Adding to

Total | |

6 | (4,2) |

5 | (3,2) (4,1) |

4 | (2,2) (3,1) (4,0) |

3 | (1,2) (2,1) (3,0) (4,-1) |

2 | (0,2) (1,1) (2,0) (3,-1) (4,-2) |

1 | (-1,2) (0,1) (1,0) (2,-1) (3,-2) |

0 | (-2,2) (-1,1) (0,0) (1,-1) (2,-2) |

-1 | (1,-2) (0,-1) (-1,0) (-2,1) (-3,2) |

-2 | (0,-2) (-1,-1) (-2,0) (-3,1) (-4,2) |

-3 | (-1,-2) (-2,-1) (-3,0) (-4,1) |

-4 | (-2,-2) (-3,-1) (-4,0) |

-5 | (-3,-2) (-4,-1) |

-6 | (-4,-2) |

Since the highest m value is 6, we expect to have a
state
which uses up one state for each m value from -6 to +6.
Now the highest m value left is 5, so a
states uses up a state at
each m value between -5 and +5.
Similarly we find a
,
, and
state.
This uses up all the states, and uses up the states at each value of m.
So we find in this case,

and that takes on every integer value between the limits. This makes sense in the vector model.

Jim Branson 2013-04-22