Normalization of a free particle quantum state

In summary, the periodic boundary condition Ψ(x)=Ψ(x+L) is used to normalize the free particle's quantum state due to the issue of non-normalizability of the free particle wavefunction. This issue can be addressed by working with wavepackets, considering only probability flux, or imposing periodic boundary conditions on free space. The latter option allows for normalization and does not affect observable results since the periodicity of free space is arbitrary.
  • #1
JK423
Gold Member
394
7
Can anyone explain to me why we use the periodic boundary condition
Ψ(x)=Ψ(x+L), in order to normalize the free particle's quantum state??

I've made 2 threads already on this some time ago, but haven't got an answer still..
I hope this time i`ll have because I am really curious about the physical significance of PBC!

John
 
Physics news on Phys.org
  • #2
Hi John,

The issue with the free particle wavefunction, [itex]e^{\pm ikx}[/itex] and linear combinations of these functions, is that they are not square integrable, and thus not normalizable if you just try to integrate over all space.

There are several ways we can get around this:

1. Work with wavepackets. We can replace the free particle's momentum space wavefunction (a delta function), with some highly peaked, non delta distribution. This will result in a position space wavefunction that is a normalizable wave packet.

2. We talk only in terms of probability flux, and thus avoid the questions of normalization and the probability of finding the particle in any finite region.

3. We can think of free space as being periodic and impose periodic boundary conditions. Doing so, allows us to normalize the wavefunction. Since the periodicity of free space is arbitrary anyway (empty space is periodic with any periodicity), observable results for the free particle should not depend on it anyway.
 
  • #3
hi
thanks
but
i am not convinced about the free particle
 
  • Like
Likes Dougias

1. What is normalization of a free particle quantum state?

Normalization is the process of ensuring that the total probability of finding a particle in a quantum state is equal to 1. In other words, it is a mathematical procedure that ensures the wavefunction of a particle is properly scaled.

2. Why is normalization important in quantum mechanics?

Normalization is important because it allows us to interpret the wavefunction as a probability amplitude. By ensuring that the total probability is equal to 1, we can accurately calculate the probability of finding a particle in a specific state.

3. How is normalization of a free particle quantum state calculated?

The normalization of a free particle quantum state is calculated by taking the integral of the wavefunction squared over all space. This integral is then set equal to 1 and solved for the normalization constant.

4. What happens if a quantum state is not normalized?

If a quantum state is not normalized, then the total probability of finding a particle in any state will be greater than 1. This violates the principles of probability and can lead to incorrect predictions in quantum mechanics.

5. Are all quantum states normalized?

No, not all quantum states are normalized. However, it is a fundamental requirement for any physically meaningful state in quantum mechanics. Non-normalized states can arise, for example, when using approximate solutions or when dealing with non-normalizable states such as the plane wave solution.

Similar threads

Replies
5
Views
43
Replies
4
Views
130
Replies
2
Views
1K
  • Quantum Physics
Replies
1
Views
229
Replies
2
Views
695
Replies
4
Views
873
  • Quantum Physics
2
Replies
57
Views
5K
Replies
5
Views
1K
  • Quantum Physics
Replies
4
Views
639
Back
Top