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Next: Magnetic Moments Up: Spin Angular Momentum Previous: Wavefunction of Spin One-Half

Rotation Operators in Spin Space

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

$\displaystyle \vert A\rangle = \langle +\vert A\rangle \vert+\rangle + \langle -\vert A\rangle \vert-\rangle$ (5.23)

in spin space. In Section 4.3, we were able to construct an operator $ R_z({\mit\Delta}\varphi)$ that rotates the system through an angle $ {\mit\Delta}\varphi$ about the $ z$ -axis in position space. Can we also construct an operator $ T_z({\mit\Delta}\varphi)$ that rotates the system through an angle $ {\mit\Delta}\varphi$ about the $ z$ -axis in spin space? By analogy with Equation (4.62), we would expect such an operator to take the form

$\displaystyle T_z({\mit\Delta}\varphi) = \exp\left(\frac{-{\rm i}\, S_z\, {\mit\Delta}\varphi}{\hbar}\right).$ (5.24)

Thus, after rotation, the ket $ \vert A\rangle$ becomes

$\displaystyle \vert A_R\rangle = T_z({\mit\Delta}\varphi)\, \vert A\rangle.$ (5.25)

To demonstrate that the operator (5.24) really does rotate the spin of the system, let us consider its effect on $ \langle S_x\rangle$ . Under rotation, this expectation value changes as follows:

$\displaystyle \langle S_x\rangle \rightarrow \langle A_R\vert \,S_x\, \vert A_R \rangle = \langle A\vert \,T_z^{\dag }\, S_x \,T_z \,\vert A\rangle.$ (5.26)

Thus, we need to compute

$\displaystyle \exp\left(\frac{\,{\rm i}\,S_z\, {\mit\Delta}\varphi}{\hbar}\right)\, S_x \, \exp\left(\frac{-{\rm i}\,S_z \,{\mit\Delta}\varphi}{\hbar}\right).$ (5.27)

This goal can be achieved in two different ways.

First, we can use the explicit formula for $ S_x$ given in Equation (5.11). We find that expression (5.27) becomes

$\displaystyle \frac{\hbar}{2} \,\exp\left(\frac{\,{\rm i}\,S_z\, {\mit\Delta}\v...
...rt\right)\, \exp\left(\frac{-{\rm i}\,S_z \,{\mit\Delta}\varphi}{\hbar}\right),$ (5.28)


$\displaystyle \frac{\hbar}{2} \left( {\rm e}^{\,{\rm i}\,{\mit\Delta}\varphi/2}...
...ert-\rangle \langle +\vert\,{\rm e}^{\,-{\rm i}\,{\mit\Delta}\varphi/2}\right),$ (5.29)

which yields

$\displaystyle \exp\left(\frac{\,{\rm i}\,S_z\, {\mit\Delta}\varphi}{\hbar}\righ...
...phi}{\hbar}\right)=S_x\,\cos{\mit\Delta}\varphi - S_y\,\sin{\mit\Delta}\varphi,$ (5.30)

where use has been made of Equations (5.11)-(5.13).

A second approach is to use the so-called Baker-Campbell-Hausdorff lemma [5,20,62]. This takes the form

$\displaystyle \exp(\,{\rm i}\, G\,\lambda)\,A\, \exp(-{\rm i} \,G \,\lambda)$ $\displaystyle \equiv A + {\rm i} \,\lambda\, [G,A] + \left(\frac{{\rm i}^{\,2} \lambda^{\,2}}{2!}\right) [G, [G,A]]$    
  $\displaystyle \phantom{=}+ \left(\frac{{\rm i}^{\,3}\lambda^{\,3}}{3!}\right)[G, [G, [G,A]]]+\cdots,$ (5.31)

where $ G$ and $ A$ are operators, and $ \lambda$ a real parameter. The proof of this lemma is left as an exercise. (See Exercise 2.) Applying the Baker-Campbell-Hausdorff lemma to expression (5.27), we obtain

  $\displaystyle S_x + \left(\frac{{\rm i}\,{\mit\Delta}\varphi}{\hbar}\right) [S_... \left(\frac{{\rm i}\,{\mit\Delta}\varphi}{\hbar}\right)^2 [S_z, [S_z, S_x]]$    
  $\displaystyle + \left(\frac{1}{3!}\right) \left(\frac{{\rm i}\,{\mit\Delta}\varphi}{\hbar}\right)^3 [S_z,[S_z, [S_z, S_x]]]+\cdots$ (5.32)

which reduces to

$\displaystyle S_x\left[ 1- \frac{({\mit\Delta}\varphi)^2}{2!} + \cdots\right] -...
... \left[{\mit\Delta}\varphi - \frac{({\mit\Delta}\varphi)^3}{3!} +\cdots\right],$ (5.33)

where use has been made of Equation (5.1). Thus,

$\displaystyle \exp\left(\frac{\,{\rm i}\,S_z\, {\mit\Delta}\varphi}{\hbar}\righ...
...phi}{\hbar}\right)=S_x\,\cos{\mit\Delta}\varphi - S_y\,\sin{\mit\Delta}\varphi.$ (5.34)

The second proof is more general than the first, because it only makes use of the fundamental commutation relation (5.1), and is, therefore, valid for systems with spin angular momentum greater than one-half.

For a spin one-half system, both methods imply that

$\displaystyle \langle S_x \rangle \rightarrow \langle S_x\rangle \,\cos{\mit\Delta}\varphi - \langle S_y\rangle\,\sin{\mit\Delta}\varphi$ (5.35)

under the action of the rotation operator (5.24). It is straightforward to show that

$\displaystyle \langle S_y \rangle \rightarrow \langle S_x\rangle\,\sin{\mit\Delta}\varphi + \langle S_y\rangle \,\cos{\mit\Delta}\varphi.$ (5.36)


$\displaystyle \langle S_z \rangle \rightarrow\langle S_z\rangle,$ (5.37)

because $ S_z$ commutes with the rotation operator. Equations (5.35)-(5.37) demonstrate that the operator (5.24) rotates the expectation value of $ {\bf S}$ through an angle $ {\mit\Delta}\varphi$ about the $ z$ -axis. In fact, the expectation value of the spin operator behaves like a classical vector under rotation. In other words,

$\displaystyle \langle S_k \rangle \rightarrow \sum_l R_{k\,l}\, \langle S_l\rangle,$ (5.38)

where the $ R_{k\,l}$ are the elements of the conventional rotation matrix for the rotation in question [92]. (Here, and in the following, $ k$ , $ l$ , et cetera, are indices that run from 1 to 3, with 1 corresponding to the $ x$ -axis, 2 to the $ y$ -axis, and so on.) It is clear, from our second derivation of the result (5.35), that this property is not restricted to the spin operators of a spin one-half system. In fact, we have effectively demonstrated that

$\displaystyle \langle J_k \rangle \rightarrow \sum_l R_{k\,l} \,\langle J_l\rangle,$ (5.39)

where the $ J_k$ are the generators of rotation, satisfying the fundamental commutation relation $ {\bf J}\times {\bf J} = {\rm i}\,\hbar\, {\bf J}$ , and the rotation operator about the $ 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 (5.24) on the state ket (5.23). It is easily seen that

$\displaystyle T_z({\mit\Delta}\varphi)\,\vert A\rangle = {\rm e}^{-{\rm i}\,{\m...
... e}^{\,{\rm i}\,{\mit\Delta}\varphi/2}\, \langle -\vert A\rangle \vert-\rangle.$ (5.40)

Consider a rotation by $ 2\pi$ radians. We find that

$\displaystyle \vert A\rangle \rightarrow T_z(2\pi)\,\vert A\rangle = -\vert A\rangle.$ (5.41)

Note that a ket rotated by $ 2\pi$ radians differs from the original ket by a minus sign. In fact, a rotation by $ 4\pi$ radians is needed to transform a ket into itself. The minus sign does not affect the expectation value of $ {\bf S}$ , because $ {\bf S}$ is sandwiched between $ \langle A\vert$ and $ \vert A\rangle$ , both of which change sign. Nevertheless, the minus sign does give rise to observable consequences, as we shall see presently.

next up previous
Next: Magnetic Moments Up: Spin Angular Momentum Previous: Wavefunction of Spin One-Half
Richard Fitzpatrick 2016-01-22