Survival Probability of a free particle in time?

Click For Summary

Homework Help Overview

The discussion revolves around the survival probability of a free particle described by the wavefunction ψ = e^(-|x|). The original poster seeks to plot this probability and is exploring the necessary mathematical framework to achieve this, particularly in relation to time evolution and normalization of the wavefunction.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss normalization of the wavefunction and the implications of the Schrödinger equation on time evolution. There are inquiries about the nature of survival probability for a free particle and the assumptions regarding separation of variables in the wavefunction.

Discussion Status

Some participants have provided guidance on normalization and the use of Fourier transforms. There is an ongoing exploration of the mathematical expressions involved, with participants questioning the validity of certain assumptions and discussing the relevance of the Dirac delta function in the context of the problem.

Contextual Notes

There are indications of confusion regarding the time dependence of the wavefunction and the conditions under which the time-independent Schrödinger equation can be applied. The original poster expresses a desire to learn and clarify these concepts as part of their project.

Poirot
Messages
87
Reaction score
2

Homework Statement


I want to plot a graph of the survival probability of the initial state ψ = e-|x| for a free particle. Hopefully this will enable me to plot some more difficult examples like the inverted oscillator etc for a project but I'm struggling fundamentally with the free particle.

Homework Equations

The Attempt at a Solution


I have tried simple things like trying to calculate the probability using the usual integral over all space but that didn't work. I think I need to use something more complex like a propagator perhaps but my knowledge is very limited. Do I need to confine my wavefunction so it can be normalised? how do I find the time dependence for this initial state?

Any help on how to approach this would be great! I'm ultimately trying to reproduce a graph in mathematica that originated from a paper on tunnelling out of a time dependent well but I am a rookie in this field and I'm trying to teach myself the physics as I go along.
 
Physics news on Phys.org
You can normalize your state as it falls off quickly enough for large x (the integral over the whole space exists).
The Schrödinger equation will tell you how the state evolves over time.

Survival of what?
 
mfb said:
You can normalize your state as it falls off quickly enough for large x (the integral over the whole space exists).
The Schrödinger equation will tell you how the state evolves over time.

Survival of what?
Thank you! I will give that a go. And it says 'survival probability of the initial state'. I'm a little confused as to why the probability for a free particle wouldn't be 1 for all time?
 
You can look at ##\langle \Psi(t) | \Psi(0) \rangle## I guess. It will start at 1 and go to zero over time. That is some sort of survival.
 
I have integrated the wavefunction and got that it's already normalised. When I introduce a factor of e^(-iwt) for the time-dependence and then take the inner product I just get 1 with no time dependence?
 
Poirot said:
When I introduce a factor of e^(-iwt) for the time-dependence
What is w?

Follow-up question: Can you really assume that?
 
mfb said:
What is w?

Follow-up question: Can you really assume that?
I meant omega as w, but I realize I made an assumption rather than calculating it.
I have plugged my function ψ(t) = c(t) e^(-|x|) into the TDSE to try and solve for c(t) this gives me an expression:
\begin{eqnarray*}
\frac{\hbar^2}{2m} c(t) e^{-|x|} + \frac{\hbar^2}{m} c(t) \delta(x) = i\hbar \frac{dc}{dt}e^{-|x|}
\end{eqnarray*}
I then equated the coeffecients of e^-|x| and integrated to get c(t) = exp[iħt/2m]
but I'm not sure if that's allowed mathematically.

I tried finding the inner product of this TD state with the initial state and again got 1 for the probability due to the e^i... part being mod^2 at the end?
 
You are assuming that you can separate the wave function into a time- and a space-dependent part. See the follow-up question from my previous post.

You cannot do that, the wave function cannot be expressed that way.
Poirot said:
but I'm not sure if that's allowed mathematically.
The ##\delta(x)## is relevant as well.
 
mfb said:
You are assuming that you can separate the wave function into a time- and a space-dependent part. See the follow-up question from my previous post.

You cannot do that, the wave function cannot be expressed that way.The ##\delta(x)## is relevant as well.
Am I right in thinking you can't assume we can use separation of variables because the TISE isn't satisfied by the wavefunction?

I haven't ever come across a wavefunction in which one can't assume separation of variables, would I have to assume c = c(x,t) and solve the differential equation for this?
Sorry to ask so many questions, I'm being pushed beyond my current knowledge for my project, and thanks for all the help!
 
  • #10
TISE = time-independent Schrödinger equation? That only works for energy eigenstates.

Poirot said:
would I have to assume c = c(x,t) and solve the differential equation for this?
Just assume ##\Psi(x,t)## (i. e. don't make an assumption) and solve that differential equation.
A Fourier transformation might help.
 
  • #11
mfb said:
TISE = time-independent Schrödinger equation? That only works for energy eigenstates.

Just assume ##\Psi(x,t)## (i. e. don't make an assumption) and solve that differential equation.
A Fourier transformation might help.
Ah okay thank you!

Do I need to assume I have a wave packet and then use a Fourier Transform? And I can't see how the dirac delta comes into play from the differential?

Thanks again
 
  • #12
You can always do a Fourier transformation. The evolution of the wave function is easier to describe if you have the transformed wave function.
Poirot said:
And I can't see how the dirac delta comes into play from the differential?
Well, it is there, and the equation has to be true everywhere, including x=0.
 
  • #13
mfb said:
You can always do a Fourier transformation. The evolution of the wave function is easier to describe if you have the transformed wave function.
Well, it is there, and the equation has to be true everywhere, including x=0.
Thanks for all your help. I think I've found the wave function with x and t dependence:
\begin{eqnarray*}
\Psi(x,t) = \frac{1}{\pi}}\int_{-\infty}^{\infty} \frac{e^{ikx - \frac{i\hbar k^2t}{2m}}}{1+k^2}
\end{eqnarray*}

I found this using Fourier transform as suggested. I have tried plugging this into mathematica to solve but it seems it isn't possible and I not entirely sure how to proceed. You mentioned the dirac delta needing to come into this but I can't see how at the moment?

Thanks in advance!

Edit: I apologise for the lack of latex, not sure what I've messed up here..
 
  • #14
The Dirac delta doesn't appear any more in the equations if you go to the momentum representation (=after Fourier transformation).
The formula doesn't parse, and the integral doesn't seem to have an integration variable (k I guess?).
I don't know if it is necessary to "solve" this integral.
 
  • #15
mfb said:
The Dirac delta doesn't appear any more in the equations if you go to the momentum representation (=after Fourier transformation).
The formula doesn't parse, and the integral doesn't seem to have an integration variable (k I guess?).
I don't know if it is necessary to "solve" this integral.
Ok thank you, and my mistake it should be integration with respect to k.

If I am trying to plot this for a 'survival probability' how would I get values out of this for the probability as a function of time?
 
  • #16
##\langle \Psi(t) | \Psi(0) \rangle## (well, the magnitude of it) - Mathematica can at least integrate it numerically, but there might be an analytic solution.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
Replies
8
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
2
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K