How Do Equations Determine Angular Momentum Eigenvalues in Quantum Mechanics?

[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...

 

[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 ...

 

[broegger]

Aarh, very clever indeed

Thanks!

 

[broegger]

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 ...

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??

 

[George Jones]

Why, you little...

Now, Now.

Physics Monkey's hints were good ones.

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

Regards,
George

 

[George Jones]

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

 

[vanesch]

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