Angular momentum raising lowering operator

In summary, to prove the commutator [L_\pm , L^2]=0, we first expand the commutator using the given equations and then use the fact that [L_{i},L_{j}]=ih\epsilon_{ijk}L_{k} to cancel out terms and arrive at the final answer of 0. Similarly, to prove [L_+,L_-]=2\hbar L_z, we can again use the commutator formula and apply the given equations to arrive at the final answer of 2\hbar L_z.
  • #1
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Homework Statement



Derive [itex] [L_\pm , L^2]=0 [/itex]



Homework Equations



[itex] L_{\pm}=L_x \pm iL_y [/itex]


The Attempt at a Solution



[itex] [L_\pm , L^2]=[L_x,L_x^2] \pm i[L_y,L_y^2]=[L_x,L_x]L_x + L_x[L_x,L_x] \pm i([L_y,L_y]L_y+L_y[L_y,L_y]) [/itex]

Is this right so far? If so, how do I proceed from this? The books and websites I'll read always just state [itex] [L_\pm , L^2]=0 [/itex] instead of proving it.
 
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  • #2
[itex]L^{2}=L^{2}_{x}+L^{2}_{y}+L^{2}_{z}[/itex]

Then

[itex][L_{±},L^{2}]=[L_{x}±iL_{y},L^{2}_{x}+L^{2}_{y}][/itex]

which is going to give you some cross terms. Using the fact that [itex][L_{i},L_{j}]=ih\epsilon_{ijk}L_{k}[/itex], you can move some stuff around and cancel terms to arrive at the final answer. I'm not sure how you got to where you are.
 
  • #3
I'm having some trouble expanding the commutators due to the +/- sign. Is it still [A,B]=AB-BA?
 
  • #4
Is it still [A,B]=AB-BA?

This is always true.

After writing the commutation relation, and expanding, the ± just becomes [itex]\mp[/itex] when multiplied by -. If you have two like terms, one of which has the sign ± and the other of which has the sign [itex]\mp[/itex], they will cancel.
 
  • #5
Is the following ok?

[itex] [L_\pm,L^2]=(L_x \pm iL_y)(L_x^2+L_y^2)-(L_x^2+L_y^2)(L_x \pm iL_y) [/itex]

[itex] [L_\pm,L^2]=L_x^3+L_xL_y^2\pm iL_x^2L_y\pm iL_y^3 -L_x^3 \mp i L_x^2 L_y -L_x L_y^2 \mp iLy^3 [/itex]

= 0
 
  • #6
I tried to do the same as what I did above to prove [itex] [L_+,L_-]=2\hbar L_z [/itex] but I got zero instead:

[itex] [L_+,L_-]=(L_x +iL_y)(L_x-iL_y)-(L_x-iL_y)(L_x+iL_y)=(L_x^2+L_y^2)-(L_x^2+L_y^2)=0 [/itex]

I found the following derivation but I don't understand how the last three equalities were arrived at:

[itex] [L_+,L_-]=[L_x+iL_y,L_x-iL_y]=i[L_y,L_x]-i[L_x,L_y]=\hbar (L_z+L_z)=2\hbar L_z [/itex]
 

1. What is the angular momentum raising lowering operator?

The angular momentum raising lowering operator is a mathematical operator used in quantum mechanics to describe the angular momentum of a particle. It is represented by the symbols L+ and L- and is used to raise or lower the value of the angular momentum of a particle by one unit.

2. How is the angular momentum raising lowering operator used in quantum mechanics?

The angular momentum raising lowering operator is used to calculate the angular momentum of a particle in a given state. It is also used to determine the probability of a particle being in a particular angular momentum state.

3. What is the relationship between the angular momentum raising lowering operator and the angular momentum operator?

The angular momentum raising lowering operator is a component of the overall angular momentum operator. It is used to change the value of the angular momentum of a particle, while the full angular momentum operator is used to calculate the total angular momentum of a particle.

4. How does the angular momentum raising lowering operator affect the energy levels of a particle?

The angular momentum raising lowering operator does not directly affect the energy levels of a particle. However, it is used in the calculation of the Hamiltonian operator, which determines the energy levels of a particle.

5. Can the angular momentum raising lowering operator be used for all types of angular momentum?

Yes, the angular momentum raising lowering operator can be used for all types of angular momentum, including orbital angular momentum and spin angular momentum. It is a fundamental tool in quantum mechanics for describing the behavior of particles with angular momentum.

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