Normalising and Probabilities of wavefunctions

Click For Summary

Homework Help Overview

The discussion revolves around the normalization of wavefunctions and the calculation of probabilities associated with eigenvalues in quantum mechanics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to normalize a wavefunction by assigning a constant to each state and questions the correctness of their approach. Some participants clarify the implications of using a constant for normalization and discuss the calculation of probabilities.

Discussion Status

Participants are actively engaging in clarifying the normalization process and the calculation of probabilities. Some guidance has been provided regarding the need to square values for probability calculations, while others question the necessity of this step based on their interpretations.

Contextual Notes

There appears to be some confusion regarding the assumptions made about the likelihood of states and the correct method for calculating probabilities, indicating a need for further exploration of these concepts.

Sekonda
Messages
201
Reaction score
0
Hey

My question is displayed below

Sc4DW.png


I think I have done this right but I wanted to check, we have to normalise the wavefunction first and I think this is done by assuming each state is equally likely and so assigning some constant 'c' to premultiply each of the 3 states.

We need multiply each state by it's Bra form such that we get 3c^{2}=1 and so c=1/√3
and provided this is correct then the probability of attaining an eigenvalue ω=1 is just

√(2/3)

Is this correct? If not what am I doing/assuming which is wrong?

Thanks,
SK
 
Physics news on Phys.org
It's correct. You are not assuming that every state is equally likely, you know it from the fact that all are multiplied by the same constant (in this case 1) in the non normalized wave function.
 
Cheers man, was thinking as I wrote that - that is was wrong... considering the question.

Thanks!
SK
 
Don't forget to square to get the probabilities :smile:
 
Ahh yes of course!, I did it wrong initally anyway - should of written 2/3 - giving (2/3)^(2) =4/9 as the probability, 44.4%

I believe this is correct...

Cheers!
Sk
 
Or is this probability just 2/3? Can someone check this

Cheers!
SK
 
The probability is 2/3. There is no need to square anything if your performed the braket.
 
Cheers, thanks for that! Good to know i am now doing it the right way.

SK
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
4K
Replies
7
Views
3K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
2
Views
2K
  • · Replies 12 ·
Replies
12
Views
1K
Replies
46
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
5
Views
3K