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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: eric gisse on 26 Mar 2010 16:50
Jay R. Yablon wrote:

>
> "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)?

I have absolutely no idea and it is not at all clear to me if that is
something that can be discerned purely based on the form of the potential
function.

I also do not see how, if at all, this is a problem. Where is it written
that each energy eigenvalue is a function of one and only one state?

Plus, is there no particular reason why you can not write the |n-2> and |
n+2> vectors in terms of ladder operators and the |n> vectors?

>
> 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: 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
From: "Juan R." González-Álvarez on 27 Mar 2010 07:56
Jay R. Yablon wrote on Thu, 25 Mar 2010 18:27:08 -0400:

> "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."

Don't seem a bad advice, is it Jay?

The problem is not in the advice but in that you continue ignoring it.

How many times did Igor, Arnold, or myself give it to you? 12? 36?

> 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
>>





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BLOG:
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From: "Juan R." González-Álvarez on 27 Mar 2010 08:05
eric gisse wrote on Fri, 26 Mar 2010 13:50:10 -0700:

> Jay R. Yablon wrote:

(...)

> I also do not see how, if at all, this is a problem. Where is it written
> that each energy eigenvalue is a function of one and only one state?

Well, nobody in this newsgroup wait you to really understand what is a
spectral decomposition or what we mean by a formula like

H = Sum_i |E_i> E_i <E_i|


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

BLOG:
http://www.canonicalscience.org/publications/canonicalsciencetoday/canonicalsciencetoday.html
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