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4.3: Rotation Operators

Consider a particle whose position is described by the spherical polar coordinates $ (r, \theta, \varphi)$ . The classical momentum conjugate to the azimuthal angle $ \varphi$ is the $ z$ -component of angular momentum, $ L_z$ . According to Section 2.5, in quantum mechanics we can always adopt the Schrödinger representation, for which ket space is spanned by the simultaneous eigenkets of the position operators $ r$ , $ \theta$ , and $ \varphi$ , and $ L_z$ takes the form

 

$\displaystyle L_z = -{\rm i}\,\hbar\, \frac{\partial}{\partial \varphi}.$ (346)

 

 

We can do this because there is nothing in Section 2.5 which specifies that we have to use Cartesian coordinates--the representation (165) works for any well-defined set of coordinates.

Consider an operator $ R({\mit\Delta}\varphi)$ that rotates the system through an angle $ {\mit\Delta}\varphi$ about the $ z$ -axis. This operator is very similar to the operator $ D({\mit\Delta} x)$ , introduced in Section 2.8, which translates the system a distance $ {\mit\Delta} x$ along the $ x$ -axis. We were able to demonstrate in Section 2.8 that

 

$\displaystyle p_x = {\rm i}\,\hbar\, \lim_{\delta x\rightarrow 0}\frac{D(\delta x)-1} {\delta x},$ (347)

 

 

where $ p_x$ is the linear momentum conjugate to $ x$ . There is nothing in our derivation of this result which specifies that $ x$ has to be a Cartesian coordinate. Thus, the result should apply just as well to an angular coordinate. We conclude that

 

$\displaystyle L_z = {\rm i}\,\hbar\, \lim_{\delta \varphi\rightarrow 0}\frac{R(\delta \varphi)-1} {\delta \varphi}.$ (348)

 

 

According to Equation (348), we can write

 

$\displaystyle R(\delta \varphi) = 1 -{\rm i}\,L_z\,\delta\varphi/\hbar$ (349)

 

 

in the limit $ \delta\varphi\rightarrow 0$ . In other words, the angular momentum operator $ L_z$ can be used to rotate the system about the $ z$ -axis by an infinitesimal amount. We say that $ L_z$ is the generator of rotations about the $ z$ -axis. The above equation implies that

 

$\displaystyle R({\mit\Delta}\varphi) = \lim_{N\rightarrow\infty} \left(1-{\rm i} \,\frac{{\mit\Delta} \varphi}{N} \frac{L_z}{\hbar}\right)^N,$ (350)

 

 

which reduces to

 

$\displaystyle R({\mit\Delta}\varphi) = \exp(-{\rm i}\,L_z \,{\mit\Delta}\varphi/\hbar).$ (351)

 

 

Note that $ R({\mit\Delta}\varphi)$ has all of the properties we would expect of a rotation operator: i.e.,

 

$\displaystyle R(0)$ $\displaystyle = 1,$ (352)
$\displaystyle R({\mit\Delta}\varphi)\,R(-{\mit\Delta}\varphi)$ $\displaystyle = 1,$ (353)
$\displaystyle R({\mit\Delta}\varphi_1)\,R({\mit\Delta}\varphi_2)$ $\displaystyle = R({\mit\Delta}\varphi_1+ {\mit\Delta}\varphi_2).$ (354)

 

 

Suppose that the system is in a simultaneous eigenstate of $ L^2$ and $ L_z$ . As before, this state is represented by the eigenket $ \vert l, m\rangle$ , where the eigenvalue of $ L^2$ is $ l\,(l+1)\,\hbar^2$ , and the eigenvalue of $ L_z$ is $ m\,\hbar$ . We expect the wavefunction to remain unaltered if we rotate the system $ 2\pi$ degrees about the $ z$ -axis. Thus,

 

$\displaystyle R(2\pi)\,\vert l,m\rangle = \exp(-{\rm i}\,L_z \,2\pi/\hbar)\,\vert l,m\rangle = \exp(-{\rm i}\, 2\pi\,m) \,\vert l,m\rangle = \vert l,m\rangle.$ (355)

 

 

We conclude that $ m$ must be an integer. This implies, from the previous section, that $ l$ must also be an integer. Thus, an orbital angular momentum can only take integer values of the quantum numbers $ l$ and $ m$ .

Consider the action of the rotation operator $ R({\mit\Delta}\varphi)$ on an eigenstate possessing zero angular momentum about the $ z$ -axis (i.e., an $ m=0$ state). We have

 

$\displaystyle R({\mit\Delta}\varphi)\,\vert l, 0\rangle = \exp(0)\,\vert l, 0\rangle = \vert l, 0\rangle.$ (356)

 

 

Thus, the eigenstate is invariant to rotations about the $ z$ -axis. Clearly, its wavefunction must be symmetric about the $ z$ -axis.

There is nothing special about the $ z$ -axis, so we can write

 

$\displaystyle R_x({\mit\Delta}\varphi_x)$ $\displaystyle = \exp(-{\rm i}\,L_x \,{\mit\Delta}\varphi_x/\hbar),$ (357)
$\displaystyle R_y({\mit\Delta}\varphi_y)$ $\displaystyle = \exp(-{\rm i}\,L_y \,{\mit\Delta}\varphi_y/\hbar),$ (358)
$\displaystyle R_z({\mit\Delta}\varphi_y)$ $\displaystyle = \exp(-{\rm i}\,L_z\, {\mit\Delta}\varphi_z/\hbar),$ (359)

 

 

by analogy with Equation (351). Here, $ R_x({\mit\Delta}\varphi_x)$ denotes an operator that rotates the system by an angle $ {\mit\Delta}\varphi_x$ about the $ x$ -axis, etc. Suppose that the system is in an eigenstate of zero overall orbital angular momentum (i.e., an $ l=0$ state). We know that the system is also in an eigenstate of zero orbital angular momentum about any particular axis. This follows because $ l=0$ implies $ m=0$ , according to the previous section, and we can choose the $ z$ -axis to point in any direction. Thus,

 

$\displaystyle R_x({\mit\Delta} \varphi_x)\, \vert,0\rangle$ $\displaystyle =\exp(0)\,\vert,0\rangle = \vert,0\rangle,$ (360)
$\displaystyle R_y({\mit\Delta} \varphi_y)\, \vert,0\rangle$ $\displaystyle =\exp(0)\,\vert,0\rangle = \vert,0\rangle,$ (361)
$\displaystyle R_z({\mit\Delta} \varphi_z) \,\vert,0\rangle$ $\displaystyle =\exp(0)\,\vert,0\rangle = \vert,0\rangle.$ (362)

 

 

Clearly, a zero angular momentum state is invariant to rotations about any axis. Such a state must possess a spherically symmetric wavefunction.

Note that a rotation about the $ x$ -axis does not commute with a rotation about the $ y$ -axis. In other words, if the system is rotated an angle $ {\mit\Delta}\varphi_x$ about the $ x$ -axis, and then $ {\mit\Delta}\varphi_y$ about the $ y$ -axis, it ends up in a different state to that obtained by rotating an angle $ {\mit\Delta}\varphi_y$ about the $ y$ -axis, and then $ {\mit\Delta}\varphi_x$ about the $ x$ -axis. In quantum mechanics, this implies that $ R_y({\mit\Delta}\varphi_y)\,R_x({\mit\Delta}\varphi_x)
\neq R_x({\mit\Delta}\varphi_x)\,R_y({\mit\Delta}\varphi_y)$ , or $ L_y \,L_x \neq L_x\, L_y$ , [see Equations (357)-(359)]. Thus, the noncommuting nature of the angular momentum operators is a direct consequence of the fact that rotations do not commute.

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