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Rotation Operators

Consider a particle whose position is described by the spherical coordinates $ (r, \,\theta,\, \varphi)$ . The classical momentum conjugate to the azimuthal angle $ \varphi$ is the $ z$ -component of angular momentum, $ L_z$ [55]. 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

$\displaystyle L_z = -{\rm i}\,\hbar\, \frac{\partial}{\partial \varphi}.$ (4.57)

Incidentally, it is legitimate to write the previous equation because there is nothing in Section 2.5 that specifies that we have to use Cartesian coordinates--the representation (2.74) works for any well-defined set of coordinates.

Consider an operator $ R_z({\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_x({\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

$\displaystyle p_x = {\rm i}\,\hbar\, \lim_{\delta x\rightarrow 0}\frac{D_x(\delta x)-1} {\delta x},$ (4.58)

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

$\displaystyle L_z = {\rm i}\,\hbar\, \lim_{\delta \varphi\rightarrow 0}\frac{R_z(\delta \varphi)-1} {\delta \varphi}.$ (4.59)

According to Equation (4.59), we can write

$\displaystyle R_z(\delta \varphi) = 1 -{\rm i}\,L_z\,\delta\varphi/\hbar$ (4.60)

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 rotation about the $ z$ -axis. The previous equation implies that

$\displaystyle R_z({\mit\Delta}\varphi) = \lim_{N\rightarrow\infty} \left(1-{\rm...
...=0,\infty}\left[\frac{(-{\rm i}\,{\mit\Delta}\varphi\,L_z/\hbar)^n}{n!}\right],$ (4.61)

which reduces to

$\displaystyle R_z({\mit\Delta}\varphi) = \exp\left(\frac{-{\rm i}\,L_z \,{\mit\Delta}\varphi}{\hbar}\right).$ (4.62)

Note that $ R_z({\mit\Delta}\varphi)$ has all of the properties we would expect a rotation operator to possess: that is,

$\displaystyle R_z(0)$ $\displaystyle =1,$ (4.63)
$\displaystyle R_z({\mit\Delta}\varphi)\,R_z(-{\mit\Delta}\varphi)$ $\displaystyle =1,$ (4.64)
$\displaystyle R_z({\mit\Delta}\varphi_1)\,R_z({\mit\Delta}\varphi_2)$ $\displaystyle = R_z({\mit\Delta}\varphi_1+ {\mit\Delta}\varphi_2).$ (4.65)

(See Exercises 7 and 8.)

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,

$\displaystyle R_z(2\pi)\,\vert l,m\rangle = \exp\left(\frac{-{\rm i}\,L_z \,2\p...
...rt l,m\rangle = \exp(-{\rm i}\, 2\pi\,m) \,\vert l,m\rangle = \vert l,m\rangle.$ (4.66)

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

$\displaystyle R_z({\mit\Delta}\varphi)\,\vert l, 0\rangle = \exp(0)\,\vert l, 0\rangle = \vert l, 0\rangle.$ (4.67)

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

$\displaystyle R_x({\mit\Delta}\varphi_x)$ $\displaystyle = \exp\left(\frac{-{\rm i}\,L_x \,{\mit\Delta}\varphi_x}{\hbar}\right),$ (4.68)
$\displaystyle R_y({\mit\Delta}\varphi_y)$ $\displaystyle = \exp\left(\frac{-{\rm i}\,L_y \,{\mit\Delta}\varphi_y}{\hbar}\right),$ (4.69)
$\displaystyle R_z({\mit\Delta}\varphi_y)$ $\displaystyle = \exp\left(\frac{-{\rm i}\,L_z\, {\mit\Delta}\varphi_z}{\hbar}\right),$ (4.70)

by analogy with Equation (4.62). Here, $ R_x({\mit\Delta}\varphi_x)$ denotes an operator that rotates the system through an angle $ {\mit\Delta}\varphi_x$ about the $ x$ -axis, et cetera. 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,

$\displaystyle R_x({\mit\Delta} \varphi_x)\, \vert,0\rangle$ $\displaystyle =\exp(0)\,\vert,0\rangle = \vert,0\rangle,$ (4.71)
$\displaystyle R_y({\mit\Delta} \varphi_y)\, \vert,0\rangle$ $\displaystyle =\exp(0)\,\vert,0\rangle = \vert,0\rangle,$ (4.72)
$\displaystyle R_z({\mit\Delta} \varphi_z) \,\vert,0\rangle$ $\displaystyle =\exp(0)\,\vert,0\rangle = \vert,0\rangle.$ (4.73)

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

Note that, in general, a rotation about the $ x$ -axis does not commute with a rotation about the $ y$ -axis. In other words, if a physical system is rotated through an angle $ {\mit\Delta}\varphi_x$ about the $ x$ -axis, and then through an angle $ {\mit\Delta}\varphi_y$ about the $ y$ -axis, it ends up in a different state to that obtained by rotating through an angle $ {\mit\Delta}\varphi_y$ about the $ y$ -axis, and then through an angle $ {\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 (4.68)-(4.70) and Exercise 3.] Thus, the noncommuting nature of the angular momentum operators is a direct consequence of the fact that rotations do not commute.

next up previous
Next: Eigenfunctions of Orbital Angular Up: Orbital Angular Momentum Previous: Eigenvalues of Orbital Angular
Richard Fitzpatrick 2016-01-22