Does This Wave Function Normalization Look Correct?

Click For Summary
SUMMARY

The discussion focuses on the normalization of the wave function \(\Psi(x,t) = A e^{-\lambda |x|}e^{-i \omega t}\) in quantum mechanics, as presented in Griffiths' textbook. The correct normalization constant \(A\) is derived to be \(A = \sqrt{\lambda}\) after evaluating the integral \(\int_{-\infty}^{\infty} |\Psi|^2 dx\). Participants confirm that the approach of integrating from 0 to infinity and multiplying by two is valid due to the symmetry of the function. The importance of understanding the treatment of absolute values in integrals is emphasized.

PREREQUISITES
  • Understanding of quantum mechanics principles
  • Familiarity with wave functions and their properties
  • Knowledge of integral calculus, specifically improper integrals
  • Experience with complex numbers and their conjugates
NEXT STEPS
  • Study the normalization of wave functions in quantum mechanics
  • Learn about the properties of even and odd functions in integrals
  • Explore the implications of symmetry in quantum systems
  • Review techniques for evaluating improper integrals
USEFUL FOR

Students of quantum mechanics, particularly those studying wave functions and normalization, as well as educators seeking to clarify concepts related to symmetry and integration in physics.

r16
Messages
42
Reaction score
0

Homework Statement


I'm starting to (trying) teach myself some quantum mechanics out of the Griffiths book, and since there are no answers in the back I have no idea if I'm on the right track or not. Could you guys look over the answer to this equation to see if it looks right?

Consider the wave function \Psi(x,t) = A e^{-\lambda |x|}e^{-i \omega t}

a Normalize \Psi

Homework Equations


1 = \int^\infty_{-\infty} | \Psi |^2 dx

The Attempt at a Solution


|\Psi(x,t)|^2 = (\Psi*)\Psi
\Psi* = A e^{-\lambda |x|}e^{i \omega t}
|\Psi(x,t)|^2 = A^2e^{-2 \lambda |x|}

1 = \int^\infty_{-\infty} A^2e^{-2 \lambda |x|} dx = \frac{A^2}{\lambda}
A = \sqrt{\lambda}

Look correct?
 
Physics news on Phys.org
That is correct. Don't forget you can check your answer by integrating the function with your normalization constant plugged in.
 
Here, the absolute value becomes a problem. The absolute value makes the integrand an even function. That means that the final function is symetric about the y axis. What you can do is drop the absolute value, Integrate from 0 to infinity, and multiply the result by two.
 
Oh wait, you did the integral already. Well, that's what you would do if you were actually doing the math rather than using a table or computer program.
 
flatmaster said:
Here, the absolute value becomes a problem. The absolute value makes the integrand an even function. That means that the final function is symetric about the y axis. What you can do is drop the absolute value, Integrate from 0 to infinity, and multiply the result by two.

i did the integral mathematically exactly the way you said! I wasn't completely sure that this was the correct way to evaluate the absolute value, but I had a feeling it was. I'm a math major in addition to being a physics major =)
 
This lies within the tips and tricks that are learned slowly and never taught explicitly. When they are taught explicitly, they are often not understood at the time.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
Replies
7
Views
2K
  • · Replies 14 ·
Replies
14
Views
5K
  • · Replies 33 ·
2
Replies
33
Views
8K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K