What's the total angular momentum operator for a system of two particles?

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SUMMARY

The total angular momentum operator for a system of two particles in two dimensions is indeed the sum of the individual angular momenta, expressed as \(\vec{L} = \vec{L}_1 + \vec{L}_2\). For each particle, the angular momentum operator is defined as \(L_j^z = -i\hbar \left[y_j\partial/\partial{x_j} - x_j\partial/\partial{y_j}\right]\), where \(j\) represents the particle index. This formulation highlights that the operator for one particle does not act on the coordinates of the other particle, maintaining the independence of their respective angular momenta.

PREREQUISITES
  • Understanding of angular momentum in quantum mechanics
  • Familiarity with polar coordinates in two dimensions
  • Knowledge of differential operators and their applications
  • Basic concepts of quantum mechanics, including the role of \(\hbar\)
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  • Study the derivation of angular momentum operators in quantum mechanics
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Suppose we're in two dimensions, and both particles have mass 1.

Particle 1's location is given by its polar coordinates [itex](r_1,\theta_1)[/itex]; likewise for Particle 2 [itex](r_2,\theta_2)[/itex].

Is it true that the total angular momentum [itex]\vec{L}[/itex] is just the sum of the individual angular momenta of the particles: [itex]\vec{L} = \vec{L}_1 + \vec{L}_2[/itex]? If that's the case, can you give me the total angular momentum operator [itex]\vec{L}[/itex] as a differential operator?
 
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yeah, just add it up: [tex]L_j^z = -i\hbar \left[y_j\partial/\partial{x_j} - {x_j}\partial/\partial{y_j}\right][/tex] where j is the particle index. Keep in mind L_2 does not act on the coordinates for the first particle.
 

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