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

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

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

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

According to Equation (4.59), we can write

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

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

which reduces to

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

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

$$R_z(0) =1,$$ (4.63)

$$R_z({\mit\Delta}\varphi)\,R_z(-{\mit\Delta}\varphi) =1,$$ (4.64)

$$R_z({\mit\Delta}\varphi_1)\,R_z({\mit\Delta}\varphi_2) = R_z({\mit\Delta}\varphi_1+ {\mit\Delta}\varphi_2).$$ (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,

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

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

$$R_x({\mit\Delta}\varphi_x) = \exp\left(\frac{-{\rm i}\,L_x \,{\mit\Delta}\varphi_x}{\hbar}\right),$$ (4.68)

$$R_y({\mit\Delta}\varphi_y) = \exp\left(\frac{-{\rm i}\,L_y \,{\mit\Delta}\varphi_y}{\hbar}\right),$$ (4.69)

$$R_z({\mit\Delta}\varphi_y) = \exp\left(\frac{-{\rm i}\,L_z\, {\mit\Delta}\varphi_z}{\hbar}\right),$$ (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,

$$R_x({\mit\Delta} \varphi_x)\, \vert,0\rangle =\exp(0)\,\vert,0\rangle = \vert,0\rangle,$$ (4.71)

$$R_y({\mit\Delta} \varphi_y)\, \vert,0\rangle =\exp(0)\,\vert,0\rangle = \vert,0\rangle,$$ (4.72)

$$R_z({\mit\Delta} \varphi_z) \,\vert,0\rangle =\exp(0)\,\vert,0\rangle = \vert,0\rangle.$$ (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.