# Fourier transform for a localized free particle

• Irid
In summary: The smaller waves are more localized and so have a lower amplitude (because they have less energy). This is analogous to the k distribution having a lower peak amplitude.
Irid

## Homework Statement

This is from Griffiths Introduction to Quantum Mechanics, Problem 2.21.

Suppose a free particle, which is initially localized in the range -a<x<a, is released at time t=0:
$$\Psi(x,0) = \begin{cases} \frac{1}{\sqrt{2a}}, & \text{if } -a<x<a,\\ 0, & \text{otherwise},\end{cases}$$
Determine the Fourier transform $$\phi(k)$$ of $$\Psi(x,0)$$ and comment on the behavior of $$\phi(k)$$ for very small and very large values of a. How does this relate to the uncertainty principle?

## Homework Equations

$$\phi(k) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty} \Psi(x,0) e^{-ikx}\, dx = \frac{1}{2\sqrt{\pi a}} \int_{-a}^{+a} e^{-ikx}\, dx = -\frac{1}{2ik\sqrt{\pi a}} \left(e^{-ika}-e^{ika}\right) = \frac{\sin (ka)}{k\sqrt{\pi a}}$$

## The Attempt at a Solution

The one Fourier transform with a greater $$a$$ has more rapid oscillations and a higher peak at k=0, when compared to the one with a smaller $$a$$. A big value of $$a$$ corresponds to a widely spread particle, while a small $$a$$ stands for a particle well localized at x=0.

My problem is that I don't see the connection between my Fourier transform and uncertainty. As far as I understand, rapid oscillations serve to pick up some specific values of energy, and the resulting wave function is close to a discrete linear combination, while slow oscillations pick up a wide spectrum of energies. But what does this have to do with uncertainty? Both seem quite uncertain to me...

Last edited:
In terms of uncertainty look at your initial wave function, what does it represent? It represnts a probability amplitude to be found at position x, if a measurement is made. The bigger a is, the more uncertainty in the initial position of the particle when a measurement is made

Now have a think about what the Fourier transfrom is 'transforming' to...ie it is shifting from description in position space to...

Then look at your solution effectively sinka/k, with a normalisation factor. Sounds like you've plotted it up but always a good place to start. Amplitude decays like 1/k and oscillates as sinka.

I wouldn't get too tied up with the oscillations at this stage they are more an artifact of the orginal position distribution shape, which is rectangular. Concentrate on the main peak for now

You're close with you comments of wide v narrow energy distribution, but why make the jump to energy? A k state for a free particle is an energy eigenstate, but work in k space for the moment. Think about the width of the peak and how this affects the range of outcomes for a measurement of k. Then consider how this relates to the uncertainty in the in the original position distribution...

Last edited:
Yes, I actually have a graph:
http://img410.imageshack.us/my.php?image=221eo7.pdf"
I only don't know if you can see it, it's a PDF.

Anyway, OK, I concentrate on the main peak. For a big $$a$$ the particle can be in a wide range of positions, so it has a high position uncertainty, but the Fourier transform in such a case gives a narrow and high peak at the center, which serves to pick up a small range of energies (roughly acts like a delta function):

$$\Psi(x,t) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty} \phi(k) e^{ikx} e^{-iE(k)t/\hbar}\, dk$$

which is the opposite of what I expect. An initially uncertain particle gets to transformed into something with a well defined energy and position (center).

Last edited by a moderator:
getting close, but you need to understand the the Fourier transfrom is taking you to momentum (k) space, so instead of a description in position (x)you now have a description in momentum (k) (ie the bottom axis on graph changes from x to k after the transform)

So once we have transformed to k space, the particle is no longer described in terms of position (the position information is inhernet in the wave function but not expressed directly)

Your observations about the realative peak widths were right, though interpretation of the k distributtion was a bit misguided. Try thinking again about these peaks in term of x & k uncertainties...

have a go thinking it through first but if you need a further hint:

Last edited:

## 1. What is a Fourier transform?

A Fourier transform is a mathematical operation that decomposes a function into its constituent frequencies. It is commonly used in signal processing and image analysis to analyze and manipulate signals and images.

## 2. How does Fourier transform work for a localized free particle?

For a localized free particle, the Fourier transform is used to analyze the wave function, which describes the position and momentum of the particle. The Fourier transform converts the wave function from the position domain to the momentum domain and vice versa.

## 3. What is the physical significance of the Fourier transform for a localized free particle?

The Fourier transform for a localized free particle allows us to understand the quantum behavior of the particle. It helps us to determine the probability of finding the particle at a certain position or with a certain momentum. It also allows us to analyze the energy states of the particle.

## 4. How is the Fourier transform related to Heisenberg's uncertainty principle?

The Fourier transform and Heisenberg's uncertainty principle are closely related. The Fourier transform of a particle's wave function is a representation of its momentum distribution, and the uncertainty principle states that there is a fundamental limit to the precision with which we can measure both the position and momentum of a particle simultaneously.

## 5. Are there any real-world applications of the Fourier transform for a localized free particle?

Yes, the Fourier transform for a localized free particle has many practical applications, particularly in quantum mechanics and quantum computing. It is also used in a variety of fields such as medical imaging, signal processing, and data compression.

Replies
0
Views
334
Replies
11
Views
1K
Replies
1
Views
1K
• Calculus and Beyond Homework Help
Replies
5
Views
461
Replies
2
Views
785
Replies
16
Views
674
Replies
1
Views
863
Replies
2
Views
4K