5.4: Rotation Operators in Spin Space

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Let us, for the moment, forget about the spatial position of the particle, and concentrate on its spin state. A general spin state $ A$ is represented by the ket

$ R_z({\mit\Delta}\varphi)$ that rotates the system through an angle $ z$ -axis in position space. Can we also construct an operator $ {\mit\Delta}\varphi$ about the $ T_z({\mit\Delta}\varphi) = \exp(-{\rm i} \,S_z\, {\mit\Delta}\varphi/\hbar).$ \ref{440}

Thus, after rotation, the ket $ \vert A_R\rangle = T_z({\mit\Delta}\varphi)\, \vert A\rangle.$

\ref{441}

To demonstrate that the operator \ref{440} really does rotate the spin of the system, let us consider its effect on $ \langle S_x\rangle \rightarrow \langle A_R\vert \,S_x\, \vert A_R \rangle = \langle A\vert \,T_z^{\dag }\, S_x \,T_z \,\vert A\rangle.$

\ref{442}

Thus, we need to compute

$ S_x$ given in Equation \ref{427}. We find that Equation \ref{443} becomes $\frac{\hbar}{2} \,\exp(\,{\rm i}\,S_z\, {\mit\Delta}\varphi/\hbar... ...rt-\rangle \langle +\vert\,)\, \exp(-{\rm i}\,S_z \,{\mit\Delta}\varphi/\hbar),$ \ref{444}

$\frac{\hbar}{2} \left( {\rm e}^{\,{\rm i}\,{\mit\Delta}\varphi/2}... ...ert-\rangle \langle +\vert\,{\rm e}^{\,-{\rm i}\,{\mit\Delta}\varphi/2}\right),$ \ref{445}

which reduces to

$ \exp(\,{\rm i}\, G\,\lambda)\,A\, \exp(-{\rm i} \,G \,\lambda)$ $ + \left(\frac{{\rm i}^3\lambda^3}{3!}\right)[G, [G, [G,A]]]+\cdots,$ \ref{447}

where $ \lambda$ a real parameter. The proof of this lemma is left as an exercise. Applying the Baker-Hausdorff lemma to Equation \ref{443}, we obtain

$S_x + \left(\frac{{\rm i}\,{\mit\Delta}\varphi}{\hbar}\right) [S_... ...\frac{{\rm i}\,{\mit\Delta}\varphi}{\hbar}\right)^2 [S_z, [S_z, S_x]] + \cdots,$ \ref{448}

which reduces to

$S_x\left[ 1- \frac{({\mit\Delta}\varphi)^2}{2!} + \frac{({\mit\De... ...{\mit\Delta}\varphi)^3}{3!}+ \frac{({\mit\Delta}\varphi)^5}{5!} +\cdots\right],$ \ref{449}

$ \langle S_x \rangle \rightarrow \langle S_x\rangle \,\cos{\mit\Delta}\varphi - \langle S_y\rangle\,\sin{\mit\Delta}\varphi$ \ref{451}

under the action of the rotation operator \ref{440}. It is straightforward to show that

$ \langle S_z \rangle \rightarrow\langle S_z\rangle,$ \ref{453}

because $ {\bf S}$ by an angle $ z$ -axis. In fact, the expectation value of the spin operator behaves like a classical vector under rotation:

$ R_{k\,l}$ are the elements of the conventional rotation matrix for the rotation in question. It is clear, from our second derivation of the result \ref{451}, that this property is not restricted to the spin operators of a spin one-half system. In fact, we have effectively demonstrated that $ J_k$ are the generators of rotation, satisfying the fundamental commutation relation $ k$ th axis is written $ R_k ({\mit\Delta}\varphi) = \exp(-{\rm i}\,J_k\, {\mit\Delta}\varphi/\hbar)$ .

Consider the effect of the rotation operator \ref{440} on the state ket \ref{439}. It is easily seen that

$T_z({\mit\Delta}\varphi)\,\vert A\rangle = {\rm e}^{-{\rm i}\,{\m... ... e}^{\,{\rm i}\,{\mit\Delta}\varphi/2}\, \langle -\vert A\rangle \vert-\rangle.$ \ref{456}

Consider a rotation by $ \vert A\rangle \rightarrow T_z(2\pi)\,\vert A\rangle = -\vert A\rangle.$

\ref{457}

Note that a ket rotated by $ 4\pi$ radians is needed to transform a ket into itself. The minus sign does not affect the expectation value of $ {\bf S}$ is sandwiched between $ \vert A\rangle$, both of which change sign. Nevertheless, the minus sign does give rise to observable consequences, as we shall see presently.

Contributors

Richard Fitzpatrick (Professor of Physics, The University of Texas at Austin)

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