Mutliple Noninteracting Spins

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Spin Polarization in Thermal Equilibrium

We now consider a simple system of \(N\) noninteracting spins of magnetic moment \(\mu\) in an external magnetic field \(B_0\) parallel to the \(z\)-axis. The Hamiltonian of this system is: \begin{equation} \hat{H}(\vec{\hat{\sigma}}(1),...,\vec{\hat{\sigma}}(N))=\hbar \omega_0 \sum_{i=1}^{N}\hat{\sigma}_z(i). \end{equation} We will calculate the average values, \(\langle \hat{X} \rangle\), and fluctuation amplitudes, \\ \(\delta X=\sqrt{\langle (\hat{X}-\langle \hat{X} \rangle)^2 \rangle}\)=\(\sqrt{ \langle \hat{X}^2 \rangle-\langle \hat{X} \rangle ^2}\), for \(\hat{X}=\hat{M}_x,\hat{M}_y,\hat{M}_z\). In thermodynamic equilibrium the statistical operator of this system is: \begin{eqnarray} \hat{\rho}(\beta)&=&\frac{e^{\beta \hbar \omega_0 \sum_{i=1}^{N}\hat{\sigma}_z(i)}}{Tr\{e^{\beta \hbar \omega_0 \sum_{i=1}^{N}\hat{\sigma}_z(i)} \}} \end{eqnarray} where \(\beta=1/kT\). The trace in the denominator can be easily calculated since the operators belonging to different spins commute \begin{eqnarray} Tr\{e^{\beta \hbar \omega_0 \sum_{i=1}^{N}\hat{\sigma}_z(i)} \}&=&Tr\{\prod_{i=1}^{N} e^{\beta \hbar \omega_0\hat{\sigma}_z(i)} \}=\prod_{i=1}^{N} Tr\{ e^{\beta \hbar \omega_0\hat{\sigma}_z(i)} \}\\ \nonumber &=&(e^{\beta \hbar \omega_0}+e^{-\beta \hbar \omega_0})^N \end{eqnarray} The average magnetization along the field is (standard NMR result): \begin{eqnarray} M_z&=&\langle \hat{M}_z \rangle=\langle \mu \sum_{i=1}^N \sigma_z(i) \rangle \\ &=&\mu \sum_{i=1}^N Tr\{\hat{\rho}(\beta) \hat{\sigma}_z(i) \}\\ &=&\mu \sum_{i=1}^N \frac{(e^{\beta \hbar \omega_0}-e^{-\beta \hbar \omega_0})(e^{\beta \hbar \omega_0}+e^{-\beta \hbar \omega_0})^{(N-1)}}{(e^{\beta \hbar \omega_0}+e^{-\beta \hbar \omega_0})^N} \\ M_z&=&\mu N \mbox{tanh} ( \beta \hbar \omega_0 )\approx \mu^2 N \beta B_0 \end{eqnarray} This is the polarization of the spins induced by the applied field.

Fluctuations of the Magnetization in Thermal Equilibrium

The average of the square of \(\hat{M}_z\) is \begin{eqnarray} \langle \hat{M}^2_z \rangle&=&\langle \mu^2 \sum_{i,j=1}^N \sigma_z(i) \sigma_z(j) \rangle \\ &=&\mu^2 \sum_{i,j=1}^N Tr\{\hat{\rho}(\beta) \hat{\sigma}_z(i) \hat{\sigma}_z(j) \}\\ &=&\mu^2 \sum_{i=j=1}^N Tr\{\hat{\rho}(\beta) \hat{\sigma}^2_z(i) \}+\mu^2 \sum_{i \neq j=1}^N Tr\{\hat{\rho}(\beta) \hat{\sigma}_z(i) \hat{\sigma}_z(j) \} \\ &=&\mu^2 N+\mu^2 N(N-1) \mbox{tanh}^2 ( \beta \hbar \omega_0 )\\ \delta M_z &=& \sqrt{\mu^2 N+\mu^2 N(N-1)} \mbox{tanh}^2 ( \beta \hbar \omega_0 )-\mu^2 N^2 \mbox{tanh}^2 ( \beta \hbar \omega_0 )\\ &\approx &\mu \sqrt{N} \end{eqnarray} We find: \begin{eqnarray} \frac{\delta M_z}{M_z} &\approx & \frac{\mbox{tanh} ( \beta \hbar \omega_0 )}{\sqrt{N}} \end{eqnarray} The average magnetization along the \(x\)-axis is: \begin{eqnarray} M_x&=&\langle \hat{M}_x \rangle=\langle \mu \sum_{i=1}^N \sigma_x(i) \rangle \\ &=&\mu \sum_{i=1}^N Tr\{\hat{\rho}(\beta) \hat{\sigma}_x(i) \}=0\\ \end{eqnarray} The average of the square of \(\hat{M}_x\) is \begin{eqnarray} \langle \hat{M}^2_z \rangle&=&\langle \mu^2 \sum_{i,j=1}^N \hat{\sigma}_x(i) \hat{\sigma}_x(j) \rangle \\ &=&\mu^2 \sum_{i,j=1}^N Tr\{\hat{\rho}(\beta) \hat{\sigma}_x(i) \hat{\sigma}_z(j) \}\\ &=&\mu^2 \sum_{i=j=1}^N Tr\{\hat{\rho}(\beta) \hat{\sigma}^2_x(i) \}+\mu^2 \sum_{i \neq j=1}^N Tr\{\hat{\rho}(\beta) \hat{\sigma}_x(i) \hat{\sigma}_x(j) \} \\ &=&\mu^2 N+0\\ \delta M_x &=&\mu \sqrt{N} \end{eqnarray} We find: \begin{eqnarray} \frac{\delta M_x}{M_z} &=& \frac{\mbox{tanh} ( \beta \hbar \omega_0 )}{\sqrt{N}} \end{eqnarray}