Prove this function satisfies the time independent schrodinger equation

Click For Summary
SUMMARY

The function Ψ = A e^(i k x) + B e^(-i k x) satisfies the time-independent Schrödinger equation when the potential V is constant. The equation is expressed as (- (ħ²) / 2m) * (d²Ψ / dx²) + VΨ = EΨ. The second-order derivative of Ψ results in (-k² A e^(i k x)) + (-k² B e^(-i k x)), leading to the equation (-(ħ²) / 2m)(-2k²) + V = E. The discussion seeks clarification on the values of k that allow Ψ to be a valid solution.

PREREQUISITES
  • Understanding of the time-independent Schrödinger equation
  • Familiarity with complex exponential functions
  • Knowledge of quantum mechanics terminology, specifically potential energy (V) and energy (E)
  • Basic calculus, particularly second-order derivatives
NEXT STEPS
  • Explore the implications of constant potential in quantum mechanics
  • Study the role of the wave number k in quantum wave functions
  • Investigate boundary conditions for solutions of the Schrödinger equation
  • Learn about normalization of wave functions in quantum mechanics
USEFUL FOR

Students and professionals in quantum mechanics, physicists analyzing wave functions, and anyone studying the properties of the time-independent Schrödinger equation.

keith river
Messages
14
Reaction score
0
Prove this function satisfies the time independent Schrödinger equation. When V is constant.
Psi = A e^(i k x) + B e^(-i k x)

Attempt:
Time independent Schrödinger equation : (- (hbar^2) / 2m) * (d^2 Psi / d x^2) + V Psi = E Psi

Second order derivative of Psi : (-k^2 A e^(i k x)) + (- k^2 B e^(-i k x))

Divide all by Psi.

I get (-(hbar^2) / 2m) (-2k^2) + V = E

From here I don't know where to go. Any pointers?
 
Physics news on Phys.org
k is a parameter in the solution. Are there any values of k for which \Psi is a solution?
 

Similar threads

Replies
7
Views
2K
Replies
29
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K