Dispersion relation (particle in a box)

Click For Summary
SUMMARY

The discussion focuses on the dispersion relation for a particle in a one-dimensional box, specifically examining the energy equations E = n²(π)²(ħ)² / 2mL² and E = (ħ)²k² / 2m. It clarifies that while momentum p is not an eigenstate, k² can be interpreted as the eigenvalue of p², linking energy eigenvalues E to momentum k. The conversation emphasizes that in solid state physics, the energy spectrum becomes independent of boundary conditions in the large volume limit, allowing for the interpretation of k as momentum under periodic boundary conditions.

PREREQUISITES
  • Understanding of solid state physics concepts, particularly energy quantization.
  • Familiarity with the Schrödinger equation and eigenstates.
  • Knowledge of boundary conditions in quantum mechanics.
  • Basic grasp of wave-particle duality and momentum-energy relationships.
NEXT STEPS
  • Explore the implications of periodic boundary conditions in quantum mechanics.
  • Study the relationship between wave vectors k and momentum p in quantum systems.
  • Investigate the concept of density of states in solid state physics.
  • Learn about the role of eigenvalues and eigenstates in quantum mechanics.
USEFUL FOR

Students and researchers in solid state physics, quantum mechanics enthusiasts, and anyone interested in the mathematical foundations of energy-momentum relationships in quantum systems.

HAMJOOP
Messages
31
Reaction score
0
I am learning some basic solid state physics idea, like density of state ...etc.

For particle in a 1D box,

E = n^2 (pi)^2 (h_bar)^2 / 2mL^2

But why it is written as
E = (h_bar)^2 k^2 /2m

does it means that energy eigenvalue E is related to momentum k ?
I guess it is not because momentum is not eigenstate.

But what is this expression talking about anyway ? what is the physical meaning of this k ?
 
Physics news on Phys.org
While p isn't an eigenstate, p^2 is. So take k^2 as eigenvalue of p^2. In solid state physics the point is that the energy spectrum in the large volume limit becomes independent of the boundary conditions. So you could also assume periodic boundary conditions and interpret k as momentum.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
9K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 0 ·
Replies
0
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K