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The quantum treatment of the spin precession does not alter essentially what we have been describing until now. However it is instructive to work out how a quantum spin behaves, and what is the equivalent of the classical vector motion.

Perhaps the most important thing is to understand the connection between this simple quantum exercise and the classical concept of nuclear magnetization expressed in a previous section by the nuclear Curie law. We shall come back to this at the end of this section.

We must start from the Zeeman Hamiltonian of an individual spin {$\mathbf{ I}$}, which is written by the correspondence principle, from the classical energy of a magnetic moment in a static field, {$U=-\mathbf{ m}\cdot\mathbf{ B} $} as

{$$ \begin{equation} H = - \gamma\hbar \mathbf{ I} \cdot \mathbf{ B} \end{equation} $$}

Let us consider for simplicity a spin {$I=\frac 1 2$}, with {$ \mathbf{ B} = B\hat z$}, as before.

The Hamiltonian may be rewritten as

{$$ H = -\gamma\hbar B I_z = \hbar \omega I_z $$}

It is useful to recall the matrix form of the three spin operator components in the representation of {$I_z,I^2$}:

{$$ I_z= \begin{bmatrix} \frac 1 2 & 0 \\ 0 & - \frac 1 2 \end{bmatrix} ,\qquad I_x= \begin{bmatrix} 0 & \frac 1 2 \\ \frac 1 2 & 0 \end{bmatrix} ,\qquad I_y = \begin{bmatrix} 0 & -\frac i 2\\ \frac i 2 & 0 \end{bmatrix} , $$}

The eigenstates of the Hamiltonian are those of {$I_z$}, which we denote {$|+\rangle,|-\rangle$} and correspond to the eigenvalues {$\pm \frac 1 2 $}. Each of these states is stationary, that is, if the spin is in one of these states it will remain forever in that state. But let us prepare the spin in a different state, such as for instance an eigenstate of {$I_x$}. In terms of {$|+\rangle,|-\rangle$} one has:

{$ (2) \qquad\qquad |\pm\rangle_x \quad = \frac 1 {\sqrt 2} \left(|+\rangle \pm |-\rangle\right) $}

If we want to determine the time evolution of the spin component perpendicular to the external field the quantum prescription is to calculate the expectation value of the spin operator, e.g. {$I_x$}, on the time evolved state (in the Schrödinger representation) or, equivalently, the expectation value of the time evolved operator on the stationary state (Heisenberg representation). The time evolution operator here is {$U(t)=e^{-iHt/\hbar}=e^{i\omega I_z t}$}. Choosing e.g. {$ |+\rangle_x $} as the state, both pictures correspond to:

{$ (3) \qquad\qquad \langle I_x(t)\rangle\quad =\quad _x\langle +|e^{i\omega I_z t} I_x e^{-i\omega I_z t}|+\rangle_x \quad $}

This quantity is more directly computed recalling the definition of the raising and lowering spin operators:

{$ I_\pm= I_x\pm i I_y $}

which have the following effect on the z-eigenstates:

{$ I_\mp |\pm\rangle= |\mp\rangle \qquad\qquad I_\pm |\pm\rangle= 0 $}

Substituting {$ I_x = (I_+ + I_-)/2 $} and the expression (2) for the state into Eq. (3) we get:

{$ (4) \qquad\qquad \begin{eqnarray} \langle I_x(t)\rangle & = & \frac 1 4 \left[ (\langle +| e^{-i\omega t/2} + \langle -| e^{i\omega t/2})(I_+ + I_-)(e^{i\omega t/2}| +\rangle + e^{-i\omega t/2}|-\rangle )\right] \\ & = & \frac 1 4 (e^{i\omega t}+e^{i\omega t}) = \frac 1 2 cos\omega t\end{eqnarray}$}

A similar calculation shows that {$\langle I_y(t)\rangle =\frac 1 2 sin\omega t$}, that is the expectation value of {$\mbox{\it \bf I}$}, precesses around the external field (being equal to {$ \frac 1 2$} in amplitude).

Considerations on the nuclear Curie Law.

We have shown here that each single spin, if we neglect couplings with its neighbours and with the lattice, precesses around the external field. The precession has a precise quantum meaning, expressed for instance by Eq. (4). In a statistical ensemble of quantum spins each will precess accordingly.

Note for instance the state {$\cos\alpha |+\rangle + \sin\alpha |-\rangle$}, cfr. Eq. (2), represents a state precessing in a cone of aperture {$2\alpha$} around the external field. That is, a continuum of classical orientations is mimicked by the coherent quantum precessions, depending on initial conditions, despite the fact that energy levels are still quantized, with only two eigenstates corresponding respectively to spin-up and to spin-down.

When the external field has been just applied to an unmagnetized ensamble the average of all these precessions will always be zero, since spins are pointing with equal probability in any direction on the sphere. However after some time the relaxation driven by the statistical quantum operator {$\exp(- H/k_B T)$} will favour slightly the spin-up state to an extent determined by the Curie susceptibility {$\chi$}.

From now on we can consider by the same argument that in all coherent manipulations (e.g. nutations induced by radio-frequency pulses) the average nuclear magnetization follows strictly the same behaviour of the single quantum spin, although rescaled by {$\chi$}.

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QuantumSpinPrecession

Considerations on the nuclear Curie Law.