Why Can't Quantum Numbers Include Half Integers?

[HiggsBoson1]

I don't understand why quantum numbers can not be divided into half integers and so on. The books I have read do not give clear explanations. Would anyone mind helping me understand this? Thanks!

 

[bhobba]

What quantum numbers are you talking about?

If you are talking about spin quantum numbers then any decent book on QM will prove from the angular momentum commutation relations why its quantized.

If you are talking about solutions to Schroedinger's equation then that is not always quantized, and when it is it depends entirely on the Hamiltonian:
http://www.physics.ox.ac.uk/Users/smithb/website/coursenotes/qi/QILectureNotes3.pdf

Thanks
Bill

 

[HiggsBoson1]

What quantum numbers are you talking about?

If you are talking about spin quantum numbers then any decent book on QM will prove from the angular momentum commutation relations why its quantized.

If you are talking about solutions to Schroedinger's equation then that is not always quantized, and when it is it depends entirely on the Hamiltonian:
http://www.physics.ox.ac.uk/Users/smithb/website/coursenotes/qi/QILectureNotes3.pdf

Thanks
Bill

That was really helpful! Thanks a lot! :)

 

[stevendaryl]

I don't understand why quantum numbers can not be divided into half integers and so on. The books I have read do not give clear explanations. Would anyone mind helping me understand this? Thanks!

The Schrodinger equation: H \psi = E \psi can be solved for arbitrary values of E, but when E is not an eigenvalue of the hamiltonian, then \psi will be unnormalizable--it will blow up at infinity, or at the origin, or somewhere. For example, a solution to the free particle Schrodinger equation with E < 0 is: \psi = e^{\hbar K x}, which corresponds to an energy of -\hbar^2 K^2/(2 m).

The differential equations for the wave function are forced to have discrete values by boundary conditions: to make sure that the wave function is single-valued in the whole domain, and to make sure that the integral of |\psi|^2 is finite.