Construction of wavepackets

  • #1
SemM
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Hi, it was suggested previously on PF by others that a way to solve a ODE where the domain of the operator in Hilbert space allowed a real solution, is through the construction of wavepackets.

The conditions for real solutions are according to Kreyszig's Functional Analysis that E, in the following equation ( HY=EY), is in the resolvent set of the Hamiltonian operator.

I would like to ask for confirmation of that this is the right approach:

1. Having ##H\psi = E\psi##, one solves the equation with E treated as a constant first. One uses no initial conditions. The true form of the energy is ## E= \frac{h^2k^2}{8\pi^2m}## . The solution is here now called ##\eta(q)##

2. Following Kreyszig's book on this, I quote

"The solution ##\eta(q)## can now be used to represent any ##\psi \in L ^2(-\infty,+\infty)## as a wave packet in the form:

\begin{equation}
\psi(q) = \frac{1}{\sqrt{2\pi}} \lim_{a\rightarrow \infty} \int_{-a}^{a}\phi(k)\eta(q)dk
\end{equation}

where


\begin{equation}
\phi(k) = \frac{1}{\sqrt{2\pi}} \lim_{a\rightarrow \infty} \int_{-a}^{a} \psi(q)\eta(q)dq
\end{equation}

Should this be exercised on the solution :

\begin{equation}
\eta(q) = e^{-2i\gamma q} + 3/4 + \frac{i\hbar}{2\gamma}
\end{equation}

How would the integral and thus the wavepacket look like? Based on what is in the book, I am not sure on what to make out of ##\phi(k)## and ##\psi(q)##, basically, how does one find these two auxiliary functions?

Hope this question was clear

Thanks
 

Answers and Replies

  • #2
vanhees71
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I've no clue what you book wants to tell me.

The most simple way to solve this problem along the line that might be what's explained in your book is to use plane waves, i.e., momentum eigenstates as a generalized basis (in the Heisenberg picture using natural units with ##\hbar=1##):
$$u_{\vec{p}}(t,\vec{x})=\langle \vec{x}|\vec{p},t \rangle=\frac{1}{(2 \pi)^{3/2}} \exp \left (-\frac{\mathrm{i} \vec{p}^2}{2m} t + \mathrm{i} \vec{p} \cdot \vec{x} \right ).$$
Then the general wave function, representing a pure state of a free particle, is given by a wave packet, i.e., it must be square integrable:
$$\psi(t,\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} u_{\vec{p}}(t,\vec{x}) \tilde{\psi}(\vec{p}),$$
where ##\tilde{\psi}(\vec{p})## is an arbitrary square-integrable function (normalized conveniently to 1, because then also the wave packet is normalized to 1).
 
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  • #3
SemM
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Thanks this was indeed very useful! I will use it further with my reading here. Thanks van Hees
 
  • #4
SemM
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Then the general wave function, representing a pure state of a free particle, is given by a wave packet, i.e., it must be square integrable:
$$\psi(t,\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} u_{\vec{p}}(t,\vec{x}) \tilde{\psi}(\vec{p}),$$
where ##\tilde{\psi}(\vec{p})## is an arbitrary square-integrable function (normalized conveniently to 1, because then also the wave packet is normalized to 1).
Does this only work for linear Hermitian Hamiltonians? I have the feeling a non-linear Hamltonian needs a different approach, because the wave is not harmonic as the linear wave is, and is rather anharmonic. Or am I mistaking?
 
Last edited:
  • #5
vanhees71
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You can expand any Hilbert-space vector with respect to any (generalized) basis you like, e.g., with respect to the momentum eigenbasis. This is independent of the Hamiltonian. Whether the choice of the basis is a good one in the sense to solve the eigenvalue problem for the Hamiltonian (stationary states) or the time evolution of the system is another question.
 
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  • #6
SemM
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You can expand any Hilbert-space vector with respect to any (generalized) basis you like, e.g., with respect to the momentum eigenbasis. This is independent of the Hamiltonian. Whether the choice of the basis is a good one in the sense to solve the eigenvalue problem for the Hamiltonian (stationary states) or the time evolution of the system is another question.
Thanks van Hees!
 

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