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AngularMomentum Quantum Numbers 
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#1
Jun911, 01:22 AM

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Is there any generally accepted or even speculative a priori reason why the orbitalangularmomentum quantum number is allowed to take on only integer values but the spinangularmomentum quantum number may have halfinteger values, or is this just empirical?



#2
Jun911, 01:48 AM

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IIRC it has to do with the fact that the angular momentum quantum numbers correspond to the numbering of the spherical harmonics used to describe the atom (the angular part of the wavefunction). Spherical harmonics only come with integer quantum numbers. Spin is not described by spherical harmonics.



#3
Jun911, 02:09 AM

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That the orbital angular momentum comes only in integer representations of the (covering of the) rotation group, SU(2) is due to the fact that orbital angular momentum is given as
[tex]\hat{\vec{L}}=\hat{\vec{x}} \times \hat{\vec{p}}.[/tex] If you apply [itex]\hat{L}_z[/itex] to, e.g., the position eigenvector [itex]\vec{x}=(0,0,z) \rangle[/itex], you'll get 0 since a rotation around the z axis doesn't change this state. You can also easily verify this by direct application of the [itex]\hat{L}_z=\hat{x} \hat{p}_y\hat{y} \hat{p}_x[/itex] to this state. Now, you can as well expand this state in orbitalangularmomentum eigenstates, and then the application of [itex]\hat{L}_z[/itex] leads to [tex]\vec{x}=(0,0,z) \rangle=\sum_l \sum_{m=l}^{l} C_{lm} r=1,l,m \rangle \; \Rightarrow \; \hat{L}_z \vec{x}=(0,0,z) \rangle = \sum_l \sum_{m=l}^{l} m C_{lm} r=z,l,m \rangle.[/tex] Thus, you must have [itex]C_{lm}=0[/itex] for all [itex]m \neq 0[/itex]. Thus only eigen vectors with [itex]m=0[/itex] contribute to the decomposition of [itex]\vec{x}=(0,0,z) \rangle[/itex], but such eigen states can exist only for integer l. Now, you can make any positioneigen vector by applying a rotation to [itex]\vec{x}=(0,0,z) \rangle[/itex]. Thus also all these vectors consist only of superpositions of integerl orbitalangularmomentum eigen vectors. Since this is a complete set of eigen vectors, l can only take integer values. 


#4
Jun911, 03:42 PM

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AngularMomentum Quantum Numbers
Because orbital angular momentum describes singlevalued functions and spin angular momentum describes spinors, which are doublevalued.



#5
Jun911, 04:15 PM

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It's also worth noting that the angular momentum of a quantum state can only ever change in integer units of hbar. This means that you have to be a little careful with statements like "spin angular momentum may have halfinteger values", because that sounds like spin may be incremented in halfinteger steps, which is not correct.
In general, a given angular momentum quantum number can take value EITHER from the set of integers OR from the set of halfintegers. As far as I know there is no way to convert between the two sets, at least not in nonrelativistic QM. 


#6
Jun911, 04:30 PM

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For a composite system it is true, the spin angular momentum addition requires the total spin to be separated by integers. So 2 1/2 spin particles added together may have spin 1, 0, or 1. This is not quite the same as Angular Momentum where you can increase a single particle's l value simply by imparting to it some angular momentum. 


#7
Jun911, 04:57 PM

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Anyway, I was more referring to something like a total angular momentum quantum number for an atom: J=L + S. L will always be an integer, but S can be either an integer or halfinteger value, so J can either be an integer OR a halfinteger in the range LS <= J <= L+S. For example, if L=1 and S=3/2, then J can take any value in the set {1/2,3/2,5/2}. If L=1 and S=1, then J can take any value from the set {0,1,2}. But you never get a J that can take values from a set like {0,1/2,1,3/2,2,5/2}. A similar condition holds for the zcomponent of angular momentum ... it can only ever increase or decrease in integer multiples of hbar (c.f. angular momentum raising and lower operators). For an electron, s_{z}=hbar/2 OR hbar/2. For a porbital (l=1), l_{z}=hbar OR 0 OR hbar. 


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