally occurring, this in contrast to the more arbitrary equations in use. Furthermore, measurements of time and length would be fundamental as opposed to the arbitrary units currently used.

How so? What are "fundamental" measurements of time and length?

 

56 minutes ago, John Henke said:

And also gravity might be explained, possibly dark matter and energy as well, probably as curvatures in the x eigenfunction. These would serve as the preeminent tests of the theory—tests, mind you, that physics as a whole has not yet passed.

You are getting waaaay ahead of yourself. How about passing the introductory tests first.

 

 

[John Henke]

30 minutes ago, swansont said:

1 hour ago, John Henke said:

 The obvious question then is why this new wavefunction is preferable. First and foremost, it is far more streamlined.

You've got perhaps a dozen variables (which you have not described in any detail). "Streamlined" is not the word that comes to mind.

I would argue starting out with something simpler leads to less opportunity at higher levels, for instance the lack of curvature.

30 minutes ago, swansont said:

Quote

Also, this model is less arbitrary. This gives a more specific description of spin and chirality, as it’s naturally occurring, this in contrast to the more arbitrary equations in use. Furthermore, measurements of time and length would be fundamental as opposed to the arbitrary units currently used.

How so? What are "fundamental" measurements of time and length?

In this model, n established a length would have a non arbitrary value assigned to it, that it would be given by nature and by no human choice or decision. For example, let's say n is a 1x1^1000, then that's going to determine a very particular value after t grows by 1. It will be completely non arbitrary and fundamental.

 

30 minutes ago, swansont said:

Quote

And also gravity might be explained, possibly dark matter and energy as well, probably as curvatures in the x eigenfunction. These would serve as the preeminent tests of the theory—tests, mind you, that physics as a whole has not yet passed.

You are getting waaaay ahead of yourself. How about passing the introductory tests first.

Of course I'm getting ahead of myself. That's what a test is. If I had the answer already then it wouldn't be a test. There are many tests, as I've mentioned, I already have the answer to, for instance, all the tests you've given me: you substitute my wavefunction or my eigenfunctions in for the wavefunctions or eigenfunctions used in the text studiot posted, and that gets the equation for the energy orbitals of hydrogen. If you take a V bar and add to its lowercase gamma component the complex conjugate of itself, then that's a 1D particle in a well as long as its a solution to Schrodinger's.

Edited April 29, 2020 by John Henke

[swansont]

1 hour ago, John Henke said:

I would argue starting out with something simpler leads to less opportunity at higher levels, for instance the lack of curvature.

In this model, n established a length would have a non arbitrary value assigned to it, that it would be given by nature and by no human choice or decision. For example, let's say n is a 1x1^1000, then that's going to determine a very particular value after t grows by 1. It will be completely non arbitrary and fundamental.

Well, then, propose a measurement to give us a value, dictated by nature, and without any arbitrary human units. Seems to me this should already be part of any proposal.

 

1 hour ago, John Henke said:

Of course I'm getting ahead of myself. That's what a test is. If I had the answer already then it wouldn't be a test.

Sure it would. Just because you've passed a test doesn't mean it wasn't a test.

Getting a head of yourself is when you claim you will be a world champion marathon runner when you haven't ever run (much less won) a 5 km race.

 

1 hour ago, John Henke said:

There are many tests, as I've mentioned, I already have the answer to, for instance, all the tests you've given me: you substitute my wavefunction or my eigenfunctions in for the wavefunctions or eigenfunctions used in the text studiot posted, and that gets the equation for the energy orbitals of hydrogen. If you take a V bar and add to its lowercase gamma component the complex conjugate of itself, then that's a 1D particle in a well as long as its a solution to Schrodinger's.

I don't know what V bar and gamma are, and if it's so straightforward, let's see the solution. You're the expert here, and this is your proposal. Show us how it's done. Show us something, anything, where we can compare your results to known QM results.

[Mordred]

Consider this challenge you have only two discrete values for the spin of an electron.

How do you produce two discrete values from a sinusoidal wave ? How do you determine the precise points on that sinusoidal wave that denotes spin 1/2 up or down ? 

You will need to answer this with the Stern Gerlach experiment in mind.

https://en.m.wikipedia.org/wiki/Stern–Gerlach_experiment

Edited April 29, 2020 by Mordred

[John Henke]

Ah I did not realize that was the case Mordred, thank you. There is a solution. You multiply by a lowercase gamma component divided by its corresponding uppercase version. 

And Swansont, I am currently away from my computer but I will do some more illustrative work on Mathematica later

[Mordred]

How does that equate to deflection from a magnet ? Ie Stern Gerlach apparatus ? If spin is determined by the classical terms your using how does that make sense ?

These questions come back to the magnetic moments involved in spin. Specifically the possible orientations of the magnetic moment due to spin. (I seriously hope you do not have the image of a spinning particle like a spinning top) when you look at the magnetic moments and the angular momentum terms for a point like particle you will find that would not make sense.

 

Edited April 29, 2020 by Mordred

[John Henke]

So are you saying angular momentum is quantized? Because multiplying by a lowercase gamma component divided by its corresponding uppercase version could quantize angular momentum as it incrementally increases the rate of rotation analagous to how multiplying by multiple gamma components with spin will increase the spin.

To be clear your multiplying a V bar, a wavefunction, by that. It is going to increase the angular momentum by increments of 1/2 with each multiplication of a lowercase gamma component divided by its uppercase version where all the components in the wavefunction have the same values for omega overbar and k overbar.

Edited April 29, 2020 by John Henke

[John Henke]

Correction: multiplying by omega components (not gamma components) increases spin by increments of 1 or 1/2 or 0 each depending on the component.

1 hour ago, Mordred said:

(I seriously hope you do not have the image of a spinning particle like a spinning top

No of course not. Look at the equations. Each point in space time is given a phase. Perhaps youve misread the graphs. The blue part is the real part and the gold the imaginary. The only thing about the graphs that has changed is the definition that the x eigenvalue of a waveparticle (this equation occuring in all of my equations) determines the position--and this is always by definition the case even when the waveparticle (denoted lambda in the article) is divided by other components such that it doesn't look like the x eigenvalue is growing over space. Given this definition of position, the graphs of the quantum wavefunction and my wavefunction should look the same provided they are both used as solutions to Schrodinger's, Dirac's, etc. Furthermore, although this gets more into my argument for the equation for a particle--these x eigenfunctions, these waveparticles, are best thought of as x axis over time (when in a single dimension). That's what I argue an x eigenfunction represents, and that's what I argue a particle essentially is, a spin zero x eigenfunction.

Edited April 29, 2020 by John Henke

[Mordred]

1 hour ago, John Henke said:

So are you saying angular momentum is quantized? 

You cannot treat particle spin strictly by its angular momentum term.

You also have it's magnetic moment. That is what you are missing and why you believe you can treat particle spin in classical terms which is wrong.

 

[John Henke]

Moreover multiple V overbars can multiply to get this into multiple spacial dimensions and time. I will continue working on my article to make this all clearer.

[Mordred]

We cross posted see above

[John Henke]

15 minutes ago, Mordred said:

You cannot treat particle spin strictly by its angular momentum term.

So the wavefunction has omega and k and my equations have omega overbar and k overbar which are basically the same except that they continue to apply when there is no spin or angular momentum. So I don't see what the wavefunction has that my equation doesn't that's allowing it to have a magnetic moment if mine doesn't.

Again, my new stance is that you're simply substituting my equation and its eigenfunctions into Schrodinger's replacing the wavefunction or its eigenfunctions, and as far as I can tell, my equations are capable of all behaviors seen in the wavefunction. But if you believe it's not I would be very curious as to why.

Edited April 29, 2020 by John Henke

[Mordred]

Nature. We detect and can measure the magnetic moment of all standard particles including the neutron. Even though it is charge neutral. You must be able to account for it in order to properly describe spin.

Ok let's take this further if you rotate a spin 1/2 particle by 360 degrees that particle will not return to its original state. You require a 720 degree rotation.

Yet a spin 1 particle only requires a 360 degree rotation 

https://en.m.wikipedia.org/wiki/Spin_(physics)

See here for reference.

Now earlier I stated three different 1/2 spin particles that will have three different gyroscopic magnetic moments.

 

 

Edited April 29, 2020 by Mordred

[John Henke]

So does this mean that the wavefunction is not a viable option either? I'm not sure how it differs such that it can express this rather abstract idea. But this is really interesting. Anyway, I will look into this further. Very interesting.

Okay, so one option, and this is admittedly improvised conjecture, is there are two wavefunctions, one for the positive x axis and one for the negative. Then you can use different forms of the wavefunction one with the positive and one with the negative curvatures I've discussed and that might serve as the dipole moments.

And actually this reminds me of another way the theory can be tested which is that almost everything is approximate. This is not so true with the wavefunction because it uses a perfect n that approaches infinity. However, with my numbers you'll find that everything is approximate, aside from certain things like Planck's constant or the speed of light.

And that model of dipole moments might fit nicely with matrices because there are four potential forms that particles with spin can take and that might help define four rows in a matrix. And again, two of the four have positive curvature and two of the four negative.

Edited April 29, 2020 by John Henke

[swansont]

Posting an example of applying your approach to some basic experiment would be illuminating, instead of describing in terms of names of variables (of which there are so many). We can speak of momentum (linear and angular) and energy, but "v overbar" and "omega" have no meaning.

2 hours ago, John Henke said:

So the wavefunction has omega and k and my equations have omega overbar and k overbar which are basically the same except that they continue to apply when there is no spin or angular momentum. So I don't see what the wavefunction has that my equation doesn't that's allowing it to have a magnetic moment if mine doesn't.

Actual solutions to physics problems is one obvious difference. Thus far, at least.

[Mordred]

Ask yourself the following questions.

1) do you need a wavefunction to describe two spin states ie spin 1/2 up or 1/2 spin down ?

2) what is the probability nature of wavefunctions in QM ? This is where you must look into. Many of QM's wavefunctions are probability functions that is described as a wavefunction. This is different from physical waveforms.

https://en.m.wikipedia.org/wiki/Wave_function

Note the part of QM stating it is a complex variable probability function

Edited April 29, 2020 by Mordred