How Big Is a Photon?

[John Clark]

I recently realized I can't give a good answer to the seemingly simple question "how big is a photon?" so I did some reading on the subject and the short answer I can now come up with is "that depends". The slightly longer answer is it depends on how the photon is produced and what you mean by “size.” In one sense photons are elementary particles with no evidence of internal structure: scattering experiments show they have no parts. When a photon strikes a photographic plate it produces a localized point, not a smear, and there is no sign of matter interacting with one portion of a photon before another. Photons are also massless and electrically neutral, so they do not have a rigid physical boundary that could be measured like the surface of a ball.

But on the other hand red light has a wavelength of about 650 nanometers, so does that mean a red photon is 650 nanometers across? No, because Heisenberg's Uncertainty Principle must be taken into account. Wavelength tells us the spacing of the crests of the underlying electromagnetic wave, not how long a single photon is in space. The actual spatial extent of a photon depends on its coherence length,  that is to say the distance over which its wave-like oscillations remain well defined. And coherence length = c × coherence time = c/(2πΔf). 

So coherence length is tied to Heisenberg’s Uncertainty Principle: ΔE × Δt ≳ ℏ/2. A photon that is emitted over a short time interval (Δt is small) has a large uncertainty in energy (ΔE is large), which translates into a broad frequency spread and therefore a short coherence length. But if the emission lasts a long time then the frequency uncertainty is small and the photon can stretch out over a very long distance. The way the photon is produced can make a big difference. 

A red photon from even a cheap laser diode has an extremely narrow frequency spread giving it a very long coherence length, several meters, and for high-quality lasers it can be several miles. By contrast, a red photon emitted from blackbody radiation, like the glow from an incandescent bulb, has a broad frequency spread, and therefore a very short coherence length, about the width of a human hair. In both cases the photons are indivisible quanta, but their wave packets have very different spatial extents.

So the “size” of a photon cannot be pinned down to a single number. In interactions, it behaves as if it has no size at all, striking at a point. But in terms of its wave-like character, it can extend over hundreds of nanometers in wavelength and, depending on how it is produced, anywhere from a fraction of a micrometer to many miles in coherence length. A photon is therefore not like a little ball of light with a fixed diameter, it's more useful to be thought of as a quantum excitation of the electromagnetic field that travels like a wave but interacts with matter like a particle, and whose spatial extent depends on both the uncertainty principle and the circumstances of its creation.

 John K Clark    See what's on my new list at  Extropolis

bph

[Alan Grayson]

If you apply the LT, a photon's length is zero. AG 

[Brent Meeker]

Thanks for the exposition.

Brent

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[Alan Grayson]

On Tuesday, September 2, 2025 at 2:16:15 PM UTC-6 Brent Meeker wrote:

Thanks for the exposition.

Brent

But you, being the expert you are, already "knew" the answer. It had to do with the wf of a photon. But oddly, you haven't a clue what differential equation would have to be solved to obtain that wf. WTF. AG

[John Clark]

On Tue, Sep 2, 2025 at 6:05 PM Alan Grayson <agrays...@gmail.com> wrote:

On Tuesday, September 2, 2025 at 2:16:15 PM UTC-6 Brent Meeker wrote:

 

>>Thanks for the exposition.
Brent

>But you, being the expert you are, already "knew" the answer. It had to do with the wf of a photon. But oddly, you haven't a clue what differential equation would have to be solved to obtain that wf. WTF. AG

As seen below, the electromagnetic wave function is described by 4 differential equations discovered by James Maxwell about 10 years before Einstein was born :

Gauss's Law for Electricity: ∇ ⋅ E = ρ / ε₀
Gauss's Law for Magnetism: ∇ ⋅ B = 0
Faraday's Law of Induction: ∇ × E = -∂B/∂t
Ampère's Law  ∇ × B = μ₀J + μ₀ε₀∂E/∂t

To this day those equations are the bread and butter of electrical engineers, they don't include any of the quantum properties of light but they don't need to in order to explain the Doppler red shift. If you want to find the quantum wave function, and not just the electromagnetic wave function, and also include relativistic effects, then things become a little more complicated. 

In the two-slit experiment, if you quantize the electromagnetic field, the field amplitude takes on the role of a photon’s probability amplitude. That amplitude behaves like a wave, so it interferes, and the squared magnitude (|E|^2) gives the probability distribution for where a photon is likely to be detected on the screen. If you send photons one at a time, each detection looks random, but over many trials the hits build up exactly into the interference pattern predicted by |E|^2.

When you use a bright source, such as a laser, you no longer just have a single photon but a huge number of them, so |E|^2 doesn’t just describe a probability distribution, it gives the average number of photons arriving at each point. And as the number of photons increases classical optics approximates quantum mechanics more and more closely. 

John K Clark    See what's on my new list at  Extropolis

tvi

[Alan Grayson]

On Wednesday, September 3, 2025 at 6:05:55 AM UTC-6 John Clark wrote:

On Tue, Sep 2, 2025 at 6:05 PM Alan Grayson <agrays...@gmail.com> wrote:

On Tuesday, September 2, 2025 at 2:16:15 PM UTC-6 Brent Meeker wrote:

 

>>Thanks for the exposition.
Brent

>But you, being the expert you are, already "knew" the answer. It had to do with the wf of a photon. But oddly, you haven't a clue what differential equation would have to be solved to obtain that wf. WTF. AG

As seen below, the electromagnetic wave function is described by 4 differential equations discovered by James Maxwell about 10 years before Einstein was born :

Gauss's Law for Electricity: ∇ ⋅ E = ρ / ε₀
Gauss's Law for Magnetism: ∇ ⋅ B = 0
Faraday's Law of Induction: ∇ × E = -∂B/∂t
Ampère's Law  ∇ × B = μ₀J + μ₀ε₀∂E/∂t

To this day those equations are the bread and butter of electrical engineers, they don't include any of the quantum properties of light but they don't need to in order to explain the Doppler red shift. If you want to find the quantum wave function, and not just the electromagnetic wave function, and also include relativistic effects, then things become a little more complicated. 

In the two-slit experiment, if you quantize the electromagnetic field, the field amplitude takes on the role of a photon’s probability amplitude. That amplitude behaves like a wave, so it interferes, and the squared magnitude (|E|^2) gives the probability distribution for where a photon is likely to be detected on the screen.

How can the amplitude give a probability if its value is generally not less than unity? AG 

[Alan Grayson]

On Wednesday, September 3, 2025 at 8:05:59 AM UTC-6 Alan Grayson wrote:

On Wednesday, September 3, 2025 at 6:05:55 AM UTC-6 John Clark wrote:

On Tue, Sep 2, 2025 at 6:05 PM Alan Grayson <agrays...@gmail.com> wrote:

On Tuesday, September 2, 2025 at 2:16:15 PM UTC-6 Brent Meeker wrote:

 

>>Thanks for the exposition.
Brent

>But you, being the expert you are, already "knew" the answer. It had to do with the wf of a photon. But oddly, you haven't a clue what differential equation would have to be solved to obtain that wf. WTF. AG

As seen below, the electromagnetic wave function is described by 4 differential equations discovered by James Maxwell about 10 years before Einstein was born :

Gauss's Law for Electricity: ∇ ⋅ E = ρ / ε₀
Gauss's Law for Magnetism: ∇ ⋅ B = 0
Faraday's Law of Induction: ∇ × E = -∂B/∂t
Ampère's Law  ∇ × B = μ₀J + μ₀ε₀∂E/∂t

To this day those equations are the bread and butter of electrical engineers, they don't include any of the quantum properties of light but they don't need to in order to explain the Doppler red shift. If you want to find the quantum wave function, and not just the electromagnetic wave function, and also include relativistic effects, then things become a little more complicated. 

In the two-slit experiment, if you quantize the electromagnetic field, the field amplitude takes on the role of a photon’s probability amplitude. That amplitude behaves like a wave, so it interferes, and the squared magnitude (|E|^2) gives the probability distribution for where a photon is likely to be detected on the screen.

How can the amplitude give a probability if its value is generally not less than unity? AG 

Another problem with this is that the amplitude is defined for multiple photons, not for a single photon, whose wf I am seeking to acquire and understand. AG 

[John Clark]

On Wed, Sep 3, 2025 at 10:06 AM Alan Grayson <agrays...@gmail.com> wrote:

How can the amplitude give a probability if its value is generally not less than unity? AG 

|E|^2 gives the relative probabilities of detecting a photon at different points on the screen, it gives you the correct curve shape but is not normalized so all the probabilities don't add up to exactly one. To make it a true probability distribution that integrates to 1 you divide |E|^2 by a constant, the total integrated intensity across the screen.

John K Clark    See what's on my new list at  Extropolis

kcj

   

[Alan Grayson]

On Wednesday, September 3, 2025 at 11:42:44 AM UTC-6 John Clark wrote:

On Wed, Sep 3, 2025 at 10:06 AM Alan Grayson <agrays...@gmail.com> wrote:

How can the amplitude give a probability if its value is generally not less than unity? AG 

|E|^2 gives the relative probabilities of detecting a photon at different points on the screen, it gives you the correct curve shape but is not normalized so all the probabilities don't add up to exactly one. To make it a true probability distribution that integrates to 1 you divide |E|^2 by a constant, the total integrated intensity across the screen.

But this E is the classical energy, and when the EM field is quantized don't we need a huge number of photons to define the curve you reference above? Is there a differential equation that when solved, gives us the wf of a RELATIVISTIC particle? Is there an equation that gives the wf of a photon? TY, AG 

[John Clark]

On Wed, Sep 3, 2025 at 6:53 PM Alan Grayson <agrays...@gmail.com> wrote:

>> |E|^2 gives the relative probabilities of detecting a photon at different points on the screen, it gives you the correct curve shape but is not normalized so all the probabilities don't add up to exactly one. To make it a true probability distribution that integrates to 1 you divide |E|^2 by a constant, the total integrated intensity across the screen.

> But this E is the classical energy, and when the EM field is quantized don't we need a huge number of photons to define the curve you reference above?

If you are able to calculate the probability distribution for where a single photon is likely to deliver its quantum of energy after passing through two slits, then you will be able to calculate how the energy will be distributed across the entire screen for any finite number of photons that passes through those two slits.

 

> Is there a differential equation that when solved, gives us the wf of a RELATIVISTIC particle?

If you start with Maxwell's Equations then you get that automatically because, although they were discovered a decade before Einstein was born, they are 100% compatible with both Special and General Relativity. Interestingly they are NOT compatible with Newtonian physics because Newton claimed that all speed was relative, but Maxwell's Equations can produce an absolute speed, the speed of light, and there are no specifications about who should observe that speed, so everybody must.  Resolving that inconsistency was the motivation Einstein had for developing relativity; he made no changes to Maxwell's Equations, he thought they were fine just as they are, instead he modified Newtonian physics.  

> Is there an equation that gives the wf of a photon? TY, AG 

As I explained before, the electromagnetic field amplitude can take on the role of a photon’s probability amplitude. 

John K Clark    See what's on my new list at  Extropolis

ijt

[Alan Grayson]

On Thursday, September 4, 2025 at 4:38:50 AM UTC-6 John Clark wrote:

On Wed, Sep 3, 2025 at 6:53 PM Alan Grayson <agrays...@gmail.com> wrote:

>> |E|^2 gives the relative probabilities of detecting a photon at different points on the screen, it gives you the correct curve shape but is not normalized so all the probabilities don't add up to exactly one. To make it a true probability distribution that integrates to 1 you divide |E|^2 by a constant, the total integrated intensity across the screen.

> But this E is the classical energy, and when the EM field is quantized don't we need a huge number of photons to define the curve you reference above?

If you are able to calculate the probability distribution for where a single photon is likely to deliver its quantum of energy after passing through two slits, then you will be able to calculate how the energy will be distributed across the entire screen for any finite number of photons that passes through those two slits.

 

> Is there a differential equation that when solved, gives us the wf of a RELATIVISTIC particle?

If you start with Maxwell's Equations then you get that automatically because, although they were discovered a decade before Einstein was born, they are 100% compatible with both Special and General Relativity. Interestingly they are NOT compatible with Newtonian physics because Newton claimed that all speed was relative, but Maxwell's Equations can produce an absolute speed, the speed of light, and there are no specifications about who should observe that speed, so everybody must.  Resolving that inconsistency was the motivation Einstein had for developing relativity; he made no changes to Maxwell's Equations, he thought they were fine just as they are, instead he modified Newtonian physics.  

But ME's describe the behavior of the EM field, not the relativistic behavior of a particle. So I pose my question above again. TY, AG 

> Is there an equation that gives the wf of a photon? TY, AG 

As I explained before, the electromagnetic field amplitude can take on the role of a photon’s probability amplitude. 

But this is within the context of the double slit experiment. If there is a wf for a photon, it should be, I think, independent of any experimental set up. AG 

[Brent Meeker]

In a context in which the photon number isn't changing, the energy wave derived from Maxwell's equation is the wf for a photon.

Brent

[Brent Meeker]

On 9/4/2025 6:24 AM, Alan Grayson wrote:

On Thursday, September 4, 2025 at 4:38:50 AM UTC-6 John Clark wrote:

On Wed, Sep 3, 2025 at 6:53 PM Alan Grayson <agrays...@gmail.com> wrote:

>> |E|^2 gives the relative probabilities of detecting a photon at different points on the screen, it gives you the correct curve shape but is not normalized so all the probabilities don't add up to exactly one. To make it a true probability distribution that integrates to 1 you divide |E|^2 by a constant, the total integrated intensity across the screen.

> But this E is the classical energy, and when the EM field is quantized don't we need a huge number of photons to define the curve you reference above?

If you are able to calculate the probability distribution for where a single photon is likely to deliver its quantum of energy after passing through two slits, then you will be able to calculate how the energy will be distributed across the entire screen for any finite number of photons that passes through those two slits.

 

> Is there a differential equation that when solved, gives us the wf of a RELATIVISTIC particle?

If you start with Maxwell's Equations then you get that automatically because, although they were discovered a decade before Einstein was born, they are 100% compatible with both Special and General Relativity. Interestingly they are NOT compatible with Newtonian physics because Newton claimed that all speed was relative, but Maxwell's Equations can produce an absolute speed, the speed of light, and there are no specifications about who should observe that speed, so everybody must.  Resolving that inconsistency was the motivation Einstein had for developing relativity; he made no changes to Maxwell's Equations, he thought they were fine just as they are, instead he modified Newtonian physics.  

But ME's describe the behavior of the EM field, not the relativistic behavior of a particle. So I pose my question above again. TY, AG 

They also describe the energy propagation of the EM field.  And given the known energy per photon that implies the probability distribution of the photons.

> Is there an equation that gives the wf of a photon? TY, AG 

As I explained before, the electromagnetic field amplitude can take on the role of a photon’s probability amplitude. 

But this is within the context of the double slit experiment. If there is a wf for a photon, it should be, I think, independent of any experimental set up. AG 

No, it is not only in the context of the double slit experiment.  The context of the double slit experiment is supplied by initial conditions.

Brent

John K Clark    See what's on my new list at  Extropolis

ijt

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[Alan Grayson]

On Thursday, September 4, 2025 at 6:54:57 PM UTC-6 Brent Meeker wrote:

On 9/4/2025 6:24 AM, Alan Grayson wrote:

On Thursday, September 4, 2025 at 4:38:50 AM UTC-6 John Clark wrote:

On Wed, Sep 3, 2025 at 6:53 PM Alan Grayson <agrays...@gmail.com> wrote:

>> |E|^2 gives the relative probabilities of detecting a photon at different points on the screen, it gives you the correct curve shape but is not normalized so all the probabilities don't add up to exactly one. To make it a true probability distribution that integrates to 1 you divide |E|^2 by a constant, the total integrated intensity across the screen.

> But this E is the classical energy, and when the EM field is quantized don't we need a huge number of photons to define the curve you reference above?

If you are able to calculate the probability distribution for where a single photon is likely to deliver its quantum of energy after passing through two slits, then you will be able to calculate how the energy will be distributed across the entire screen for any finite number of photons that passes through those two slits.

 

> Is there a differential equation that when solved, gives us the wf of a RELATIVISTIC particle?

If you start with Maxwell's Equations then you get that automatically because, although they were discovered a decade before Einstein was born, they are 100% compatible with both Special and General Relativity. Interestingly they are NOT compatible with Newtonian physics because Newton claimed that all speed was relative, but Maxwell's Equations can produce an absolute speed, the speed of light, and there are no specifications about who should observe that speed, so everybody must.  Resolving that inconsistency was the motivation Einstein had for developing relativity; he made no changes to Maxwell's Equations, he thought they were fine just as they are, instead he modified Newtonian physics.  

But ME's describe the behavior of the EM field, not the relativistic behavior of a particle. So I pose my question above again. TY, AG 

They also describe the energy propagation of the EM field.  And given the known energy per photon that implies the probability distribution of the photons.

> Is there an equation that gives the wf of a photon? TY, AG 

As I explained before, the electromagnetic field amplitude can take on the role of a photon’s probability amplitude. 

But this is within the context of the double slit experiment. If there is a wf for a photon, it should be, I think, independent of any experimental set up. AG 

No, it is not only in the context of the double slit experiment.  The context of the double slit experiment is supplied by initial conditions.

Brent

What initial conditions could yield a similar result that we get for a free non-relativistic particle; namely, that the wf spreads over time so its position probability becomes progressively uncertain?  So No, I don't agree that applying the double slit will yield a wf for a photon which supports a similar result as in the case just mentioned. And I am beginning to suspect that we have no equation, the solution of which gives us the wf of a RELATIVISTIC particle, or a PHOTON. If such equations exist, plesae post them. TY, AG

[Brent Meeker]

The initial condition that the wave function is spherically symmetric.

So No, I don't agree that applying the double slit will yield a wf 

The double slit doesn't "yield a wf".  It implies an initial condition

for a photon which supports a similar result as in the case just mentioned. And I am beginning to suspect that we have no equation, the solution of which gives us the wf of a RELATIVISTIC particle, or a PHOTON. If such equations exist, plesae post them. TY, AG

You want me to waste 15m typing in a lot of display math just so you can see a bunch of equations?  Go look'em up.

Brent

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[Alan Grayson]

I tried that, but nothing came up. If they have names, please name the names. They must have names. AG 

[Alan Grayson]

Ever hear of COPY AND PASTE? AG 

[Alan Grayson]

What are the special initial conditions of the double slit which results in a wf which spreads for free particles? AG 

[John Clark]

On Thu, Sep 4, 2025 at 11:10 PM Alan Grayson <agrays...@gmail.com> wrote:

> I am beginning to suspect that we have no equation, the solution of which gives us the wf of a RELATIVISTIC particle, 

The Dirac Equation was discovered way back in 1928, and it describes the quantum wave function of a RELATIVISTIC electron.

(iγμ∂μ−m)ψ=0

ψ is a four-component wave function that represents the quantum state of a fermion

m is the mass of the particle

γμ is called the gamma matrix, it encode spin and relativistic effects

∂μ=(∂/∂t, ∂/∂x, ∂/∂y, ∂/∂z),  it's called the four-gradient and it represents spacetime derivatives 

And i of course is √-1

>or a PHOTON.

Maxwell’s Equations already are relativistic and they already describe the correct degrees of freedom. So quantization of the electromagnetic field automatically results in photons. So there’s no need for a separate photon wavefunction equation.

  John K Clark    See what's on my new list at  Extropolis

010

[Alan Grayson]

On Friday, September 5, 2025 at 4:18:23 AM UTC-6 John Clark wrote:

On Thu, Sep 4, 2025 at 11:10 PM Alan Grayson <agrays...@gmail.com> wrote:

> I am beginning to suspect that we have no equation, the solution of which gives us the wf of a RELATIVISTIC particle, 

The Dirac Equation was discovered way back in 1928, and it describes the quantum wave function of a RELATIVISTIC electron.

(iγμ∂μ−m)ψ=0

ψ is a four-component wave function that represents the quantum state of a fermion

m is the mass of the particle

γμ is called the gamma matrix, it encode spin and relativistic effects

∂μ=(∂/∂t, ∂/∂x, ∂/∂y, ∂/∂z),  it's called the four-gradient and it represents spacetime derivatives 

And i of course is √-1

>or a PHOTON.

I was aware of Dirac's equation and that it described the electron, but I didn't know that the variable function solved for, was the electron's wf. Are you sure its solution is the wf of a relativistic electron? I infer there is no equation for the PHOTON. AG

Maxwell’s Equations already are relativistic and they already describe the correct degrees of freedom. So quantization of the electromagnetic field automatically results in photons. So there’s no need for a separate photon wavefunction equation.

But any initial condition for a double slit experiment cannot result in the interference pattern for a free particle, where the probability spreads over time. Regardless of the initial condition, we always get an interference pattern on the screen, which although fixed on the screen and changing along the screen, doesn't yield a spreading probability. So I have no idea what Brent was referring to, when he suggested the initial condition would shed light on this issue. AG

[John Clark]

On Fri, Sep 5, 2025 at 6:45 AM Alan Grayson <agrays...@gmail.com> wrote:

> I was aware of Dirac's equation and that it described the electron, but I didn't know that the variable function solved for, was the electron's wf. Are you sure its solution is the wf of a relativistic electron? 

 

Yes. And for bringing Quantum Mechanics and Special Relativity into harmony is why Dirac received the Nobel Prize, and why he is considered to be one of the best physicists of the 20th century. 

 

> Regardless of the initial condition, we always get an interference pattern on the screen,

No we do not, not if one of the initial conditions is that there's a detector near one of the slits that enables us to know which slit the photon (or electron) went through. 

 John K Clark    See what's on my new list at  Extropolis

4r4

[Alan Grayson]

I was assuming we don't know which-way. But even if we do, there's no way to get the result for a free particle, with the probability spreading in time. I think we get a central peak and nothing time variant if we know which-way. AG

[John Clark]

On Fri, Sep 5, 2025 at 7:45 AM Alan Grayson <agrays...@gmail.com> wrote:

> there's no way to get the result for a free particle,

 A "free particle" is a particle that is not subject to ANY external force or have any potential energy, so its total energy is kinetic. That makes things simpler and as a result the quantum wave function is just a plane wave. But in the real world things are more complicated because "free particles" do not exist.   

 John K Clark    See what's on my new list at  Extropolis

tmc

[Alan Grayson]

On Friday, September 5, 2025 at 6:50:49 AM UTC-6 John Clark wrote:

On Fri, Sep 5, 2025 at 7:45 AM Alan Grayson <agrays...@gmail.com> wrote:

> there's no way to get the result for a free particle,

 A "free particle" is a particle that is not subject to ANY external force or have any potential energy, so its total energy is kinetic. That makes things simpler and as a result the quantum wave function is just a plane wave. But in the real world things are more complicated because "free particles" do not exist.   

But we can imagine a free particle in empty space and we do that often. So, IMO, there's no way to get a particle's wf from a double slit experiment, even a rough approximation. Consequently, ME's are not enough to get a particle's wf by observing distribution patterns on the screen. And to correct my previous statement, if we don't know which-way in double slit experiment, we get two overlapping peaks, centered on the screen, opposite the space between the slits. AG

[Alan Grayson]

CORRECTION: ... If we KNOW which-way in double slit experiment ... 

[Alan Grayson]

On Friday, September 5, 2025 at 5:19:00 AM UTC-6 John Clark wrote:

On Fri, Sep 5, 2025 at 6:45 AM Alan Grayson <agrays...@gmail.com> wrote:

> I was aware of Dirac's equation and that it described the electron, but I didn't know that the variable function solved for, was the electron's wf. Are you sure its solution is the wf of a relativistic electron? 

 

Yes. And for bringing Quantum Mechanics and Special Relativity into harmony is why Dirac received the Nobel Prize, and why he is considered to be one of the best physicists of the 20th century. 

Has the solution wf for the electron been solved? If so, by whom? Dirac? AG 

[John Clark]

On Fri, Sep 5, 2025 at 9:44 AM Alan Grayson <agrays...@gmail.com> wrote:

>>  A "free particle" is a particle that is not subject to ANY external force or have any potential energy, so its total energy is kinetic. That makes things simpler and as a result the quantum wave function is just a plane wave. But in the real world things are more complicated because "free particles" do not exist.   

> But we can imagine a free particle in empty space

If you're forcing a particle to go through two slips in a wall before it reaches a detector screen then it is not a free particle. 

>and we do that often. 

College professors often do that because it makes a problem much simpler, so simple he can give it as a homework problem and his students might actually have a fighting chance of actually being able to solve it.  But nobody does it because we often (or ever) see a "free particle" in nature. 

> So, IMO, there's no way to get a particle's wf from a double slit experiment,

Perhaps you should consider changing your opinion. In his book "The Feynman Lectures on Physics", Richard Feynman said "the two slit experiment contains all the mysteries in Quantum Mechanics, a phenomenon impossible to explain in classical terms". 

> Has the solution wf for the electron been solved? If so, by whom? Dirac?

As I've mentioned before, Dirac found a way for Quantum Mechanics and Special Relativity to live together in harmony, and he did it 97 years ago. To this day nobody has been able to do the same thing for General Relativity, it's the holy grail of physics. 

  John K Clark    See what's on my new list at  Extropolis

nws

[Alan Grayson]

On Friday, September 5, 2025 at 12:59:54 PM UTC-6 John Clark wrote:

On Fri, Sep 5, 2025 at 9:44 AM Alan Grayson <agrays...@gmail.com> wrote:

>>  A "free particle" is a particle that is not subject to ANY external force or have any potential energy, so its total energy is kinetic. That makes things simpler and as a result the quantum wave function is just a plane wave. But in the real world things are more complicated because "free particles" do not exist.   

> But we can imagine a free particle in empty space

If you're forcing a particle to go through two slips in a wall before it reaches a detector screen then it is not a free particle. 

Right, and that's why the double slit experiment cannot exactly reproduce the wf results for particles, inclusive of photons. AG 

>and we do that often. 

College professors often do that because it makes a problem much simpler, so simple he can give it as a homework problem and his students might actually have a fighting chance of actually being able to solve it.  But nobody does it because we often (or ever) see a "free particle" in nature. 

Einstein and other notables also do that, habitually. And that's the reason S's equation is used in teaching QM, because it is easily solved. I've never seen a solution to Dirac's equation. AG 

> So, IMO, there's no way to get a particle's wf from a double slit experiment,

Perhaps you should consider changing your opinion. In his book "The Feynman Lectures on Physics", Richard Feynman said "the two slit experiment contains all the mysteries in Quantum Mechanics, a phenomenon impossible to explain in classical terms". 

> Has the solution wf for the electron been solved? If so, by whom? Dirac?

As I've mentioned before, Dirac found a way for Quantum Mechanics and Special Relativity to live together in harmony, and he did it 97 years ago. To this day nobody has been able to do the same thing for General Relativity, it's the holy grail of physics. 

But did Dirac actually SOLVE his equation? What does the solution look like? This raises another important issue IMO. He needed a special equation for the electron, implying, unlike S's equation, that all particle wf's can be determined as solutions to a single, the same differential equation.  AG 

[John Clark]

On Fri, Sep 5, 2025 at 3:57 PM Alan Grayson <agrays...@gmail.com> wrote:

 > I've never seen a solution to Dirac's equation

Well now you have:  

Solving the Dirac Equation

  John K Clark    See what's on my new list at  Extropolis

kfs

n

[Alan Grayson]

On Friday, September 5, 2025 at 2:05:36 PM UTC-6 John Clark wrote:

On Fri, Sep 5, 2025 at 3:57 PM Alan Grayson <agrays...@gmail.com> wrote:

 > I've never seen a solution to Dirac's equation

Well now you have:  

Solving the Dirac Equation

TY. To the best of your information and knowledge, has anyone ever compared the wf results of the Dirac equation for NON-RELATIVISTIC motion, with the results from S's equation? AG

  John K Clark    See what's on my new list at  Extropoliss

n