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From: "Juan R." González-Álvarez on 25 Mar 2010 17:33
Jay R. Yablon wrote on Thu, 25 Mar 2010 17:19:48 -0400:

> "Robert Israel" <israel(a)math.MyUniversitysInitials.ca> wrote in message
> news:rbisrael.20100325203421$0c4a(a)news.acm.uiuc.edu...
>> "Jay R. Yablon" <jyablon(a)nycap.rr.com> writes:
>>
>>> Following up on some recent discussion in sci.physics.research about
>>> momentum and position operators, I am trying to clarify how one should
>>> think generally about the energy eigenvalues of the Hamiltonian. This
>>> is laid out in the three page file linked below:
>>>
>>> http://jayryablon.files.wordpress.com/2010/03/energy-eigenvalues.pdf
>>>
>>> Any help you can provide in getting my thinking straightened out about
>>> this, is appreciated.
>>
>> The free Hamiltonian has continuous spectrum. There are no eigenvalues
>> or eigenvectors (in the sense that mathematicians give to those terms).
>
> Thanks Robert.
>
> Is there some general set of conditions which tell us the circumstances
> under which there will be a continuous spectrum, versus those (such as
> the harmonic oscillator) under which there is a discrete spectrum of
> eigenvalues?

Opening a textbook, Jay, you would know what is the spectrum for a free
particle and the different techniques to quantize it.

Opening a textbook, Jay, you would know what is the spectrum for an
oscillator and you would know under what circunstances it can be considered
continuous :-D

The advice is the same:

"Jay, the 90% of your questions are solved by *opening a textbook*.
Igor Khavkine has recently done a similar remark to you in another
thread. I repeat what he said you because it seems you did not read:

"Doing this kind of exercise is great and absolutely necessary to get
a good grasp on the subject. However, that by itself does not put you
any closer to discovery than any first year physics graduate student.

Here's a very general rule of thumb, that I believe I've tried to
express in the past, but it bears repeating. Just because you've
stumbled onto an original idea (any idea, that you've arrived at on
your own), that does not mean that you are the first to do so. In
fact, very few ideas anyone has are completely new. Other people
have already learned what you've learned and they've thought the same
things that you've thought. It takes quite a bit of work to make it
plausible that any particular idea or result is new. It takes an
intimate familiarity with the relevant research literature.
Obtaining such a familiarity can take a very long time, years is not
uncommon. If you are just starting to learn a subject, immmediately
obtaining new results should not be your goal. Understanding to
personal satisfaction and obtaining a familiarity with the relevant
literature are much better, not to mention more concrete, goals."

Regards.


--
http://www.canonicalscience.org/

BLOG:
http://www.canonicalscience.org/publications/canonicalsciencetoday/canonicalsciencetoday.html
From: eric gisse on 25 Mar 2010 17:43
Jay R. Yablon wrote:

> Following up on some recent discussion in sci.physics.research about
> momentum and position operators, I am trying to clarify how one should
> think generally about the energy eigenvalues of the Hamiltonian. This
> is laid out in the three page file linked below:
>
> http://jayryablon.files.wordpress.com/2010/03/energy-eigenvalues.pdf
>
> Any help you can provide in getting my thinking straightened out about
> this, is appreciated.

So when you apply the ladder operators to the free space hamiltonian, you
get a superposition of multiple particle states. Makes sense, if you think
about it.

Start from the beginning. Why would you expect there to be a discrete
spectrum of particle states in vacuum? I thought it was well understood that
the free space solutions to QM were all continuous. That's what I recall
being taught, at least.

>
> Jay
> ____________________________
> Jay R. Yablon
> Email: jyablon(a)nycap.rr.com
> co-moderator: sci.physics.foundations
> Weblog: http://jayryablon.wordpress.com/
> Web Site: http://home.roadrunner.com/~jry/FermionMass.htm

From: Jay R. Yablon on 25 Mar 2010 18:27

"eric gisse" <jowr.pi.nospam(a)gmail.com> wrote in message
news:hogld6$bj4$4(a)news.eternal-september.org...
> Jay R. Yablon wrote:
>
>> Following up on some recent discussion in sci.physics.research about
>> momentum and position operators, I am trying to clarify how one
>> should
>> think generally about the energy eigenvalues of the Hamiltonian.
>> This
>> is laid out in the three page file linked below:
>>
>> http://jayryablon.files.wordpress.com/2010/03/energy-eigenvalues.pdf
>>
>> Any help you can provide in getting my thinking straightened out
>> about
>> this, is appreciated.
>
> So when you apply the ladder operators to the free space hamiltonian,
> you
> get a superposition of multiple particle states. Makes sense, if you
> think
> about it.
>
> Start from the beginning. Why would you expect there to be a discrete
> spectrum of particle states in vacuum? I thought it was well
> understood that
> the free space solutions to QM were all continuous. That's what I
> recall
> being taught, at least.

Yes, Eric,

But the point I am trying to explore, is that for some V(q-hat) one will
get a continuous spectrum with a superposition of states, and for at
least one V(q-hat) (the harmonic oscillator) one gets a discrete
spectrum.

How does one know as a rule of thumb what to expect in any given
situation, based on the chosen V(q-hat)?

Now, I'll save Juan the trouble of his next post: "Jay, just read the
textbook."

For anyone who does not mind having this discussion, I look forward.

Jay

>
>>
>> Jay
>> ____________________________
>> Jay R. Yablon
>> Email: jyablon(a)nycap.rr.com
>> co-moderator: sci.physics.foundations
>> Weblog: http://jayryablon.wordpress.com/
>> Web Site: http://home.roadrunner.com/~jry/FermionMass.htm
>

From: "Juan R." González-Álvarez on 25 Mar 2010 20:07
eric gisse wrote on Thu, 25 Mar 2010 14:38:22 -0700:

> Juan R. González-Álvarez wrote:
>
> [...]
>
> Hee-hee, remember all those day ago when you were claiming I was banned
> from posting to sci.physics.foundations?

Where?

> Can you say 'projection'? I knew you could.





--
http://www.canonicalscience.org/

BLOG:
http://www.canonicalscience.org/publications/canonicalsciencetoday/canonicalsciencetoday.html
From: Darwin123 on 26 Mar 2010 15:40
On Mar 25, 11:38 am, "Jay R. Yablon" <jyab...(a)nycap.rr.com> wrote:
> Following up on some recent discussion in sci.physics.research about
> momentum and position operators, I am trying to clarify how one should
> think generally about the energy eigenvalues of the Hamiltonian.  This
> is laid out in the three page file linked below:
>
> http://jayryablon.files.wordpress.com/2010/03/energy-eigenvalues.pdf
>
> Any help you can provide in getting my thinking straightened out about
> this, is appreciated.
You asked in your link how you should interpret equation 21. You
said it is not a real eignenvalue because certain properties weren't
fulfilled. However, I think your question reveals fundamental
misunderstandings about the Dirac notation you are using. I think it
is good you are trying to understand the problem physically. However,
you need to grasp linear algebra better in order to understand the
physical picture.
The physically real quantities in quantum mechanics are not the
vectors, but the matrix elements. "Eigenvalues" are derived by
diagonalizing the matrix elements. Somewhere in many quantum mechanics
problems are the steps of forming a matrix and diagonalizing it. These
are two separate step.
The function |n> is not an eigenvalue, or even a function per
se. Therefore, equation 21 can not be a real eigenvalue because it is
a ket vector. An eigenvalue is a sort of scalar.
In order to get scalar values, the ket vector has to be
multiplied by a bra vector. Examples of bra vectors are <n|, <n+1| and
<2|. Examples of ket vectors are |n>, |n+1> and |2>.
I think that you should start thinking of the quantum mechanical
notation as a generalization of linear algebra. The treatment of bra
vectors is entirely analogous to the treatment of row vectors in
linear algebra, and the treatment of ket vectors is entirely analogous
to the treatment of column vectors. You could think of a bra vector as
a row vector with an infinite number of elements, and a ket vector as
a column vector with an infinite number of elements. The numbers
inside the vector a merely labels.
Form the matrix first, and then diagonalize it. Then, try to
picture what physical processes you mathematical operations correspond
to.
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