Finding fourier transfrom of the following wavefunction

In summary, the conversation discusses the wavefunction of a free particle, specifically the function Psi(x,0)=E^(ik0x) when x=(-a/2,a/2) and zero elsewhere. The question is whether this can be a wavefunction for a free particle, and the answer is yes because any function of x can be expressed as a wavepacket. The conversation also touches on calculating the probability of finding the particle between two positions, as well as the Fourier transform and the uncertainty principle. The conclusion is that while k may be certain, we know nothing about the position of the particle.
  • #1
Feynmanfan
129
0
Let Psi(x,0)=E^(ik0x) when x=(-a/2,a/2) and zero elsewhere.

Can this be a wavefunction of a free particle. I believe it is so because every function of x can be expressed as a wavepacket. Is this correct?

If I want to calculate P(x,0), probability to find the particle between x, x+dx it's just the square of the modulus. But what about P(k,0)? I'm having trouble calculating it's Fourier transform, I think that the delta function must show somewhere but I don't know how.

k seems to be certain k=k0 , right? However, P(x,0)=1/a is the same everywhere.
 
Physics news on Phys.org
  • #2
Feynmanfan said:
Let Psi(x,0)=E^(ik0x) when x=(-a/2,a/2) and zero elsewhere.

Can this be a wavefunction of a free particle. I believe it is so because every function of x can be expressed as a wavepacket. Is this correct?

If I want to calculate P(x,0), probability to find the particle between x, x+dx it's just the square of the modulus. But what about P(k,0)? I'm having trouble calculating it's Fourier transform, I think that the delta function must show somewhere but I don't know how.

k seems to be certain k=k0 , right? However, P(x,0)=1/a is the same everywhere.

If k_0 is one definite value, then the particle has one precise momentum. What does the uncertainty principle say about the position of such a particle?
 
  • #3
Is the wavefunction normalizable...?It is a generalized eigenfunction of the Hamiltonian...?

Daniel.
 
  • #4
Yes it is normalizable. But it's not a generalized function of the Hamiltonian, is it? It's a particular case where k=ko.

I'm asked to draw P(x,0) and P(k,0) and find out delta(x) and delta(k) and justify it using Heisenberg's uncertainty principle.

By doing Psi's Fourier transform I get a complicated function and I don't know if that's the way I can justify the following: we know nothing about the position (cause all probabilities are the same) but k is certain.
 

1. What is a Fourier transform?

A Fourier transform is a mathematical tool used to convert a function of time or space into a function of frequency, and vice versa. It is commonly used in mathematics, physics, and engineering to analyze and understand the frequency components of a signal or function.

2. Why is it important to find the Fourier transform of a wavefunction?

The Fourier transform allows us to break down a complex wavefunction into its individual frequency components. This can help us better understand the behavior and properties of the wavefunction, and can also be used to solve differential equations and other mathematical problems.

3. How do you find the Fourier transform of a wavefunction?

To find the Fourier transform of a wavefunction, you first need to express the wavefunction as a function of time. Then, you can apply the Fourier transform formula, which involves an integral over time. The resulting function will be a representation of the wavefunction in terms of frequency.

4. What are some common applications of Fourier transforms?

Fourier transforms have many applications in various fields, including signal processing, image processing, data compression, and quantum mechanics. They are also used in solving differential equations, analyzing vibrations and waves, and studying the properties of materials and structures.

5. Is there a specific method for finding the Fourier transform of a wavefunction?

Yes, there is a specific formula for finding the Fourier transform, which involves taking an integral over time. However, the exact method or technique used may vary depending on the specific properties and behavior of the wavefunction. It may also require some mathematical manipulation and understanding of the properties of Fourier transforms.

Similar threads

  • Introductory Physics Homework Help
Replies
6
Views
278
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
9
Views
686
  • Introductory Physics Homework Help
Replies
14
Views
2K
  • Quantum Physics
Replies
4
Views
801
  • Introductory Physics Homework Help
Replies
28
Views
361
  • Calculus and Beyond Homework Help
Replies
5
Views
342
Replies
2
Views
263
  • Introductory Physics Homework Help
Replies
8
Views
750
  • Introductory Physics Homework Help
Replies
1
Views
1K
Back
Top