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From: Jay R. Yablon on 27 Mar 2010 11:46

"Darwin123" <drosen0000(a)yahoo.com> wrote in message
news:9ddfd86a-0e89-472e-8d53-7b4fb7fd3af0(a)r27g2000yqn.googlegroups.com...
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.

Darwin,

Your three posts are very helpful and appreciated. One question for
right now:

Referring to: http://en.wikipedia.org/wiki/Diagonalizable_matrix which
talks about how to diagonalize a matrix, don't you need a non-vanishing
determinant? For example, the x-hat and p-hat matrices do not have
eigenvalues, witness the zeros that run all the way down the main
diagonal, so these cannot be diagonalized so far as I can tell. Or, can
they?

You say "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."

Can you please explain in a little more detail, using acsii if you wish
(I was able to follow what you wrote;-), how you are doing this
diagonalization of, e.g., (15) in
http://jayryablon.files.wordpress.com/2010/03/energy-eigenvalues.pdf.

Thanks,

Jay




From: Jay R. Yablon on 27 Mar 2010 13:45

"Darwin123" <drosen0000(a)yahoo.com> wrote in message
news:f010606b-666c-4ff6-9f86-b8035fd0781d(a)r27g2000yqn.googlegroups.com...
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.

[Jay]
No, I understand all of that. I was just thinking in terms of
A-hat|a>=a|a> yielding eigenvalues a of A-hat, because (A-hat - a)|a> =
0. But, if A-hat is not diagonal, the vectors |a> one obtains will
contain non-zero elements in more than one position.

[Darwin]
Form the matrix first, and then diagonalize it. Then, try to
picture what physical processes you mathematical operations correspond
to.

[Jay]
Agreed. On to your next posts, which talk more about diagonalizing.

Jay

From: Jay R. Yablon on 27 Mar 2010 14:19

"Darwin123" <drosen0000(a)yahoo.com> wrote in message
news:f0ef81c2-7ba1-4e7c-9aa5-5d443f335fcd(a)33g2000yqj.googlegroups.com...
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.

[Jay]
I agree completely. But, it was not unconscious. I am quite conscious
of this, and that is what bothers me, and that is at the root of what I
have been trying to straighten out in my posts of the past couple of
weeks.

[Darwin]
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.

[Jay]
Agreed. Which we can also write as:

H_0=P^2+Q^2 = (P+iQ)(P-iQ)

[Darwin]
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.

[Jay]
Agreed. Go on...

[Darwin]
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.

[Jay]
That is very interesting, but not quite what I was subconsciously
asking. Let me give it another try:

H_0 = P^2 + Q^2 (1)

as you define it above is the Hamiltonian for a Harmonic oscillator. I
can, in relation to this, define operators P, Q, a and a* and number N
in the usual way.

Now, suppose I want to know about the energy spectrum of any Hamiltonian

H = P^2 + V(Q) (2)

The V(Q) = 0 discussed above is one example, but my interest is in the
general case just above, where V(Q) <> Q^2, that is, where my potential
is NOT the oscillator potential but something -- anything -- else.

Does it even make sense for me to discuss the energy spectrum of using
the P, Q, a and a* and number N that I defined using the Harmonic
potential (1)? Or, do I need to define a different set of P, Q, a and
a* and number N for whatever V(Q) interests me in (2).

It seems to me that the answer to this plays out in one of two ways:

1) No, it does not make sense to use the P, Q, a and a* and number N
defined in relation to harmonic potential, and I need to define a
different set of P, Q, a and a* and number N operators for my V(Q) <>
Q^2. If this is so, then my question is: what is the general procedure
for defining P, Q, a and a* and number N for a potential V(Q) <> Q^2?
And, might this procedure be based on using:

H_0=P^2+V(Q) = (P+i sqrt(V(Q))(P-i sqrt(V(Q)) (3)

and, if so, then what happens to:

[Q,P] = i h-bar?

(Which I presume has to survive or we ruin the translation operation.)

2) Yes, it *does* make sense to use the P, Q, a and a* and number N
defined in relation to harmonic potential, but now the Hamiltonian will
in general no longer be diagonal so I then need to diagonalize the
Hamiltonian and in that situation I will end up with a continuous
spectrum of eigenvalues for the energy. In that case, my questions are
these:

a) I'd like to see exactly how that diagonalization takes place which
brings one from a discrete to a continuous spectrum (which I already
asked in reply to your third post). AND

b) What makes the Harmonic potential so darned special, that I can use
it to define operators P, Q, a, a* and N, and then use these very same
operators as the basis for calculating energy spectra for ANY potential
V(Q) <> Q^2. I may be wrong about this comment, and if so would
appreciate a solid rebuttal, but in using the harmonic oscillator, are
we not guilty of being the drunk who looks for the quarter under the
(harmonic oscillator) lamppost because that is where the light is? My
son does point out that in the series expansion for a potential with a
minimum at zero, the zero and first order terms drop out and the
coefficient of the leading term is (1/2!) d^2V/dx^2 |_x=0, which means
that the harmonic oscillator will be the leading order for most or all
physical potentials of interest and the "omega" which shows up defines
the "stiffness" of the equilibrium state. So, I can perhaps buy that
the Harmonic oscillator gives us a first order term for most situations
of interest (and would like to see how we do this with a coulomb 1/r
potential). But I am still troubled by the fact that one would be able
to use operators P, Q, a, a* and N established for a harmonic oscillator
potential as the basis for calculating valid results for any other
potential V(Q) <> Q^2. What makes the harmonic oscillator that
everybody uses all the time almost as an assumption without discussion,
so darned special? Why are we not all being drunks under the lamppost
when we just invoke a harmonic oscillator and then start calculating?
*How do we justify using the harmonic operators as a foundation for
doing calculations for non-harmonic potentials?* In doing so, are we
limiting ourselves to getting results that will only be valid in the
lowest order?

Jay

From: Darwin123 on 27 Mar 2010 19:08
On Mar 27, 11:46 am, "Jay R. Yablon" <jyab...(a)nycap.rr.com> wrote:
> "Darwin123" <drosen0...(a)yahoo.com> wrote in message
>
> news:9ddfd86a-0e89-472e-8d53-7b4fb7fd3af0(a)r27g2000yqn.googlegroups.com...
> 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:
>

> Can you please explain in a little more detail, using acsii if you wish
> (I was able to follow what you wrote;-), how you are doing this
> diagonalization of, e.g., (15) inhttp://jayryablon.files.wordpress.com/2010/03/energy-eigenvalues.pdf.
>
> Thanks,
>
> Jay
Use the orthonormality of the basis vectors, and the
operational definition.
We can assume that the "number" space of the harmonic
oscillator that the basis is orthonormal, so the value of the product
<m|n> is given by:
1) <m|n>=0 if |n-m|>0
2) <m|n>=1 if n=m.
We also use the formulas for the raising and lowering
operators, which have already been derived for the harmonic
oscillator. I will refer to "a" as the lowering operator and trans(a)
as the raising operator. When doing the ordinary "hramonic oscillator"
problem, using the commutation relations, we found that,
3) trans(a)|n>=sqrt(n+1)|n+1>
for n=0,1,2,3,...
4) a|0>=0|0>
5) a|n>=sqrt(n)|n-1>
for n=1,2,3,4,5,...

Your free particle Hamiltonian is presented in your link as
equation 15. The askii gets a little complicated. However, I think I
can tell you what to do better in English sentences.
The important thing to note is that your Hamiltonian, H, is the
sum of an "tranpose(a)a" term, an "aa" term, and an "trans(a)trans(a)"
term.
We can derive the matrix elements for all m,n=0,1,2,3,... from
your equation 15 and my equations 1-5.
It is easy to derive from equations 1-5 if |m-n|>2, <m|H|n>=0. So
we already have a large simplification. Note
6) <m|aa|n>=sqrt(n)sqrt(n-1)<m|n> for n=1,2,...
7) <m|trans(a)trans(a)|n>=sqrt(n+1)sqrt(n+2)<m|n> for n=0,1,2,...
8) <m|trans(a)a|n>=n<m|n> for n=0,1,2,...
You get the idea. Figure out all the elements of the matrix <m|H|
n> using equations 6-8 and your equation 15. I will call this the
matrix {H}. {H} will have both on-diagonal and off diagonal elements.
However, |m-n|<3 for all nonzero elements of {H}. So it actually is a
rather sparse matrix.
Define the eigenvectors of {H} to be the column vector {v(m')},
where m' is an eienvalue of {H}. This is an infinite dimensional
column vector. However, it is still a column vector.
By definition of the eigenvalue problem,
9) {H}{v(m')}=m'{v(m')}
You only have to write out a few terms of equation 9 to get an
idea of what is going on. The mathematical problem is well defined,
despite there being an infinite number of elements of v. However,
linear algebra methods often used for finite dimensional matrices
won't work. The "determinate" of {H} is not well defined. However, you
don't need the determinate of {H} to solve the problem.
What you do is use equation 9 to set up a recursion equation for
any value m'. The n+2th element of {v} is defined in terms of the n
+1,n,n-1 and n-2 element of {v}. You will find that there is a
solvable recursion equation for each value of m'>0.
I think I may have lost you with the last two paragraphs. I have
reached the limits of my ascii communication skills. However, maybe
this will be enough to get you started.
From: Jay R. Yablon on 27 Mar 2010 19:55

"Jay R. Yablon" <jyablon(a)nycap.rr.com> wrote in message
news:8170i5Fep2U1(a)mid.individual.net...
>
> "Darwin123" <drosen0000(a)yahoo.com> wrote in message
> news:f0ef81c2-7ba1-4e7c-9aa5-5d443f335fcd(a)33g2000yqj.googlegroups.com...
> On Mar 25, 11:38 am, "Jay R. Yablon" <jyab...(a)nycap.rr.com> wrote:
>
.. . .
> [Jay]
> That is very interesting, but not quite what I was subconsciously
> asking. Let me give it another try:
>
> H_0 = P^2 + Q^2 (1)
>
> as you define it above is the Hamiltonian for a Harmonic oscillator.
> I can, in relation to this, define operators P, Q, a and a* and number
> N in the usual way.
>
> Now, suppose I want to know about the energy spectrum of any
> Hamiltonian
>
> H = P^2 + V(Q) (2)
>
> The V(Q) = 0 discussed above is one example, but my interest is in the
> general case just above, where V(Q) <> Q^2, that is, where my
> potential is NOT the oscillator potential but something -- anything --
> else.
>
> Does it even make sense for me to discuss the energy spectrum of using
> the P, Q, a and a* and number N that I defined using the Harmonic
> potential (1)? Or, do I need to define a different set of P, Q, a and
> a* and number N for whatever V(Q) interests me in (2).
>
> It seems to me that the answer to this plays out in one of two ways:
>
> 1) No, it does not make sense to use the P, Q, a and a* and number N
> defined in relation to harmonic potential, and I need to define a
> different set of P, Q, a and a* and number N operators for my V(Q) <>
> Q^2. If this is so, then my question is: what is the general
> procedure for defining P, Q, a and a* and number N for a potential
> V(Q) <> Q^2? And, might this procedure be based on using:
>
> H_0=P^2+V(Q) = (P+i sqrt(V(Q))(P-i sqrt(V(Q)) (3)
>
> and, if so, then what happens to:
>
> [Q,P] = i h-bar?
>
> (Which I presume has to survive or we ruin the translation operation.)
>
> 2) Yes, it *does* make sense to use the P, Q, a and a* and number N
> defined in relation to harmonic potential, but now the Hamiltonian
> will in general no longer be diagonal so I then need to diagonalize
> the Hamiltonian and in that situation I will end up with a continuous
> spectrum of eigenvalues for the energy. In that case, my questions
> are these:
>
> a) I'd like to see exactly how that diagonalization takes place which
> brings one from a discrete to a continuous spectrum (which I already
> asked in reply to your third post). AND
>
> b) What makes the Harmonic potential so darned special, that I can use
> it to define operators P, Q, a, a* and N, and then use these very
> same operators as the basis for calculating energy spectra for ANY
> potential V(Q) <> Q^2. I may be wrong about this comment, and if so
> would appreciate a solid rebuttal, but in using the harmonic
> oscillator, are we not guilty of being the drunk who looks for the
> quarter under the (harmonic oscillator) lamppost because that is where
> the light is? My son does point out that in the series expansion for
> a potential with a minimum at zero, the zero and first order terms
> drop out and the coefficient of the leading term is (1/2!) d^2V/dx^2
> |_x=0, which means that the harmonic oscillator will be the leading
> order for most or all physical potentials of interest and the "omega"
> which shows up defines the "stiffness" of the equilibrium state. So,
> I can perhaps buy that the Harmonic oscillator gives us a first order
> term for most situations of interest (and would like to see how we do
> this with a coulomb 1/r potential). But I am still troubled by the
> fact that one would be able to use operators P, Q, a, a* and N
> established for a harmonic oscillator potential as the basis for
> calculating valid results for any other potential V(Q) <> Q^2. What
> makes the harmonic oscillator that everybody uses all the time almost
> as an assumption without discussion, so darned special? Why are we
> not all being drunks under the lamppost when we just invoke a harmonic
> oscillator and then start calculating? *How do we justify using the
> harmonic operators as a foundation for doing calculations for
> non-harmonic potentials?* In doing so, are we limiting ourselves to
> getting results that will only be valid in the lowest order?
>
> Jay
>
I posted a file at:

http://jayryablon.files.wordpress.com/2010/03/square-root-potential-3.pdf

which better explains what I am getting at, in the event that 1) above
is a valid answer, and we can do the development generally from

H_0=P^2+V(Q) = (P+i sqrt(V(Q))(P-i sqrt(V(Q)) (3)

Don't hold me to this; I am simply exploring.

Jay

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