# Angular momentum eigenvalues

1. Oct 31, 2005

### broegger

Hi,

How can you infer from these equations,

$$a = b_{max}(b_{max}+\hbar) \quad \text{and} \quad a = b_{min}(b_{min}-\hbar),$$​

that $$b_{max} = -b_{min}$$? It is used in the derivation of the angular momentum eigenvalues...

2. Oct 31, 2005

### Physics Monkey

Set them equal. You should find $$b^2_{max} + \hbar b_{max} = b^2_{min} - \hbar b_{min}$$. This implies that $$b^2_{max} - b^2_{min} = - (b_{max} + b_{min} )\hbar$$. Try factoring the left hand side ...

3. Oct 31, 2005

### broegger

Aarh, very clever indeed

Thanks!

4. Nov 2, 2005

### broegger

Why, you little....

That doesn't work since:

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

What am I missing here??

5. Nov 2, 2005

### George Jones

Staff Emeritus
Now, Now.

Physics Monkey's hints were good ones.

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

Regards,
George

6. Nov 3, 2005

### George Jones

Staff Emeritus
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

7. Nov 3, 2005

### vanesch

Staff Emeritus
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