If Lx, Ly and Lz are respectively the x, y and z components of angular momentum operator L, the commutator [Lx, Ly, Lz] is equal to
A. $$i\hbar \left( {L_x^2 + L_y^2} \right)$$
B. $$2i\hbar {L_z}$$
C. $$i\hbar \left( {L_x^2 - L_y^2} \right)$$
D. zero
Answer: Option C
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Related Questions on Quantum Mechanics
A. In the ground state, the probability of finding the particle in the interval $$\left( {\frac{L}{4},\,\frac{{3L}}{4}} \right)$$ is half
B. In the first excited state, the probability of finding the particle in the interval $$\left( {\frac{L}{4},\,\frac{{3L}}{4}} \right)$$ is half This also holds for states with n = 4, 6, 8, . . . .
C. For an arbitrary state $$\left| \psi \right\rangle ,$$ the probability of finding the particle in the left half of the well is half
D. In the ground state, the particle has a definite momentum
A. (e-ax1 - e-ax2)
B. a(e-ax1 - e-ax2)
C. e-ax2 (e-ax1 - e-ax2)
D. None of the above

Isn't Lx^2=Ly^2??