Question 11.1: (a) Use the transformation between Cartesian and spherical c......
(a) Use the transformation between Cartesian and spherical coordinates
x=r\sin (\theta ) \cos (\phi ),
y=r\sin (\theta )\sin (\phi ),
z=r\cos (\theta ),
to obtain the expression \hat{L}_{z} =-i\hbar \frac{\partial}{\partial \phi }.
(b) Calculate \left[\phi ,\hat{L}_{z} \right].
(c) Show that \hat{L}^{2} =-\hbar ^{2} \left(\frac{1}{\sin (\theta )}\frac{\partial}{\partial \theta } \left(\sin (\theta )\frac{\partial}{\partial \theta }\right)+\frac{1}{\sin^{2}(\theta ) } \frac{\partial}{\partial \phi ^{2} } \right).
(d) Derive \left[\hat{L}_{y},\hat{L}_{z} \right] =i\hbar \hat{L}_{x} using the fact that \left[\hat{x},\hat{p} _{x} \right] =i\hbar.
(e) If \hat{L}_{\pm } =\hat{L}_{x}\pm i \hat{L}_{y} show that \left[\hat{L}^{2}, \hat{L}_{\pm }\right] =0,\left[\hat{L}_{+ },\hat{L}_{-}\right] =2\hbar \hat{L}_{z}, and \left[\hat{L}_{z },\hat{L}_{\pm }\right]=\pm \hbar \hat{L}_{\pm }
(f) Show that \hat{L}^{2} =\hat{L}_{\pm }\hat{L}_{\mp } +\hat{L}^{2}_{z} \mp \hbar \hat{L}_{z} and \hat{L}^{2} =\frac{1}{2}(\hat{L}_{+}\hat{L}_{-}+\hat{L}_{-}\hat{L}_{+})+\hat{L} ^{2}_{z}.
(a) For a particle at position r with momentum p, the classical angular momentum is L = r×p and so the z-component of angular momentum is L_{z} =xp_{y}-yp_{x} . The corresponding quantum mechanical operator in Cartesian coordinates is
\hat{L}_{z} =\hat{x} \hat{p}_{y} -\hat{y} \hat{p}_{x}=\hat{x} \left(-i\hbar \frac{\partial}{\partial y} \right) -\hat{y} \left(-i\hbar \frac{\partial}{\partial x} \right) =-i\hbar \left(\hat{x}\frac{\partial}{\partial y}-\hat{y} \frac{\partial}{\partial x} \right)Using the transformation to spherical coordinates
x=r\sin (\theta ) \cos (\phi ),
y=r\sin (\theta )\sin (\phi ),
z=r\cos (\theta ),
we need to find the expressions for the partial derivatives appearing in the operator \hat{L}_{z}.
First, we can verify the solution given, \hat{L}_{z} =-ih\frac{\partial}{\partial \phi }, by taking the derivatives
\frac{\partial x}{\partial \phi } =-r\sin (\theta )\sin (\phi )=-y\frac{\partial y}{\partial \phi } =r\sin (\theta )\cos (\phi )=x
\frac{\partial z}{\partial \phi } =0
so that
\hat{L}_{z} =-i\hbar \frac{\partial}{\partial \phi } =-i\hbar \left(\frac{\partial x\partial }{\partial \phi \partial x }+\frac{\partial y\partial }{\partial \phi \partial y}+\frac{\partial z\partial }{\partial \phi \partial z} \right) =-i\hbar \left(-\hat{y}\frac{\partial}{\partial x} +\hat{x} \frac{\partial}{\partial y} +0 \right) =-i\hbar \left(\hat{x}\frac{\partial}{\partial y}-\hat{y}\frac{\partial}{\partial x} \right)However, in the absence of the given solution, we need to do a bit more work to find the derivatives \frac{\partial}{\partial x} and \frac{\partial}{\partial y} in spherical coordinates.
Starting with the radial position, r=\left(x^{2} +y^{2}+z^{2} \right)^{1/2}, we find the partial derivative \frac{\partial r}{\partial x} =\frac{2x}{2\sqrt{(x^{2}+y^{2}+z^{2} )} } =\frac{x}{r}, and similarly \frac{\partial r}{\partial y} =\frac{y}{r} and \frac{\partial r}{\partial z} =\frac{z}{r}.
We now note that \cos (\theta )=\frac{z}{r}. Taking the partial derivative with respect to x results in -\sin (\theta ) \frac{\partial \theta }{\partial x} =z\left(\frac{-1\partial r}{r^{2}\partial x } \right) and \frac{\partial \theta }{\partial x} =\frac{z}{r^{2} \sin (\theta )} \frac{x}{r}=\frac{\cos (\theta )\cos (\phi )}{r}. Likewise, the partial derivative with respect to y yields -\sin (\theta )\frac{\partial \theta }{\partial y} =z\left(\frac{-1\partial r}{r^{2} \partial y} \right) and \frac{\partial \theta }{\partial y} =\frac{z}{r^{2} \sin (\theta )}\frac{y}{r} =\frac{\cos (\theta )\sin (\phi )}{r}. And for z we obtain -\sin (\theta )\frac{\partial \theta }{\partial z} =\frac{1}{r} -\frac{z}{r^{2} } \frac{\partial r}{\partial z} =\frac{1}{r} -\frac{z}{r^{2} }\left(\frac{z}{r} \right) =\frac{1-\cos ^{2}(\theta ) }{r} =\frac{\sin ^{2} (\theta )}{r} and \frac{\partial \theta }{\partial z} =\frac{-\sin (\theta )}{r}.
Since \tan (\theta )=\frac{y}{x} we can take the partial derivative with respect to x to obtain \sec ^{2} (\phi )\frac{\partial \phi }{\partial x} =y\left(\frac{-1}{x^{2} } \right) and \frac{\partial \phi }{\partial x} =\frac{-y}{x^{2} } \cos ^{2} (\phi ). Likewise, the partial derivative with respect to y yields \sec^{2} (\phi )\frac{\partial \phi }{\partial y} =\frac{1}{x} and \frac{\partial \phi }{\partial x} =\frac{1}{x}\cos^{2} (\phi ). And the partial derivative with respect to z gives \sec ^{2} (\phi )\frac{\partial \phi }{\partial z} =0 and \frac{\partial \phi }{\partial z} =0. Therefore
Since
\left(\frac{xy}{r}- \frac{yx}{r} \right) \frac{\partial}{\partial r} =0and
\left(x\frac{\cos (\theta )\sin (\phi )}{r} -y\frac{\cos (\theta )\sin (\phi )}{r} \right)=(\sin (\theta )\cos (\theta )\sin (\phi )\cos (\theta )-\sin (\theta )\cos (\theta )\sin (\phi )\cos (\theta ))=0
we are left with
\hat{L}_{z} =-i\hbar \left(\left(\cos^{2}(\phi )+\frac{y^{2} }{x^{2} }\cos ^{2} (\phi ) \right)\frac{\partial}{\partial \phi } \right) =-i\hbar \left(\cos^{2}(\phi ) (1+\tan^{2}(\phi ) )\frac{\partial}{\partial \phi } \right)\hat{L}_{z} =-i\hbar\left(\cos^{2}(\phi )\sec^{2} (\phi ) \frac{\partial}{\partial \phi } \right) =-i\hbar \frac{\partial}{\partial \phi }
(b) The calculation of \left[\phi ,\hat{L}_{z}\right] proceeds as
\left[\hat{\phi } ,\hat{L}_{z}\right] \psi =\left(\hat{\phi } \hat{L}_{z}- \hat{L}_{z}\hat{\phi } \right) \psi =\hat{\phi } \left(-i\hbar \frac{\partial}{\partial \phi } \right)\psi -\left(-i\hbar \frac{\partial}{\partial \phi }\right) \hat{\phi }\psi =i\hbar \left(\frac{\partial}{\partial \phi } (\hat{\phi }\psi )-\hat{\phi }\frac{\partial}{\partial \phi }\psi \right)\left[\hat{\phi } ,\hat{L}_{z}\right] \psi =i\hbar \left(\psi \left(\frac{\partial}{\partial \phi }\hat{\phi } \right)+\hat{\phi }\left(\frac{\partial}{\partial \phi } \psi \right)-\hat{\phi }\left(\frac{\partial}{\partial \phi }\psi \right) \right) =i\hbar \psi \frac{\partial}{\partial \phi } \hat{\phi }=i\hbar \psi
so that
\left[\hat{\phi } ,\hat{L}_{z}\right] =i\hbar(c) Because \hat{L}^{2} =\hat{L}^{2}_{x} +\hat{L}^{2}_{y}+\hat{L}^{2}_{z} we need to find expressions for the x, y, and z components of \hat{L}. Starting with \hat{L}_{x} in Cartesian coordinates
\hat{L}_{x}=-i\hbar\left(\left(\frac{yz}{r}-\frac{zy}{r} \right)\frac{\partial}{\partial r} +\left(\frac{-y\sin (\theta )}{r} -\frac{z\cos (\theta )\sin (\phi )}{r} \right) \frac{\partial}{\partial \theta }+\left(0-\frac{z}{x} \cos^{2} (\phi )\right)\frac{\partial}{\partial \phi } \right)
\hat{L}_{x}=-i\hbar\left(-(\sin ^{2}(\theta )\sin (\phi )+\cos^{2}(\theta ) \sin (\phi )) \frac{\partial}{\partial \theta }+(-\cot (\theta )\cos (\phi )) \frac{\partial}{\partial \phi } \right)
\hat{L}_{x}=i\hbar\left(\sin (\phi )\frac{\partial}{\partial \theta } +\cot (\theta )\cos (\phi )\frac{\partial}{\partial \phi } \right)
Similarly for \hat{L}_{y} we have
\hat{L}_{y}=\hat{z}\hat{p} _{x} -\hat{x}\hat{p}_{z} =-i\hbar \left(\hat{z}\frac{\partial}{\partial x} -\hat{x}\frac{\partial}{\partial z} \right)\hat{L}_{y}=-i\hbar \left(\hat{z} \left(\frac{\partial r\partial }{\partial x\partial r} +\frac{\partial \theta \partial }{\partial x\partial \theta }+\frac{\partial \phi \partial }{\partial x\partial \phi }\right)-\hat{x} \left(\frac{\partial r\partial }{\partial z\partial r}+\frac{\partial \theta \partial }{\partial z\partial \theta }+\frac{\partial \phi \partial }{\partial x\partial \phi }\right) \right)
\hat{L}_{y}=-i\hbar\left(\left(\frac{zx}{r}-\frac{xz}{r} \right)\frac{\partial}{\partial r}+\left(\frac{z\cos (\theta )\sin (\phi )}{r}+\frac{x\sin (\theta )}{r} \right) \frac{\partial}{\partial \theta } +\left(\frac{-zy}{x^{2} }\cos ^{2} (\phi )-0 \right) \frac{\partial}{\partial \phi } \right)
\hat{L}_{y}=-i\hbar\left(-(\cos^{2}(\theta )\cos (\phi ) +\sin^{2} (\theta) \cos (\phi))\frac{\partial}{\partial \theta } +\left(\frac{-r^{2} \sin (\theta )\cos (\theta )\sin (\phi )}{r^{2}\sin^{2}(\theta ) \cos ^{2}(\phi ) } \right)\frac{\partial}{\partial \phi } \right)
\hat{L}_{y}=i\hbar\left(\cos (\phi )\frac{\partial}{\partial \theta } -\cot (\theta )\sin (\phi )\frac{\partial}{\partial \phi } \right)
and we already have the expression for \hat{L}_{z} from part (a). We can now find \hat{L}^{2} using \hat{L}^{2} =\hat{L}^{2}_{x} +\hat{L}^{2}_{y} +\hat{L}^{2}_{z}.
For \hat{L}^{2}_{x} we have
For \hat{L}^{2}_{y} we have
\hat{L}^{2}_{y}=-\hbar ^{2} \left(\cos^{2}(\phi )\frac{\partial^{2} }{\partial \theta ^{2} } +\cot^{2} (\theta )\sin ^{2} (\phi )\frac{\partial^{2} }{\partial \phi ^{2} }-\cos (\phi )\frac{\partial}{\partial \theta }\cot (\theta )\sin (\phi )\frac{\partial}{\partial \phi } -\cot (\theta )\sin (\phi ) \frac{\partial}{\partial \phi }\cos (\phi )\frac{\partial}{\partial \theta } \right)\hat{L}^{2}_{x}=-\hbar ^{2} \left(\cos^{2}(\phi )\frac{\partial^{2} }{\partial \theta ^{2} } +\cot^{2} (\theta )\sin ^{2} (\phi )\frac{\partial^{2} }{\partial \phi ^{2} } + \sin (\phi )\frac{\partial}{\partial \theta }\cot (\theta ) \cos (\phi )\frac{\partial}{\partial \phi } +\cot (\theta )\sin (\phi )(-\sin (\phi ))\frac{\partial}{\partial \theta }+\cot (\theta )\sin (\phi )\frac{\partial \partial }{\partial \phi \partial \theta } \right)
Hence
\hat{L}^{2}=-\hbar ^{2}\left(\frac{\partial^{2} }{\partial \theta ^{2} }+\cot^{2}(\theta ) \frac{\partial ^{2} }{\partial \phi^{2} }+\cot (\theta )(\cos^{2} (\phi )+\sin^{2}(\phi ) ) \frac{\partial}{\partial \theta }+\frac{\partial^{2} }{\partial \phi ^{2} } \right)\hat{L}^{2}=-\hbar ^{2}\left(\frac{\partial^{2} }{\partial \theta ^{2} }+(1+\cot ^{2}(\theta ) )\frac{\partial^{2} }{\partial \phi^{2} } +\cot (\theta )\frac{\partial}{\partial \theta } \right)
Noting that 1+\cot²(θ) = \csc²(θ) = 1/ \sin²(θ), we obtain
\hat{L}^{2}=-\hbar ^{2}\left(\frac{\partial^{2} }{\partial \theta ^{2} }+\cot (\theta )\frac{\partial}{\partial \theta }+\frac{1}{\sin ^{2}(\theta ) } \frac{\partial^{2} }{\partial \phi^{2} } \right)or, equivalently,
\hat{L}^{2}=-\hbar ^{2}\left(\frac{1}{\sin (\theta )}\frac{\partial}{\partial \theta } \left(\sin (\theta )\frac{\partial}{\partial \theta } \right)+\frac{1}{\sin ^{2} (\theta )} \frac{\partial^{2} }{\partial \phi^{2} } \right)(d) We can use \left[\hat{x},\hat{p}_{x} \right] =i\hbar to find
\left[\hat{L}_{y} ,\hat{L}_{z} \right] =\left[\hat{z}\hat{p}_{x} -\hat{x}\hat{p}_{z},\hat{x}\hat{p}_{y} -\hat{y}\hat{p}_{x}\right] =\left[\hat{z}\hat{p}_{x},\hat{x}\hat{p}_{y}\right] -\left[\hat{z}\hat{p}_{x},\hat{y}\hat{p}_{x}\right] -\left[\hat{x}\hat{p}_{z},\hat{x}\hat{p}_{y}\right] +\left[\hat{x}\hat{p}_{z},\hat{y}\hat{p}_{x}\right]The terms \left[\hat{z}\hat{p}_{x},\hat{y}\hat{p}_{x}\right]=\left[\hat{x}\hat{p}_{z},\hat{x}\hat{p}_{y}\right]=0 so that
\left[\hat{L}_{y},\hat{L}_{z}\right]=\hat{z}\left[\hat{p}_{x},\hat{x} \right] \hat{p}_{y} +\hat{y} \left[\hat{x},\hat{p}_{x} \right] \hat{p}_{z} =i\hbar \hat{y} \hat{p}_{z} -i\hbar \hat{z} \hat{p}_{y}=i\hbar \left(\hat{y}\hat{p}_{z} -\hat{z}\hat{p}_{y} \right) =i\hbar \hat{L} _{x}(e) We are given \hat{L}_{\pm } =\hat{L} _{x} \pm i\hat{L}_{y}. Because \hat{L}^{2} commutes with \hat{L} _{x} and \hat{L} _{y}, it follows that
\left[\hat{L}^{2},\hat{L}_{\pm } \right] =\left[\hat{L}^{2},\left(\hat{L}_{x} \pm i\hat{L}_{y} \right) \right] =\left[\hat{L}^{2},\hat{L}_{x} \right]+(\pm i)\left[\hat{L}^{2},\hat{L}_{y} \right] =0To show that \left[\hat{L}_{+},\hat{L} _{-} \right] =2\hbar \hat{L}_{z} we expand the commutator and make use of the commutation relation \left[\hat{L}_{x},\hat{L} _{y} \right] =i\hbar \hat{L}_{z} so that
\left[\hat{L}_{+},\hat{L} _{-} \right] =\left[\hat{L}_{x}+i\hat{L}_{y},\hat{L}_{x}-i\hat{L}_{y} \right] =\left[\hat{L}_{x},\hat{L} _{x}\right]+\left[\hat{L}_{y},\hat{L} _{y} \right]+\left[\hat{L}_{x},-i\hat{L} _{y} \right]+\left[i\hat{L}_{y},\hat{L} _{x} \right]Obviously, \left[\hat{L}_{x},\hat{L} _{x}\right]=\left[\hat{L}_{y},\hat{L} _{y} \right]=0 and the terms \left[\hat{L}_{x},-i\hat{L} _{y}\right]=\left[i\hat{L}_{y},\hat{L} _{x} \right]=\hbar \hat{L}_{z}, so that \left[\hat{L}_{+},\hat{L} _{-} \right]=2\hbar \hat{L}_{z}
The last relationship we are asked to find is
\left[\hat{L}_{z},\hat{L} _{\pm } \right]=\left[\hat{L}_{z},\hat{L} _{x} \right]\pm i\left[\hat{L}_{z},\hat{L} _{y } \right]=i\hbar \hat{L} _{y }\pm i(-i\hbar \hat{L} _{x})=\pm \hbar (\hat{L} _{x}\pm i\hat{L} _{y })=\pm \hbar \hat{L} _{\pm }(f) Starting from
\hat{L} _{\pm } \hat{L} _{\mp }=(\hat{L} _{x}\pm i\hat{L} _{y})(\hat{L} _{x}\mp i\hat{L}_{y} )=\hat{L}^{2}_{x} +\hat{L}^{2}_{y}\pm i(\hat{L}_{x} \hat{L}_{y} -\hat{L}_{y}\hat{L}_{x})=\hat{L}^{2} -\hat{L}^{2}_{z}\pm i(i(\hbar \hat{L}_{z} ))which can be rearranged to give
\hat{L}^{2}=\hat{L} _{\pm } \hat{L} _{\mp }+\hat{L} ^{2}_{z} \mp \hbar \hat{L} _{z}It follows that
2\hat{L}^{2}=\hat{L}_{+} \hat{L}_{-}+\hat{L}^{2}_{z} -\hbar \hat{L}_{z} +\hat{L}_{-} \hat{L}_{+}+\hat{L}^{2}_{z}+\hbar \hat{L}_{z}=\hat{L}_{+} \hat{L}_{-}+\hat{L}_{-} \hat{L}_{+}+2\hat{L}^{2}_{z}so that
\hat{L}^{2}=\frac{1}{2} (\hat{L}_{+} \hat{L}_{-}+\hat{L}_{-} \hat{L}_{+})+\hat{L} ^{2}_{z}