Proving Parity in Normalized Solutions of the Schrodinger Equation

  • Thread starter Thread starter ProfessorChaos
  • Start date Start date
  • Tags Tags
    Proof Schrödinger
Click For Summary
SUMMARY

The discussion focuses on proving that normalized solutions to the time-independent Schrödinger equation exhibit definite parity when the potential V(x) is an even function, specifically V(x) = V(-x). Participants emphasize that if u(x) is a solution, then it must satisfy the condition u(x) = ±u(-x). The approach involves substituting y = -x into the equation and analyzing the resulting solutions. The consensus is that this method effectively demonstrates the parity of the solutions.

PREREQUISITES
  • Understanding of the time-independent Schrödinger equation
  • Knowledge of quantum mechanics concepts, particularly potential energy functions
  • Familiarity with the concept of parity in physics
  • Basic mathematical skills for manipulating equations and functions
NEXT STEPS
  • Study the properties of even and odd functions in quantum mechanics
  • Learn about normalization techniques in quantum wave functions
  • Explore the implications of parity in quantum systems
  • Investigate the role of potential functions in determining solution characteristics
USEFUL FOR

Students and researchers in quantum mechanics, physicists interested in wave function properties, and anyone studying the implications of symmetry in physical systems.

ProfessorChaos
Messages
2
Reaction score
0
Hopefully this is in the correct section.

Struggling with this question though I don't think it should be particularly difficult:

Show that if V(x) = V(-x) normalised solutions to the time-independent Schrödinger equation have definite parity - that is, u(x) = +-u(-x)

(+- means plus or minus. Sorry for poor formatting - on phone)

Thanks in advance.
 
Physics news on Phys.org
Just read the "how to ask for help" sticky.

My attempt at a solution involved subbing y = -x. Obtaining a solution, then trying to normalise it. However I don't get anywhere really, and my line of argument through my current page of working reads unconvincing (I'm just guessing).

Cheers
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 17 ·
Replies
17
Views
3K
  • · Replies 5 ·
Replies
5
Views
5K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
2K
Replies
4
Views
2K