(a) Calculate the Poisson bracket between the x and y components of the classical orbital angular momentum.
(b) Calculate the commutator between the x and y components of the orbital angular momentum operator.
(c) Compare the results obtained in (a) and (b).
(a) Using the definition (3.113)
\left\{A,B\right\} =\sum\limits_{j} \left(\frac{∂A}{∂qj}\frac{∂B}{∂pj}-\frac{∂A}{∂pj}\frac{∂B}{∂qj} \right). (3.113)
we can write the Poisson bracket \left\{l_{x},l_{y} \right\} as
\left\{l_{x},l_{y} \right\}=\sum\limits_{j=1}^{3} \left (\frac {∂l_{x}}{∂q_{j} }\frac{∂l_{y}}{∂p_{j} }-\frac{∂l_{x}}{∂p_{j} }\frac{∂l_{y}}{∂q_{j} } \right), (3.227)
where q_{1}=x,q_{2}=y,q_{3}=z, p_{1}=p_{x},p_{2}=p_{y}, and p_{3}=p_{z}. Since l_{x}=yp_{z}-zp_{y},l_{y}=zp_{x}-xp_{z},l_{z}=xp_{y}-yp_{x}, the only partial derivatives that survive are ∂l_{x}/∂z=-p_{y},∂l_{y}/∂p_{z}=-x,∂l_{x}/∂p_{z}=y, and ∂l_{y}/∂z=p_{x}. Thus, we have
\left\{l_{x},l_{y} \right\}=\frac{∂l_{x}}{∂z}\frac{∂l_{y}}{∂p_{z}} -\frac{∂l_{x}}{∂p_{z}}\frac{∂l_{y}}{∂z}=xp_{y}-yp_{x} =l_{z}. (3.228)
(b) The components of \hat{\vec{L}} are listed in (3.26) to (3.28): \hat{L}_{x}=\hat{Y}\hat{P}_{z}-\hat{Z}\hat{P}_{y} ,\hat{Z}\hat{P}_{x}-\hat{X}\hat{P}_{z}, and \hat {L}_{z} =\hat{X}\hat{P}_{y}-\hat{Y}\hat{P}_{x}. Since \hat{X},\hat{Y} and \hat{Z} mutually commute and so do \hat{P}_{x}, \hat{P}_{y} and \hat{P}_{x}, \hat{P}_{y},
\hat{L}_{x}=\hat{Y}\hat{P}_{z}-\hat{Z}\hat{P}_{y}=-i\hbar \left(\hat{Y}\frac{∂}{∂z}-\hat{Z}\frac{∂}{∂y} \right), (3.26)
\hat{L}_{y}=\hat{Z}\hat{P}_{x}-\hat{X}\hat{P}_{z}=-i\hbar \left(\hat{Z}\frac{∂}{∂x}-\hat{X}\frac{∂}{∂Z} \right), (3.27)
\hat{L}_{z}=\hat{X}\hat{P}_{y}-\hat{Y}\hat{P}_{x}=-i\hbar \left(\hat{X}\frac{∂}{∂y}-\hat{Y}\frac{∂}{∂x} \right). (3.28)
we have
[\hat{L}_{x},\hat{L}_{y}]=[\hat{Y}\hat{P}_{z}-\hat{Z}\hat {P}_{y},\hat{Z}\hat{P}_{x}-\hat{X}\hat{P}_{z}]
=[\hat{Y}\hat{P}_{z},\hat{Z}\hat{P}_{x}]-[\hat{Y}\hat {P}_{z},\hat{X}\hat{P}_{z}]-[\hat{Z}\hat{P}_{y}, \hat{Z}\hat {P}_{x}]+[\hat{Z}\hat{P}_{y},\hat{X}\hat{P}_{z}][/latex]
=\hat{Y}[\hat{P}_{z},\hat{Z}]\hat{P}_{x}+\hat{X}[\hat {Z},\hat{P}_{z}]\hat{P}_{y}=i\hbar (\hat{X}\hat{P}_{y}-\hat{Y} \hat{P}_{x})
=i\hbar \hat{L}_{z}. (3.229)
(c) A comparison of (3.228) and (3.229) shows that
\left\{l_{x},l_{y} \right\}=l_{z}\longrightarrow [\hat{L}_{x}, \hat{L}_{y}]=i\hbar \hat{L}_{z}. (3.230)