Sample Space for Free Particle in the general case

Click For Summary
SUMMARY

The discussion focuses on the interpretation of Schrödinger's Equation for a free particle, specifically the wave function represented as ψ(X,T) = e^{i/\hslash ( px - Et)}. It is established that this wave function is not normalizable, leading to the conclusion that a real free particle must be represented by a wave packet instead. The participants clarify that while the provided wave function is a partial solution, it remains useful in certain applications, such as scattering problems involving finite square barriers, where it yields satisfactory results despite its limitations.

PREREQUISITES
  • Understanding of Schrödinger's Equation in quantum mechanics
  • Familiarity with wave functions and probability density functions
  • Knowledge of wave packets and their significance in quantum mechanics
  • Basic concepts of scattering theory in quantum mechanics
NEXT STEPS
  • Study the concept of wave packets in quantum mechanics
  • Learn about normalizable wave functions and their properties
  • Explore scattering problems using wave packets and compare results with plane wave solutions
  • Investigate the implications of momentum representation in quantum mechanics
USEFUL FOR

Students and enthusiasts of quantum mechanics, particularly those interested in wave functions, scattering theory, and the foundational concepts of quantum particle behavior.

IronHamster
Messages
28
Reaction score
0
I am a beginner to quantum mechanics and am trying to make sense of Schrödinger's Equation. I am attempting to find probabilities in the case of a free particle in the general case.

It is my understanding that the solution to Schrödinger's Equation in the general case of a free particle is as follows:

\psi(X,T) = e^{i/\hslash ( px - Et)}

The modulus square of this is 1, which means the probability density function is uniform.

Two questions:
1. Over what values of x is this pdf defined? Can we eliminate all values of x > ct?
2. Am I correct to interpret x as the distance from the (known) starting position of the particle at t = 0?

Thanks.
 
Physics news on Phys.org
Notice that that wave-function is not normalizable. The integral of the modulus square of that wave-function over all space is infinite. A free particle wave function cannot actually be what you gave, but must be a wave-packet.
 
Matterwave said:
Notice that that wave-function is not normalizable. The integral of the modulus square of that wave-function over all space is infinite. A free particle wave function cannot actually be what you gave, but must be a wave-packet.

So are you saying that the solution I mentioned does not describe a free particle wave? I'm not sure how that could be, I have read from multiple sources that it is.

Is there a different approach that needs to be taken to achieve a normalizable function?
 
A real free particle cannot be represented by that function because that function is not normalizable. A real free particle is represented by a wave packet. That function is a function of a particle with exactly 1 momentum (p), but really a particle is represented by a wave packet which has a range of momenta.

You can say that your equation is only a "partial" solution. It hasn't been fixed up yet.

Still, that function is useful for many applications. For example, if we are doing a scattering problem off of a finite square barrier, we tend to just use that function. The central results you obtain by using that function (the transmission and reflection coefficients) is surprisingly good to the results you would get if you made wave packets; however, doing a scattering problem with wave packets is a nightmare.
 
Oh ok that makes sense. Thanks!
 

Similar threads

  • · Replies 25 ·
Replies
25
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 27 ·
Replies
27
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 17 ·
Replies
17
Views
3K