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Addition of angular momenta

  1. Jan 11, 2009 #1
    hi everybody.

    [tex]\textbf{J}_1[/tex] and [tex]\textbf{J}_2[/tex] are angular momentum (vector-)operators.
    In many textbooks [tex]\left[\textbf{J}_1,\textbf{J}_2\right] = 0[/tex] is stated to be a condition to show that [tex]\textbf{J}=\textbf{J}_1+\textbf{J}_2[/tex] is also an angular momentum (vector-)operator. But what is meant with [tex]\left[\textbf{J}_1,\textbf{J}_2\right] = 0[/tex]. When i show that [tex]\textbf{J}[/tex] is an angular momentum operator (i.e. [tex]\left[J_x,J_y\right]=iJ_z[/tex] ...) i always need the condition [tex]\left[(\textbf{J}_1)_x,(\textbf{J}_2)_x\right][/tex] and the like. So the components of [tex]\textbf{J}_1[/tex] and [tex]\textbf{J}_2[/tex] should mutually commute. Is this the meaning of [tex]\left[\textbf{J}_1,\textbf{J}_2\right] = 0[/tex]? For me it looks like [tex](\textbf{J}_1)_x(\textbf{J}_2)_x+(\textbf{J}_1)_y(\textbf{J}_2)_y+(\textbf{J}_1)_z(\textbf{J}_2)_z-(\textbf{J}_2)_x(\textbf{J}_1)_x-(\textbf{J}_2)_y(\textbf{J}_1)_y-(\textbf{J}_2)_z(\textbf{J}_1)_z=0[/tex] and this does not imply the conditions i need (as far as i see).
    I know Operators acting on different spaces commute and this fact is often used but i want to know how to treat the situation above only with the formal condition [tex]\left[\textbf{J}_1,\textbf{J}_2\right] = 0[/tex].

    thanks and greetings tommy.
     
  2. jcsd
  3. Jan 11, 2009 #2

    strangerep

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    Science Advisor

    Huh? You have two separate copies of the so(3) Lie algebra, so that commutation relation
    just says that every generator from copy #1 commutes with every generator from copy #2.

    Or did I misunderstand the question?
     
  4. Jan 12, 2009 #3
    Hi. Thank you for your answer. As i mentioned i know that every component of [tex]\textbf{J}_1[/tex] has to commute with every component of [tex]\textbf{J}_2[/tex] to show that [tex]\textbf{J}_1+\textbf{J}_2[/tex] is an angular momentum operator (another generator of the group) and because they act on different subspaces of the system. But the explicit form of [tex]\left[\textbf{J}_1,\textbf{J}_2\right]=0[/tex] (which is the only given condition) is [tex](\textbf{J}_1)_x(\textbf{J}_2)_x+(\textbf{J}_1)_y( \textbf{J}_2)_y+(\textbf{J}_1)_z(\textbf{J}_2)_z-(\textbf{J}_2)_x(\textbf{J}_1)_x-(\textbf{J}_2)_y(\textbf{J}_1)_y-(\textbf{J}_2)_z(\textbf{J}_1)_z=0
    [/tex]. So is [tex]\left[\textbf{J}_1,\textbf{J}_2\right]=0[/tex] just an abbreviation for [tex]\left[(\textbf{J}_1)_i,(\textbf{J}_2)_j\right]=0 \forall i,j[/tex] or does the explicit form of the commutator imply this?
    greetings.
     
  5. Jan 12, 2009 #4
    Yes, the abbrevation you wrote is correct (there is no scalar product involved in the commutator, statement is about each element of the vectors [tex]\mb{J}_1,\mb{J}_2[/tex]). If it helps you can write the elements of what is basically [tex]\mathfrak{su}(2)\otimes\mathfrak{su}(2)[/tex] algebra as [tex](J_1)_i=J_i\otimes 1, [/tex] and [tex](J_2)_i=1\otimes J_i[/tex]. Then it becomes more obvious that the two copies of the underlying algebra commute.

    Hope this helps
     
  6. Jan 12, 2009 #5
    many thanks!
     
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