In article ,
ganesh wrote:
hi,
In response to everyone who replied to my querry....

The fermionic wavefunction changes sign when we interchange two fermions:
this can be used to explain the pauli exclusion principle...And why do
the wavefunction changes sign?? because the relevant operators (talking of
QFT) anticommute instead of commuting.
So now the anticommutation principle becomes the axion instead of
pauli's exclusion principle....This probably may be a slightly deeper
axion,...but it still doesnt seem to fundamentally answer some Qs.

Axiom. An axion is a hypothetical particle proposed to explain certain
symmetries of the strong force.

When you go to the relativistic theory, if you choose the wrong
commutation rules you can violate causality. So the anticommutation of
fermions is forced on us from that angle.
--
"Let us learn to dream, gentlemen, then perhaps we shall find the
truth... But let us beware of publishing our dreams before they have been
put to the proof by the waking understanding." -- Friedrich August Kekulé

Bjoern Feuerbacher wrote:
John Sefton wrote:

Gregory L. Hansen wrote:

In article ,
ganesh wrote:


One basic Q: Is pauli's exclusion principle an axiom?? Or can it be
derived using the electromagnetic force + .....??

ganesh



Identical particles in quantum mechanics are not identical in the sense of
two cue balls with the same mass, shape, color, and so on. They're
identical in the strongest possible sense -- you can't say particle A is
here and particle B is there because particle B is also here and A is also
there. When you work out the two-particle wavefunction you must compute
both possibilities and add them together. If the properties are indicated
by A and B and our assumption of which is which is indicated by position,
we have

psi(A,B) + psi(B,A)

Fermions, particles with half-integral spin (1/2, 3/2, 5/2,...)
anti-commute, which means there's a minus sign.

psi(A,B) - psi(B,A)

If A and B are the same (two particles in identical quantum states) we
have

psi(A,A) - psi(A,A) = 0

The probability goes to zero, hence, no two fermions can share the same
state.

For instance, in the two-point, one pathway standing
wave below
http://rapfast.petcom.com/~john/He.GIF
the first five positions a
0,0,80 0,0,-80
5.9726,-14.4192,78.4128 -5.9726,14.4192,-78.4128
21.6478,-21.6478,73.9104 -21.6478,21.6478,-73.9104
41.0624,-17.0086,66.5176 -41.0624,17.0086,-66.5176
56.5685,0,56.5685 -56.5685,0,-56.5685
The Galaxy pattern orbital is produced when a point is
rotating round a circle at a fixed radius from a center
and the circle is precessing at twice the frequency of
the rotation. If there are more than two points on the
circle, ONLY points exactly opposite each other will follow
the same path. THIS is why the Pauli Exclusion Principle
holds.


Does this explain also why the electrons inside a metal obey the
Fermi-Dirac statistic?


Why ask a crackpot?
You know everything.

John Sefton wrote:

Bjoern Feuerbacher wrote:
John Sefton wrote:

Gregory L. Hansen wrote:

In article ,
ganesh wrote:


One basic Q: Is pauli's exclusion principle an axiom?? Or can it be
derived using the electromagnetic force + .....??

ganesh



Identical particles in quantum mechanics are not identical in the sense of
two cue balls with the same mass, shape, color, and so on. They're
identical in the strongest possible sense -- you can't say particle A is
here and particle B is there because particle B is also here and A is also
there. When you work out the two-particle wavefunction you must compute
both possibilities and add them together. If the properties are indicated
by A and B and our assumption of which is which is indicated by position,
we have

psi(A,B) + psi(B,A)

Fermions, particles with half-integral spin (1/2, 3/2, 5/2,...)
anti-commute, which means there's a minus sign.

psi(A,B) - psi(B,A)

If A and B are the same (two particles in identical quantum states) we
have

psi(A,A) - psi(A,A) = 0

The probability goes to zero, hence, no two fermions can share the same
state.

For instance, in the two-point, one pathway standing
wave below
http://rapfast.petcom.com/~john/He.GIF
the first five positions a
0,0,80 0,0,-80
5.9726,-14.4192,78.4128 -5.9726,14.4192,-78.4128
21.6478,-21.6478,73.9104 -21.6478,21.6478,-73.9104
41.0624,-17.0086,66.5176 -41.0624,17.0086,-66.5176
56.5685,0,56.5685 -56.5685,0,-56.5685
The Galaxy pattern orbital is produced when a point is
rotating round a circle at a fixed radius from a center
and the circle is precessing at twice the frequency of
the rotation. If there are more than two points on the
circle, ONLY points exactly opposite each other will follow
the same path. THIS is why the Pauli Exclusion Principle
holds.


Does this explain also why the electrons inside a metal obey the
Fermi-Dirac statistic?


Why ask a crackpot?

Because it may turn out after all that he isn't a crackpot? (although I
think this is highly unlikely in most cases)
Or because the question may give the crackpot something to think about -
it may tell him that there are other phenomena which are already
explained by standard physics, but not his theories? (although I again
think it is highly unlikely that such questions will make any crackpot
think about his claims)


You know everything.

Nonsense. I never claimed this. On the contrary, I know very little. I
have knowledge in particle physics, QFT, cosmology and some other
fields, but there are lots of people out there who know far more than me
about these topics, and I don't know much about most other topics
(chemistry, biology, medicine, law, theology, ...........).

OTOH, I do know enough to recognize a crackpot when I see his claims.


Bye,
Bjoern

(I'm right now
going back to pre QFT, since QFT doesnt talk of spacetime as an observable

It doesn't use hermitean operators for position and time, right. So
what? Even QM doesn't has time as an observable! You shouldn't confuse
the use of the noun "observable" in quantum theory with the question if
something can be observed or not!


and this is something I really dont feel comfortable with).

Well, that's your problem. Physics doesn't care about if you feel
comfortable with it or not.

Well, its like this....If I can perform an experiment, then I expect the
theory to explain the results of an experiment. (I guess this is v. much
the problem of physics!). Now, if I can perform an experiment and determine
the position of a particle (within the limits of uncertainity principle),
then certainly do expect QFT to give me a mechanism to predict it (given
initial conditions, the environment, blah blah blah....)
The second reason I feel uncomfortable is how does one reconcile
heisenberg's momentum-position uncertainity with QFT??



Since spin commutes with space,

What is this supposed to mean, specifically? That spin commutes with the
position operator or what?

yeah...why?? is it incorrect??



Yes, definitely. What have you read about QM and QFT so far?

I acknowledge not much.....

ganesh

ganesh wrote:

(I'm right now
going back to pre QFT, since QFT doesnt talk of spacetime as an observable

It doesn't use hermitean operators for position and time, right. So
what? Even QM doesn't has time as an observable! You shouldn't confuse
the use of the noun "observable" in quantum theory with the question if
something can be observed or not!


and this is something I really dont feel comfortable with).

Well, that's your problem. Physics doesn't care about if you feel
comfortable with it or not.

Well, its like this....If I can perform an experiment, then I expect the
theory to explain the results of an experiment.

Well, that's a sensible expectation, I agree. Did you notice that I
mentioned (at least) two times that QFT *can* explain the exclusion
principle? (it can be proven that the ladder operators for fermions have
to anticommute)


(I guess this is v. much
the problem of physics!). Now, if I can perform an experiment and determine
the position of a particle (within the limits of uncertainity principle),
then certainly do expect QFT to give me a mechanism to predict it (given
initial conditions, the environment, blah blah blah....)

Right. Now, where is the reason why you feel uncomfortable??


The second reason I feel uncomfortable is how does one reconcile
heisenberg's momentum-position uncertainity with QFT??

Why do you think there is a problem? Could you please be more specific?



Since spin commutes with space,

What is this supposed to mean, specifically? That spin commutes with the
position operator or what?

yeah...why?? is it incorrect??

No, it's corrected. It's merely a very strange way to express this,
IMHO.



Yes, definitely. What have you read about QM and QFT so far?

I acknowledge not much.....

Could you be more specific, please? I have to know your level of
knowledge in order to be able to help you with your problem.


Bye,
Bjoern

Bjoern Feuerbacher wrote in message ...
ganesh wrote:

(I'm right now
going back to pre QFT, since QFT doesnt talk of spacetime as an observable

It doesn't use hermitean operators for position and time, right. So
what? Even QM doesn't has time as an observable! You shouldn't confuse
the use of the noun "observable" in quantum theory with the question if
something can be observed or not!


and this is something I really dont feel comfortable with).

Well, that's your problem. Physics doesn't care about if you feel
comfortable with it or not.

Well, its like this....If I can perform an experiment, then I expect the
theory to explain the results of an experiment.

Well, that's a sensible expectation, I agree. Did you notice that I
mentioned (at least) two times that QFT *can* explain the exclusion
principle? (it can be proven that the ladder operators for fermions have
to anticommute)

I agree u mentioned that QFT can explain the exclusion principle. I was
thinking about what it...I'll probably get back to it after thinking more..
(if necessary).


(I guess this is v. much
the problem of physics!). Now, if I can perform an experiment and determine
the position of a particle (within the limits of uncertainity principle),
then certainly do expect QFT to give me a mechanism to predict it (given
initial conditions, the environment, blah blah blah....)

Right. Now, where is the reason why you feel uncomfortable??

Well, the spacetime operators are not hermitian. Had the spacetime operator
\phi been a number operator, I could have said look "the eigenvalues of
\phi(x) gives me the number of particles at spacetime pt 'x' = these
particles are situated in spacetime pt 'x' ". However, this interpretation
is invalid, since \phi(x) is not hermitian. Given this, how does the
theory (QFT) predict the position (which I can get by doing some experiment..
again within some uncertainity range)??


The second reason I feel uncomfortable is how does one reconcile
heisenberg's momentum-position uncertainity with QFT??

Why do you think there is a problem? Could you please be more specific?

If QFT were to predict the position (within a certain range as in QM),
then one can calculate or estimate \delta_k*\delta_x. If the position
is not predicted by the theory, then what does the uncertainity relation
mean?? Or does the uncertainity relation take some other form??




Yes, definitely. What have you read about QM and QFT so far?

I acknowledge not much.....

Could you be more specific, please? I have to know your level of
knowledge in order to be able to help you with your problem.

I'm an elec engg by profession interested in
physics....I'm currently reading "Quantum theory of fields" by Wienberg.
Earlier read QFT by A. Mandl, and found it tooo brief and not much
physical insight... Am currently trying to understand the basic
formulation of QFT...its interpretation and why it is the way it is...

ganesh

ganesh wrote:

Bjoern Feuerbacher wrote in message ...
ganesh wrote:

(I'm right now
going back to pre QFT, since QFT doesnt talk of spacetime as an observable

It doesn't use hermitean operators for position and time, right. So
what? Even QM doesn't has time as an observable! You shouldn't confuse
the use of the noun "observable" in quantum theory with the question if
something can be observed or not!


and this is something I really dont feel comfortable with).

Well, that's your problem. Physics doesn't care about if you feel
comfortable with it or not.

Well, its like this....If I can perform an experiment, then I expect the
theory to explain the results of an experiment.

Well, that's a sensible expectation, I agree. Did you notice that I
mentioned (at least) two times that QFT *can* explain the exclusion
principle? (it can be proven that the ladder operators for fermions have
to anticommute)

I agree u mentioned that QFT can explain the exclusion principle. I was
thinking about what it...I'll probably get back to it after thinking more..
(if necessary).

I finally found the reference:
W. Pauli, The Connection between Spin and Statistics, Physical Review 58
(1940),
page 716.

Have fun with it! :-) (there is some heavy stuff in it, like the
properties
of the Lorentz group; I hope you will understand enough to see what he
is
doing there...)



(I guess this is v. much
the problem of physics!). Now, if I can perform an experiment and determine
the position of a particle (within the limits of uncertainity principle),
then certainly do expect QFT to give me a mechanism to predict it (given
initial conditions, the environment, blah blah blah....)

Right. Now, where is the reason why you feel uncomfortable??

Well, the spacetime operators are not hermitian.

What do you mean by "spacetime operators"?


Had the spacetime operator
\phi been a number operator,

Phi is usually used to denote a scalar field, not a "spacetime
operator", whatever
this is supposed to mean.


I could have said look "the eigenvalues of
\phi(x)

Phi(x)? So you indeed talk about a scalar field? Then what is this stuff
about
"spacetime operators" supposed to mean???

And, BTW, the operators phi(x) *are* hermitean, so what are you talking
about?


gives me the number of particles at spacetime pt 'x'

Why should it???

The eigenvalues of phi(x) (provided that you are really talking about a
scalar
field) are the possible field strengths at point x. This has *nothing*
to do
with the number of particles there!


= these
particles are situated in spacetime pt 'x' ". However, this interpretation
is invalid, since \phi(x) is not hermitian.

It is - provided that you talk about a scalar field operator.


Given this, how does the
theory (QFT) predict the position (which I can get by doing some experiment..
again within some uncertainity range)??

If you want to have the probability amplitude that if you start with a
particle
at point x at time t and find it later at point x' at time t', this is
given by
the propagator (also called 2-point Green's function).

OTOH, if you want to answer a question like "what is the probability
amplitude
that a will find a particle, for which I know the momentum to be p, at
point x?",
you have to calculate, just as in ordinary QM, the overlap between the
states |p
and |x. You can get these states with suitable ladder operators from
the vacuum.


The second reason I feel uncomfortable is how does one reconcile
heisenberg's momentum-position uncertainity with QFT??

Why do you think there is a problem? Could you please be more specific?

If QFT were to predict the position (within a certain range as in QM),
then one can calculate or estimate \delta_k*\delta_x.

Yes. So what? Why can't this be reconciled with Heisenberg's uncertainty
relation?


If the position
is not predicted by the theory, then what does the uncertainity relation
mean??

It is used mainly in ordinary QM, but can be used in QFT, too. However,
if you want to calculate the position, you first have to say "position
of *what*". In other words, you have to say which particle you talk
about
- in which state it is, for example. Just like in QM...


Or does the uncertainity relation take some other form??

In QFT, one talks about fields (often about particles, too, but the
fields
are the more fundamental objects). Hence the uncertainty relation is
usually
expressed for fields - well, what one actually writes down are the
commutation
relations, from which the uncertainty relation can be derived. For a
scalar
field, it's something like
[phi(x,t), pi(y,t)] = i delta(x-y),
where x and y are two different points in space, t is the *same* time in
both,
pi is the "momentum density" of the field phi, derived from the
Lagrangian,
and delta on the right means the Dirac delta "function".



Yes, definitely. What have you read about QM and QFT so far?

I acknowledge not much.....

Could you be more specific, please? I have to know your level of
knowledge in order to be able to help you with your problem.

I'm an elec engg by profession interested in
physics....I'm currently reading "Quantum theory of fields" by Wienberg.

Oh, I don't think this is a good starting point - I would say that
Weinberg
is better suited for advanced readers. The books I liked were from 1)
Peskin and
Schroeder and 2) Rhyder (sp?).


Earlier read QFT by A. Mandl,

Sorry, I don't know this.


and found it tooo brief and not much
physical insight... Am currently trying to understand the basic
formulation of QFT...its interpretation and why it is the way it is...

Try Peskin and Schroeder - I think they are doing a good job.


What knowledge do you have about QM? Specifically, are you familiar with
the Dirac notation? Although, are you familiar with Special Relativity,
specifically four-vector notation?


Bye,
Bjoern

I finally found the reference:
W. Pauli, The Connection between Spin and Statistics, Physical Review 58
(1940),
page 716.

Have fun with it! :-) (there is some heavy stuff in it, like the
properties
of the Lorentz group; I hope you will understand enough to see what he
is
doing there...)
I'll try to get this paper.... and I guess the lorentz group is inevitable
if I want to get into QFT...

??

Well, the spacetime operators are not hermitian.

What do you mean by "spacetime operators"?

oops,...made a mess....actually wanted to mean \phi(x) is not a position
operator.....but ended up with something twisted...
I meant \phi(x) as the field operator only....and ok mistake corrected...
its hermitian.


Given this, how does the
theory (QFT) predict the position (which I can get by doing some experiment..
again within some uncertainity range)??

If you want to have the probability amplitude that if you start with a
particle
at point x at time t and find it later at point x' at time t', this is
given by
the propagator (also called 2-point Green's function).

ok, wud read about it.....

OTOH, if you want to answer a question like "what is the probability
amplitude
that a will find a particle, for which I know the momentum to be p, at
point x?",
you have to calculate, just as in ordinary QM, the overlap between the
states |p
and |x. You can get these states with suitable ladder operators from
the vacuum.

So, let me try to get what u r saying. I use the creation operaton a'(k)
on the vacum and finally get |p. Then I get |x if I know p|x ...since
|x = |pp|x

But then how do I find the values of x|p?? from which eqn??

And one more doubt....lets say I've |x, and let c be correspondig probability
amplitude, then shouldnt c satisfy the following continuity eqn of some sort

d/dt(c*c) + \grad.S = 0

where S is whatever.....(I'm not really bothered)

And from what I've read, in the continuity eqn
d/dt(P) + \grad.S = 0 that we get from QFT,
this guy P is not positive definite,....and so P cannot be interpreted
as postition probability density





If the position
is not predicted by the theory, then what does the uncertainity relation
mean??

It is used mainly in ordinary QM, but can be used in QFT, too. However,
if you want to calculate the position, you first have to say "position
of *what*". In other words, you have to say which particle you talk
about
- in which state it is, for example. Just like in QM...


Or does the uncertainity relation take some other form??

In QFT, one talks about fields (often about particles, too, but the
fields
are the more fundamental objects).
Ah, this is one of the greatest botheration to me,...and wud be really
grateful if someone can tell me in simple stuff, what the hell forced mankind
to think about fields instead of particles?? I guess it cant just be lorentz
invariance, because, the original dirac eqn was lorentz invariant and still
talked about single particles and not fields.



What knowledge do you have about QM? Specifically, are you familiar with
the Dirac notation? Although, are you familiar with Special Relativity,
specifically four-vector notation?

yeah, but cant really comment on the depth of knowledge....

ganesh wrote:


I finally found the reference:
W. Pauli, The Connection between Spin and Statistics, Physical Review 58
(1940),
page 716.

Have fun with it! :-) (there is some heavy stuff in it, like the
properties
of the Lorentz group; I hope you will understand enough to see what he
is
doing there...)

I'll try to get this paper.... and I guess the lorentz group is inevitable
if I want to get into QFT... ??

Some books about QFT manage to explain it without using the Lorentz
group.
IIRC, Ryder is one of them Bjorken and Drell may be one other.




Well, the spacetime operators are not hermitian.

What do you mean by "spacetime operators"?

oops,...made a mess....actually wanted to mean \phi(x) is not a position
operator.....but ended up with something twisted...
I meant \phi(x) as the field operator only....and ok mistake corrected...
its hermitian.

Well, at least as long it describes a real field... ;-)

[snip]


OTOH, if you want to answer a question like "what is the probability
amplitude
that a will find a particle, for which I know the momentum to be p, at
point x?",
you have to calculate, just as in ordinary QM, the overlap between the
states |p
and |x. You can get these states with suitable ladder operators from
the vacuum.

So, let me try to get what u r saying. I use the creation operaton a'(k)
on the vacum and finally get |p. Then I get |x if I know p|x

No.


...since
|x = |pp|x

No, the last equation is wrong. Where did you get it from?


But then how do I find the values of x|p?? from which eqn??

You first have to construct |x by a suitable creation operator, then
you can calculate x|p.


And one more doubt....lets say I've |x, and let c be correspondig
probability amplitude,

Corresponding to what?


then shouldnt c satisfy the following continuity eqn of some sort

d/dt(c*c) + \grad.S = 0

where S is whatever.....(I'm not really bothered)

If c is a non-relativistic wavefunction, then it satisfies such an
equation. If it's a relativistic wave function, it doesn't satisfy
such an equation necessarily (because particle number is not
conserved).


And from what I've read, in the continuity eqn
d/dt(P) + \grad.S = 0 that we get from QFT,
this guy P is not positive definite,....and so P cannot be interpreted
as postition probability density

Are you talking about the Klein-Gordon equation here, where it's not
clear how to define the probability density?



If the position
is not predicted by the theory, then what does the uncertainity relation
[snip]


Or does the uncertainity relation take some other form??

In QFT, one talks about fields (often about particles, too, but the
fields are the more fundamental objects).

Ah, this is one of the greatest botheration to me,...and wud be really
grateful if someone can tell me in simple stuff, what the hell forced
mankind to think about fields instead of particles??

You have to ask Faraday, IIRC he brought the idea up. ;-)

Electrodynamics is best described using fields; even now when we have
the "particle" interpretation of it (real and virtual photons), in
actual calculations about (macroscopic) electrodynamics phenomena,
it's *still* *MUCH* easier to use fields rather than particles.


I guess it cant just be lorentz
invariance, because, the original dirac eqn was lorentz invariant and still
talked about single particles and not fields.

Well, one can argue that the wave function itself is a field. ;-)
(a "field" in physics is (sometimes?) defined simply as a function
which depends on all three space coordinates and is defined on all
space)


What knowledge do you have about QM? Specifically, are you familiar with
the Dirac notation? Although, are you familiar with Special Relativity,
specifically four-vector notation?

yeah, but cant really comment on the depth of knowledge....

It's enough if you know what these terms mean and how to use them in
simple calculations.


Bye,
Bjoern

ah, thank you, I guess, I got quite a bit of pointers and directions
to proceed...... will read something and then come back to USENET....

gans

Bjoern Feuerbacher wrote in message ...
ganesh wrote:


I finally found the reference:
W. Pauli, The Connection between Spin and Statistics, Physical Review 58
(1940),
page 716.

Have fun with it! :-) (there is some heavy stuff in it, like the
properties
of the Lorentz group; I hope you will understand enough to see what he
is
doing there...)

I'll try to get this paper.... and I guess the lorentz group is inevitable
if I want to get into QFT... ??

Some books about QFT manage to explain it without using the Lorentz
group.
IIRC, Ryder is one of them Bjorken and Drell may be one other.




Well, the spacetime operators are not hermitian.

What do you mean by "spacetime operators"?

oops,...made a mess....actually wanted to mean \phi(x) is not a position
operator.....but ended up with something twisted...
I meant \phi(x) as the field operator only....and ok mistake corrected...
its hermitian.

Well, at least as long it describes a real field... ;-)

[snip]


OTOH, if you want to answer a question like "what is the probability
amplitude
that a will find a particle, for which I know the momentum to be p, at
point x?",
you have to calculate, just as in ordinary QM, the overlap between the
states |p
and |x. You can get these states with suitable ladder operators from
the vacuum.

So, let me try to get what u r saying. I use the creation operaton a'(k)
on the vacum and finally get |p. Then I get |x if I know p|x

No.


...since
|x = |pp|x

No, the last equation is wrong. Where did you get it from?


But then how do I find the values of x|p?? from which eqn??

You first have to construct |x by a suitable creation operator, then
you can calculate x|p.


And one more doubt....lets say I've |x, and let c be correspondig
probability amplitude,

Corresponding to what?


then shouldnt c satisfy the following continuity eqn of some sort

d/dt(c*c) + \grad.S = 0

where S is whatever.....(I'm not really bothered)

If c is a non-relativistic wavefunction, then it satisfies such an
equation. If it's a relativistic wave function, it doesn't satisfy
such an equation necessarily (because particle number is not
conserved).


And from what I've read, in the continuity eqn
d/dt(P) + \grad.S = 0 that we get from QFT,
this guy P is not positive definite,....and so P cannot be interpreted
as postition probability density

Are you talking about the Klein-Gordon equation here, where it's not
clear how to define the probability density?



If the position
is not predicted by the theory, then what does the uncertainity relation
[snip]


Or does the uncertainity relation take some other form??

In QFT, one talks about fields (often about particles, too, but the
fields are the more fundamental objects).

Ah, this is one of the greatest botheration to me,...and wud be really
grateful if someone can tell me in simple stuff, what the hell forced
mankind to think about fields instead of particles??

You have to ask Faraday, IIRC he brought the idea up. ;-)

Electrodynamics is best described using fields; even now when we have
the "particle" interpretation of it (real and virtual photons), in
actual calculations about (macroscopic) electrodynamics phenomena,
it's *still* *MUCH* easier to use fields rather than particles.


I guess it cant just be lorentz
invariance, because, the original dirac eqn was lorentz invariant and still
talked about single particles and not fields.

Well, one can argue that the wave function itself is a field. ;-)
(a "field" in physics is (sometimes?) defined simply as a function
which depends on all three space coordinates and is defined on all
space)


What knowledge do you have about QM? Specifically, are you familiar with
the Dirac notation? Although, are you familiar with Special Relativity,
specifically four-vector notation?

yeah, but cant really comment on the depth of knowledge....

It's enough if you know what these terms mean and how to use them in
simple calculations.


Bye,
Bjoern