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Are they out of phase from each other?

How could we tell, by looking at their wave-functions, the difference between an electron and a positron?

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- Thread starter JDude13
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Are they out of phase from each other?

How could we tell, by looking at their wave-functions, the difference between an electron and a positron?

- #2

tiny-tim

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People usually think of the wave-function of an electron as being

It

Essentially, it's a constant spinor times a "wavy" complex part.

The spinor is usually written as 4 components (it's not a vector, it just

The top two represent an electron, the bottom two represent a positron …

(the rest of it represents the spin …

eg an electron with spin up along the z-axis is (1,0,0,0), but a positron is (0,0,1,0))

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People usually think of the wave-function of an electron as beingcomplex-valued.

Itisn't, it'sspinor-valued.

Essentially, it's a constant spinor times a "wavy" complex part.

The spinor is usually written as 4 components (it's not a vector, it justlookslike one).

The top two represent an electron, the bottom two represent a positron …that'show the wave-functions look different from each other!

(the rest of it represents the spin …

eg an electron with spin up along the z-axis is (1,0,0,0), but a positron is (0,0,1,0))

*yelps as previous comment whizzes right over his head*

I'm 16 here...

Umm... Forgive me if I'm not getting this... The wave function of a particle operates in 4 spacial dimensions?

- #4

tiny-tim

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*yelps as previous comment whizzes right over his head*

I'm 16 here...

Yes, I

The wave function of a particle operates in 4 spacial dimensions?

Sort of, but they're not the usual space dimensions, they're extra ones.

Best to think of the wave function of an electron as a variable complex number (the "wavy" part) mutiplied by a constant black box labelled "charge and spin-direction" (the constant "spinor" part, with 4 components) …

the label tells you all you need to know.

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Sort of, but they're not the usual space dimensions, they're extra ones.

Best to think of the wave function of an electron as a variable complex number (the "wavy" part) mutiplied by a constant black box labelled "charge and spin-direction" (the constant "spinor" part, with 4 components) …

the label tells you all you need to know.

Okay... So what about the Planck relation? Doesn't that specify a real, measurable quantity? And photons are very measurable... Do photons act on fewer dimensions than electrons, positrons and other elementary particles?

How would you represent the wave function of the simples particle you know how?... For curiosity's sake...

- #6

jtbell

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Dirac started by writing down his equation based on what he thought the properties of its solutions should be, and then he realized that the only way to solve it was to make the wave function into a 4-component spinor and some of the coefficients in the Dirac equation into matrices instead of plain old numbers.

You might try Googling for "Dirac equation" and see what turns up.

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Have you considered that maybe, you should be more curious about simple, basic physics

I know that maybe, finding a forum like this might feel like being a kid in a candy store. But at some point, you need to learn to grab only things that are at your height rather than trying to aim for those on the top shelf. We all learn things based on a foundation of things we

These answers are over your head because they ARE over your head. People understand them after a painful series and sequential process of understanding simpler things. This is something that you appear to not realize.

Zz.

- #8

tiny-tim

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Okay... So what about the Planck relation? Doesn't that specify a real, measurable quantity? And photons are very measurable

I'm not following you.

I never said anything wasn't measurable …

multiply an electron wave function (black box and all) by its complex conjugate, and you get a measurable probability (an ordinary number), just as expected.

... Do photons act on fewer dimensions than electrons, positrons and other elementary particles?

Bosons like photons have a simpler mathematical representation than fermions like electrons. Their black boxes need fewer numbers to describe them.

How would you represent the wave function of the simples particle you know how?... For curiosity's sake...

A scalar spin-0 boson that didn't interact with anything would just be a complex number.

- #9

A. Neumaier

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People usually think of the wave-function of an electron as beingcomplex-valued.

Itisn't, it'sspinor-valued.

Essentially, it's a constant spinor times a "wavy" complex part.

No. It is two independent wave functions, whereas your recipe gives two components that are multiples of each other.

The spins at any two spatial positions can (in principle) be independently assigned.

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A. Neumaier

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I'm 16 here...

Umm... Forgive me if I'm not getting this... The wave function of a particle operates in 4 spacial dimensions?

Wave functions for a single particle always consist of a number m of components, each of which is a function of a 4-dimensional argument (space and time). What m is depends on the particle.

For a heavy particle one can ignore spin and then has m=1 - just one component.

The wave function of a nonrelativistic electron has two components - one for the spin up amplitude, one for the spin down amplitude.

The wave function of a Dirac particle has 4 components (that this equals the space-time dimension is an accident without any deeper meaning) - one for each combination of spin direction and charge sign. It describes in general arbitrary superpositions of an electron (with two possible values of the spin) and a positron (also with two possible values of the spin). We have a pure electron if the two positron components vanish, and a pure positron when the two electron components vanish.

The wave function of a nucleon has 8 components - four Dirac components for the proton/antiproton and four Dirac components for the neutron/antineutron. It describes in general arbitrary superpositions of a proton, an antiproton, a neutron, and an antineutron, each with two possible values of the spin. We have a pure proton if the six non-proton components vanish, etc..

I hope you got the general recipe. To check you understanding, figure out how many components has a quark state (which exists in 3 colors, 6 flavors, two spin states, and two values +-1/3 of the baryon number - which distinguish between the quark and the antiquark).

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Are they out of phase from each other?

How could we tell, by looking at their wave-functions, the difference between an electron and a positron?

You're only 16. At that age I didn't know anything about quantum physics, I had never heard of wavefunctions, I was not aware of positrons. I like to think of myself I wasn't retarded back then, though. :D

My advice: take things easily. Finish your high-school first and then, if still interested, go to university and discover theoretical physics. A good school will help you get your answers to the questions you ask today. Posting questions on a forum may result in getting (hopefully correct) answers whose understanding is waay over your head. Simply because you haven't been taught real physics so far.

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tiny-tim

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Hi Arnold!

But surely if we use a frame in which the electron is stationary, one of those components is zero, and so in any other frame the components will have the same "wavy part" as the stationary electron*and as each other?*

btw, I've just seen your interesting "http://www.mat.univie.ac.at/~neum/physfaq/physics-faq.html" [Broken]" (which you referred to in another thread) …

wouldn't it be easier to read in some more modern font than (I think) Courier?

No. It is two independent wave functions, whereas your recipe gives two components that are multiples of each other.

The spins at any two spatial positions can (in principle) be independently assigned.

But surely if we use a frame in which the electron is stationary, one of those components is zero, and so in any other frame the components will have the same "wavy part" as the stationary electron

btw, I've just seen your interesting "http://www.mat.univie.ac.at/~neum/physfaq/physics-faq.html" [Broken]" (which you referred to in another thread) …

wouldn't it be easier to read in some more modern font than (I think) Courier?

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- #13

A. Neumaier

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But surely if we use a frame in which the electron is stationary, one of those components is zero, and so in any other frame the components will have the same "wavy part" as the stationary electronand as each other?

Why should one component have to be zero? The components have nothing to do with space-time components. The Hilbert space of a nonrelativistic electron is parameterized

by the spectrum of a maximally commuting set of observables, e.g., position and the z-component of spin, and any superposition of (position, spin) eigenstates is a valid state.

btw, I've just seen your interesting "http://www.mat.univie.ac.at/~neum/physfaq/physics-faq.html" [Broken]" (which you referred to in another thread) …

wouldn't it be easier to read in some more modern font than (I think) Courier?

Every year in the Christmas break I spend some time to modernize the presentation of the FAQ. During the year I usually only update content. Ultimately I want to convert everything to html with appropriate cross-links between the topics, and working external links. But I have many other things to do, and this has no high priority.

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tiny-tim

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Why should one component have to be zero? The components have nothing to do with space-time components.

Yes, but they transform between Lorentz frames in accordance with a representation of the Lorentz group (is that the correct terminology? ) …

if the "lower part" is zero in one frame, then in any other frame won't the "lower part" have the same "wavy function" as the "upper part"?

The Hilbert space of a nonrelativistic electron is parameterized by the spectrum of a maximally commuting set of observables, e.g., position and the z-component of spin, and any superposition of (position, spin) eigenstates is a valid state.

Yes of course, but by "an electron" I was thinking of an eigenstate … "

Every year in the Christmas break …

You need

Why not get some geeky students to do it for you?

- #15

A. Neumaier

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Yes, but they transform between Lorentz frames in accordance with a representation of the Lorentz group.

if the "lower part" is zero in one frame, then in any other frame won't the "lower part" have the same "wavy function" as the "upper part"?

No, because with a Lorentz transformation, you can force the lower part to zero only at a single point in space.

Yes of course, but by "an electron" I was thinking of an eigenstate … "an" electron.

Do you really think that an electron in a superposition of eigenstates is no longer an electron??? An electron remains an electron no matter which state it is in. (In second quantization, the electron number remains exactly once!)

You needelves!

Why not get some geeky students to do it for you?

If I let anyone else do it using the usual automatic tools, the result would be something

that I couldn't maintain easily. (Download the source of one of my html pages and some random html pages from elsewhere, and look at the difference!)

My main goal is to get the information across, and the present pages do this well enough.

- #16

tiny-tim

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No, because with a Lorentz transformation, you can force the lower part to zero only at a single point in space.

ooh, I never knew that!

I've always assumed that producing an electron (in say an electron-positron pair production) resulted in something which in its rest frame was pure electron.

- #17

A. Neumaier

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I've always assumed that producing an electron (in say an electron-positron pair production) resulted in something which in its rest frame was pure electron.

Well, you mix two things. I was referring to the 2-component electron, where you claimed the two components would have to be constant multiples of each other.

But now you are referring to a four component state, which describes the superposition of an electron and a positron state of the same particle. Here the Lorentz transform acts separately on the top two and the bottom two components. So if you have a pure electron, the bottom two components are identically zero, and remain so after a Lorentz transformation.

Thus an electron remains an electron, no matter in which frame you look at it.

But the top two components change, and you can make one of the components zero at a single chosen point but not everywhere.

- #18

tiny-tim

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Well, you mix two things. I was referring to the 2-component electron, where you claimed the two components would have to be constant multiples of each other.

But now you are referring to a four component state, which describes the superposition of an electron and a positron state of the same particle.

I'm not sure we're talking about the same thing.

I've intended all along to mean a 4-component spinor.

So eg if a stationary electron with spin-up along the z-direction is (1,0,0,0), in a different frame it's Lorentz-transformed to a multiple of (1,0,0,

(I can't quickly find an on-line version of this, so I'm basing it on pp 77-82 of my Aitchison and Hey, 2nd edition)

what I'm referring to as an "upper part" that's essentially electron-ish and a "lower part" that's essentially positron-ish …

the actual wave-function (in the first, rest-frame) will be (1,0,0,0) times a "wavy part", and that "wavy part" will therefore be the same in both upper and lower parts of the Lorentz-transformed wave-function.

Here the Lorentz transform acts separately on the top two and the bottom two components. So if you have a pure electron, the bottom two components are identically zero, and remain so after a Lorentz transformation.

In my version, it crosses-over, much as a pure electric field becomes partly magnetic.

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Have you considered that maybe, you should be more curious about simple, basic physicsfirstbefore asking about stuff that are clearly way over your head? I mean, antimatter wavefunction? Tachyons?

I know that maybe, finding a forum like this might feel like being a kid in a candy store. But at some point, you need to learn to grab only things that are at your height rather than trying to aim for those on the top shelf. We all learn things based on a foundation of things wealready understood. If not, the new things that we acquire will be just dangling there in mid air, without any context or connection to other things. This is not physics and it isn't a way to learn.

These answers are over your head because they ARE over your head. People understand them after a painful series and sequential process of understanding simpler things. This is something that you appear to not realize.

Zz.

My understanding and intended use of this information is irrelavent. I am asking a friendly community to give me a little help in understanding quantum mechanics. If I can't "reach the candy on the top shelf", I'm sure someone will stumble across it and take it after I'm gone.

I don't like to not know what's going on... Being intentionally naive about the universe whilst I learn the basic mathematics is not something I want to do... But that's also irrelavent. If I don't use this information someone else will...

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My understanding and intended use of this information is irrelavent. I am asking a friendly community to give me a little help in understanding quantum mechanics. If I can't "reach the candy on the top shelf", I'm sure someone will stumble across it and take it after I'm gone.

I don't like to not know what's going on... Being intentionally naive about the universe whilst I learn the basic mathematics is not something I want to do... But that's also irrelavent. If I don't use this information someone else will...

Then you are not interested in learning or understanding anything. All you want is to be able to compete in Jeopardy, where all you will have are disjointed pieces of information without any connection to other things.

And you are also asking the rest of us to put a tremendous amount of effort, because each answer we give, we will have to keep on backtracking ourselves because you have no ability to understand the answers we gave. That is what has been going on in practically ALL the threads that you have created.

It was a friendly advice. Take it or leave it. But if you think you are actually understanding something from the way you are doing this, you are mistaken.

Zz.

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Then you are not interested in learning or understanding anything. All you want is to be able to compete in Jeopardy, where all you will have are disjointed pieces of information without any connection to other things.

And you are also asking the rest of us to put a tremendous amount of effort, because each answer we give, we will have to keep on backtracking ourselves because you have no ability to understand the answers we gave. That is what has been going on in practically ALL the threads that you have created.

It was a friendly advice. Take it or leave it. But if you think you are actually understanding something from the way you are doing this, you are mistaken.

Zz.

Your advice is noted... I would like to remind you that people don't

I urge anyone on this community who feel far superior to everyone else (inclding ZapperZ) to stop responding to inferior minds as it's probably not worth your valuable, valuable time.

I'd apologise for wasting your time, Zapper, but I didn't; you did...

And fyi: When I made this thread I thought I had a large grasp on wavefunction. The more I find out, the weaker my grasp appears. This community has shown me that there is a lot more to learn when it comes to mathematics and physics...

But seriously, if you don't want to respond, don't... SIMPLE!

- #22

A. Neumaier

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I've intended all along to mean a 4-component spinor.

So eg if a stationary electron with spin-up along the z-direction is (1,0,0,0), in a different frame it's Lorentz-transformed to a multiple of (1,0,0,σ·p/E+m)

Probably you meant (1,

A stationary electron can have spin-up along the z-direction in position x but spin-down along the z-direction in position y, indeed any combination of spins in arbitrary directions at different positions. Your assumption that psi(x)=(1,0,0,0) independent of x is therefore

very special, since it requires the spin to be up everywhere in space.

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