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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: 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.
From: Darwin123 on 26 Mar 2010 16:13
On Mar 25, 11:38 am, "Jay R. Yablon" <jyab...(a)nycap.rr.com> wrote:

> Any help you can provide in getting my thinking straightened out about
> this, is appreciated.
I gave a general recommendation in another post. Here, I will try
to be a little more specific. However, Google doesn't provide good
mathematical notation and I don't know how to make a link like you
did. So I will use some ascii notation and hope it is understandable.
You UNCONSCIOUSLY started out with a basis that was derived from
the Hamiltonian of a harmonic oscillator. I will express everything in
terms of normalized coordinates where m=1/2 and k=2. So the harmonic
oscillator Hamiltonian you started
H_0=P^2+Q^2
where P is the momentum operator and Q is the displacement operator.
Note H_0 is an operator.
Why did I say unconscious? Because you defined your basis vectors
as the ket vectors, |n>. Here "n" is the eigenvalue of H_0. It is not
an eigenvalue of the Hamiltonian you decided to use. The Hamiltonian
you decide was interesting was the free space Hamiltonian:
H=P^2
It is important to note that the eigenvectors of H are not the
eigenvectors of H_0. I think that is your problem. The particle in a
harmonic potential is not a free particle. You think an eigenvalue has
a physical meaning independent of the Hamiltonian.
What you asked for was a little muddled, so I will make a
conjecture as to what you really meant (subconsciously?).
You were asking what linear combination of harmonic oscillator
states will produce an eigenvector of harmonic oscillator states. You
are asking:
Suppose I have a harmonic oscillator that is in a stationary
state with V(q) as given. I cut off the potential energy contribution
so that v=0. What will be the wave function of the free particle that
has been released from the harmonic oscillator potential?
To do this, first form a matrix. Let h_mn be a matrix element
defined by:
h_mn=<m|H|n>
where
H=P^2
where,
m=1,2,3,...
and
n=1,2,3,...
This matrix will have infinite but coutable dimensionality. Note
that this matrix will have both diagonal and of-diagonal terms. Thus,
neither |m> nor |n> are eignevectors of H.
Now, diagonalize the matrix <m|H|n>. You will get a series of
eigenvalues. I will call these eignenvalues m' and n'. So you will get
a new set of eigenvectors |m'> where are linear combinations of |m>.
|m'>=a_mn|n>.
Many problems come clear if you think about the diagonalizing
process. The physical analogue?
Imagine a classical harmonic oscillator. A weight bob attached to
a horizontal spring. Start the harmonic oscillator going. Start it
with a total energy corresponding to m. That is the eigenvector of
H_0. Now, cut the spring. The weight bob sails away. Measure the
kinetic energy of the bob. Do it many times. Take an average of all
the momenta measured. The average kinetic will correspond to the
eigenvalue of H.

From: Darwin123 on 26 Mar 2010 17:03
On Mar 26, 4:13 pm, Darwin123 <drosen0...(a)yahoo.com> wrote:
> On Mar 25, 11:38 am, "Jay R. Yablon" <jyab...(a)nycap.rr.com> wrote:
I just did more of the problem. I found something which I think
relates to one of your questions in another post.
When you diagonalized <m|H|n>, you will find the the eigenvalue
"m'" is continuous. Although m=1,2,3,4,..., the eigenvalues of H can
be any value. For every value of m', there is an eigenvector |m'>. m'
can be anything at all.
So the diagonalization process has turned a discrete set of
eigenvalues into a continuous sent of eigenvalues.
Mathematically, at least, we we why the free particle has a
continuous set of energy levels and the harmonic oscillator has a
discrete set of eigenvalues. The nonzero off-diagonal terms correspond
to a broadening of the line spectrum.
To figure out how a free particle relates to a bound state, one
has to look at the corresponding linear algebra problem.
Here is the relevant part of what I wrote before:

>       To do this, first form a matrix. Let h_mn be a matrix element
> defined by:
> h_mn=<m|H|n>
> where
> H=P^2
> where,
> m=1,2,3,...
> and
> n=1,2,3,...
>     This matrix will have infinite but coutable dimensionality. Note
> that this matrix will have both diagonal and of-diagonal terms. Thus,
> neither |m> nor |n> are eignevectors of H.
>      Now, diagonalize the matrix <m|H|n>. You will get a series of
> eigenvalues. I will call these eignenvalues m' and n'. So you will get
> a new set of eigenvectors |m'> where are linear combinations of |m>.
> |m'>=a_mn|n>.
>        Many problems come clear if you think about the diagonalizing
> process. The physical analogue?
>     Imagine a classical harmonic oscillator. A weight bob attached to
> a horizontal spring. Start the harmonic oscillator going. Start it
> with a total energy corresponding to m. That is the eigenvector of
> H_0. Now, cut the spring. The weight bob sails away. Measure the
> kinetic energy of the bob. Do it many times. Take an average of all
> the momenta measured. The average kinetic will correspond to the
> eigenvalue of H.

I
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