Harmonic Oscillator: Evaluating Ground State Probability

Click For Summary
SUMMARY

The discussion focuses on evaluating the probability of finding a harmonic oscillator in the ground state beyond classical turning points using the wave function \(\psi(x) = Ae^{-ax^{2}}\). The normalization constant \(A\) is defined as \(A = \left(\frac{m\kappa}{\pi^{2}\hbar^{2}}\right)^{1/8}\) with \(a = \sqrt{m\kappa}/2\hbar\). Participants clarify the interpretation of the problem, emphasizing the need to calculate probabilities at both tails of the distribution, specifically \(\int^{0}_{-\infty}\psi^{2}(x)\;dx\) and \(1 - \int^{0.1}_{-\infty}\psi^{2}(x)\;dx\).

PREREQUISITES
  • Understanding of quantum mechanics principles, particularly harmonic oscillators
  • Familiarity with wave functions and normalization constants
  • Knowledge of calculus, specifically integration techniques
  • Basic understanding of physical constants such as mass, force constant, and Planck's constant
NEXT STEPS
  • Study the derivation of the harmonic oscillator wave functions in quantum mechanics
  • Learn about the implications of normalization in quantum states
  • Explore the concept of probability density functions in quantum mechanics
  • Investigate the role of classical turning points in quantum systems
USEFUL FOR

Students and professionals in physics, particularly those studying quantum mechanics, as well as educators looking for insights into teaching harmonic oscillators and probability distributions.

glebovg
Messages
156
Reaction score
0

Homework Statement



Using the normalization constant A and the value of a, evaluate the probability to find an oscillator in the ground state beyond the classical turning points ±x0. Assume an electron bound to an atomic-sized region (x0 = 0.1 nm) with an effective force constant of 1.0 eV/nm2.

Homework Equations



\psi(x)=Ae^{-ax^{2}}, where A=(\frac{m\kappa}{\pi^{2}\hbar^{2}})^{1/8} and a=\sqrt{m\kappa}/2\hbar

The Attempt at a Solution



How do I find m?
 
Last edited:
Physics news on Phys.org
The mass of the electron? You look it up.
 
or you could weigh one …

if you have one on you o:)
 
I will try to read the question more carefully next time.
 
Am I supposed to find probabilities at both tails of the distribution?
 
Yes.
 
Apparently there is an inconsistency in the question and there are two interpretations of the question. What would be the other one?
 
Beats me.
 
My interpretation: find probabilities at both tails of the distribution i.e. \int^{0}_{-\infty}\psi^{2}(x)\;dx and 1-\int^{0.1}_{-\infty}\psi^{2}(x)\;dx.
 
  • #10
I'd say the problem is asking for one number, P(|x|>x0). The upper limit on your first integral is wrong.
 
  • #11
Do you mean it is wrong according to your interpretation or it is generally wrong? Should it be 0.1?
 
Last edited:
  • #12
It is generally wrong. Why would the upper limit be 0?
 
  • #13
It should be -0.1, right?
 
Last edited:

Similar threads

  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
2K
Replies
11
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
1
Views
2K
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
6K