Probability of Staying in x<0 for Superposition of 2 Gaussians

Click For Summary

Discussion Overview

The discussion revolves around calculating the probability of a superposition of two Gaussian wavefunctions remaining in the region where x < 0. Participants explore the mathematical approach to integrating the squared wavefunction and the implications of normalization, particularly in the context of quantum mechanics and probability current.

Discussion Character

  • Homework-related, Mathematical reasoning, Conceptual clarification, Debate/contested

Main Points Raised

  • One participant outlines the need to calculate the integral of the squared wavefunction over the range from -∞ to 0 to find the probability of x < 0.
  • Another participant suggests that the integrals of the two Gaussian components can be calculated separately and summed.
  • Concerns are raised about the possibility of obtaining a probability greater than 1 if both integrals are taken from -∞ to ∞, given that the Gaussians are separately normalized.
  • There is uncertainty about the interpretation of "staying in x < 0," with some suggesting it may refer to the probability at all times t, while others question if it is meant for a single time value.
  • A participant mentions the need for proper normalization constants and the importance of knowing the proportions of the Gaussians before combining them.
  • Another participant discusses the requirement for an analytic expression for probability current and the evolution of probability over time, referencing a specific paper for context.
  • There is a query about whether equal weighting of the Gaussians is necessary, with a suggestion that the coefficients could vary as long as the overall wavefunction is normalized.

Areas of Agreement / Disagreement

Participants express differing views on the normalization of the wavefunctions and the interpretation of the probability calculation. There is no consensus on how to proceed with the calculations or the meaning of "staying in x < 0."

Contextual Notes

Participants note that the wavefunctions are separately normalized, which complicates the integration process. There are also discussions about the implications of normalization on the total probability and the need for additional information regarding the combination of the Gaussians.

WWCY
Messages
476
Reaction score
15

Homework Statement


I am supposed to find probability of staying in x < 0 for a superposition of two Gaussians. The wavefunction is something along the lines of:

Screen Shot 2017-11-05 at 5.43.08 PM.png


Homework Equations

The Attempt at a Solution


Usually, the step involved in finding probabilities for 1 particle is just to perform the integral of ##|\Psi|^2## between 2 points (##-\infty## to 0 i believe) . However, I believe this wavefunction is the sum of 2 separately normalized gaussians. I'm not sure how I should proceed.

Advice is greatly appreciated!
 

Attachments

  • Screen Shot 2017-11-05 at 5.43.08 PM.png
    Screen Shot 2017-11-05 at 5.43.08 PM.png
    5.1 KB · Views: 997
Physics news on Phys.org
You can still calculate this integral. Just calculate the integral of both separately and add them.
 
mfb said:
You can still calculate this integral. Just calculate the integral of both separately and add them.

Thanks for the response!

But if I choose to do both integrals from -infinity to infinity, and sum them, does that not give me a probability of 2?
 
WWCY said:

Homework Statement


I am supposed to find probability of staying in x < 0 for a superposition of two Gaussians. The wavefunction is something along the lines of:

View attachment 214386

Homework Equations

The Attempt at a Solution


Usually, the step involved in finding probabilities for 1 particle is just to perform the integral of ##|\Psi|^2## between 2 points (##-\infty## to 0 i believe) . However, I believe this wavefunction is the sum of 2 separately normalized gaussians. I'm not sure how I should proceed.

Advice is greatly appreciated!

The probability that ##x < 0## at time ##t## is ##\Pr(x < 0|t) = P(t) = \int_{-\infty}^0 |\psi(x,t)|^2 \, dx##, so you need to write out the four terms of ##|\psi|^2 = \psi^* \, \psi## and then integrate them separately. It will be complicated and unpleasant.

However, I am not quite sure what the statement "staying in ##x < 0##" means. Interpreted literally, it is asking for the probability that ##x<0## for all ##t##, and this is very different from asking that ##x < 0## for any single value of ##t##. Is that really what is wanted?
 
Ray Vickson said:
The probability that ##x < 0## at time ##t## is ##\Pr(x < 0|t) = P(t) = \int_{-\infty}^0 |\psi(x,t)|^2 \, dx##, so you need to write out the four terms of ##|\psi|^2 = \psi^* \, \psi## and then integrate them separately. It will be complicated and unpleasant.

Will this not give me something unphysical? In the problem, these Gaussians were normalized separately, with A being their normalization constants, if I chose to do the integral from ##-\infty## to ##\infty##, would it not give me something > 1?

Ray Vickson said:
However, I am not quite sure what the statement "staying in ##x < 0##" means. Interpreted literally, it is asking for the probability that ##x<0## for all ##t##, and this is very different from asking that ##x < 0## for any single value of ##t##. Is that really what is wanted?

Yep, I was asked to investigate how the probability for the region ##-\infty## to 0 evolved with time, so I do need an expression for all ##t##. To provide some context, all of this is part of a project about quantum backflow, and I'm tasked to do the "simpler" calculations like the ones above.

Thanks for the assistance!
 
Ray Vickson said:
staying in x<0" means. Interpreted literally, it is asking for the probability that x<0 for all t
Quite. I wonder if it was intended to ask for the probability at t=∞.
WWCY said:
these Gaussians were normalized separately
Then you do not have enough information. You need to know the proportions in which to combine them before renormalising.
 
If the given wavefunction is properly normalized, then the normalization constants take care that the total integral over the squared wave function is 1. This means ##|A_1|^2 + |A_2|^2 = 1##. An integral over a smaller x range will then give a value smaller than 1.

If the question asks for ##t \to \infty##, you'll need a different approach. Hint: Look at the momentum distribution.
 
Hi all, thanks for the responses, I think it's best for me to upload the paper I'm referring to.

https://arxiv.org/pdf/1301.4893.pdf

All of the stuff that I'm referring to is on pages 4 and 5 of the document. In particular, I'm looking to recover analytic expressions for Figures 1. (Probability Current) and 2. (Probability from ##-\infty<x<0##).

For probability current, I'm assuming that it's just a case of plugging the wavefunction into the Current equation.

I'd really appreciate it if someone could look through it and advise me on the interpretation!
 
  • Like
Likes   Reactions: mfb
haruspex said:
Then you do not have enough information. You need to know the proportions in which to combine them before renormalising.

Do you mean something like ##\frac{1}{\sqrt{2}} \psi_1 + \frac{1}{\sqrt{2}} \psi_2##?
 
  • #10
I initially made a post in the HW section about this topic but I believe that my question was poorly formulated, hence I'm attempting to ask a better one here, since it isn't exactly HW.

I am studying the probability current, and probability of a sum of 2 gaussians. The type of wavefunction is as shown :
Screen Shot 2017-11-05 at 5.43.08 PM.png

where ##\hbar = m = 1##

The ultimate goals are to:
a) Obtain an analytic expression for Probability Current at x = 0
b) Obtain an analytic expression for probability from ##-\infty < x < 0## for all ##t## - to study how the probability of the two wavepackets remaining at ##x<0## changes with time.
c) To obtain plots similar to Figures 1 (Probability Current) and 2 (Probability) in this paper (pages 4 and 5 respectively): https://arxiv.org/pdf/1301.4893.pdf

What I intend to do:

I was given an initial Gaussian of: ##\Psi(x,0) = e^{ipx}e^{\frac{-x^2}{2\sigma ^2}}##.

Normalizing, finding ##\widetilde{\psi}## and then ##\Psi(x,t)## for 1 Gaussian gives,
$$\Psi_k = \frac{\sqrt{\sigma}}{\pi^{1/4} \sqrt{\sigma ^2 + it}}exp(ip_k (x - \frac{p_k}{2}t) + \frac{(x - p_k t)^2}{2(\sigma ^2 + it)}) $$
then summing over 2 Gaussians will give,
$$\Psi_k =\sum_{k=1,2} \frac{\sqrt{\sigma}}{\pi^{1/4} \sqrt{\sigma ^2 + it}}exp(ip_k (x - \frac{p_k}{2}t) + \frac{(x - p_k t)^2}{2(\sigma ^2 + it)}) $$
which is not dissimilar to the one above, except for the fact that I should have started with ##\Psi = e^{ipx}e^{\frac{-x^2}{4\sigma ^2}}##, I believe.

For a: I intend to place the Sum of Gaussians into the Probability Current equation, and chug through the algebra to derive the expression, evaluating the expression at x = 0.

For b: This is the part where I am not too sure about what to do. This is a sum of separately normalized wavefunctions, simply performing ##\int_{-\infty}^{0}|\Psi(x,t)|^2 dx## does not seem the right thing to do as I believe such functions don't seem to be orthogonal, and more fundamentally ##\int_{-\infty}^{\infty}|\Psi(x,t)|^2 dx## might ##>1## (?).

In my earlier post, some commenters seemed to question my interpretation of a) and b). I too am slightly unsure and I hope someone can take the time to skim through the noted sections of the paper and provide me with some advice, or some notes I should go through before attempting to solve the problem.

Any assistance is greatly appreciated!
 

Attachments

  • Screen Shot 2017-11-05 at 5.43.08 PM.png
    Screen Shot 2017-11-05 at 5.43.08 PM.png
    6.1 KB · Views: 437
  • #11
Why do you feel each Gaussian must have equal weighting? The coefficients for each term could quite general as long as the whole this is normalised.
 

Similar threads

  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 21 ·
Replies
21
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 10 ·
Replies
10
Views
3K