Using Symmetry Arguments to Solve Angular Momentum Problems

[iamalexalright]

btw, we can use symmetry arguments on all of these...

a. For a particle in a state with known values of $$L^{2}$$ and $$L_{z}$$ with quantum numbers l and m, what is the expectation value of the angular momentum _vector_.

I have a feeling this should be very easy and I'm not thinking hard enough but I can't think of anything... any hints?

b. What is the expectation of $$<L_{x}^{2} + L_{y}^{2}$$ for the particle in part a?

$$<L_{x}^{2} + L_{y}^{2}> =$$
$$<L^{2} - L_{z}^{2} =$$
$$<\varphi_{lm}|L^{2}|\varphi_{lm}> - <\varphi_{lm}|L_{z}^{2}|\varphi_{lm}> =$$
$$\hbar^{2}l(l + 1)<\varphi_{lm}|\varphi_{lm}> - <\varphi_{lm}|L_{z}L_{z}|\varphi_{lm}> =$$
$$\hbar^{2}l(l + 1) - \hbar^{2}m^{2} =$$
$$\hbar^{2}(l(l + 1) - m^{2})$$

c. If we assume $$<L_{x}^{2}> = <L_{y}^{2}>$$, what is the uncertainty in $$L_{x}$$?

[$$(\Delta L_{x})^{2} = <L_{x}^{2}> - <L_{x}>^{2}$$

Okay, L_x is 0 since it is in a state with known L_z and
$$<L_{x}^{2}> =$$
$$(1/2)<L_{x}^{2} + L_{y}^{2}> =$$
$$(1/2)\hbar^{2}(l(l + 1) - m^{2})$$

so
$$\Delta L_{x} = \hbar \sqrt{\frac{1}{2}(l(l + 1) - m^{2})$$

look correct?

 

[mark726]

I would like to confirm that your calculations and reasoning are correct. Using symmetry arguments is a powerful tool in solving problems in quantum mechanics. In this case, we can use the fact that the expectation value of <L_{x}^{2}> is equal to the average of <L_{x}^{2}> and <L_{y}^{2}> due to the symmetry of the system. This allows us to simplify the calculation and arrive at the same result as you did. Additionally, your calculation for the uncertainty in L_{x} is also correct. Keep up the good work!