How Do Equations Determine Angular Momentum Eigenvalues in Quantum Mechanics?

Click For Summary

Discussion Overview

The discussion revolves around the derivation of angular momentum eigenvalues in quantum mechanics, specifically focusing on the equations relating to the variables \( b_{max} \) and \( b_{min} \). Participants explore the implications of these equations and the conditions under which certain relationships hold.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant questions how the equations \( a = b_{max}(b_{max}+\hbar) \) and \( a = b_{min}(b_{min}-\hbar) \) lead to the conclusion that \( b_{max} = -b_{min} \).
  • Another participant suggests setting the equations equal to derive \( b^2_{max} + \hbar b_{max} = b^2_{min} - \hbar b_{min} \), leading to a factorization approach.
  • A later reply challenges the validity of the factorization, arguing that it leads to an absurd conclusion, questioning what might be missing in the reasoning.
  • One participant points out the use of a reductio ad absurdum argument, asserting that the assumption \( (b_{max} + b_{min}) \neq 0 \) leads to a contradiction, thus suggesting \( (b_{max} + b_{min}) = 0 \).
  • Another participant humorously reflects on the concept of dividing by zero, sharing a personal anecdote about a math lesson that illustrates the pitfalls of such reasoning.

Areas of Agreement / Disagreement

Participants express differing views on the validity of the mathematical manipulations and the conclusions drawn from them. There is no consensus on the correct interpretation of the equations or the implications for \( b_{max} \) and \( b_{min} \).

Contextual Notes

Some assumptions regarding the definitions of \( b_{max} \) and \( b_{min} \) are not explicitly stated, and the discussion includes unresolved mathematical steps that affect the conclusions drawn.

broegger
Messages
257
Reaction score
0
Hi,

How can you infer from these equations,

[tex]a = b_{max}(b_{max}+\hbar) \quad \text{and} \quad a = b_{min}(b_{min}-\hbar),[/tex]​

that [tex]b_{max} = -b_{min}[/tex]? It is used in the derivation of the angular momentum eigenvalues...
 
Physics news on Phys.org
Set them equal. You should find [tex]b^2_{max} + \hbar b_{max} = b^2_{min} - \hbar b_{min}[/tex]. This implies that [tex]b^2_{max} - b^2_{min} = - (b_{max} + b_{min} )\hbar[/tex]. Try factoring the left hand side ...
 
Aarh, very clever indeed :rolleyes:

Thanks!
 
Physics Monkey said:
Set them equal. You should find [tex]b^2_{max} + \hbar b_{max} = b^2_{min} - \hbar b_{min}[/tex]. This implies that [tex]b^2_{max} - b^2_{min} = - (b_{max} + b_{min} )\hbar[/tex]. Try factoring the left hand side ...

Why, you little... :mad:

That doesn't work since:


[tex]b^2_{max} - b^2_{min} = - (b_{max} + b_{min} )\hbar \quad \Leftrightarrow \quad b_{max} - b_{min} = -\hbar \quad \Leftrightarrow \quad \text{nonsense}[/tex]

What am I missing here??
 
broegger said:
Why, you little... :mad:

Now, Now.

Physics Monkey's hints were good ones.

All kinds of nonsense can be "proved" by dividing by zero.

Regards,
George
 
You have used a reductio ad absurdum argument to prove the required result, i.e., assume that (b_max + b_min) is not equal to zero, divide by (b_max + b_min), arrive at nonsense.

Thus the assumption is false and (b_max + b_min) = 0.

Regards,
George
 
George Jones said:
All kinds of nonsense can be "proved" by dividing by zero.

That's true. One day, my math teacher told us how we could convince our parents to double our pocket money by showing that A = 2 A (using again some division by zero). My dad's comment was: since A = 2 A, I'll give you half of what you get normally, that shouldn't then make a difference for you :redface:
 

Similar threads

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