n particle wave function

[Gerard Westendorp]

Hi,

I'm trying to interpret the time-dependence part of the n-particle
Schroedinger equation, and its wave function solutions.

i d_t PSI(x1,x2,x3...,t)
= -1/2m (d2_x1 + d2_x3 + d2_x3+...+ V) PSI(x1,x2,x3,...,t)

It seems that the more particles you add, the higher the frequency
of the wave function solutions.

w= k1^2 + k2^2 + k3^2 +... -V

I'm having some trouble with this. Is the frequency really higher,
or is this some kind of unobservable thing?


thanks,

Gerard

[Igor Khavkine]

On Jan 13, 6:51 pm, Gerard Westendorp <west...@xs4all.nl> wrote:
> Hi,
>
> I'm trying to interpret the time-dependence part of the n-particle
> Schroedinger equation, and its wave function solutions.
>
> i d_t PSI(x1,x2,x3...,t)
> = -1/2m (d2_x1 + d2_x3 + d2_x3+...+ V) PSI(x1,x2,x3,...,t)
>
> It seems that the more particles you add, the higher the frequency
> of the wave function solutions.
>
> w= k1^2 + k2^2 + k3^2 +... -V

On a pedantic note, this only works if V is a constant.

> I'm having some trouble with this. Is the frequency really higher,
> or is this some kind of unobservable thing?

The frequency is directly proportional to the energy. In as much as
(non-relativistically) the absolute energy is only defined up to a
constant, the magnitude of the frequency has no meaning. However,
energy differences are well defined (are observable quantities,
through spectroscopy for example), and so are the energy differences.

A wave function with n sets of independent coordinates represents a
collective state of n particles. The frequency of the wave function is
then proportional to the total energy of the n-particle system, which
is just the sum of the energies of the n particles. The energy of each
particle is its kinetic energy (the k^2 term) and its potential energy
(that's contained in the V term). So, what you observed is perfectly
correct and not in any way unusual.

BTW, one should keep in mind that wave function frequencies are well
defined only for eigenstates of the Hamiltonian (which are plane waves
for a system of free particles). All other wave functions are quantum
superpositions of states of definite frequency (energy).

Hope this helps.

Igor

[Arnold Neumaier]

Gerard Westendorp schrieb:
> Hi,
>
> I'm trying to interpret the time-dependence part of the n-particle
> Schroedinger equation, and its wave function solutions.
>
> i d_t PSI(x1,x2,x3...,t)
> = -1/2m (d2_x1 + d2_x3 + d2_x3+...+ V) PSI(x1,x2,x3,...,t)
>
> It seems that the more particles you add, the higher the frequency
> of the wave function solutions.
>
> w= k1^2 + k2^2 + k3^2 +... -V
>
> I'm having some trouble with this. Is the frequency really higher,
> or is this some kind of unobservable thing?

The only frequencies that have a physical meaning are the differences
of eigenvalues of the Hamiltonian, multiplied by Planck's constant.

All Hamiltonians have an unbounded spectrum, so that most solutions have
arbitrarily high frequencies. But observable are usually only the low
frequencies; the high frequency part is treated as noise (as in
classical mechanics.


Arnold Neumaier

[Gerard Westendorp]

Igor Khavkine wrote:
[..]

> A wave function with n sets of independent coordinates represents a
> collective state of n particles. The frequency of the wave function is
> then proportional to the total energy of the n-particle system, which
> is just the sum of the energies of the n particles. The energy of each
> particle is its kinetic energy (the k^2 term) and its potential energy
> (that's contained in the V term). So, what you observed is perfectly
> correct and not in any way unusual.

Thanks.
This seems indeed to be what the n-particle Schroedinger equation is saying.

But take a 1 MHz radio wave. The 1 MHz can be be observed directly on an
oscilloscope.
This radio wave consists of a large number of 1 MHz photons. It seems
that both the collective of these particles and the individual particles
have the same frequency, seemingly in contradiction with the idea that N
particles oscillate at N times the frequency of a single one.

Or better, lets make it a resonant cavity of 1 MHz. Then we might think
of this system as a quantum oscillator with frequency (f). Every time it
changes energy, it adsorbs or emits a photon of the same frequency (f).
According to the Schroedinger equation of the quantum oscillator, the
frequency of the oscillator is actually N*f.

I find it hard to believe that there is something oscillating at N*f.

Gerard

[Igor Khavkine]

On Jan 15, 3:09 am, Gerard Westendorp <west...@xs4all.nl> wrote:

> But take a 1 MHz radio wave. The 1 MHz can be be observed directly on an
> oscilloscope.
> This radio wave consists of a large number of 1 MHz photons. It seems
> that both the collective of these particles and the individual particles
> have the same frequency, seemingly in contradiction with the idea that N
> particles oscillate at N times the frequency of a single one.

> I find it hard to believe that there is something oscillating at N*f.

Not all frequencies are created equal. Just because "something"
oscillates at frequency N*f does not mean that "everything else" must
oscillate with the same frequency. When you look at a wave form on an
oscilloscope, you are essentially measuring the superposition of the
wave forms of individual photons the many that compose the state of
the electromagnetic field. The frequency that you see on the
oscilloscope is the average frequency, which will be f, no matter what
N is. N is encoded in the amplitude of the field's wave form. To
measure N, you need to calculate the energy contained in the intensity
of the field. Now, energy is not measured in Hz, but if you divide
that energy by hbar, you'll get N*f. Ts expected, since the wave
function frequency N*f measures the total energy of the state
(relative say to the ground state of the field).

Hope this helps.

Igor

[Arnold Neumaier]

Gerard Westendorp schrieb:
> Igor Khavkine wrote:
>
>> A wave function with n sets of independent coordinates represents a
>> collective state of n particles. The frequency of the wave function is
>> then proportional to the total energy of the n-particle system, which
>> is just the sum of the energies of the n particles. The energy of each
>> particle is its kinetic energy (the k^2 term) and its potential energy
>> (that's contained in the V term). So, what you observed is perfectly
>> correct and not in any way unusual.
>
> Thanks.
> This seems indeed to be what the n-particle Schroedinger equation is saying.
>
> But take a 1 MHz radio wave. The 1 MHz can be be observed directly on an
> oscilloscope.
> This radio wave consists of a large number of 1 MHz photons. It seems
> that both the collective of these particles and the individual particles
> have the same frequency, seemingly in contradiction with the idea that N
> particles oscillate at N times the frequency of a single one.

> Or better, lets make it a resonant cavity of 1 MHz. Then we might think
> of this system as a quantum oscillator with frequency (f). Every time it
> changes energy, it adsorbs or emits a photon of the same frequency (f).
> According to the Schroedinger equation of the quantum oscillator, the
> frequency of the oscillator is actually N*f.
>
> I find it hard to believe that there is something oscillating at N*f.

A 1 MHz radio wave is not an N-particle system but (in the relevant
approximation) an ensemble of single-particle systems prepared in
the same state.

This explains the lack of oscillations at N*f.


Arnold Neumaier