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Angular momentum operator for 2-D harmonic oscillator

  • #1
1. The problem statement

I want to write the angular momentum operator ##L## for a 2-dimensional harmonic oscillator, in terms of its ladder operators, ##a_x##, ##a_y##, ##a_x^\dagger## & ##a_y^\dagger##, and then prove that this commutes with its Hamiltonian.

The Attempt at a Solution



I get

##L = (x + y) × (p_x + p_y) = -i \hbar ((a_x^\dagger +a_y^\dagger) × (a_x + a_y) + a_x × a_x + a_y × a_y)##.

Does this seem to make any sense?

As for the commutator, with Hamiltonian ##H = \hbar \omega (a_x^\dagger a_x + a_y^\dagger a_y + 1) ##, I consequently get all sorts of exotic product terms, e.g.

## \left ( (a_x^\dagger +a_y^\dagger) × (a_x + a_y) \right ) a_x^\dagger a_x ##

and the like, and I'm not comfortable with handling these. I want all of them to go away, for said operators to commute, but am not sure how to do/see that.

Any comments would be appreciated.
 

Answers and Replies

  • #2
hilbert2
Science Advisor
Insights Author
Gold Member
1,343
415
You should rather show that the operator ##L^2##, which is the same as ##L_{z}^{2}## for motion on the xy-plane, commutes with the Hamiltonian. This actually happens only when the "spring constants" are the same in x and y directions and the potential energy ##V(r,\theta )## depends only on the radial coordinate ##r##.
 
  • #3
Chandra Prayaga
Science Advisor
649
148
You can first work out the commutator of Lz with the Hamiltonian. In your attempt, the way you wrote L is wrong. You should work out the cross product of the position vector and the momentum vector.
 

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