Quantum mechanics and operators


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Topic: Science > Physics
User: "cfgauss"
Date: 15 Nov 2004 01:31:54 AM
Object: Quantum mechanics and operators
I'm trying to find out more about where the operators in quantum mechanics
come from. From what I see, the justification for using them is that using
them with the wave function, which is already known, gives you the
expectation value for that operator, i.e., <E> = Integral[ phi* (-i hbar d/dt)
phi*]. But then I see the justification for the wave function as that if
you put the operators in the equation E = p^2/2m, you get the Schrodinger
equation. Well, that's great, but that argument just goes in a big circle.
If you didn't know the wave equation, the solution to the wave equation, or
the operators, how would you be able to come up with them? And what is the
justification for what the operators are, other than "because it works." Do
they "come" from anywhere?
Also, if we look at the relativistic correct wave equation, my quantum
mechanics books tell me that it comes from putting the operators into the
relativistic energy equation. How can you justify doing this? How do you
know the operators are relativisticly correct? How do we know that we don't
have anything like an operator for mass that we have to put into the
equations?
I'm also interested in the use of complex numbers to write the equation. My
quantum mechanics book mentions that they are used to simplify the
equations, and that it doesn't physically mean anything. It mentions that
you can write, for example, the equations describing electricity and
magnetism in a form like that by saying F = E + icB, then you can write
Maxwell's equations in terms of this F instead of separate equations for E
and for B. So, you could also write the wave equation as the sum of two
other "things," phi = phi_1 + i phi_2. Is there any kind of representation
for these "fields" phi_1 and phi_2? Are they physically meaningful? Also,
is there a physically meaningful representation of the field F = E + icB?
I've tried to look through several quantum mechanics books for answers to
these questions, but I haven't seen any argument that doesn't seem circular
to me.
Thanks very much!
- Jeremy
.

User: "John T Lowry"

Title: Re: Quantum mechanics and operators 15 Nov 2004 12:18:05 PM
"cfgauss" <cfgauss@u.washington.edu> wrote in message
news:9e2e7039.0411142331.6c0a8309@posting.google.com...

I'm trying to find out more about where the operators in quantum
mechanics
come from. From what I see, the justification for using them is that
using
them with the wave function, which is already known, gives you the
expectation value for that operator, i.e., <E> = Integral[ phi* (-i
hbar d/dt)
phi*]. But then I see the justification for the wave function as that
if
you put the operators in the equation E = p^2/2m, you get the
Schrodinger
equation. Well, that's great, but that argument just goes in a big
circle.
If you didn't know the wave equation, the solution to the wave
equation, or
the operators, how would you be able to come up with them? And what
is the
justification for what the operators are, other than "because it
works." Do
they "come" from anywhere?

Also, if we look at the relativistic correct wave equation, my quantum
mechanics books tell me that it comes from putting the operators into
the
relativistic energy equation. How can you justify doing this? How do
you
know the operators are relativisticly correct? How do we know that we
don't
have anything like an operator for mass that we have to put into the
equations?

I'm also interested in the use of complex numbers to write the
equation. My
quantum mechanics book mentions that they are used to simplify the
equations, and that it doesn't physically mean anything. It mentions
that
you can write, for example, the equations describing electricity and
magnetism in a form like that by saying F = E + icB, then you can
write
Maxwell's equations in terms of this F instead of separate equations
for E
and for B. So, you could also write the wave equation as the sum of
two
other "things," phi = phi_1 + i phi_2. Is there any kind of
representation
for these "fields" phi_1 and phi_2? Are they physically meaningful?
Also,
is there a physically meaningful representation of the field F = E +
icB?

I've tried to look through several quantum mechanics books for answers
to
these questions, but I haven't seen any argument that doesn't seem
circular
to me.

Thanks very much!
- Jeremy

Here's a provocative statement, to your quest, in L.E. Ballentine's
Quantum Mechanics: A Modern Development, p. 77: "The dynamics of a free
particle are invariant under the full Galilei group of space-time
transformations, and this turns out to be sufficient to completely
identify the operators for its dynamical variables. The method is based
on a paper by T.F. Jordan (1975)."
I suspect you'll find quite a bit of what you're looking for in
Ballentine's very excellent book.
John Lowry
Flight Physics
.
User: "Edward Green"

Title: Re: Quantum mechanics and operators 15 Nov 2004 06:59:55 PM
"John T Lowry" <jlowry100@earthlink.net> wrote in message news:<xp6md.4498$G36.2430@newsread3.news.pas.earthlink.net>...

Here's a provocative statement, to your quest, in L.E. Ballentine's
Quantum Mechanics: A Modern Development, p. 77: "The dynamics of a free
particle are invariant under the full Galilei group of space-time
transformations, and this turns out to be sufficient to completely
identify the operators for its dynamical variables.

ITI "the full Galilei group" means the group of transformations
characterizing the fabled "Gallilean invariance", and that this
therefore applies to non-relativistic quantum mechanics.
Intriguing!

The method is based
on a paper by T.F. Jordan (1975)."

I suspect you'll find quite a bit of what you're looking for in
Ballentine's very excellent book.

John Lowry
Flight Physics

.


User: "Edward Green"

Title: Re: Quantum mechanics and operators 15 Nov 2004 09:18:14 PM
(cfgauss) wrote in message news:<9e2e7039.0411142331.6c0a8309@posting.google.com>...

I'm trying to find out more about where the operators in quantum mechanics
come from. From what I see, the justification for using them is that using
them with the wave function, which is already known, gives you the
expectation value for that operator, i.e., <E> = Integral[ phi* (-i hbar d/dt)
phi*]. But then I see the justification for the wave function as that if
you put the operators in the equation E = p^2/2m, you get the Schrodinger
equation. Well, that's great, but that argument just goes in a big circle.
If you didn't know the wave equation, the solution to the wave equation, or
the operators, how would you be able to come up with them? And what is the
justification for what the operators are, other than "because it works." Do
they "come" from anywhere?

Also, if we look at the relativistic correct wave equation, my quantum
mechanics books tell me that it comes from putting the operators into the
relativistic energy equation. How can you justify doing this? How do you
know the operators are relativisticly correct? How do we know that we don't
have anything like an operator for mass that we have to put into the
equations?

I'm also interested in the use of complex numbers to write the equation. My
quantum mechanics book mentions that they are used to simplify the
equations, and that it doesn't physically mean anything. It mentions that
you can write, for example, the equations describing electricity and
magnetism in a form like that by saying F = E + icB, then you can write
Maxwell's equations in terms of this F instead of separate equations for E
and for B. So, you could also write the wave equation as the sum of two
other "things," phi = phi_1 + i phi_2. Is there any kind of representation
for these "fields" phi_1 and phi_2? Are they physically meaningful? Also,
is there a physically meaningful representation of the field F = E + icB?

I've tried to look through several quantum mechanics books for answers to
these questions, but I haven't seen any argument that doesn't seem circular
to me.

It seems to me you need a healthy dose of Castor Philosophy of
Science!
"Justification" only applies to deduction, which is to say, the
elaboration of consequences from existing rule sets. How then to
generate a new rule set, not implied by existing rule sets? It's
called "creativity" or "genius", which is a signal we don't really
know, but one fruitful method seems to involve "kind of sort of"
arguments, which would be laughed out of court in deduction: look at
this situation over here -- doesn't it kind of look like that one over
there? What if we tried to build some analogous structure?
There is no "justification" for such procedures, except, in the end,
that they work. That is, some do: no doubt most such attempts are
dead ends. Given that the new rule set parsimoniously describes a
broad swath of experiment, either broader or at least more
parsimoniously than before, it's no longer subject to any need to
justify the logical scaffolding used in its erection.
This is a point not often grasped by would-be apple cart upsetters:
for example, a gentleman posting here thinks that since an argument
used by Einstein to motivate GR cannot be "justified" in your sense
("pulled out of his *****" might suggest itself to less refined
commentators), that he has discovered a fatal flaw in GR, which must
collapse! Which is nonsense. Such arguments are motivational and
marketing tools for a new theory: cosmetics, not structure.
And one more thing: "physically meaningful" is a cant phrase.
Theories are physically meaningful when they map to experiment. If
they have internal machinery which cannot be directly mapped to
experiment, yet apparently cannot be pulled out without breaking the
black box, then they are "physically meaningful". Saying otherwise is
like pointing to a particular gear inside a watch and saying its not
"temporally meaningful" because it doesn't connect directly to the
hands! If its necessary for the watch to keep time, its meaningful.
But it's certainly still a good idea to constantly reexamine logical
structures which work to try to understand just what their necessary
content is -- just because some highly trained people can turn the
cranks, doesn't mean they see everything there which is worth seeing.
.

User: "zigoteau"

Title: Re: Quantum mechanics and operators 15 Nov 2004 10:18:47 AM
(cfgauss) wrote in message news:<9e2e7039.0411142331.6c0a8309@posting.google.com>...
Hi, Jeremy,

I'm trying to find out more about where the operators in quantum mechanics
come from. From what I see, the justification for using them is that using
them with the wave function, which is already known, gives you the
expectation value for that operator, i.e., <E> = Integral[ phi* (-i hbar d/dt)
phi*]. But then I see the justification for the wave function as that if
you put the operators in the equation E = p^2/2m, you get the Schrodinger
equation. Well, that's great, but that argument just goes in a big circle.
If you didn't know the wave equation, the solution to the wave equation, or
the operators, how would you be able to come up with them? And what is the
justification for what the operators are, other than "because it works." Do
they "come" from anywhere?

Ultimately, there is no better answer than "because it works". Do you
have a problem with that? Way back in the time of Nebuchadnezzar,
people didn't know any physics. All we've learnt since is guesswork
confirmed by experiment.
That being said, there was a path to the Schrödinger equation. Newton
thought about whether light was composed of particles or waves.
Maxwell and Fresnel seemed to prove conclusively that light was a
wave. Then Planck came along and showed that light sometimes behaved
like particles. Energy E and momentum p are properties of particles,
frequency in time (f) and space (q) are properties of waves. Einstein
confirmed the Planck relationship E = hf. Subsequently de Broglie
showed that p = hq. The time-frequency operator is i/2/pi*d/dt.
Schrödinger was the first to write down an equation for electron
waves, but he wasn't just guessing - a lot of the spadework had been
done by others before him. That is however not the reason that it is
still taught in textbooks. It is still taught "because it works".

Also, if we look at the relativistic correct wave equation, my quantum
mechanics books tell me that it comes from putting the operators into the
relativistic energy equation. How can you justify doing this?

"Because it works" seems a good enough reason.

How do you
know the operators are relativisticly correct?

There are a number of relativistic wave equations: the Klein-Gordon
equation and the Dirac equation, for starters. You choose the one
which gives results corresponding to experiment. It's a damn sight
more convincing than gazing at your navel, or taking
Aristotle/Mohammed/the Pope's word for it.

How do we know that we don't
have anything like an operator for mass that we have to put into the
equations?

How do we know anything?

I'm also interested in the use of complex numbers to write the equation. My
quantum mechanics book mentions that they are used to simplify the
equations, and that it doesn't physically mean anything.

Complex numbers come in whenever you have waves. In electronics and
electrical engineering, the voltage on a wire is always real, but it
is represented by a complex number. Complex numbers have a real part
and an imaginary part, which are a lot easier to write down on a piece
of paper than a voltage at all possible instants of time, of which
there are infinitely many.
The name "imaginary" for the square root of -1 suggests something a
little weird. But when you write down a complex number on a piece of
paper it just sits there, and does not start to smoke or disappear.
It's just a mathematical object that you can calculate with according
to well-defined rules. If you prefer to separate the real and
imaginary part and to work with them as distinct real numbers, that's
fine, but you will soon start to see that those two numbers always go
together, and only get combined in certain symmetrical ways.

It mentions that
you can write, for example, the equations describing electricity and
magnetism in a form like that by saying F = E + icB, then you can write
Maxwell's equations in terms of this F instead of separate equations for E
and for B. So, you could also write the wave equation as the sum of two
other "things," phi = phi_1 + i phi_2. Is there any kind of representation
for these "fields" phi_1 and phi_2? Are they physically meaningful?
Also, is there a physically meaningful representation of the field F = E + icB?

What do you mean by "physically meaningful"?

I've tried to look through several quantum mechanics books for answers to
these questions, but I haven't seen any argument that doesn't seem circular
to me.

Quantum mechanics can be very confusing for a beginner. Relax, you're
in very good company. Even the greatest physicists, including Einstein
and Feynman, had the sort of problems with quantum mechanics that you
are having. The main problem for people writing the textbooks is that
*nobody* knows what it means.
However ultimately, the stuff in the textbooks is there "because it
works".
Cheers,
Zigoteau.
.
User: "Bob Cain"

Title: Re: Quantum mechanics and operators 19 Nov 2004 01:35:12 AM
zigoteau wrote:

Complex numbers come in whenever you have waves. In electronics and
electrical engineering, the voltage on a wire is always real, but it
is represented by a complex number. Complex numbers have a real part
and an imaginary part, which are a lot easier to write down on a piece
of paper than a voltage at all possible instants of time, of which
there are infinitely many.

Isn't it fair to say that complex numbers are just neat ways
to represent two valued variables and functions where the
values are orthogonal?
With the harmonic content of signals it is magnitude and
phase. In fact they are usually in that context.
Bob
--
"Things should be described as simply as possible, but no
simpler."
A. Einstein
.
User: "zigoteau"

Title: Re: Quantum mechanics and operators 20 Nov 2004 05:45:48 AM
Bob Cain <arcane@arcanemethods.com> wrote in message news:<cnk7mb0hg0@enews2.newsguy.com>...
Hi, Bob,

Complex numbers come in whenever you have waves. In electronics and
electrical engineering, the voltage on a wire is always real, but it
is represented by a complex number. Complex numbers have a real part
and an imaginary part, which are a lot easier to write down on a piece
of paper than a voltage at all possible instants of time, of which
there are infinitely many.


Isn't it fair to say that complex numbers are just neat ways
to represent two valued variables and functions where the
values are orthogonal?

I have to go along with Robert Kolker on that one. Yes, you're right,
but complex numbers have additional mathematical structure.

With the harmonic content of signals it is magnitude and
phase. In fact they are usually in that context.

I can see that that would be the case for an audio enthusiast like
you. I come across complex numbers in all sorts of other contexts.
Cheers,
Zigoteau.
.
User: "Bob Cain"

Title: Re: Quantum mechanics and operators 20 Nov 2004 07:50:14 PM
zigoteau wrote:

Isn't it fair to say that complex numbers are just neat ways
to represent two valued variables and functions where the
values are orthogonal?


I have to go along with Robert Kolker on that one. Yes, you're right,
but complex numbers have additional mathematical structure.

But in analysis, especially Fourier analysis, of signals,
which is a pretty general problem classification, is that
full structure not utilized?

With the harmonic content of signals it is magnitude and
phase. In fact they are usually in that context.


I can see that that would be the case for an audio enthusiast like
you.

Actually it's from being an E.E. but you're forgiven the
diminution. :-)

I come across complex numbers in all sorts of other contexts.

No doubt. I was thinking specifically of QM where it pretty
much is about magnitude and phase, right? Is not its use in
the Schrodinger equation of that same nature?
I've just always felt that too much intrinsic meaning has
been given the use of the complex numbers when they are just
a computational or notational convenience for certain kinds
of problems.
Bob
--
"Things should be described as simply as possible, but no
simpler."
A. Einstein
.
User: ""

Title: Re: Quantum mechanics and operators 21 Nov 2004 05:09:46 AM
In article <cnos6a02b3a@enews2.newsguy.com>,
Bob Cain <arcane@arcanemethods.com> wrote:



zigoteau wrote:

Isn't it fair to say that complex numbers are just neat ways
to represent two valued variables and functions where the
values are orthogonal?


I have to go along with Robert Kolker on that one. Yes, you're right,
but complex numbers have additional mathematical structure.


But in analysis, especially Fourier analysis, of signals,
which is a pretty general problem classification, is that
full structure not utilized?

With the harmonic content of signals it is magnitude and
phase. In fact they are usually in that context.


I can see that that would be the case for an audio enthusiast like
you.


Actually it's from being an E.E. but you're forgiven the
diminution. :-)

I come across complex numbers in all sorts of other contexts.


No doubt. I was thinking specifically of QM where it pretty
much is about magnitude and phase, right? Is not its use in
the Schrodinger equation of that same nature?

I've just always felt that too much intrinsic meaning has
been given the use of the complex numbers when they are just
a computational or notational convenience for certain kinds
of problems.

<splutter> [emoticon wipes oatmeal off TTY screen] Now
think about computers' floating point and decimal arithmetic.
Would you care to strengthen your use of the
word "convenience"? ;-)
/BAH
Subtract a hundred and four for e-mail.
.
User: "Steve Harris"

Title: Re: Quantum mechanics and operators 21 Nov 2004 10:08:34 PM
wrote in message news:<v5mdncyLEMjv4j3cRVn-1w@rcn.net>...

In article <cnos6a02b3a@enews2.newsguy.com>,
Bob Cain <arcane@arcanemethods.com> wrote:



zigoteau wrote:

Isn't it fair to say that complex numbers are just neat ways
to represent two valued variables and functions where the
values are orthogonal?


I have to go along with Robert Kolker on that one. Yes, you're right,
but complex numbers have additional mathematical structure.


But in analysis, especially Fourier analysis, of signals,
which is a pretty general problem classification, is that
full structure not utilized?

With the harmonic content of signals it is magnitude and
phase. In fact they are usually in that context.


I can see that that would be the case for an audio enthusiast like
you.


Actually it's from being an E.E. but you're forgiven the
diminution. :-)

I come across complex numbers in all sorts of other contexts.


No doubt. I was thinking specifically of QM where it pretty
much is about magnitude and phase, right? Is not its use in
the Schrodinger equation of that same nature?

I've just always felt that too much intrinsic meaning has
been given the use of the complex numbers when they are just
a computational or notational convenience for certain kinds
of problems.


<splutter> [emoticon wipes oatmeal off TTY screen] Now
think about computers' floating point and decimal arithmetic.
Would you care to strengthen your use of the
word "convenience"? ;-)

/BAH

COMMENT
It's all the same to a computer. The convenience of "i" is for the guy
with the blackboard. By the time you're doing numerical solutions to
complex equations, the i's are long gone. It all comes out in the wash
to positive and negative strings of 1's and 0's. When you want to
reduce quantum mechanics to numerical solutions, you convert all those
i's to sins and cosines (or maybe you're using a high level language
that does that for you) and inside the computer they go though
whatever series algorithm the machine uses to calculate trig functions
and exponentials and it goes round and round and it comes out a
real-numbered answer. Even if it was QED or QCD to begin with. So are
the i's really there in nature? No more than the trig functions and
integrations and exponentiations are "really" there. At some level,
all physics can be expressed as strings of rules for adding and
subtracting 1's and 0's. Which level you view as "reality" is a matter
of taste. And of course, of your speed with adding and substracting
1's and 0's.
SBH
.
User: ""

Title: Re: Quantum mechanics and operators 22 Nov 2004 06:15:10 AM
In article <79cf0a8.0411212008.51537cac@posting.google.com>,
(Steve Harris sbharris@ROMAN9.netcom.com) wrote:

jmfbahciv@aol.com wrote in message

news:<v5mdncyLEMjv4j3cRVn-1w@rcn.net>...

In article <cnos6a02b3a@enews2.newsguy.com>,
Bob Cain <arcane@arcanemethods.com> wrote:



zigoteau wrote:

Isn't it fair to say that complex numbers are just neat ways
to represent two valued variables and functions where the
values are orthogonal?


I have to go along with Robert Kolker on that one. Yes, you're right,
but complex numbers have additional mathematical structure.


But in analysis, especially Fourier analysis, of signals,
which is a pretty general problem classification, is that
full structure not utilized?

With the harmonic content of signals it is magnitude and
phase. In fact they are usually in that context.


I can see that that would be the case for an audio enthusiast like
you.


Actually it's from being an E.E. but you're forgiven the
diminution. :-)

I come across complex numbers in all sorts of other contexts.


No doubt. I was thinking specifically of QM where it pretty
much is about magnitude and phase, right? Is not its use in
the Schrodinger equation of that same nature?

I've just always felt that too much intrinsic meaning has
been given the use of the complex numbers when they are just
a computational or notational convenience for certain kinds
of problems.


<splutter> [emoticon wipes oatmeal off TTY screen] Now
think about computers' floating point and decimal arithmetic.
Would you care to strengthen your use of the
word "convenience"? ;-)

/BAH



COMMENT

It's all the same to a computer. The convenience of "i" is for the guy
with the blackboard. By the time you're doing numerical solutions to
complex equations, the i's are long gone. It all comes out in the wash
to positive and negative strings of 1's and 0's. When you want to
reduce quantum mechanics to numerical solutions, you convert all those
i's to sins and cosines (or maybe you're using a high level language
that does that for you) and inside the computer they go though
whatever series algorithm the machine uses to calculate trig functions
and exponentials and it goes round and round and it comes out a
real-numbered answer. Even if it was QED or QCD to begin with. So are
the i's really there in nature? No more than the trig functions and
integrations and exponentiations are "really" there. At some level,
all physics can be expressed as strings of rules for adding and
subtracting 1's and 0's. Which level you view as "reality" is a matter
of taste. And of course, of your speed with adding and substracting
1's and 0's.

I had the very rare pleasure of watching two people do the work
that you so blithely dismiss as mere arrangements and manipulations
of zeros and one. With all due respect, you don't know that
you're talking about.
My comment had to do with the level of the understatement
committed by the poster when he said convenience.
/BAH
Subtract a hundred and four for e-mail.
.
User: "Ken Muldrew"

Title: Re: Quantum mechanics and operators 22 Nov 2004 01:41:12 PM
wrote:

I had the very rare pleasure of watching two people do the work
that you so blithely dismiss as mere arrangements and manipulations
of zeros and one. With all due respect, you don't know that
you're talking about.

I know what Steve is talking about (briefly: math was developed when
arithmetic was hard to do, so a lot of the structures, techniques, and
the notation are designed to minimize the amount of arithmetic that
must be done) but I don't know what you're talking about.

My comment had to do with the level of the understatement
committed by the poster when he said convenience.

OK, getting computers to do math wasn't easy. Neither was it easy to
produce tables of logarithms for an earlier method of simplifying
arithmetic. But once the tables are there, and similarly once
computers can get the job done, it's better to simply take those
developments as foundational and see what can be done with them. To do
otherwise is to become paralyzed by a neverending reverance for the
past. Progress is made by internalizing what our predecessors have
done and using those tools to their full potential. Understatement is
a necessary part of the ritual.
Ken Muldrew
kmuldrezw@ucalgazry.ca
(remove all letters after y in the alphabet)
.
User: ""

Title: Re: Quantum mechanics and operators 23 Nov 2004 07:22:23 AM
In article <41a23ecd.335855162@news.ucalgary.ca>,
(Ken Muldrew) wrote:

jmfbahciv@aol.com wrote:

I had the very rare pleasure of watching two people do the work
that you so blithely dismiss as mere arrangements and manipulations
of zeros and one. With all due respect, you don't know that
you're talking about.


I know what Steve is talking about (briefly: math was developed when
arithmetic was hard to do, so a lot of the structures, techniques, and
the notation are designed to minimize the amount of arithmetic that
must be done) but I don't know what you're talking about.

My comment had to do with the level of the understatement
committed by the poster when he said convenience.


OK, getting computers to do math wasn't easy.

Exactly.

..Neither was it easy to
produce tables of logarithms for an earlier method of simplifying
arithmetic. But once the tables are there, and similarly once
computers can get the job done, it's better to simply take those
developments as foundational and see what can be done with them.

Sure. However, the difference between hardcopy numbers and
computer-stored numbers is that all computer storage isn't
alike. With each architecture development, the bit arrangement
work has to be done all over again. What's more, you can
different results. This is a serious aspect of using
computers and is not a matter of "done once, and never done
again."

..To do
otherwise is to become paralyzed by a neverending reverance for the
past. Progress is made by internalizing what our predecessors have
done and using those tools to their full potential. Understatement is
a necessary part of the ritual.

Of course. But I'm seeing the myth that computers can add
automagically becoming the common assumption, even among
people who should know better.
It is my habit to stomp out smoking ashes before they start
a forest fire.
/BAH
Subtract a hundred and four for e-mail.
.



User: "Keith F. Lynch"

Title: Re: Quantum mechanics and operators 10 Dec 2004 08:06:49 PM
Steve Harris
<sbharris@ix.netcom.com> wrote:

It's all the same to a computer. The convenience of "i" is for
the guy with the blackboard. By the time you're doing numerical
solutions to complex equations, the i's are long gone.

Agreed.

It all comes out in the wash to positive and negative strings of 1's
and 0's.

Typically there isn't a separate sign bit, at least for integers.
Instead, if the leading bit is 1 the number is treated as negative.
That way there's no need for subtraction, just addition and
complementing.
The base ten analog of this would be to call the year 1 BC 9999 AD
instead. One year before that was 9998, etc. Five years after 9998
AD was 10003 AD, or, discarding the leading 1, 3 AD. So long as you
never go back further than 4999 BC or forward further than 4999 AD, it
works well.
(Yes, I know that there wasn't actually a year 0, but that 1 BC was
immediately followed by 1 AD -- a complication I am deliberately
ignoring.)

At some level, all physics can be expressed as strings of rules for
adding and subtracting 1's and 0's.

Just adding, and complementing, no need for subtraction.
(Complementing consists of replacing all 0s with 1s and vice versa,
then adding 1.) And addition just consists of excusive-oring bits and
carrying.

Which level you view as "reality" is a matter of taste.

Yes. Most people prefer something like a story, or a picture, and not
just a long list of calculations.
On a separate subject, did you get my email of Monday the 6th?
Thanks.
--
Keith F. Lynch - http://keithlynch.net/
Please see http://keithlynch.net/email.html before emailing me.
.
User: "Lewis Mammel"

Title: Re: Quantum mechanics and operators 12 Dec 2004 12:40:45 AM
"Keith F. Lynch" wrote:


Steve Harris

<sbharris@ix.netcom.com> wrote:

It all comes out in the wash to positive and negative strings of 1's
and 0's.


Typically there isn't a separate sign bit, at least for integers.
Instead, if the leading bit is 1 the number is treated as negative.
That way there's no need for subtraction, just addition and
complementing.

Sure, but I think it's ironic to accept the 1's and 0's as a concrete
reality which dispels the abstract illusion of "i". The 1's and 0's
are an abstract interpretation of the electronic states of a carefully
engineered silicon substrate, and this engineering is firmly grounded
in the quantum mechanics of the solid state, an understanding of which
would be impossible without ... yeah, you got it, "i" .
Lew Mammel, Jr.
.
User: "Edward Green"

Title: Re: Quantum mechanics and operators 12 Dec 2004 07:40:25 AM
Lew Mammel, Jr. wrote:

Sure, but I think it's ironic to accept the 1's and 0's as a concrete
reality which dispels the abstract illusion of "i". The 1's and 0's
are an abstract interpretation of the electronic states of a

carefully

engineered silicon substrate, and this engineering is firmly grounded
in the quantum mechanics of the solid state, an understanding of

which

would be impossible without ... yeah, you got it, "i" .

Well, that's a rather roundabout and convoluted defense of "i", but I
agree with you in spirit.
You might say that any feasible calculation involves the manipulation
of a finite symbol set, which could be 0,1; 0,....,9; 0,1,-1,i,-i; or
*,^,7,q . The names or glyphs are irrelevant, only the number of
symbols and their interrelations under defined operations. As the 0,1
convention shows, any sufficient set can be used to represent any
other set.
("Sufficient" is a hedge to make this vague assertion true, but one
gets the idea. ;-)
.







User: "robert j. kolker"

Title: Re: Quantum mechanics and operators 19 Nov 2004 02:00:49 AM
Bob Cain wrote:

Isn't it fair to say that complex numbers are just neat ways to
represent two valued variables and functions where the values are
orthogonal?

No sir. The complex numbers have an algebraic structure. It is the
structure you get when you adjoin the root of x^2 + 1 = 0 to the real
number field.
What you described is the cartesian product R X R where R is the real
numbers or if you prefer E^2, the euclidean plane.


With the harmonic content of signals it is magnitude and phase. In fact
they are usually in that context.

Bob Kolker
.



User: ""

Title: Re: Quantum mechanics and operators 15 Nov 2004 10:18:09 AM
(cfgauss) wrote in message news:<9e2e7039.0411142331.6c0a8309@posting.google.com>...
[a lot of confused ramblings about quantum mechanics]
I can answer all of your questions about quantum mechanics
*IF* you can answer the corresponding questions about *any*
physical theory.
Socks
.

User: "Alfred Einstead"

Title: Re: Quantum mechanics and operators 23 Nov 2004 06:50:43 PM
(cfgauss) wrote:

I'm trying to find out more about where the operators in quantum mechanics
come from.

Von Neumann answerer that question decisively with the combined
classical-quantum formalism based on the prior notion of states and
state spaces. It proceeds along the following stages -- which apply
generally to classical or quantum physics, up to a certain point.
I'll generalize it slightly below and indicate where von Neumann
comes into the picture:
(A) The observables
===================
(1) A system is described by a set of coordinates q1,q2,...
(2) As a function of time they obey a 2nd order system
q''(t) = a(q(t), q'(t))
(q(t) is the vector of all the q's)
(3) The q's form an algebra in which
qi(t) qj(t) = qj(t) qi(t) for all i,j = 0,1,2,...
(4) Dispersion relations:
for all functions A(q), B(q) of the coordinates
A(q(t)) B(q(t+dt)) - B(q(t+dt)) A(q(t)) = i W_AB(t) dt
where the W_AB is a functional of A, B.
For classical physics W_AB = 0. For quantum physics
W_AB is a function of a parameter h-bar such that
W_{qi qj}/h-bar -> Wij as h-bar -> 0
with W a non-singular matrix.
In the quantum case, this places severe constraints on which
equations of motion are possible, since one needs to have
d/dt (qi(t) qj(t) - qj(t) qi(t)) = 0
and similar relations constructed form the other commutators,
as well as the Jacobi relation. The constraints, in fact,
force the equations of motion in the classical limit to be
those deriveable from a Lagrangian. Then the matrix W is
actually the inverse of the "mass" matrix:
W = M^{-1}
M_ij = d^2 L(q,q') / d qi' d qj'
(This result is a theorem that wasn't known until around 1990).
This is also the condition which determines when a Lagrangian
system has a Hamiltonian (i.e., when the matrix M has an inverse).
So, the system in the case of quantum mechanics has a Hamiltonian
system as its classical limit -- all quantum systems as defined
above are therefore quantizations of classical Hamiltonian systems.
Now, this is where von Neumann comes into the picture.
(B) The states
==============
Since the system's equations of motion are solved uniquely by the
initial values (q(0), q'(0)), then the system is effectively described
by this collection. Classically, this would comprise the phase
space. All variables q(t), q'(t) are functions of the initial values
q(0) and q'(0), so that one only needs to focus attention on these
values. An algebra is defined solely by the commutator relations
for q(0) and q'(0) -- i.e., equal-time commutator relations at a
specific time, time 0.
A state is defined solely by the set of expectation values it
yields for all functions A(q(0), q'(0)). If W is the state then
the corresponding expectation value would be W[A(q(0),q'(0))].
The requirements are that
W[1] = 1; W[A^2] >= 0;
W[A1],W[A2],W[A3]...->W[A], if A1,A2,A3,...->A.
States can be combined in mixtures. If W1, W2 are states then the
combination
p W1 + q W2; p + q = 1
also satisfies the definition requirements for a state. When W1,
W2 are different it's called a mixed state. Those states not
decomposable in this fashion are then called pure.
The key theorem proven is:
For commutative algebra the space of pure states is
exactly the same as the phase space. Each pure state of
the form
W_{qv}[A(q(0),q'(0))] = A(q,v) corresponding to a point
(q,v) of the phase space.
For a non-commutative algebra, such as that satisfying the
relations above for the quantum case (the theorem proven
applies more generally), every pure state W_v corresponds
to a normed vector (up to a complex number z with |z|=1)
in a Hilbert space H, with observables corresponding to
operators over H such that:
W_v[A(q(0),q'(0))] = <v| A(Q(0),Q'(0)) |v>
where Q(0), Q'(0) are the respective vectors of operators
corresponding to the vectors q(0) and q'(0).
That theorem is where the operator representation ultimately
emerges from.
.

User: "Gregory L. Hansen"

Title: Re: Quantum mechanics and operators 17 Nov 2004 08:36:33 PM
In article <9e2e7039.0411142331.6c0a8309@posting.google.com>,
cfgauss <cfgauss@u.washington.edu> wrote:

I'm trying to find out more about where the operators in quantum mechanics
come from. From what I see, the justification for using them is that using
them with the wave function, which is already known, gives you the
expectation value for that operator, i.e., <E> = Integral[ phi* (-i hbar d/dt)
phi*]. But then I see the justification for the wave function as that if
you put the operators in the equation E = p^2/2m, you get the Schrodinger
equation. Well, that's great, but that argument just goes in a big circle.
If you didn't know the wave equation, the solution to the wave equation, or
the operators, how would you be able to come up with them? And what is the
justification for what the operators are, other than "because it works." Do
they "come" from anywhere?

Operators do what they're designed to do.
First recall that in the position representation, X psi(x) = x*psi(x), and
in the momentum representation, P psi(p) = p*psi(x). These are really
just the first moment, as when you calculate a center of mass with
x*density(x). The wavefunction |psi> is a vector in a Hilbert space. A
classical analogy would be a coordinate in phase space with a matrix that
picks it out,
(x1 0 0 0...)(0)
(0 x2 0 0 )(1) = x2
(0 0 x3 0 )(0)
...
And you might make the vector represent position probabilities, or the
frequencies of occurance in an ensemble. A rotation matrix acting on a
vector is also an operator. Position vector, vector in a Hilbert space,
it's all linear algebra.
With the definitions P psi(p) = p psi(p) and X psi(x) = x psi(x), I
thought I could Fourier transform one of them, integrate by parts, and
derive the position operator in the momentum representation. I haven't
figured that out, and I'm not sure why it doesn't seem to work.
But it's more usual to define momentum as the generator of translations.
Fourier transform a translated wavefunction,
psi(x+a) = psi(x) + a (d/dx) psi + 1/2 a^2 (d/dx)^2 psi + ...
= exp(a d/dx) psi
And it's certainly true that you can boost a wavefunction with an operator
of that type,
psi' = exp(ipx/hbar) psi
(or something like thta). For an infinitesmal translation e,
psi(x + e) = psi(x) + e (d/dx) psi(x)
When you know that psi(x) is a wavefunction and p=h/lambda, you can fix
the constants.
Someone more clever than me might be able to derive it from the
commutation relations, or the definition of momentum from the Lagrangian.
And I thought I'd seen that somewhere, and it disturbs me that I can't
seem to find it now.

Also, if we look at the relativistic correct wave equation, my quantum
mechanics books tell me that it comes from putting the operators into the
relativistic energy equation. How can you justify doing this? How do you
know the operators are relativisticly correct? How do we know that we don't
have anything like an operator for mass that we have to put into the
equations?

By definition. -i*hbar*d/dt is the energy operator, i*hbar*d/dx is the
momentum operator in the position representation. Mass is a property
intrinsic to a particle, but there is a number operator.


I'm also interested in the use of complex numbers to write the equation. My
quantum mechanics book mentions that they are used to simplify the
equations, and that it doesn't physically mean anything. It mentions that
you can write, for example, the equations describing electricity and
magnetism in a form like that by saying F = E + icB, then you can write
Maxwell's equations in terms of this F instead of separate equations for E
and for B. So, you could also write the wave equation as the sum of two
other "things," phi = phi_1 + i phi_2. Is there any kind of representation
for these "fields" phi_1 and phi_2? Are they physically meaningful? Also,
is there a physically meaningful representation of the field F = E + icB?

I've tried to look through several quantum mechanics books for answers to
these questions, but I haven't seen any argument that doesn't seem circular
to me.

Thanks very much!
- Jeremy

--
"When the fool walks through the street, in his lack of understanding he
calls everything foolish." -- Ecclesiastes 10:3, New American Bible
.
User: "Edward Green"

Title: Re: Quantum mechanics and operators 19 Nov 2004 05:36:01 PM
(Gregory L. Hansen) wrote in message news:<cnh1rg$t3s$1@hood.uits.indiana.edu>...

With the definitions P psi(p) = p psi(p) and X psi(x) = x psi(x), I
thought I could Fourier transform one of them, integrate by parts, and
derive the position operator in the momentum representation. I haven't
figured that out, and I'm not sure why it doesn't seem to work.

Hmm... I was able to make something like that work on my first
blundering attempt, though I'm not sure if I have grown in wisdom by
it.
Consider a pair of functions related by:
f(x) = Int[+/-oo] dk exp(ikx)g(k)
then...
f'(x) = Int[+/-oo] dk exp(ikx)(ik)g(k)
Apparently differentiating one side of the transform pair is
equivalent to multiplying the other side by i times the argument.
(And may I say by carefully writing "P" vs. "p" and "X" vs. "x" above
you are making significant progress in your fight with notational
abuse, and may someday look forward to unsupervised visits? ;-)
.
User: "Gregory L. Hansen"

Title: Re: Quantum mechanics and operators 19 Nov 2004 07:23:24 PM
In article <eca320d0.0411191536.f3bd022@posting.google.com>,
Edward Green <spamspamspam3@netzero.com> wrote:

glhansen@steel.ucs.indiana.edu (Gregory L. Hansen) wrote in message
news:<cnh1rg$t3s$1@hood.uits.indiana.edu>...

With the definitions P psi(p) = p psi(p) and X psi(x) = x psi(x), I
thought I could Fourier transform one of them, integrate by parts, and
derive the position operator in the momentum representation. I haven't
figured that out, and I'm not sure why it doesn't seem to work.


Hmm... I was able to make something like that work on my first
blundering attempt, though I'm not sure if I have grown in wisdom by
it.

Consider a pair of functions related by:

f(x) = Int[+/-oo] dk exp(ikx)g(k)

then...

f'(x) = Int[+/-oo] dk exp(ikx)(ik)g(k)

Apparently differentiating one side of the transform pair is
equivalent to multiplying the other side by i times the argument.

I was about to say that assumes a particular basis, but the Fourier
transform itself assumes a particular basis. You don't get momentum
eigenstates if your potential isn't flat.
I wanted to find some approach that's basis-agnostic. I suppose that
means either using the commutator, or the Lagrangian. But in the brief
time I gave to the problem, I wasn't clever enough to figure it out.
Even my little trick with Taylor-expanding psi(x+a) uses knowledge of
the plane wave to spruce up the constants, but at least it shows where the
d/dx comes from. exp(k*d/dx) is just how you add something if you can
only multiply.


(And may I say by carefully writing "P" vs. "p" and "X" vs. "x" above
you are making significant progress in your fight with notational
abuse, and may someday look forward to unsupervised visits? ;-)

I'm taking it one day at a time, Edward... One day at a time.
--
"A nice adaptation of conditions will make almost any hypothesis agree
with the phenomena. This will please the imagination but does not advance
our knowledge." -- J. Black, 1803.
.
User: "Edward Green"

Title: Re: Quantum mechanics and operators 20 Nov 2004 10:38:16 AM
(Gregory L. Hansen) wrote in message news:<cnm6ac$kkb$1@hood.uits.indiana.edu>...

In article <eca320d0.0411191536.f3bd022@posting.google.com>,
Edward Green <spamspamspam3@netzero.com> wrote:

Consider a pair of functions related by:

f(x) = Int[+/-oo] dk exp(ikx)g(k)

then...

f'(x) = Int[+/-oo] dk exp(ikx)(ik)g(k)

Apparently differentiating one side of the transform pair is
equivalent to multiplying the other side by i times the argument.


I was about to say that assumes a particular basis, but the Fourier
transform itself assumes a particular basis. You don't get momentum
eigenstates if your potential isn't flat.

Hmm... that's an interesting comment. It makes one think. I was
about to say "is that a particular basis, or a particular potential",
but maybe on second thought, I want to say "does it matter"?
A wave function may be expressed in the positional representation
regardless of the potential, may it not? Psi(x) is Psi(x), for a
hydrogen atom or for a free particle. And I think the same thing may
be said of the momentum representation...
[I was about to write:
"f(x) = Int[+/-oo] dk exp(ikx)g(k)"
as above, indicating a certain prejudice in favor of position, since I
didn't find it necessary to express the momentum representation in
terms of the position representation; I just wrote "Psi(x)"!]
Anyway, what I'm saying -- I didn't know I knew this ;-) -- is that
the position and momentum representations are potential independent:
they _mean_ "decompose the wave function on these particular bases",
and they don't care about no stinkin' potential. In a general these
bases will not be eigenstates of the Hamiltonian, and it turns out to
be screwy to ask if this is "basis independent", since the
representation amounts to a particular choice of basis -- independent
of the potential.

I wanted to find some approach that's basis-agnostic. I suppose that
means either using the commutator, or the Lagrangian.

Oh, you and your damn commutators and Lagrangians.

But in the brief
time I gave to the problem, I wasn't clever enough to figure it out.
Even my little trick with Taylor-expanding psi(x+a) uses knowledge of
the plane wave to spruce up the constants, but at least it shows where the
d/dx comes from. exp(k*d/dx) is just how you add something if you can
only multiply.

Well, that's my story and I'm sticking to it: A position eigenstate
is a position eigenstate and a momentum eigenstate is a momentum
eigenstate, and the meaning of these terms do not change depending on
the Hamiltonian -- they are eigenstates of the position and momentum
operators, not of the Hamiltonian.
.
User: "Gregory L. Hansen"

Title: Re: Quantum mechanics and operators 20 Nov 2004 03:54:01 PM
In article <eca320d0.0411200838.411b814e@posting.google.com>,
Edward Green <spamspamspam3@netzero.com> wrote:

glhansen@steel.ucs.indiana.edu (Gregory L. Hansen) wrote in message
news:<cnm6ac$kkb$1@hood.uits.indiana.edu>...

In article <eca320d0.0411191536.f3bd022@posting.google.com>,
Edward Green <spamspamspam3@netzero.com> wrote:


Consider a pair of functions related by:

f(x) = Int[+/-oo] dk exp(ikx)g(k)

then...

f'(x) = Int[+/-oo] dk exp(ikx)(ik)g(k)

Apparently differentiating one side of the transform pair is
equivalent to multiplying the other side by i times the argument.


I was about to say that assumes a particular basis, but the Fourier
transform itself assumes a particular basis. You don't get momentum
eigenstates if your potential isn't flat.


Hmm... that's an interesting comment. It makes one think. I was
about to say "is that a particular basis, or a particular potential",
but maybe on second thought, I want to say "does it matter"?

A wave function may be expressed in the positional representation
regardless of the potential, may it not? Psi(x) is Psi(x), for a
hydrogen atom or for a free particle. And I think the same thing may
be said of the momentum representation...

[I was about to write:

"f(x) = Int[+/-oo] dk exp(ikx)g(k)"

as above, indicating a certain prejudice in favor of position, since I
didn't find it necessary to express the momentum representation in
terms of the position representation; I just wrote "Psi(x)"!]

Anyway, what I'm saying -- I didn't know I knew this ;-) -- is that
the position and momentum representations are potential independent:
they _mean_ "decompose the wave function on these particular bases",
and they don't care about no stinkin' potential. In a general these
bases will not be eigenstates of the Hamiltonian, and it turns out to
be screwy to ask if this is "basis independent", since the
representation amounts to a particular choice of basis -- independent
of the potential.

Oh yeah, that's right. Just because momentum isn't an eigenstate of the
Hamiltonian doesn't mean momentum ceases to exist.
That reminds me, I wanted to try QED with bases other than plane waves.


I wanted to find some approach that's basis-agnostic. I suppose that
means either using the commutator, or the Lagrangian.


Oh, you and your damn commutators and Lagrangians.

Commutators are the REAL way to do QM. You can crank out your
differential equations if you want. You'll be done long before I figure
out how to do it algebraically, but my derivation will eventually be
shorter.


But in the brief
time I gave to the problem, I wasn't clever enough to figure it out.
Even my little trick with Taylor-expanding psi(x+a) uses knowledge of
the plane wave to spruce up the constants, but at least it shows where the
d/dx comes from. exp(k*d/dx) is just how you add something if you can
only multiply.


Well, that's my story and I'm sticking to it: A position eigenstate
is a position eigenstate and a momentum eigenstate is a momentum
eigenstate, and the meaning of these terms do not change depending on
the Hamiltonian -- they are eigenstates of the position and momentum
operators, not of the Hamiltonian.

--
"I fart for joy and I laugh more than if I had cast my old age, as a
serpent does its skin." -- Aristophanes, Peace, 421 BC
.





User: "ben ito"

Title: QM 19 Nov 2004 06:28:23 PM
QUANTUM MECHANICS IS A DECEPTION. QUANTUM MECHANICS IS A WAVE THEORY.
QUANTUM MECHANICS USES A WAVE STRUCTURE AND ONLY USES A SEGMENT OF
THE WAVE TO FORM A QUANTUM. THEREFORE, QUANTUM MECHANICS VIOLATES
THE LAW OF CONSERVATION OF ENERGY.. THATS WHY WE HAVE THE GREEN
HOUSE EFFECT BECAUSE PHYSICIST ARE DISREGARD THE LAW OF CONSERVATION
OF ENERGY IN ALL THEIR THEORIES. MATHATICIANS AND PHYSICIST ARE
RESPONSIBLE FOR THE END OF THE WORLD WITH THEIR ***** THEORIES.
*-----------------------*
Posted at:
www.GroupSrv.com
*-----------------------*
.
User: "Gregory L. Hansen"

Title: Re: QM 19 Nov 2004 07:28:47 PM
In article <419e8fa7$2_1@Usenet.com>,
ben ito <benito20044@yahoo-dot-com.no-spam.invalid> wrote:

QUANTUM MECHANICS IS A DECEPTION. QUANTUM MECHANICS IS A WAVE THEORY.
QUANTUM MECHANICS USES A WAVE STRUCTURE AND ONLY USES A SEGMENT OF
THE WAVE TO FORM A QUANTUM. THEREFORE, QUANTUM MECHANICS VIOLATES
THE LAW OF CONSERVATION OF ENERGY.. THATS WHY WE HAVE THE GREEN
HOUSE EFFECT BECAUSE PHYSICIST ARE DISREGARD THE LAW OF CONSERVATION
OF ENERGY IN ALL THEIR THEORIES. MATHATICIANS AND PHYSICIST ARE
RESPONSIBLE FOR THE END OF THE WORLD WITH THEIR ***** THEORIES.

*-----------------------*
Posted at:
www.GroupSrv.com
*-----------------------*

Someone had better fix theoretical physics fast, before the ice caps melt!
--
"Suppose you were an idiot... And suppose you were a member of
Congress... But I repeat myself." - Mark Twain
.
User: "William Creighton"

Title: Re: QM 19 Nov 2004 07:36:13 PM
On Sat, 20 Nov 2004, Gregory L. Hansen wrote:

In article <419e8fa7$2_1@Usenet.com>,
ben ito <benito20044@yahoo-dot-com.no-spam.invalid> wrote:

QUANTUM MECHANICS IS A DECEPTION. QUANTUM MECHANICS IS A WAVE THEORY.
QUANTUM MECHANICS USES A WAVE STRUCTURE AND ONLY USES A SEGMENT OF
THE WAVE TO FORM A QUANTUM. THEREFORE, QUANTUM MECHANICS VIOLATES
THE LAW OF CONSERVATION OF ENERGY.. THATS WHY WE HAVE THE GREEN
HOUSE EFFECT BECAUSE PHYSICIST ARE DISREGARD THE LAW OF CONSERVATION
OF ENERGY IN ALL THEIR THEORIES. MATHATICIANS AND PHYSICIST ARE
RESPONSIBLE FOR THE END OF THE WORLD WITH THEIR ***** THEORIES.

*-----------------------*
Posted at:
www.GroupSrv.com
*-----------------------*


Someone had better fix theoretical physics fast, before the ice caps melt!

Don't be alarmed it turns out it just needed an oil change.
.


User: "Sam Wormley"

Title: Re: QM 19 Nov 2004 07:26:13 PM
ben ito wrote:

QUANTUM MECHANICS IS A DECEPTION. QUANTUM MECHANICS IS A WAVE THEORY.
QUANTUM MECHANICS USES A WAVE STRUCTURE AND ONLY USES A SEGMENT OF
THE WAVE TO FORM A QUANTUM. THEREFORE, QUANTUM MECHANICS VIOLATES
THE LAW OF CONSERVATION OF ENERGY.. THATS WHY WE HAVE THE GREEN
HOUSE EFFECT BECAUSE PHYSICIST ARE DISREGARD THE LAW OF CONSERVATION
OF ENERGY IN ALL THEIR THEORIES. MATHATICIANS AND PHYSICIST ARE
RESPONSIBLE FOR THE END OF THE WORLD WITH THEIR ***** THEORIES.

There has NEVER been a prediction of QM that was contradicted by an
experiment or observation. NEVER!
.


User: "Dead Parrot"

Title: Re: Quantum mechanics and operators 15 Nov 2004 11:45:02 AM
The operators are transformations performed on the Hamiltonian
operator the eigenvalues of these transformations represent observable
quantities.
cfgauss@u.washington.edu (cfgauss) wrote in message news:<9e2e7039.0411142331.6c0a8309@posting.google.com>...

I'm trying to find out more about where the operators in quantum mechanics
come from. From what I see, the justification for using them is that using
them with the wave function, which is already known, gives you the
expectation value for that operator, i.e., <E> = Integral[ phi* (-i hbar d/dt)
phi*]. But then I see the justification for the wave function as that if
you put the operators in the equation E = p^2/2m, you get the Schrodinger
equation. Well, that's great, but that argument just goes in a big circle.
If you didn't know the wave equation, the solution to the wave equation, or
the operators, how would you be able to come up with them? And what is the
justification for what the operators are, other than "because it works." Do
they "come" from anywhere?

Also, if we look at the relativistic correct wave equation, my quantum
mechanics books tell me that it comes from putting the operators into the
relativistic energy equation. How can you justify doing this? How do you
know the operators are relativisticly correct? How do we know that we don't
have anything like an operator for mass that we have to put into the
equations?

I'm also interested in the use of complex numbers to write the equation. My
quantum mechanics book mentions that they are used to simplify the
equations, and that it doesn't physically mean anything. It mentions that
you can write, for example, the equations describing electricity and
magnetism in a form like that by saying F = E + icB, then you can write
Maxwell's equations in terms of this F instead of separate equations for E
and for B. So, you could also write the wave equation as the sum of two
other "things," phi = phi_1 + i phi_2. Is there any kind of representation
for these "fields" phi_1 and phi_2? Are they physically meaningful? Also,
is there a physically meaningful representation of the field F = E + icB?

I've tried to look through several quantum mechanics books for answers to
these questions, but I haven't seen any argument that doesn't seem circular
to me.

Thanks very much!
- Jeremy

.
User: "lunatic"

Title: Re: Quantum mechanics and operators 16 Nov 2004 07:33:03 PM

cfgauss@u.washington.edu (cfgauss) wrote in message
news:<9e2e7039.0411142331.6c0a8309@posting.google.com>...

I'm trying to find out more about where the operators in quantum
mechanics
come from. From what I see, the justification for using them is that
using
them with the wave function, which is already known, gives you the
expectation value for that operator, i.e., <E> = Integral[ phi* (-i hbar
d/dt)
phi*]. But then I see the justification for the wave function as that if
you put the operators in the equation E = p^2/2m, you get the Schrodinger
equation. Well, that's great, but that argument just goes in a big
circle.
If you didn't know the wave equation, the solution to the wave equation,
or
the operators, how would you be able to come up with them? And what is
the
justification for what the operators are, other than "because it works."
Do
they "come" from anywhere?

Also, if we look at the relativistic correct wave equation, my quantum
mechanics books tell me that it comes from putting the operators into the
relativistic energy equation. How can you justify doing this? How do
you
know the operators are relativisticly correct? How do we know that we
don't
have anything like an operator for mass that we have to put into the
equations?

I'm also interested in the use of complex numbers to write the equation.
My
quantum mechanics book mentions that they are used to simplify the
equations, and that it doesn't physically mean anything. It mentions
that
you can write, for example, the equations describing electricity and
magnetism in a form like that by saying F = E + icB, then you can write
Maxwell's equations in terms of this F instead of separate equations for
E
and for B. So, you could also write the wave equation as the sum of two
other "things," phi = phi_1 + i phi_2. Is there any kind of
representation
for these "fields" phi_1 and phi_2? Are they physically meaningful?
Also,
is there a physically meaningful representation of the field F = E + icB?

I've tried to look through several quantum mechanics books for answers to
these questions, but I haven't seen any argument that doesn't seem
circular
to me.

Thanks very much!
- Jeremy

Why operators?
I'd like to answer this question if I may in two ways. Firstly in my normal
voice and then in a kind of silly high-pitched whine...... ...no, sorry.
Firstly from a historical viewpoint and then from a modern axiomatic way.
(historically)
This is a 25 year long story beginning in 1900 so I cannot even begin to go
into details here.
Using insight and results from names no less than Planck, Rutherford, Bohr,
Einstein, Ehrenfest and Kramers, it was Werner Heisenberg who became
responsible for getting operators into business. When working on the then
evergreen problem of (atomic) spectral lines and their intensities, he tried
to keep the known dynamical equation of motion x''+f(x) = 0 but to reject
the interpretation of x as simply position as a function of time.
Now normally x would be expanded into a Fourier series, when thinking of an
electron of an atom interacting with light, absorbing and emitting at
certain freqencies: x(t) = sum a_n exp(inwt) /read w as omega/ a_n are the
Fourier coefficients. However, instead of a_n he made use of a(n,n-m) , a
quantity depending on two indices, to introduce 'transition amplitudes' into
his formula (electron jumping from energy level n to level n-m). This seems
quite bold, and yes it was, but there were good reasons to do this.
Transition amplitudes were already introduced by Einsteins paper from 1916 -
he had A_nm and such everywhere, where m and n denoted energy levels of the
bound electron. You really need to see preceding papers on dispersion theory
and the correspondence principle which was guiding all these 'guesses'.
Heisenberg very strongly felt that the reason for classical dynamics not
working in the atomic domain was to be found in the underlying kinematics.
He clearly expressed this in the introduction for his paper. So he tried to
change that.
As I see this was the very turning point, as now suddenly all kinds of
kinematical quantities got two indices with the formulae remaining the same!
Multiplication of these quantities obeyed the rules of multiplications of
matrices, as a consequence of the Fourier analogy. Bohr and Jordan
formalized these results very neatly the same year, so that no one could
deny any more that there were matrices (operators) in work. They worked out
pq-qp=h/2pi.i (By another sideline, Shrödingers work, wave equations, were
shown to be equivalent with the matrix method at a later time.)
So this was how it all began....
I find it very strange that all the phenomena involved in the calculations
which led to the new formalism is related to either black body radiation,
discrete spectra or dispersion theory - so in the end related to the
interaction of a photon with a bound electron. Matrices came because of
discrete energy levels and transitons between them! Noone seemed to care
about how a photon or free electron moves or what really happens when they
don't interact. Yet the new formalism was applied to these, later, without
the least of hesitation.
(axiomatically)
The logic of a mechanical system is an orthomodular lattice. (OK, logic is
the way how all the events of a given phyisical system relate to each
other). Physical quantities are mappings so that every statement about a
physical quatity is associated with an element of the logic, an 'event'. We
assume that such statements constitute a set algebra.
1 Now if our system is quantum-mechanical, its logic is (isomorphic to) a
lattice of projections of a complex Hilbert space.
2 Components of coordinate and momentum in a given frame are real
quantities, so statements regarding them correspond to Borel sets of R
(reals), constituting the set algebra in this case.
Because of 1 and 2 our physical quantities are now mappings from the Borel
set to the lattice of projections. These are called projection-valued
measures. For every such measure there is a unique self-adjoint operator on
the Hilbert space whose spectral extension is the given projection-valued
measure, that is physical quantity! Backwards: this operator can be
constructed by integrating the identity function on the reals with respect
to the projection-valued measure. So in this sense the physical quatity can
be identified with this self-adjoint operator.
Fun, isn't it?
-lunatic-
.


User: "Mark Martin"

Title: Re: Quantum mechanics and operators 15 Nov 2004 10:35:13 AM
cfgauss wrote:

I'm trying to find out more about where the operators in quantum

mechanics

come from. From what I see, the justification for using them is that

using

them with the wave function, which is already known, gives you the
expectation value for that operator, i.e., <E> = Integral[ phi* (-i

hbar d/dt)

phi*]. But then I see the justification for the wave function as

that if

you put the operators in the equation E = p^2/2m, you get the

Schrodinger

equation. Well, that's great, but that argument just goes in a big

circle.

If you didn't know the wave equation, the solution to the wave

equation, or

the operators, how would you be able to come up with them? And what

is the

justification for what the operators are, other than "because it

works." Do

they "come" from anywhere?

Also, if we look at the relativistic correct wave equation, my

quantum

mechanics books tell me that it comes from putting the operators into

the

relativistic energy equation. How can you justify doing this? How

do you

know the operators are relativisticly correct? How do we know that

we don't

have anything like an operator for mass that we have to put into the
equations?

I'm also interested in the use of complex numbers to write the

equation. My

quantum mechanics book mentions that they are used to simplify the
equations, and that it doesn't physically mean anything. It mentions

that

you can write, for example, the equations describing electricity and
magnetism in a form like that by saying F = E + icB, then you can

write

Maxwell's equations in terms of this F instead of separate equations

for E

and for B. So, you could also write the wave equation as the sum of

two

other "things," phi = phi_1 + i phi_2. Is there any kind of

representation

for these "fields" phi_1 and phi_2? Are they physically meaningful?

Also,

is there a physically meaningful representation of the field F = E +

icB?


I've tried to look through several quantum mechanics books for

answers to

these questions, but I haven't seen any argument that doesn't seem

circular

to me.

Thanks very much!
- Jeremy

An operator is a device which turns one function into another function.
There's a before-and-after-ness to the relationship. In mechanics a
dynamic system undergoes a change of state when there's an interaction
of some sort. It goes from one mode to some subsequent mode. An
operator is used to stand for how the system changes in a
characteristic way.
-Mark Martin
.

User: "Rene Tschaggelar"

Title: Re: Quantum mechanics and operators 15 Nov 2004 01:22:49 PM
cfgauss wrote:

I'm trying to find out more about where the operators in quantum mechanics
come from. From what I see, the justification for using them is that using
them with the wave function, which is already known, gives you the
expectation value for that operator, i.e., <E> = Integral[ phi* (-i hbar d/dt)
phi*]. But then I see the justification for the wave function as that if
you put the operators in the equation E = p^2/2m, you get the Schrodinger
equation. Well, that's great, but that argument just goes in a big circle.
If you didn't know the wave equation, the solution to the wave equation, or
the operators, how would you be able to come up with them? And what is the
justification for what the operators are, other than "because it works." Do
they "come" from anywhere?

Consider an operator to be an active mathematical object that
lives in a confined mathematical space. As it works on functions
and produces another set of functions, it may have eigenfunctions.
Eg the fourier transform is an operator living in the schwartz space
where functions vanish faster than any polynominal. Its eigen-
functions are the gaussians.
To start with, the fact that they just work should be sufficient.
Then you need to have a closer look at the properties.
Rene
--
Ing.Buero R.Tschaggelar - http://www.ibrtses.com
& commercial newsgroups - http://www.talkto.net
.


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