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Angular momentum of pair of particles

  1. Feb 7, 2012 #1
    1. The problem statement, all variables and given/known data
    There are two noninteracting, spinless, distinguishable particles an isotropic oscillator potential. One particle is in the state [itex]|1\rangle =\frac{1}{\sqrt{2}}(|1,1,-1\rangle + |1,1,0\rangle )[/itex] and the other is in the state [itex]|2\rangle = \frac{1}{\sqrt{2}}(|1,1,0\rangle - |1,1,1\rangle )[/itex] where $|a,b,c\rangle$ corresponds to [itex]n=a,l=b,m=c[/itex]. What is the expectation value of [itex]L^{2}[/itex]?


    3. The attempt at a solution
    We know that the angular momentum of the system is equal to the sum of the individual angular momenta ([itex]L=L_{1}+L_{2}[/itex]), and since the particles are independent, [itex]|\varphi \rangle =|1\rangle |2\rangle[/itex] where [itex]|\varphi \rangle[/itex] is the state of the entire system. We also know that [itex]L^{2}=L_{x}^{2}+L_{y}^{2}+L_{z}^{2}[/itex]. Now taking [itex]\langle L^2 \rangle = \langle \varphi | L^2 | \varphi \rangle[/itex], we end up with [itex]\langle L^2 \rangle = \langle L_{1}^2 \rangle + \langle L_{2}^2 \rangle + 2 \langle L_{1}L_{2} \rangle[/itex]. The [itex]\langle L_{1}^2 \rangle[/itex] and [itex]\langle L_{2}^2 \rangle[/itex] are easy to calculate, but I don't see how to calculate [itex]\langle L_{1}L_{2} \rangle[/itex]. Perhaps there's a way to show that [itex]\langle L_{1}\rangle=\langle L_{2} \rangle[/itex] so that this would reduce to [itex]\langle L^2 \rangle = 4 \langle L_{x}^2 \rangle[/itex], but I don't see a way to do that.
     
  2. jcsd
  3. Feb 8, 2012 #2

    vela

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    Since you have the states, you could just explicitly calculate what the expectation value of $$L_1\cdot L_2 = {L_1}_x{L_2}_x + {L_1}_y{L_2}y + {L_1}_z{L_2}_z$$ is by expressing Lx and Ly in terms of the ladder operators, L+ and L-.

    Another approach would be to write the ##|1\rangle|2\rangle## state in terms of the eigenstates of L2, L12, and L22. Have you studied how to add angular momenta?
     
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