B Do wave packets really represent a free particle?

referframe

Gold Member
Given a source of electrons, like from an electron gun. Physicists call these freely traveling particles and often use a Gaussian wave packet to represent them with the group velocity being precisely defined as the velocity of the center of the packets. But if we do not measure the position of the particle then we have no idea where the "center of the packet" is and if we do measure the position of the particle then the wave function, because of dispersion, becomes an expanding spherical wave with basically a zero or undefined group velocity.

So does nature really know about traveling, well-defined wave packets?

ftr
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Mentz114

Gold Member
Given a source of electrons, like from an electron gun. Physicists call these freely traveling particles and often use a Gaussian wave packet to represent them with the group velocity being precisely defined as the velocity of the center of the packets. But if we do not measure the position of the particle then we have no idea where the "center of the packet" is and if we do measure the position of the particle then the wave function, because of dispersion, becomes an expanding spherical wave with basically a zero or undefined group velocity.

So does nature really know about traveling, well-defined wave packets?

Is there really such a thing as a 'free' particle ?

A wave packet in a harmonic potential is very lile a classical particle in same.

Demystifier

2018 Award
But if we do not measure the position of the particle then we have no idea where the "center of the packet" is and if we do measure the position of the particle then the wave function, because of dispersion, becomes an expanding spherical wave with basically a zero or undefined group velocity.
There is also a third possibility, that we measure both position and momentum, but neither with perfect precision. That's consistent with Heisenberg uncertainty relations, and that's what actually happens in nature. In this way the wave function can be a Gaussian with a finite width in both position and momentum space.

sweet springs

Electrons from a gun controlled by applied voltage have centered velocity and centered time of ejection. CRT or Braun tubes prove it.
Something quantum leaving such and such time travelling with such and such velocity do not seem funny to me.

atyy

A free particle is defined as a particle governed by a Hamiltonian in which the potential is 0.

Wave packets do represent free particles, but are not the only wave functions of a free particle.

ftr

A free particle is defined as a particle governed by a Hamiltonian in which the potential is 0.

Wave packets do represent free particles, but are not the only wave functions of a free particle.
So what is it in the model that destroys the wave-packet when in a potential, for example.

sweet springs

For example, diffusing Gaussian wave packet could represent a free particle and a standing Gaussian wave packet represents a particle in the ground state of harmonic oscillator.

kith

So what is it in the model that destroys the wave-packet when in a potential, for example.
The statement was that a particle is called free if the potential is zero, not that wavepackets occur only for free particles. A particle which oscillates in a harmonic potential is also represented by a wavepacket.

Gold Member

ftr

The statement was that a particle is called free if the potential is zero, not that wavepackets occur only for free particles. A particle which oscillates in a harmonic potential is also represented by a wavepacket.
I think I misunderstood his sentence, he usually writes in clear english, maybe he was in a hurry.

ftr

For some funny simulations of Schrödinger wave packets in simple step and box potentials, see
So it seems we can take the whole wave to represent the "particle" at least sometimes, right? What is the analog in QFT.

vanhees71

Gold Member
It's not clear to me what you mean when you say `we can take the whole wave to represent the "particle"'. The clear meaning of a single-particle wave function in non-relativistic QT is that its modulus squared is the probability distribution for the position of the particle at a given time.

In relativistic QT wave packets make only sense for free particle, defining asymptotic free states. That's why one needs QFT to define relativistic QT of interacting particles in a proper way. The reason is that at relativistic scattering energies you always have the possibility that particles get created or destroyed in the process, i.e., you need a formalism that describes reactions, where the particle number changes, and the most convenient description is thus in terms of QFT.

ftr

That is what I mean, why not think about the wave in ordinary QM the same as relativistic free particle.

vanhees71

Gold Member
As I said, if you have relativistic particles there's always the chance to create and destroy particles in scattering processes. That's the reason, why already a proper interpretation of a single-particle wave function in an external potential for relativistic wave equations is difficult if not impossible. That's why you need a theory taking the production and destruction of particles in scattering processes into account, and the most simple way to formulate such a theory is quantum field theory.

Mark Harder

Gold Member
So what is it in the model that destroys the wave-packet when in a potential, for example.
An example of a particle in a bound state (i.e. not a free particle) is an electron constrained by the coulombic potential of an atomic nucleus. The probability densities of position are given by more complicated distributions than a simple Gaussian packet. The clouds of electrons in atoms are called orbitals. Their structure is given by solutions of the Schroedinger equation in terms of special functions that have nodes and maxima at positions about the nucleus.

jtbell

Mentor
For some funny simulations of Schrödinger wave packets in simple step and box potentials, see

http://theory.gsi.de/~vanhees/faq/quant/node33.html

and the following pages.
Also see here for a Gaussian wave packet spreading while bouncing back and forth between the walls of an infinite square well:

http://www.optics.rochester.edu/~stroud/animations/swdecay.html

And a circular orbit wave packet in a Coulomb potential, constructed from several hydrogen energy levels:

http://www.optics.rochester.edu/~stroud/animations/decay.html

ftr

An example of a particle in a bound state (i.e. not a free particle) is an electron constrained by the coulombic potential of an atomic nucleus. The probability densities of position are given by more complicated distributions than a simple Gaussian packet. The clouds of electrons in atoms are called orbitals. Their structure is given by solutions of the Schroedinger equation in terms of special functions that have nodes and maxima at positions about the nucleus.
Of course, That is well known. My question was in a response to a misunderstanding to what atyy said.

"Do wave packets really represent a free particle?"

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