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Dimensionality of total angular momentum space

  1. Feb 9, 2012 #1
    1. The problem statement, all variables and given/known data

    There are 2 electrons, one with n=1, l=0 and the other with n=2, l=1. The question asks what is the dimensionality of total angular momentum space.

    2. Relevant equations
    (2[itex]j_{1}[/itex]+1)(2[itex]j_{2}[/itex]+1)


    3. The attempt at a solution
    I know for 2 electrons (spin 1/2 each) the possible values of total spin are s=0 or s=1.
    the total angular momentum is l=0 + l=1 = 1 (right?)
    So does this mean that j1 = 1+0 and j2 = 1+1
    which gives a dimensionality of (2(1) +1)(2(2) +1) = 15??
    The number seems a little off to me, perhaps the equation I have for dimensionality is incorrect?
     
  2. jcsd
  3. Feb 9, 2012 #2

    vela

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    Take the s=0, l=1 combination. Your formula says there are (2s+1)(2l+1) = 3 total states. Those states are
    \begin{align*}
    &\vert s=0,\ m_s=0;\ l=1,\ m_l=1\rangle \\
    &\vert s=0,\ m_s=0;\ l=1,\ m_l=0\rangle \\
    &\vert s=0,\ m_s=0;\ l=1,\ m_l=-1\rangle.
    \end{align*}The other way to look at it is to sum the angular momenta together, ##\vec{J} = \vec{S} + \vec{L}##. According to the rules of addition of angular momenta, there is only a single allowed value for j, namely j=1, so there are 2j+1 = 3 states, corresponding to mj = 1, 0, and -1. Either way you get three states.

    Now you do the s=1, l=1 combination. What are the allowed values of j?
     
  4. Feb 9, 2012 #3
    So if I understand correctly, j= l+s = 2 for l=1, s=1 which means that the possible values of [itex]m_{j}[/itex] are -2,-1,0,1,2 which gives the possible states of:
    \begin{align*}
    &\vert s=1,\ m_s=-1;\ l=1,\ m_l=1\rangle \\
    &\vert s=1,\ m_s=-1;\ l=1,\ m_l=0\rangle \\
    &\vert s=1,\ m_s=-1;\ l=1,\ m_l=-1\rangle \\
    &\vert s=1,\ m_s=0;\ l=1,\ m_l=1\rangle \\
    &\vert s=1,\ m_s=0;\ l=1,\ m_l=0\rangle \\
    &\vert s=1,\ m_s=0;\ l=1,\ m_l=-1\rangle \\
    &\vert s=1,\ m_s=1;\ l=1,\ m_l=1\rangle \\
    &\vert s=1,\ m_s=1;\ l=1,\ m_l=0\rangle \\
    &\vert s=1,\ m_s=1;\ l=1,\ m_l=-1\rangle.
    \end{align*}

    which gives 9 states total... (including the 3 states in the middle there which are the same as the ones given by s=0)
     
    Last edited: Feb 9, 2012
  5. Feb 9, 2012 #4

    vela

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    No, that's not correct. I think you need to go back and study the addition of angular momenta as you seem to have some basic misunderstandings about angular momentum in quantum mechanics.
     
  6. Feb 9, 2012 #5
    I just re-edited my 2nd post there, i confused [itex]m_{j}[/itex] with [itex]m_{s}[/itex], but I know that for s=1 [itex]m_{s}[/itex] is -1,0,1 which combined with [itex]m_{l}[/itex] =-1,0,1 gives 9 states in total including the 3 given by s=0 right?
     
  7. Feb 9, 2012 #6

    vela

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    Yes, there are nine states. (You meant ms and ml, not mj.)

    But think about this. If j=2, then mj can be -2, -1, 0, 1, 2. That's five states. What are the other four in the |j mj> basis?
     
  8. Feb 9, 2012 #7
    Would those be the case where only one spin is taken into account aka s=1/2, -1/2
    which gives [itex]m_{j}[/itex]= -3/2, -1/2, 1/2, 3/2

    also, just to clarify, the only possible values for j are 1 and 2 right?
     
    Last edited: Feb 9, 2012
  9. Feb 9, 2012 #8

    vela

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    No.

    No.
     
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