Wave packet

  • #1
3,507
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In a QM introductory book , I have read that the wave packet is not a solution of the Schrodinger equation, is this true in some context or is it just an mistake of the author?
 

Answers and Replies

  • #2
368
12
It depends on the potential. If there is a potential, then the wave packet is not a solution. If there is no potential, then the Hamiltonian describes a free field, in which case a wave packet is a solution, and the Schrodinger equation describes how the packet moves and spreads out through time. The easiest way to see this is to break the wave packet apart into different energy eigenstates (this is just a Fourier expansion), which are trivial solutions of the SE.
 
  • #3
649
2
It depends on the potential. If there is a potential, then the wave packet is not a solution. If there is no potential, then the Hamiltonian describes a free field, in which case a wave packet is a solution, and the Schrodinger equation describes how the packet moves and spreads out through time. The easiest way to see this is to break the wave packet apart into different energy eigenstates (this is just a Fourier expansion), which are trivial solutions of the SE.
Why? The Schrödinger equation is linear also in the presence of a potential, so you can still create a wave packet solution by superposing an infinity of solutions (all solutions if the system with potential of course).

Or does the "wave packet" terminology imply a gaussian weighted integral over free states? As far as I'm concerned, any localized travelling wave solution is a "wave packet".
 
  • #4
tom.stoer
Science Advisor
5,766
161
Look at the time-dep. S.eq.

[tex]i\partial_t u_E(x,t) = Hu_E(x,t) = Eu_E(x,t)[/tex]

which is solved by

[tex]u_E(x,t) = e^{-iEt}u_E(x)[/tex]

with some dispersion relation

[tex]E=E(k) = k^2/2m[/tex]

where the last equality holds for the free particle.

A wave packet is defined as

[tex]\psi(x,t) = \int dk\,e^{-iEt}u_E(x)\,a(k)[/tex]

Applying the operator

[tex][i\partial_t - H][/tex]

this vanishes identically which shows that the wave packet[tex] \psi(x,t)[/tex] is indeed a solution.

For a non-vanishing potential one has to use the solution [tex]u_E(x,t)[/tex]; the factorization [tex]u_E(x,t) = e^{-iEt}u_E(x)[/tex] derived for vanishing potential is no longer valid.
 
  • #5
368
12
Why? The Schrödinger equation is linear also in the presence of a potential, so you can still create a wave packet solution by superposing an infinity of solutions (all solutions if the system with potential of course).
Well, I guess it depends on what you mean by "wave packet". If you define it as "some Gaussian-looking thing that moves around and spreads out, but always looks like a Gaussian", then I'm pretty sure it's only true if you don't have a potential. Of course you can always set up something that looks like a Gaussian at a specific point in time, the key is whether it will continue to look like one after time evolution. I can't remember exactly what will happen if you try that in a potential--one would assume that if you did it in a very very weak potential field it would still basically look like a Gaussian, but in general I don't think it's going to stay very well localized if you've got something strong like a big square well or an SHO or something. In the particular case of a square well, I know it will slosh around between the walls, and interfere with itself, and eventually settle down to pretty much occupying the entire square well evenly. So in that case, it's probably valid to say that it's no longer a wave packet.
 

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