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Using Schwinger's Model to find J=J1+J2=0 state

  1. Oct 31, 2012 #1
    hey guys, i have been trying to derive an expression for the total angular momentum zero state from the addition of two angular momenta using schwinger's model of angular momentum.

    i.e. the state

    [itex]\left| {{j}_{1}},{{m}_{1}} \right\rangle \otimes \left| {{j}_{2}},{{m}_{2}} \right\rangle =\left| {{j}_{1}},{{m}_{1}};{{j}_{2}},{{m}_{2}} \right\rangle[/itex]

    where [itex] J = J_1 + J_2 = 0 [/itex]

    so

    [itex]{{J}^{2}}\left| j,m \right\rangle ={{\hbar }^{2}}j(j+1)\left| j,m \right\rangle = 0[/itex]

    and

    [itex] j={{j}_{1}}+{{j}_{2}},...,|{{j}_{1}}-{{j}_{2}}|=0 [/itex]

    which means [itex] j_1 = j_2 [/itex]

    also

    [itex] {{J}_{z}}\left| j,m \right\rangle =m\left| j,m \right\rangle =0\,\,\,where\,\,m={{m}_{1}}+{{m}_{2}}=0[/itex]

    so

    [itex] m_1 = -m_2 [/itex]

    so the state will be

    [itex] \left| {{j}_{1}},{{m}_{1}};{{j}_{1}},{{-m}_{1}} \right\rangle [/itex]

    Now using the Schwinger model of angular momentum you can write any state in terms of annihilation and creation operators in the following way, (this is for one angular momenta not the addition of two)

    [itex]\left| j,m \right\rangle =\frac{{{({{a}^{\dagger }})}^{j+m}}{{({{b}^{\dagger }})}^{j-m}}}{\sqrt{(j+m)!}\sqrt{(j-m)!}}\left| 0,0 \right\rangle[/itex]

    for two momenta, each momenta will have two ladder operators associated, so is it correct that the state i am looking for can be written

    [itex]\,\,\left| {{j}_{1}},{{m}_{1}};{{j}_{2}},{{m}_{2}} \right\rangle =\frac{{{({{a}_{1}}^{\dagger })}^{{{j}_{1}}+{{m}_{1}}}}{{({{b}_{1}}^{\dagger })}^{{{j}_{1}}-{{m}_{1}}}}}{\sqrt{({{j}_{1}}+{{m}_{1}})!}\sqrt{({{j}_{1}}-{{m}_{1}})!}}\frac{{{({{a}_{2}}^{\dagger })}^{{{j}_{1}}+{{m}_{2}}}}{{({{b}_{2}}^{\dagger })}^{{{j}_{1}}-{{m}_{2}}}}}{\sqrt{({{j}_{1}}+{{m}_{2}})!}\sqrt{({{j}_{1}}-{{m}_{2}})!}}\left| 0,0;0,0 \right\rangle[/itex]

    and for the total angular momentum = 0 state

    [itex]\left| {{j}_{1}},{{m}_{1}};{{j}_{1}},-{{m}_{1}} \right\rangle =\frac{{{({{a}_{1}}^{\dagger })}^{{{j}_{1}}+{{m}_{1}}}}{{({{b}_{1}}^{\dagger })}^{{{j}_{1}}-{{m}_{1}}}}{{({{a}_{2}}^{\dagger })}^{{{j}_{1}}-{{m}_{1}}}}{{({{b}_{2}}^{\dagger })}^{{{j}_{1}}+{{m}_{1}}}}}{({{j}_{1}}+{{m}_{1}})!({{j}_{1}}-{{m}_{1}})!}\left| 0,0;0,0 \right\rangle[/itex]

    does that look correct for the total angular momentum = 0 state that is a result of the addition of two momenta?
     
  2. jcsd
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