how are hilbert space and schrodinger equation related ?
Craig Joyce
Schrodinger equation? Is the set of wave functions a vector space /or
something similar ? Please help. Thanks,
Craig
Franz Heymann
"Craig Joyce" <naught...@yahoo.com> wrote in message
news:7e5f4e39.02112...@posting.google.com...
Briefly and loosely, but not bereft of meaning, here it is:
The totality of the eigenfunctions of Schrodinger's equation for a
particular system are a "complete orthogonal set". That means that
*any* (reasonable) function can be expressed as a series expansion in
these eigenfunctions. In particular, any solution of Schrodinger's
equation can be so expressed. Fourier analysis is but one example of
this procedure.
Now, if you know the eigenfunctions in terms of which you expressed
the solution with which you are playing, you can write them in
sequence on an old envelope and stick it in your inside pocket. In
your actual calculations, all you need to do is to keep tabs on the
relative amplitudes of the terms in the expansion. (Rather like
expressing a complicated periodic electrical signal in terms of
sinusoidal harmonics, and thenceforth simply characterising the signal
in terms of the amplitudes of the harmonics) The "basis vectors" of
the Hilbert space you have chosen to work in are simply the actual set
of eigenvalues. The components of the vector is the set of
amplitudes. The nice thing now is that the quantum mechanical
operators, which are the objects which enable you to study the
observables in your system, can be written in such a way that you only
need to operate on the vector in Hilbert space (i.e. the set of
amplitudes) rather than invoking the complete wave function.
A hardy theorist will probably lay hell out of me for this
explanation, but it is nevertheless a fairly close introduction to the
strict truth.
Franz Heymann
Dirk Van de moortel
"Franz Heymann" <Franz....@btopenworld.com> wrote in message news:arvmhd$kf4$1...@helle.btinternet.com...
>
> "Craig Joyce" <naught...@yahoo.com> wrote in message
> news:7e5f4e39.02112...@posting.google.com...
> > Hi! Can anyone explain the relation between Hilbert Space and the
> > Schrodinger equation? Is the set of wave functions a vector space
> /or
> > something similar ? Please help. Thanks,
>
> Briefly and loosely, but not bereft of meaning, here it is:
>
> The totality of the eigenfunctions of Schrodinger's equation for a
> particular system are a "complete orthogonal set". That means that
> *any* (reasonable) function can be expressed as a series expansion in
> these eigenfunctions. In particular, any solution of Schrodinger's
> equation can be so expressed. Fourier analysis is but one example of
> this procedure.
> Now, if you know the eigenfunctions in terms of which you expressed
> the solution with which you are playing, you can write them in
> sequence on an old envelope and stick it in your inside pocket.
Excellent visualization!
One for my son this evening.
He's just working with eigenvalues/vectors/spaces :-)
> In
> your actual calculations, all you need to do is to keep tabs on the
> relative amplitudes of the terms in the expansion. (Rather like
> expressing a complicated periodic electrical signal in terms of
> sinusoidal harmonics, and thenceforth simply characterising the signal
> in terms of the amplitudes of the harmonics) The "basis vectors" of
> the Hilbert space you have chosen to work in are simply the actual set
> of eigenvalues.
of eigenfunctions.
> The components of the vector is the set of
> amplitudes. The nice thing now is that the quantum mechanical
> operators, which are the objects which enable you to study the
> observables in your system, can be written in such a way that you only
> need to operate on the vector in Hilbert space (i.e. the set of
> amplitudes) rather than invoking the complete wave function.
>
> A hardy theorist will probably lay hell out of me for this
> explanation, but it is nevertheless a fairly close introduction to the
> strict truth.
Hey Franz, I'm not a hardy theorist but nevertheless gotcha!
Sorry, couldn't resist ;-)
Dirk Vdm
Arkadiusz Jadczyk
wrote:
There are two ways of answering this question. The two ways are to some
extent equivalent (but there is a sense in which they are not quite
equivalent, especially when one thinks of ways to generalize quantum
mechanics)
1) The set of solutions of (time-dependent) Schrodinger equation forms a
Hilbert space. One can define a scalra product between two solutions. To
compute this scalar product one chooses arbitrary time t, and integrates
over space at this time. Then one proves that the intrgral does not
depend on the chosen time.
2) Initial (Cauchy) data for Schrodinger equation form a Hilbert space.
So we have a family of Hilbert spaces, parametrized by time. Sometimes
we identify these Hilbert spaces, sometimes we don't. It all depends
whether one is an "advanced user" or not.
ark
--
Arkadiusz Jadczyk
http://www.cassiopaea.org/quantum_future/homepage.htm
--
Franz Heymann
"Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com>
wrote in message news:ohJE9.7$5M3....@news.cpqcorp.net...
Hell's delight, I said something crappy there!
Mea culpa.
>
> > The components of the vector is the set of
> > amplitudes. The nice thing now is that the quantum mechanical
> > operators, which are the objects which enable you to study the
> > observables in your system, can be written in such a way that you
only
> > need to operate on the vector in Hilbert space (i.e. the set of
> > amplitudes) rather than invoking the complete wave function.
> >
> > A hardy theorist will probably lay hell out of me for this
> > explanation, but it is nevertheless a fairly close introduction to
the
> > strict truth.
>
> Hey Franz, I'm not a hardy theorist but nevertheless gotcha!
> Sorry, couldn't resist ;-)
To my eternal shame.
It is better that you pointed it out, rather than one of our resident
morons.
{:-)
Franz Heymann
Edward Green
> news:7e5f4e39.02112...@posting.google.com...
> > Hi! Can anyone explain the relation between Hilbert Space and the
> > Schrodinger equation? Is the set of wave functions a vector space
> /or
> > something similar ? Please help. Thanks,
>
> Briefly and loosely, but not bereft of meaning, here it is:
>
> The totality of the eigenfunctions of Schrodinger's equation for a
> particular system are a "complete orthogonal set". That means that
> *any* (reasonable) function can be expressed as a series expansion in
> these eigenfunctions. In particular, any solution of Schrodinger's
> equation can be so expressed. Fourier analysis is but one example of
> this procedure.
<snip rest of beautiful synopsis>
I've heard recently a comment to the effect that quantum mechanics is
only linear by the trick of being decomposed into an infinite
dimensional vector space -- and that any non-linear system can be so
"linearized".
Obviously I don't have this quite right, because the Schroedinger
equation is indeed linear as it stands -- but any idea what the
comment means?
Franz Heymann
"Edward Green" <null...@aol.com> wrote in message
news:2a0cceff.02112...@posting.google.com...
I have not got a clue what that might mean. Surely it is only a
linear equation that has the property that any solution can be written
as a sum over a number of other solutions. And that is the essence of
bringing in a concept like Hilbert space.
[...]
I took the liberty of snipping what Ed claims too have snipped.
Franz Heymann
Dirk Van de moortel
"Franz Heymann" <Franz....@btopenworld.com> wrote in message news:as062b$a7m$2...@venus.btinternet.com...
>
> "Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com>
> wrote in message news:ohJE9.7$5M3....@news.cpqcorp.net...
> >
> > "Franz Heymann" <Franz....@btopenworld.com> wrote in message
> news:arvmhd$kf4$1...@helle.btinternet.com...
> > >
[snip]
> > > Now, if you know the eigenfunctions in terms of which you
> > > expressed
> > > the solution with which you are playing, you can write them in
> > > sequence on an old envelope and stick it in your inside pocket.
> >
> > Excellent visualization!
> > One for my son this evening.
> > He's just working with eigenvalues/vectors/spaces :-)
[snip]
> > Hey Franz, I'm not a hardy theorist but nevertheless gotcha!
> > Sorry, couldn't resist ;-)
>
> To my eternal shame.
> It is better that you pointed it out, rather than one of our resident
> morons.
> {:-)
My pleasure.
My son loved the pocket picture :-)
Dirk Vdm