Solving Gaussian Distribution Problem: Find (Δx)^2 and Uncertainty δp

In summary, the wavefunction is just \phi = \frac{1}{a^{1/2} (2\pi)^{1/4}} \cdot \exp(\frac{-(x-x_0)^2}{4a^2})\exp(\frac{i p_0 x}{\hbar})\exp(-i \omega_0 t)
  • #1
Domnu
178
0
Problem
Let us define a wave function [tex]\phi = A \exp \(\frac{-(x-x_0)^2}{4a^2}\) \exp \(\frac{i p_0 x}{\hbar}\) \exp(-i \omega_0 t)[/tex]. Show that [tex](\Delta x)^2 = a^2[/tex]. Also, calculate the uncertainty [tex]\delta p[/tex] for a particle in the given state.

Attempt at a solution
I honestly have no idea as to how to proceed... could someone give me a hint without giving away the answer?
 
Physics news on Phys.org
  • #2
Well what exactly is giving you problems?
Do you know how to normalize the wavefunction?
Do you know in theory how to calculate the expectation values, and uncertainties?
Or is all that fine and the mathematics is giving you problems?
 
  • #3
Okay, well I know how to normalize the function... we just have

[tex]\int_{-\infty}^{\infty} A^2 e^{\frac{-(x-x_0)^2}{2a^2}} = 1[/tex].

So if we let [tex]X = (x-x_0)/a[/tex], then we get

[tex]A^2 \cdot a \cdot \int_{-\infty}^{\infty} e^{-X^2/2} dX = 1[/tex],

and this just yields

[tex]A^2 \cdot a \cdot \sqrt{2\pi} = 1 \iff A = \frac{1}{a^{1/2} (2\pi)^{1/4}}[/tex].

So the wavefunction is just

[tex]\phi = \frac{1}{a^{1/2} (2\pi)^{1/4}} \cdot \exp \(\frac{-(x-x_0)^2}{4a^2}\) \exp \(\frac{i p_0 x}{\hbar}\) \exp(-i \omega_0 t)[/tex]

From here, I tried to set [tex](\Delta x)^2 = \langle x^2 \rangle - \langle x \rangle ^2[/tex], but nothing happened from there...
 
  • #4
What did you find for <x> and <x^2>? It's not clear what you mean by nothing happened from there.
 
  • #5
Well, I tried to calculate <x^2>, but it ended up that I got an integral of the form

[tex]\int_{-\infty}^{\infty} x^2 e^{x^2} dx[/tex]

However, <x> was pretty easy to find... it's just [tex]x_0[/tex]. I'm just wondering... maybe I set up the integral the wrong way... to find <x^2>, is it just

[tex]\int_{-\infty}^{\infty} \phi^* \hat{x}^2 \phi dx = \int_{-\infty}^{\infty} \phi^* x^2 \phi dx[/tex]?
 
  • #6
Yes that's what it should be. To do an integral of the form
[tex]\int{x^2e^{-x^2}}dx[/tex], here is a trick.
Let [tex]I(a) = \int_{-\infty}^{\infty}e^{-ax^2}dx[/tex]. Then [tex]I'(a)=\int_{-\infty}^{\infty}-x^2e^{-ax^2}dx [/tex]. You should already know I(a), so there shouldn't be a problem.
 
  • #7
Wow that's clever! Thanks very much =)
 

1. What is a Gaussian distribution?

A Gaussian distribution, also known as a normal distribution, is a probability distribution that is commonly used in statistics to represent a continuous set of data. It is symmetrical and bell-shaped, with the majority of data points falling near the mean and a few outliers on either side.

2. Why is it important to solve Gaussian distribution problems?

Gaussian distribution problems are important because they allow us to analyze and understand a set of data in a meaningful way. By finding the standard deviation and uncertainty of the data, we can make predictions and draw conclusions about the population from which the data was collected.

3. What is Δx and how is it related to Gaussian distribution?

Δx (pronounced "delta x") is a symbol used to represent the standard deviation of a Gaussian distribution. It measures the spread of data points around the mean and is an important factor in calculating the uncertainty of the data.

4. How do you find (Δx)^2 in a Gaussian distribution problem?

To find (Δx)^2, you need to first calculate the variance of the data set. This is done by taking the sum of the squared differences between each data point and the mean, divided by the total number of data points. The variance is equal to (Δx)^2, so you can simply take the square root of the variance to find Δx.

5. What is the uncertainty δp in a Gaussian distribution problem?

The uncertainty, δp, is a measure of how much we can expect the data points to deviate from the mean. It is calculated by multiplying the standard deviation, Δx, by the square root of the number of data points, n. This value represents the range within which we can expect the true mean of the population to fall.

Similar threads

  • Calculus and Beyond Homework Help
Replies
1
Views
806
Replies
2
Views
1K
  • High Energy, Nuclear, Particle Physics
Replies
2
Views
322
  • Advanced Physics Homework Help
Replies
3
Views
1K
Replies
2
Views
525
  • Advanced Physics Homework Help
Replies
1
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
1K
  • Advanced Physics Homework Help
Replies
3
Views
1K
  • Advanced Physics Homework Help
Replies
5
Views
2K
Replies
16
Views
21K
Back
Top