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Offline QuantumClue

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What is the Dirac equation?
« on: 26/12/2010 22:27:54 »
This will be my first part. I will post the second part to the Dirac Equation later this week. Here we will learn many of the basics which quantum theory has to say about the Dirac Equation. Note I cannot make proper matrix algebra here, so I will need to skip many parts of it.

The Dirac Equation

Note (there are many parts of this ''lecture'' which cannot be represented here. Since matrix mathematics cannot be represented, we will unfortunately have to skip many parts, but the reader will still get a reasonable insight into the world of Diracs Equation).

We first begin with the psi-function which represents symbolically the wave function of quantum mechanics.

What is the wave function?

The wave function is a mathematical absraction represented by Ψ.

The wave function is a probability distribution over all space and all time, providing the possibilities of any physical system to appear anywhere at any time.

The abstraction itself is used frequently to make sense of particle distributions provided through wave interferences.

We will begin with a well-known function given as:

Ψ = ei(kx-ωt)

This function is one which has been given its exponential function e. k and w are in fact the wave number and theangular frequency, where the angular frequency can be thought of relativistic terms.

This function is very reliable for mathematical reasons, which we physicists tend to think we know much about. The equation which makes sense of our function is:

ω/k=Vp

where Vp is the phase velocity, which has an important meaning in physics, and in many senses, reconfigures our understanding of particle velocities which are luminal and sub-luminal.

Interestingly, the value ω/k can take values which are either +1 or -1. The meaning of the two becomes more valid under more research.

The equation of interest now comes in the form:

∂ΨR/∂ x + ∂ΨR/∂ t = 0

If we pull the derivative down in respect to x in our function Ψ = ei(kx-ωt) gives ik, and if we pull down our derivative in respect to t gives us iω, and if we take the two from each other we have:

iω - ik = 0

What does this mean?

Interestingly this makes k and ω as on equal footing:

k=ω

Now, understanding this, and understanding a graph which cannot be physically or visually shown here today, positive energy and negative energy have very interesting properties.

We become to realize that we have a concept known as right moving and left moving particles.

What are right moving and left moving particles

Physically, all we need to know is that it represents a physical system. What this physical system is will become clear eventually.

Right moving particles are expressed as:

∂ψR/∂ t = -∂ψR/∂x

which represent right moving particles (ω/k = +1).

Left moving particles are represented by:

∂ψL/ ∂t = +∂ψL /∂ x

These two physical quantities now have a specific meaning in our equations, which take the form of particles moving off to the left and particles moving off to the right.

They physically represent particles and antiparticles.

Represented as column vectors, we have:

φR
φL = ψ

Which is nothing but a convenient *(mathematical matrix) way to respresent our functions.

Squaring the value of ω gives us:

ω2=m2+k2

Which is a relativistic equation where k is the same mathematically as momentum p.



If there are any mistakes, I welcome corrections/
« Last Edit: 29/12/2010 18:57:55 by QuantumClue »


 

Offline QuantumClue

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What is the Dirac equation?
« Reply #1 on: 28/12/2010 17:21:05 »
I've started my second writeup. Personally I don't want to finish unless the above made sense. Do I need to define the variables more?
 

Offline yor_on

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What is the Dirac equation?
« Reply #2 on: 29/12/2010 18:37:08 »
You might expand on this one.
show how it is thought to be done
As you jumped directly from ω/k=Vp

"

∂ΨR/∂ x + ∂ΨR/∂ t = 0

If we pull the derivative down in respect to x in our function gives ik, and if we pull down our derivative in respect to t gives us iω, and if we take the two from each other we have:"


 

Offline QuantumClue

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What is the Dirac equation?
« Reply #3 on: 29/12/2010 18:56:17 »
You might expand on this one.
show how it is thought to be done
As you jumped directly from ω/k=Vp

"

∂ΨR/∂ x + ∂ΨR/∂ t = 0

If we pull the derivative down in respect to x in our function gives ik, and if we pull down our derivative in respect to t gives us iω, and if we take the two from each other we have:"




Oh right, you pull those derivatives from the function Ψ = ei(kx-ωt). The wave equation is a simple partial equation which for mathematical terms, is part of the understanding of the function. I will change this. Thanks.
 

Offline JP

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What is the Dirac equation?
« Reply #4 on: 30/12/2010 03:56:26 »
Nice post, QC, and nice idea for posting about interesting topics. 

Would you be ok with me splitting this post so that your discussion of the Dirac equation gets its own thread?  It might make it easier for interested parties to find it for discussion (I didn't realize this thread included a discussion of the Dirac equation until I got half way through it).
 

Offline QuantumClue

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What is the Dirac equation?
« Reply #5 on: 30/12/2010 19:41:02 »
Yes of course. I'm writing up the second part which is a bit more lengthly.
 

Offline QuantumClue

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What is the Dirac equation?
« Reply #6 on: 30/12/2010 22:46:33 »
The Dirac Sea

So we have managed to understand that we have left movers and right moving wave, and in order for Dirac to stop real particles loosing energy and simply falling back into the vacuum, he required what he called a sea of negative particles, both contained with an equal proportion of left movers and right movers so it is symmetric with respect to left and right so there is no longer such a huge momentum in one particular directionality. The spin states would cancel out particles quite simply, and whenever we create a particle from the vacuum, this leaves a hole - the hole is the antiparticle of the electron, the positron.

So last time we more or less ended with our standard matrix notation expressed in column vectors:

φR
φL = ψ

This kind of notation is very handy to say the least; if one was to calulate this, we would find a matrix we will introduce soon. Usually we would place a dot above the expressions φR and φL to express that we are taking the time derivative of the quantities.

Such an equation would be read as:

(∂/∂dt) ψ = α (∂ψ/∂x)

We have a newcomer, and his name is alpha α and he is a matrix. The importance of alpha will become clear as we proceed.

If we take the square root of the equation

ω2=m2+k2

we naturally have ω=√m2+k2

What we would like to do now, is rewrite this equation (∂/∂dt) ψ = α (∂ψ/∂x) in a more compact form. When you take a wave and hit it with ∂/∂dt it may be remembered that this pulls down -iω. We then look at ∂ψ/∂x and this pulls down a -ik. The equation now has the more simplified form of:

-iω=-α-ik

Cancelling out the minus signs and the i's we finish with:

ω=αk

The alpha here acts like an instruction to express k either positively or negatively. Because of our relativistic relationship ω=√m2+k2 we would like to consider a mass and a new matrix which is called beta:

ω=αk+mβ

Anticommutation in a Clifford Algebra

A requirement on our matrices is that whatever α and β are, they must satisfy that the square of ω is equal to m2+k2. So let's square it:

ω22k2+m2β2

Decomposing this equation, we find a usual notation:

ω2=(αk+mβ)(αk+mβ)

we are now going to expand our terms - we get:

(αk+mβ)(αk+mβ)=α2k2+m2β2+αβKm+βαkm

we can rewrite the superlfuous quantities αβKm+βαkm here, knowing that k is simply momentum and m is the mass as:

(αβ+βα)km

why are they superfluous? Remember what I said, the matrices require that omega squared is equal to k2+m2 so looking at our terms, we can see that the first lot of them ie. α2k2+m2β2 already satisfy the omega squared part. Because of this, (αβ+βα)km can be seen as an unecessery left over.

Whatever alpha is (we know its a matrix) but whatever it would have been to the imagination before this, we would know that through matrix calculations, α2=1. Likewise, whatever beta would be also requires β2=1. Because (αβ+βα)km has no place in our formula, this means that we need α and β to satisfy an algebraic property called anticommutation in that (αβ+βα)=0. This is actually a clifford algebra. With a little matrix calculation on the side, we must have alpha obide by the algebraic properties of

10
0-1 = α

so α2 is

10
01 = α2

β cannot be the same as α because if it where, αβ+βα would just be twice alpha squared, so that doesn't work. So:

01
10 = β

If you are any good at metrix algebra, you can check that by squaring it.

Once we know alpha and beta, we can explicitely rewrite everything to produce the famous Dirac Equation.

The Dirac Equation in a Finalized Form

After some mathematics expressed in matrix notation, one can arrive at the equation:

R'=-i∂xψR+mψL which are right moving waves
L'=+i∂xψL+mψR which are left moving waves

Where the dash A' of an object here just means taking the mathematical object by its time derivative, as an alternative to upper dot notation. And ∂x is standard partial derivative notation is just ∂/∂x. Because beta interchanges the sign of the wave from psi-left and psi-right, the equation:

iψ'=-iα∂xψ+mβ

becomes a coupled equation which just means our left movers and right movers become coupled and this is what a mass term does acting as a scale factor of β. Spin enters the Dirac equation when Pauli Matrices are taken into consideration.

The Dirac Equation explicitely describes fermions with an intrinsic spin, and if you wanted to persue an equation which is void of spin, the Klein-Gorden Equation would satisfy.

In a compact notation, the theory of spin would arise from two specific matrices:

0I
I0 = β

Where ''I'' is the unit matrix

σj0
j = α

Each entry here is a 2X2 matrix and σj is the presence of the pauli matrix where j=1,2,3.
« Last Edit: 31/12/2010 04:05:38 by QuantumClue »
 

Offline MissD

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What is the Dirac equation?
« Reply #7 on: 31/12/2010 00:55:23 »
Hi I'm so in awe of your conversation.  I have no maths or physics to use to discuss with you.  My layman's interest is:  1)  How can something come from nothing?
                          2)  If there is nothing, why do we think we are here?
So looking forward to your thoughts without equations.

Di xx
 

Offline QuantumClue

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What is the Dirac equation?
« Reply #8 on: 31/12/2010 01:03:59 »
Depends on what context question 1) comes under. If this is the overpopularized ''particle explanation'' of the casimir effect where we are said to be getting something from nothing the statement is not actually true. We don't get something from nothing, because that would imply that the vacuum is nothing when real matter is not present - the true representation of the vacuum is that there is in fact still something even when you remove all the matter and energy in that particular part of spacetime.

Unless question 1 is resorting to the before the big bang concept where nothing existed, and sporadically there appears a universe. To be honest, physics cannot answer this question, the laws of physics break down before we can even postulate pre-planckian events. On a personal interpretation, I feel there was not nothing per se - there was in fact everything already present to create the meaningful universe we have around us today. So this ''nothingness'' we speak of actually had every ingredient necessery for the birth of a universe.

Question 2 is like the philosophical idiom ''I think, therefore I am''. If you think you exist, then there is meaning, and therefore there is.
« Last Edit: 31/12/2010 01:19:08 by QuantumClue »
 

Offline QuantumClue

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What is the Dirac equation?
« Reply #9 on: 31/12/2010 04:07:53 »
I guess I could expand a little on how spin enters the equation. It's not too difficult, unless everyone is happy so far?
 

Offline Foolosophy

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What is the Dirac equation?
« Reply #10 on: 31/12/2010 06:34:59 »
What is the Dirac Equation??

 

Offline Geezer

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What is the Dirac equation?
« Reply #11 on: 31/12/2010 07:22:07 »
Yeah! Brilliant post! (Way beyond me of course)
 

Offline Foolosophy

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« Reply #12 on: 31/12/2010 13:37:46 »
Yeah! Brilliant post! (Way beyond me of course)

Mathematics is NOT a science - its a philosophical game of proofs based on delusional axioms
 

Offline lightarrow

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What is the Dirac equation?
« Reply #13 on: 31/12/2010 14:23:34 »
The wave function is a probability distribution over all space and all time, providing the possibilities of any physical system to appear anywhere at any time.

Not exactly. Ψ is not a probability distribution, |Ψ|2 is the probability distribution. Ψ(x,y,z,t) is a mathematical function which describe the "state" of the quantum system and which physical meaning is that |Ψ(x,y,z,t)|2*dV is the probability to find the particle in the region of space with volume dV around the point (x,y,z) at the time t.

Quote
ω/k=Vp

where Vp is the phase velocity,
....
Interestingly, the value ω/k can take values which are either +1 or -1.

In which case? If one choose c as unit of speed, what you wrote seems to be valid for photons only. I don't understand.

Quote

The meaning of the two becomes more valid under more research.

The equation of interest now comes in the form:

∂ΨR/∂ x + ∂ΨR/∂ t = 0

How did you get this from the Dirac equation? Can you explain?
 

Offline QuantumClue

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« Reply #14 on: 31/12/2010 16:10:55 »
Lightarrow

Not exactly. Ψ is not a probability distribution, |Ψ|2 is the probability distribution. Ψ(x,y,z,t) is a mathematical function which describe the "state" of the quantum system and which physical meaning is that |Ψ(x,y,z,t)|2*dV is the probability to find the particle in the region of space with volume dV around the point (x,y,z) at the time t.


Yes, the probability density. I'll fix it.

In which case? If one choose c as unit of speed, what you wrote seems to be valid for photons only. I don't understand.

When learning about the Dirac Equation in a class, you must be introduced to the phase velocity in this form. There is no point jumping equations because then nothing is learned. If we ask how fast a particle goes, with this equation here ω2=m2+k2  then we need to compute the group velocity ∂ω/∂k=k/√m2+k2. And this tells us the velocity of a wave.

How did you get this from the Dirac equation? Can you explain?

Do you know anything about partial derivates? Ψ = ei(kx-ωt) is what we need. ∂ΨR/∂ x + ∂ΨR/∂ t = 0 whenever you take a wave form like this against the derivative in respect to position it pulls down an ik - that is the derivative in respect to ∂ΨR/∂ x is the same as ik. Likewise the derivative with respect to t is iω so what you have is a standard wave equation just written in a form we have already covered. 




« Last Edit: 31/12/2010 16:39:10 by QuantumClue »
 

Offline lightarrow

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« Reply #15 on: 31/12/2010 21:21:49 »

In which case? If one choose c as unit of speed, what you wrote seems to be valid for photons only. I don't understand.

When learning about the Dirac Equation in a class, you must be introduced to the phase velocity in this form. There is no point jumping equations because then nothing is learned.

Sorry, but I haven't understood. I don't know anything about Dirac eq., so what ω stands for? And k? Because, if they had the same meaning they have in non-relativistic QM, the phase velocity cannot be just 1 or -1 (case of photons), but it is (1/k)*sqrt(m2+k2). Do you mean that the equation you wrote is the one valid for photons only? I'm just trying to understand.

Quote
If we ask how fast a particle goes, with this equation here ω2=m2+k2  then we need to compute the group velocity ∂ω/∂k=k/√m2+k2. And this tells us the velocity of a wave.
Yes, certainly, infact is the inverse of the phase velocity (their product is c2 for a free particle, but here c =1).

Quote
How did you get this from the Dirac equation? Can you explain?

Do you know anything about partial derivates?

Yes.

Quote
Ψ = ei(kx-ωt) is what we need. ∂ΨR/∂ x + ∂ΨR/∂ t = 0 whenever you take a wave form like this

Ah, yes, certainly. Only that I haven't understood why "you take a wave form like this".
 

Offline yor_on

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What is the Dirac equation?
« Reply #16 on: 31/12/2010 22:04:34 »
I'm pleased reading you, you explain your equations and where they differ. That's a good sign of understanding them. And as a guess, this is not an easy one to explain, so when Lightarrow ask then it is because he actually sees the meaning, but needs to see the significance and difference of the symbols used more clearly.

This gotta be a pretty good way for us all to see how to break down a equation, or theorem, no matter where we stand mathematically. If we continue like this there will be a danger of us all starting to understand :)

And Foolosophy: "its a philosophical game of proofs based on delusional axioms" It have everything to do with your definitions I agree :) choosing the right one is very important, and sometimes that is a intuitive concept as I think. I liked it :) it was funny, and even true in some cases.
« Last Edit: 31/12/2010 22:06:52 by yor_on »
 

Offline Geezer

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« Reply #17 on: 01/01/2011 05:26:57 »
Yeah! Brilliant post! (Way beyond me of course)

Mathematics is NOT a science - its a philosophical game of proofs based on delusional axioms

Well, if you say so.

(Hmmmm? I thought I had a monopoly on PBLA on TNS. Just goes to show, you never know.)
 

Offline QuantumClue

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« Reply #18 on: 01/01/2011 19:21:53 »

In which case? If one choose c as unit of speed, what you wrote seems to be valid for photons only. I don't understand.

When learning about the Dirac Equation in a class, you must be introduced to the phase velocity in this form. There is no point jumping equations because then nothing is learned.

Sorry, but I haven't understood. I don't know anything about Dirac eq., so what ω stands for? And k? Because, if they had the same meaning they have in non-relativistic QM, the phase velocity cannot be just 1 or -1 (case of photons), but it is (1/k)*sqrt(m2+k2). Do you mean that the equation you wrote is the one valid for photons only? I'm just trying to understand.

Quote
If we ask how fast a particle goes, with this equation here ω2=m2+k2  then we need to compute the group velocity ∂ω/∂k=k/√m2+k2. And this tells us the velocity of a wave.
Yes, certainly, infact is the inverse of the phase velocity (their product is c2 for a free particle, but here c =1).

Quote
How did you get this from the Dirac equation? Can you explain?

Do you know anything about partial derivates?

Yes.

Quote
Ψ = ei(kx-ωt) is what we need. ∂ΨR/∂ x + ∂ΨR/∂ t = 0 whenever you take a wave form like this

Ah, yes, certainly. Only that I haven't understood why "you take a wave form like this".

Sorry, but I haven't understood. I don't know anything about Dirac eq., so what ω stands for? And k? Because, if they had the same meaning they have in non-relativistic QM, the phase velocity cannot be just 1 or -1 (case of photons), but it is (1/k)*sqrt(m2+k2). Do you mean that the equation you wrote is the one valid for photons only? I'm just trying to understand.

Then you might be really interested when I tell you that k and ω are actually 2 by 2 matrices?


Ah, yes, certainly. Only that I haven't understood why "you take a wave form like this".



Oh well, this is just a matter of rewriting equations where our terms are equivalent.


Happy new year!

 

Offline QuantumClue

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« Reply #19 on: 01/01/2011 19:44:50 »
Sorry ...

ω is the angular frequency and k is the wave number.
 

Offline QuantumClue

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« Reply #20 on: 01/01/2011 19:46:07 »
Mathematics is not dilusional; it's logical.
 

Offline yor_on

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« Reply #21 on: 01/01/2011 20:22:54 »
In itself it's not (delusional I meant) I agree QC. Although some conclusions following might be, if compared to our SpaceTime. Like parallel tracks always meeting. Saying that it could be true 'somewhere' becomes a philosophical question, and possible too. If that was the truth though it would still make those tracks meeting a impossibility inside SpaceTime. If we are right in assuming that they don't :)
« Last Edit: 01/01/2011 20:47:53 by yor_on »
 

Offline lightarrow

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« Reply #22 on: 01/01/2011 20:44:29 »

Sorry, but I haven't understood. I don't know anything about Dirac eq., so what ω stands for? And k? Because, if they had the same meaning they have in non-relativistic QM, the phase velocity cannot be just 1 or -1 (case of photons), but it is (1/k)*sqrt(m2+k2). Do you mean that the equation you wrote is the one valid for photons only? I'm just trying to understand.

Then you might be really interested when I tell you that k and ω are actually 2 by 2 matrices?
It was you to write:

<<If we ask how fast a particle goes, with this equation here ω2=m2+k2  then we need to compute the group velocity ∂ω/∂k=k/√m2+k2. And this tells us the velocity of a wave>>.

Do we need matrices to deduce from ω2 = m2 + k2 that ∂ω/∂k = k/sqrt(m2 + k2)? It doesn't seem so.

Quote
Ah, yes, certainly. Only that I haven't understood why "you take a wave form like this".
Oh well, this is just a matter of rewriting equations where our terms are equivalent.
And I asked you how. In the net I see the Dirac equation written as:

ih∂ψ/∂t = -ihc α•gradψ + βmc2ψ

where α and β are some matrixes related to the Dirac matrixes.

Even considering just 1 spatial dimension, in which way this equation becomes the one you have written? How can the term βmc2ψ disappear?


Thanks and Happy New Year to you too!
« Last Edit: 01/01/2011 20:46:58 by lightarrow »
 

Offline QuantumClue

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« Reply #23 on: 02/01/2011 01:42:54 »
Right, ok. What we have done is set everything into natural units. Equations like E2=p2c2+M2c4 reduce quite naturally to E2=M2+p2. The term βmc2ψ does not disappear as such. You will notice our simplified Dirac Equation iψ'=-iα∂xψ+mβ has not lost its mass term, nor has the beta matrix vanished. What has vanished are all units which can be set to 1. This was acheived very early on when considering one of our old enemies ω2 (which is in fact equal to energy) and this is in fact simply equal to ω=k.

Do we need matrices to deduce from ω2 = m2 + k2 that ∂ω/∂k = k/sqrt(m2 + k2)? It doesn't seem so.

No of course not, but that doesn't become clear until one investigates it. Bottom line is that k can take on a negative or a positive number, which is easily expressable in the mathematics. The thing which dictates whether takes on a value of 1+ or -1 depends on alpha interestingly enough. Which is just as important as learning that beta dictates which waves are considered in a matrix.

« Last Edit: 02/01/2011 01:45:29 by QuantumClue »
 

Offline QuantumClue

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« Reply #24 on: 02/01/2011 01:47:09 »
In itself it's not (delusional I meant) I agree QC. Although some conclusions following might be, if compared to our SpaceTime. Like parallel tracks always meeting. Saying that it could be true 'somewhere' becomes a philosophical question, and possible too. If that was the truth though it would still make those tracks meeting a impossibility inside SpaceTime. If we are right in assuming that they don't :)
I forget who said it was delusional now... I think foolosophy :)
 

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