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I faced some problems while learning the time evolution of Gaussian wavepacket . I started the part with a wavefunction in momentum space which was like $ \ \phi(p) = c.e^\frac{-(p-p_0)^2}{2.\sigma_p^2} \ $ . Then found the position space wavefunction by Inverse Fourier transform . And found a $\psi(x) \ $. I was asked to find out the $\psi(x,t)$. I thought that like the bound state problems I will multiply the $ \ \psi(x) \ $by $ \ e^\frac{-iEt}{\hslash} \ $and find the time evolution of the wavepacket. But in the solution they invoked a solution with integrals like : $ \ \psi(x,t) = (2\pi \hslash)^\frac{1}{2}.\int_{-\infty}^{+\infty} e^{(kx-\omega t)}.\phi(p) \ dp \ $ and solved this to find the time dependent wavefunction . Can anyone tell me that why they followed this approach ? I am suspecting that the Gaussian case is not always bounded so they didn't followed the $ \ e^\frac{-iEt}{\hslash} \ $ approach . Can the Gaussian wavepacket represent a free particle ?

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  • $\begingroup$ The gaussian wavepacket is a superposition of infinitely many energy eigenstates, so each frequency component has to be weighted separately to yield the time evolution of the whole packet. Regarding the physicality of the wave packet, it is useful to represent realistic free particles that occur in experiments in which there is a spread in the momentum distribution (we can't have a delta function in Nature). $\endgroup$ Jun 7 at 20:23

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The time evolution of a system is actually given by $e^{-iHt/\hbar}$ operating on the initial condition, where $H$ is the Hamiltonian describing the system. You have to know how the Hamiltonian operates on the state your system is in. IF your system is in an energy eigenstate, then the $H \phi_E = E \phi_E$ and you get the term you were expecting. However if its not, then you need expand the state in the basis of energy eigenstates. This can involve an integral over all the energy eigenstates. IF the momentum eigenstates are also energy eigenstates, and you know the relationship between the momentum in a given eigenstate and the energy in that eigenstate, then that integral can become an integration over all possible momenta.

Try to figure out what the momentum eigenstates are for your system. Can you write your initial condition in terms of an integral over all the momentum eigenstates? Do you know what the energy is when you operate your Hamiltonian on a single momentum eigenstate?

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  • $\begingroup$ Yes I understood your point . And I think it's a correct approach . In this problem as you can see I can find the space wavefunction by inverse transform easily and hence the momentum also can be found . But the potential energy term is not described anywhere . It can be a free particle or it's not impossible to be a bound particle . As potential energy term is unknown how the hamiltonian H can be calculated for a single momentum eigenstate ? Only I can work with the kinetic energy term . $\endgroup$
    – M Sagnik .
    Jun 9 at 2:13
  • $\begingroup$ The fact that they gave you an initial wavefunction in momentum space does imply that they probably meant it to be a free particle. One can choose to write a wave function in momentum space even if there is some potential that means momentum is not conserved, but you probably wouldn't. But the point is you need to write your wave function in terms of the energy eigenstates to evaluate the time evolution, which happen to also be the momentum eigenstates if you have a constant potential. $\endgroup$ Jun 9 at 21:03

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