Normalization of wave function (Griffiths QM, 2.5)

Click For Summary
SUMMARY

The discussion focuses on normalizing the wave function of a particle in an infinite square well, specifically the initial wave function given by Ψ(x,0) = A[ψ1(x) + ψ2(x)]. The key conclusion is that the stationary states ψ1 and ψ2 are already normalized, which simplifies the normalization process. The integral |A|² ∫ (|ψ1|² + |ψ2|²) dx = 1 leads to the realization that the integral evaluates to 2 over the interval, confirming the normalization condition. This highlights the importance of exploiting the orthonormality of the stationary states in quantum mechanics.

PREREQUISITES
  • Understanding of quantum mechanics principles, specifically wave functions.
  • Familiarity with the concept of orthonormality in quantum states.
  • Knowledge of the infinite square well model in quantum mechanics.
  • Basic proficiency in integral calculus.
NEXT STEPS
  • Study the normalization of wave functions in quantum mechanics.
  • Explore the properties of orthonormal functions in Hilbert spaces.
  • Learn about the infinite square well potential and its implications in quantum mechanics.
  • Investigate the implications of superposition in quantum states.
USEFUL FOR

Students of quantum mechanics, physics educators, and anyone seeking to understand wave function normalization and the properties of quantum states in the context of the infinite square well model.

Saraphim
Messages
46
Reaction score
0

Homework Statement


A particle in the infinite square well has its initial wave function an even mixture of the first two stationary states:

\Psi(x,0) = A\left[ \psi_1(x) + \psi_2(x) \right]

Normalize \Psi(x,0). Exploit the orthonormality of \psi_1 and \psi_2

Homework Equations


\psi_n(x) = \sqrt{\frac{2}{a}} \sin \left( \frac{n\pi}{a}x\right),
where a is the width of the infinite square well.

The Attempt at a Solution


I've managed to eliminate the orthogonal parts of my integral, so I'm now left with
|A|^2 \int |\psi_1|^2 + |\psi_2|^2 dx = 1

I have the feeling that I now have to exploit the fact that they are both normalized, but why is that so? What's the logic here?

EDIT: I had written a wrong expression for \psi_n. Sorry! :(
 
Last edited:
Physics news on Phys.org
Solution: The expression for \psi_n is already normalized. I should have realized this. Therefore, the integral yields 2 over the interval
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 22 ·
Replies
22
Views
2K
Replies
1
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 29 ·
Replies
29
Views
3K
Replies
4
Views
2K