Is the Size of a Wave Packet Constant in Quantum Physics?

[luisgml_2000]

Hello:

I need some help with a homework problem that was taken from Quantum Physics by Gasiorowicz.

The problem goes like this: You have a beam of electrons and know the size of the wave packet, and by the uncertainty principle you can estimate the dispersion in p at t=0. The problem is to know the size of the wave packet after the beam has crossed 10^4km in two cases: i) when the K.E. of the beam is 13.6 eV and ii) 100MeV.

It is clear for me that the wave packet does not have a constant size since the dispersion relation does "disperse"; I also know that in the first case the problem can be treated non-relativistically, while the second case is relativistic.

I don't want assume the wave packet to be gaussian.

However, I do not know how to combine these ideas to solve the problem.

 

[meopemuk]

I don't want assume the wave packet to be gaussian.

The shape of the wave packet was not specified, so, I think, the Gaussian form assumption is justified. You are looking for a rough estimate, anyway. So, why not make your task easier?

Eugene.

 

[luisgml_2000]

If I assume the wave packet to be gaussian, the problem would be solved, since the dynamics of such a wave packet can be calculated.

The theoretical facts I want to use are the uncertainty principle and the dispersion relation E=p^2/2m.

 

[nrqed]

If I assume the wave packet to be gaussian, the problem would be solved, since the dynamics of such a wave packet can be calculated.

The theoretical facts I want to use are the uncertainty principle and the dispersion relation E=p^2/2m.

I would simply calculate the distance traveled by an electron if it had the energy given in the problem. Then I would calculate the distance traveled by an electron with momentum p_0 + \delta p where delta p is given by the uncertainty principle. The difference between the two results is a rough estimate of the dispersion of the wave packet

 

[luisgml_2000]

That's an excellent idea. Thank you very much for your help.

 

[Anglea]

The shape of the wave packet was not specified, so, I think, the Gaussian form assumption is justified. You are looking for a rough estimate, anyway. So, why not make your task easier?

Eugene.

I've been struggling with that question long time, which is what is the advantage of assuming the wavepacket as Gaussian ??please help me out

 

[meopemuk]

I've been struggling with that question long time, which is what is the advantage of assuming the wavepacket as Gaussian ??please help me out

In order to solve your problem you need to

1. Assume some initial form of the wave function at t=0.
2. Apply the non-stationary Schroedinger equation to find the wave function at t>0.

If you choose the wave function in 1. to be Gaussian, then the solution of 2. is given by an analytically calculable integral. Moreover, Gaussian wavepackets have the advantage that it is easy to go from position representation to the momentum representation and back.

 

[Anglea]

In order to solve your problem you need to

1. Assume some initial form of the wave function at t=0.
2. Apply the non-stationary Schroedinger equation to find the wave function at t>0.

If you choose the wave function in 1. to be Gaussian, then the solution of 2. is given by an analytically calculable integral. Moreover, Gaussian wavepackets have the advantage that it is easy to go from position representation to the momentum representation and back.

appreciate your time, so you are saying such assumption lead to analytically calculable integral, by using the useful Gaussian integral relation, which make the calculation more easy. my basic understanding of the advantage of the Gaussian wavepacket is to ensure the localization and then the Normalization. how does ti sound? please enlighten me more??

 

[meopemuk]

appreciate your time, so you are saying such assumption lead to analytically calculable integral, by using the useful Gaussian integral relation, which make the calculation more easy. my basic understanding of the advantage of the Gaussian wavepacket is to ensure the localization and then the Normalization. how does ti sound? please enlighten me more??

Of course, your wavepacket should be localized and normalized. The former condition follows from the fact that you are trying to represent a classical particle, which is localized. The latter condition is just a rule of quantum mechanics. So, localization and normalization are necessary requirements: you can't do without them. However, these requirements can be satisfied with many different functional forms of the packet. The only advantage of the Gaussian form is that it is very easy to calculate integrals with Gaussian functions, like \exp(-ar^2).

 

[Anglea]

In order to solve your problem you need to

1. Assume some initial form of the wave function at t=0.
2. Apply the non-stationary Schroedinger equation to find the wave function at t>0.

You help much appreciated...
what du you mean by non-stationary? Is the word wavepacket means that(localization+Normalization)??

Gaussian wavepackets have the advantage that it is easy to go from position representation to the momentum representation and back.

please can you kindly elaborate this sentence by an example? cos I didn't understand it...
when we say Gaussian ensembles?Is this same as Gaussian wavepackets??

 

[Anglea]

As an example on localization and normalization, herr the bound and scattered state...

a scattered state is one that is influenced by a scattering potential but has nontrivial |ψ|2 at r -> infinity, and a bound state is one that has |ψ|2 -> 0 at r -> infinity.

I s this means that the bound state is localized and normalized, and the scattered state is not?if so why is that as I think it is the opposite as the bound state usually given by exponential function and ,e.g bound particle always have an exponential wave function?so this means is not narmalised, please show me where is my confusion

 

[meopemuk]

You help much appreciated...
what du you mean by non-stationary? Is the word wavepacket means that(localization+Normalization)??

The non-stationary Schroedinger equation is

i \hbar \frac{d \Psi(r,t)}{dt} = H \Psi(r,t)

If you know the wavefunction (or wavepacket) at t=0 you can solve this Schroedinger equation and find the wavefunction at any other time.

please can you kindly elaborate this sentence by an example? cos I didn't understand it...
when we say Gaussian ensembles?Is this same as Gaussian wavepackets??

By Gaussian wavepacket centered at position \mathbf{r}_0 I mean function

\Psi(\mathbf{r}) = N \exp(-a (\mathbf{r} - \mathbf{r}_0)^2)

where N is a normalization constant chosen to make sure that the probability distribution associated with this wavefunction is normalized to 1. I don't know about "Gaussian ensembles".

 

[Anglea]

The non-stationary Schroedinger equation is

i \hbar \frac{d \Psi(r,t)}{dt} = H \Psi(r,t)
By Gaussian wavepacket centered at position \mathbf{r}_0 I mean function

\Psi(\mathbf{r}) = N \exp(-a (\mathbf{r} - \mathbf{r}_0)^2)

where N is a normalization constant chosen to make sure that the probability distribution associated with this wavefunction is normalized to 1. I don't know about "Gaussian ensembles".

Thanks, Is the non-stationary Schroedinger equation is
\imath\hbar\frac{\partial}{\partial t}\psi(\textbf{r},t)=\left(\frac{-\hbar^2}{2m}\nabla_r^2+V(r) \right)\psi(\textbf{r},t)
if so, what is the formula for stationary Schroedinger equation?
Q2 I've read this sentence, and it is really confused me

there is a contradiction between the normalization of the members of the wavebacket and the Gaussian distribution of the wavefunction

please enlighten me

 

[meopemuk]

Thanks, Is the non-stationary Schroedinger equation is
\imath\hbar\frac{\partial}{\partial t}\psi(\textbf{r},t)=\left(\frac{-\hbar^2}{2m}\nabla_r^2+V(r) \right)\psi(\textbf{r},t)
if so, what is the formula for stationary Schroedinger equation?

Yes, this is the non-stationary (or time-dependent) Schroedinger equation. The stationary Schroedinger equation is used to find out the energy spectrum E_n of the system

\left(\frac{-\hbar^2}{2m}\nabla_r^2+V(r) \right)\psi_n(\textbf{r}) = E_n \psi_n(\textbf{r})


About the confusing sentence you should ask its author. It is just as confusing to me.

 

[Anglea]

can anyone tell me what one means by these bracket \langle\ psi(r)\psi^*(r) \rangle , is it integral or what?and is there any idea how to calculate this integral \int\int{\mid\langle\psi(r)\psi^*(r')\rangle\mid}^2 dr dr'

 

[Anglea]

Is the regular system means non-chaotic one? if so what is the irregular system, can anyone give an examples to both systems?please help me