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Next: Pauli Two-Component Formalism Up: Spin Angular Momentum Previous: Magnetic Moments

Spin Precession

The Hamiltonian for an electron at rest in a $ z$ -directed magnetic field, $ {\bf B}=
B\,{\bf e}_z$ , is

$\displaystyle H = -$   $\displaystyle \mbox{\boldmath$\mu$}$$\displaystyle \cdot {\bf B} = \left(\frac{e}{m_e}\right) {\bf S} \cdot {\bf B} = \omega\, S_z,$ (463)


$\displaystyle \omega = \frac{e\,B}{m_e}.$ (464)

According to Equation (247), the time evolution operator for this system is

$\displaystyle T(t,0) = \exp(-{\rm i} \,H\, t/\hbar) = \exp(-{\rm i} \,S_z\, \omega\, t/\hbar).$ (465)

It can be seen, by comparison with Equation (440), that the time evolution operator is precisely the same as the rotation operator for spin, with $ {\mit\Delta}\varphi$ set equal to $ \omega \,t$ . It is immediately clear that the Hamiltonian (463) causes the electron spin to precess about the $ z$ -axis with angular frequency $ \omega$ . In fact, Equations (451)-(453) imply that

$\displaystyle \langle S_x\rangle_t$ $\displaystyle = \langle S_x\rangle_{t=0} \cos(\omega \,t) - \langle S_y\rangle_{t=0} \sin(\omega\, t),$ (466)
$\displaystyle \langle S_y\rangle_t$ $\displaystyle = \langle S_y\rangle_{t=0} \cos(\omega\, t) + \langle S_x\rangle_{t=0} \sin(\omega\, t),$ (467)
$\displaystyle \langle S_z\rangle_t$ $\displaystyle = \langle S_z\rangle_{t=0}.$ (468)

The time evolution of the state ket is given by analogy with Equation (456):

$\displaystyle \vert A, t\rangle = {\rm e}^{-{\rm i}\,\omega \,t/2}\, \langle +\...
...+ {\rm e}^{\,{\rm i}\,\omega \,t/2}\, \langle -\vert A, 0\rangle \vert-\rangle.$ (469)

Note that it takes time $ t= 4\pi/\omega$ for the state ket to return to its original state. By contrast, it only takes times $ t=2\pi/\omega$ for the spin vector to point in its original direction.

We now describe an experiment to detect the minus sign in Equation (457). An almost monoenergetic beam of neutrons is split in two, sent along two different paths, $ A$ and $ B$ , and then recombined. Path $ A$ goes through a magnetic field free region. However, path $ B$ enters a small region where a static magnetic field is present. As a result, a neutron state ket going along path $ B$ acquires a phase-shift $ \exp(\mp{\rm i}\, \omega \,T/2)$ (the $ \mp$ signs correspond to $ s_z = \pm 1/2$ states). Here, $ T$ is the time spent in the magnetic field, and $ \omega$ is the spin precession frequency

$\displaystyle \omega = \frac{g_n\, e\,B}{m_p}.$ (470)

This frequency is defined in an analogous manner to Equation (464). The gyromagnetic ratio for a neutron is found experimentally to be $ g_n = -1.91$ . (The magnetic moment of a neutron is entirely a quantum field effect). When neutrons from path $ A$ and path $ B$ meet they undergo interference. We expect the observed neutron intensity in the interference region to exhibit a $ \cos( \pm \omega\, T/2 + \delta)$ variation, where $ \delta$ is the phase difference between paths $ A$ and $ B$ in the absence of a magnetic field. In experiments, the time of flight $ T$ through the magnetic field region is kept constant, while the field-strength $ B$ is varied. It follows that the change in magnetic field required to produce successive maxima is

$\displaystyle {\mit\Delta} B = \frac{4\pi \,\hbar}{e\, g_n\, \lambdabar\, l},$ (471)

where $ l$ is the path-length through the magnetic field region, and $ \lambdabar$ is the de Broglie wavelength over $ 2\pi$ of the neutrons. The above prediction has been verified experimentally to within a fraction of a percent. This prediction depends crucially on the fact that it takes a $ 4\pi$ rotation to return a state ket to its original state. If it only took a $ 2\pi$ rotation then $ {\mit\Delta} B$ would be half of the value given above, which does not agree with the experimental data.

next up previous
Next: Pauli Two-Component Formalism Up: Spin Angular Momentum Previous: Magnetic Moments
Richard Fitzpatrick 2013-04-08