Prev: There is no "pull" of gravity, only the PUSH of flowing ether!
Next: Unified Field Solutions
From: George Hammond on 28 Mar 2010 22:38
On Thu, 25 Mar 2010 18:27:08 -0400, "Jay R. Yablon"
<jyablon(a)nycap.rr.com> wrote:

>

>
>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)?
>
>Jay
>
>
[Hammond]
It makes a difference if V is a (positive) even or an odd
function of q. Generally an even function of q constitutes
a potential "well" wheras an odd function of q constitutes a
potential "wall" (oo barrier). The SHO= q^2 is the
simplest potential well. Only a potential "well" will
produce discrete states because it traps the particle which
then bounces back and forth inside the well.... an infinite
barrier (such as q^7 wch. you mentioned) is just a barrier
at the origin so you will have continuous states for
negative q.
Of course you can have V=k-q^2 which is an upside down
well (a "hill") or finite barrier which will produce
continuous states also for E>V, etc.
Discrete states are caused only when the particle is
"trapped" inside a potential well... otherwise the states
are continuous, such as for V=0.
As for why the vacuum has discrete particle states, I
dunno, that must be caused by the (ad hoc particle physics)
characteristics of the creation and annihilation operators
or something rather than the effects of a potential...?
And perhaps that is so because the "vacuum" containing an
infinite number of virtual particles is actually a
"potential well"... sort of like Dirac's "negative energy
sea of electrons" or something?
From: Darwin123 on 29 Mar 2010 16:16
On Mar 29, 1:56 pm, Darwin123 <drosen0...(a)yahoo.com> wrote:
> On Mar 27, 7:55 pm, "Jay R. Yablon" <jyab...(a)nycap.rr.com> wrote:
>
> > "Jay R. Yablon" <jyab...(a)nycap.rr.com> wrote in messagenews:8170i5Fep2U1(a)mid.individual.net...
> >  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.
>
>     As long as you don't hold me to my criticisms, as I am only trying
> to help you out. So far, I have two problems with this.

I told you not to hold me too it. I made a slight boo-boo. I forgot
the factor of 1/2 in front of the k. So if,
w^2=k/u
where k is the spring constant and "u" is the mass, then,
H=(1/2u)P^2+(k/2)Q^2
Let us use a harmonic oscillator with a frequency 2w. Thus k goes to
4k. So,
this case:
H'=(1/2u)P^2+2kQ^2
One can map the eigenvalues of H onto the eigenevalues of H'.
One can make a transformation on P and Q to find a P' and Q' such
that:
H'=(1/2u)P'^2+(k/2)Q'^2
The factor of 1/2 is unimportant for the point I was trying to
make. However, I felt obliged to correct myself.
The point. One can redefine creation operators "trans(a)" to
"trans(a')", annihilation operators "a", position operators P to P'
and momentum operators Q to Q' to satisfy the new Hamiltonian. One
problem is mapped to the other.
Physicists often solve a complex problem by "mapping" it onto a
simpler problem. I think the harmonic oscillator problem is important
only because it is one of the simplest problems. Many complicated
problems are mapped onto this problem.
Another way to solve a complex problem is to subtract the results
of a simple problem from the complex solution. The difference may be a
simple solution. This sort of thing is called "perturbation theory."
Again, the importance of the harmonic oscillator problem may be based
on the fact that the solution of the harmonic oscillator is relatively
simple.

First  |  Prev  | 
Pages: 1 2 3 4 5
Prev: There is no "pull" of gravity, only the PUSH of flowing ether!
Next: Unified Field Solutions