Schrodinger probability problem

Click For Summary
SUMMARY

The discussion centers on evaluating the probability of finding an oscillator in the ground state beyond classical turning points using the normalization constant A=(mw/h*pi)^(1/4) and a = mw/2h. The user calculated A^2 as 1.0735 nm^-1 and a as 1.8102 nm^-2, leading to a probability of 0.394 for finding the oscillator beyond A0=0.1 nm. The community suggests that the large probability may stem from unit inconsistencies and emphasizes the importance of ensuring all units are compatible, particularly for h and m, and verifying the normalization constant and integration process.

PREREQUISITES
  • Quantum mechanics fundamentals, particularly harmonic oscillators
  • Understanding of wave functions and their normalization
  • Familiarity with integration techniques in physics
  • Knowledge of unit conversions and dimensional analysis
NEXT STEPS
  • Review the derivation of the normalization constant A for quantum harmonic oscillators
  • Learn about dimensional analysis in quantum mechanics
  • Study the implications of kinetic energy in quantum systems
  • Explore advanced integration techniques relevant to quantum probability calculations
USEFUL FOR

Students and researchers in quantum mechanics, particularly those studying harmonic oscillators and wave function normalization, as well as anyone involved in theoretical physics calculations.

eku_girl83
Messages
89
Reaction score
0
Here's my problem:
Using the normalization constant A=(mw/h*pi)^(1/4) and a = mw/2h, evaluate the probability to find an oscillator in the ground state beyond the classical turning points -A0 and A0. Assume A0= .1nm an k = 1 eV/square nm. The h variables actually represent h/2pi.

The wave function for the ground state is Ae^(-a*x^2). So the square of the wave function is A^2*e^(-2*a*x^2), which I will integrate as x ranges from .1 to infinity.
I found A^2 to be 1.0735 nm^-1 and a = 1.8102 nm^-2.
Subsituting these values and integrating yields .394 (probability of finding the oscillator beyond A0). This seems awfully large. What am I doing wrong? Could I possibly have a problem with units?

Any help appreciated!
 
Physics news on Phys.org
Assuming you have calculated correctly, such large value is possible as you are trying to squeeze to wavefunction in a small space; atomic order. So a wavefunction with such a large curvature leads to a large kinetic energy. As kinetic energy is twice the derivative of "x". With a large kinetic energy this may be possible.
 


Your calculation seems correct, but it is possible that you have a problem with units. It is important to make sure that all units are consistent in order to get the correct result. Here are a few things to check:

1. Make sure that all units are in the correct form. In this problem, the units for A0 and k are both in nm, so make sure that the units for h and m are also in nm.

2. Check that the units for the normalization constant A are correct. The formula for A given in the problem is (mw/h*pi)^(1/4), but the units for this should be (nm^-1)^1/4, not (nm)^1/4. This could be a simple mistake in writing out the formula.

3. Double check your calculation for A^2 and a. If these values are incorrect, it will affect your final result.

4. Make sure that the integration is done correctly. It is important to use the correct limits of integration and to perform the integration accurately.

If all of these things are correct and you are still getting a large probability, it could be a problem with the given values of A0 and k. Make sure that these values make sense in the context of the problem and that they are in the correct units. If you are still having trouble, it may be helpful to consult with your teacher or a classmate to see if they can spot any errors in your calculation.
 

Similar threads

  • · Replies 17 ·
Replies
17
Views
3K
  • · Replies 32 ·
2
Replies
32
Views
3K
  • · Replies 12 ·
Replies
12
Views
5K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 0 ·
Replies
0
Views
2K