Magnetic Ordering in a System of Identical Particles with Arbitrary Spin

Abstract

The Wigner–Eckart theorem is used to consider collective effects associated with the ordering of spins in systems of identical particles in Ferro- and antiferromagnetic electronic systems, as well as magnetic effects arising in high-spin systems. The Hamiltonian obtained by Heisenberg, Dirac, and Van Vleck was written in the spin representation used to describe spin ordering in systems of particles with spin 1/2. This form is not suitable for describing systems of particles with a spin other than 1/2. “High-spin” particles in the spin representation should be described by other forms of the exchange interaction Hamiltonian in the spin representation. The Hamiltonian for high-spin particles is derived from the first principles. This chapter discusses the effects of magnetic ordering in systems of identical particles with arbitrary spin.

Appendices

Appendix 1: Variational Heisenberg Model for the Spin-1/2 System

A system of identical particles with spins 1/2 is described by the HDV Hamiltonian Eq. (64)

$$\hat{H}_{{{\text{int}} 1/2}} = - \sum\limits_{k < l} {J_{kl} {\hat{\mathbf{s}}}_{k} \cdot {\hat{\mathbf{s}}}_{l} } .$$

(A1)

Suppose each pair of particles has the same coupling constant $J_{kl} = J$. In the system of N particles, N₊ particles have spin “up” and N_- - spin “down”. Then average value of spin for each particle is

$$\overline{s} = \frac{1}{2}\frac{{N_{ + } - N_{ - } }}{N},$$

(A2)

or

$$N_{ \pm } = \frac{1}{2}N\left( {1 \pm 2\overline{s}} \right).$$

(A3)

The total energy of interaction in this case is

$$\begin{gathered} \overline{{\hat{H}_{{{\text{int}} 1/2}} }} = - \sum\limits_{k < l} {\overline{{J_{kl} {\hat{\mathbf{s}}}_{k} \cdot {\hat{\mathbf{s}}}_{l} }} } = - J\sum\limits_{k < l} {\overline{{{\hat{\mathbf{s}}}_{k} \cdot {\hat{\mathbf{s}}}_{l} }} } = - J\overline{s}^{2} \frac{{N\left( {N - 1} \right)}}{2}, \hfill \\ C_{N}^{2} = \frac{N!}{{2!\left( {N - 2} \right)!}} = \frac{{N\left( {N - 1} \right)}}{2}. \hfill \\ \end{gathered}$$

(A4)

Then the average energy of interaction per one particle is

$$\overline{\varepsilon }_{{\text{int}}} = - J\overline{s}^{2} \frac{{\left( {N - 1} \right)}}{2} \approx - \frac{J}{2}\overline{s}^{2} N.$$

(A5)

We used here that N >> 1.

The statistical sum of such state with the energy $\overline{\varepsilon }_{{\text{int}}}$ per particle is

$$z = \frac{N!}{{N_{ + } !N_{ - } !}}e^{{ - \overline{\varepsilon }_{{\text{int}}} /T}} = \frac{N!}{{N_{ + } !N_{ - } !}}\exp \left( {\frac{{J\overline{s}^{2} N}}{2T}} \right).$$

(A6)

We take the temperature in the energy scale (T = k_BT^o). The free energy for this case is

$$\begin{gathered} F = - T\ln z = - T\left\{ {N_{ + } \ln \frac{N}{{N_{ + } }} + N_{ - } \ln \frac{N}{{N_{ - } }} + \frac{{J\overline{s}^{2} N}}{2T}} \right\} = \hfill \\ = - T\left\{ {\frac{1}{2}N\left( {1 + 2\overline{s}} \right)\ln \frac{N}{{\frac{1}{2}N\left( {1 + 2\overline{s}} \right)}} + } \right. \hfill \\ + \left. {\frac{1}{2}N\left( {1 - 2\overline{s}} \right)\ln \frac{N}{{\frac{1}{2}N\left( {1 - 2\overline{s}} \right)}} + \frac{{J\overline{s}^{2} N}}{2T}} \right\} = \hfill \\ = - TN\left\{ { - \frac{{\left( {1 + 2\overline{s}} \right)}}{2}\ln \frac{{\left( {1 + 2\overline{s}} \right)}}{2} - \frac{{\left( {1 - 2\overline{s}} \right)}}{2}\ln \frac{{\left( {1 - 2\overline{s}} \right)}}{2} + \frac{{J\overline{s}^{2} }}{2T}} \right\}. \hfill \\ \end{gathered}$$

(A7)

The Stirling formula was used for the factor representation:

$$\begin{gathered} N! \approx \sqrt {2\pi N} \left( \frac{N}{e} \right)^{N} , \hfill \\ \ln N! \approx \frac{1}{2}\ln \left( {2\pi N} \right) + \left( {N\ln N - N} \right) = \left( {N\ln N - N} \right), \hfill \\ \frac{1}{2}\ln \left( {2\pi N} \right) \ll N. \hfill \\ \end{gathered}$$

(A8)

To find the optimal value of the mean spin, $\overline{s}$, we take the derivative of the free energy F, defined in (A7), with respect to $\overline{s}$.

$$\begin{gathered} \frac{\partial }{{\partial \overline{s}}}F = - TN\left\{ { - \ln \frac{{\left( {1 + 2\overline{s}} \right)}}{2} - 1 + \ln \frac{{\left( {1 - 2\overline{s}} \right)}}{2} + 1 + \frac{{J\overline{s}}}{T}} \right\} = 0, \Rightarrow \hfill \\ \ln \left( {\frac{{1 - 2\overline{s}}}{{1 + 2\overline{s}}}} \right) + \frac{{J\overline{s}}}{T} = 0, \hfill \\ \left( {\frac{{1 + 2\overline{s}}}{{1 - 2\overline{s}}}} \right) = e^{{\frac{{J\overline{s}}}{T}}} , \Rightarrow \hfill \\ 2\overline{s} = \frac{{e^{{\frac{{J\overline{s}}}{T}}} - 1}}{{e^{{\frac{{J\overline{s}}}{T}}} + 1}}, \Rightarrow \hfill \\ \end{gathered}$$

(A9)

We come to the following transcendental equation for $\overline{s}$.

$$2\overline{s} = \tanh \left( {\frac{{J\overline{s}}}{2T}} \right),$$

(A10)

or in the form

$$\left( {2\overline{s}} \right) = \tanh \left( {\frac{{J\left( {2\overline{s}} \right)}}{4T}} \right).$$

(A11)

The possible solutions to the equation Eq. (A11) are shown in Fig. 4. It is clear that in the case of parameter $\frac{J}{4T} > 1$, the equation has a nontrivial solution for $\left( {2\overline{s}} \right)$, which means the excitation of the spontaneous polarization in the system with the total spin $\Sigma = N\overline{s}$ (It is shown in Fig. 4a). The case b) shows the absence of the spontaneous spin polarization in the system (the parameter $\frac{J}{4T} < 1$. The case c) on the picture shows a state which is close to the phase transformation in the spin system. It means that the critical temperature of the phase transition is $T* = \frac{J}{4}$. It is a second-order phase transition from paramagnetic to ferromagnetic state.

Fig. 4

Graphs of the functions $Y = \tanh \left( {\frac{{J\left( {2\overline{s}} \right)}}{4T}} \right)$, $f = \left( {2\overline{s}} \right)$ and their intersections as a solution to the transcendental equation (A11). The case (a) the parameter $\frac{J}{4T} = 2$, the case (b) the parameter $\frac{J}{4T} = 0.9$, and (c) the parameter $\frac{J}{4T} = 1.1$

Appendix 2: Ising Model for the Spin-1/2 Chain

We consider a spin chain in the magnetic field. HDV Hamiltonian with respect to the interaction of eigen magnetic momentums of spin-1/2 particles has the form

$$\hat{H}_{{{\text{int}} 1/2}} = - \sum\limits_{k < l} {J_{kl} {\hat{\mathbf{s}}}_{k} \cdot {\hat{\mathbf{s}}}_{l} } - 2\mu_{B} {\mathbf{H}} \cdot \sum\limits_{k} {{\hat{\mathbf{s}}}_{k} } ,$$

(A12)

where ${\mathbf{H}}$ is a magnetic field strength.

As it was shown in Eq. (35)

$$\left( {\widehat{{{\mathbf{j}}_{1} }} \cdot \widehat{{{\mathbf{j}}_{2} }}} \right) = \frac{1}{2}\left( {\widehat{{j_{1}^{ + } }}\widehat{{j_{2}^{ - } }} + \widehat{{j_{1}^{ - } }}\widehat{{j_{2}^{ + } }} + 2\widehat{{j_{1z} }}\widehat{{j_{2z} }}} \right) \to \widehat{{j_{1z} }}\widehat{{j_{2z} }},$$

In this model (Ising Model), the first two terms are neglected. In the nearest-neighbor case (with periodic or free boundary conditions), an exact solution is available. The Hamiltonian of the one-dimensional Ising model on a lattice of L sites with periodic boundary conditions is

$$\begin{gathered} \hat{H}_{I\sin g} = - \sum\limits_{k} {J_{kk + 1} \hat{s}_{zk} \hat{s}_{zk + 1} } - 2\mu_{B} H \cdot \sum\limits_{k} {\hat{s}_{zk} } = \hfill \\ = - J\sum\limits_{k} {\hat{\sigma }_{zk} \hat{\sigma }_{zk + 1} } - \mu_{B} H \cdot \sum\limits_{k} {\hat{\sigma }_{zk} } , \hfill \\ \end{gathered}$$

(A13)

where $\hat{\sigma }_{zk}$-are Pauli matrix for k-th spin projection on the z-axis, $J_{kk + 1} = 4J$.

$$\begin{gathered} \hat{H}_{I\sin g} = - J\sum\limits_{k} {\hat{\sigma }_{zk} \hat{\sigma }_{zk + 1} } - \mu_{B} H \cdot \sum\limits_{k} {\hat{\sigma }_{zk} } = \hfill \\ = \left( { - J\hat{\sigma }_{z1} \hat{\sigma }_{z2} - \frac{{\mu_{B} H}}{2}\left( {\hat{\sigma }_{z1} + \hat{\sigma }_{z2} } \right)} \right) + \hfill \\ + \left( { - J\hat{\sigma }_{z2} \hat{\sigma }_{z3} - \frac{{\mu_{B} H}}{2}\left( {\hat{\sigma }_{z2} + \hat{\sigma }_{z3} } \right)} \right) + \hfill \\ + ... + \left( { - J\hat{\sigma }_{zN} \hat{\sigma }_{z1} - \frac{{\mu_{B} H}}{2}\left( {\hat{\sigma }_{zN} + \hat{\sigma }_{z1} } \right)} \right). \hfill \\ \end{gathered}$$

(A14)

Statistical sum z is

$$\begin{gathered} z = Sp\exp \left( { - \frac{{\hat{H}_{I\sin g} }}{T}} \right) = \hfill \\ = Sp\left\{ {e^{{\frac{{2J\hat{\sigma }_{z1} \hat{\sigma }_{z2} + \mu_{B} H\left( {\hat{\sigma }_{z1} + \hat{\sigma }_{z2} } \right)}}{2T}}} \cdot e^{{\frac{{2J\hat{\sigma }_{z2} \hat{\sigma }_{z3} + \mu_{B} H\left( {\hat{\sigma }_{z2} + \hat{\sigma }_{z3} } \right)}}{2T}}} \cdot ... \cdot e^{{\frac{{2J\hat{\sigma }_{zN} \hat{\sigma }_{z1} + \mu_{B} H\left( {\hat{\sigma }_{zN} + \hat{\sigma }_{z1} } \right)}}{2T}}} } \right\} = \hfill \\ = Sp\hat{V}_{12} \hat{V}_{23} ...\hat{V}_{N1} , \hfill \\ \hat{V}_{kk + 1} = e^{{\frac{{2J\hat{\sigma }_{zk} \hat{\sigma }_{zk + 1} + \mu_{B} H\left( {\hat{\sigma }_{zk} + \hat{\sigma }_{zk + 1} } \right)}}{2T}}} = \left( {\begin{array}{*{20}c} {e^{{\frac{{J + \mu_{B} H}}{T}}} } & {e^{{ - \frac{J}{T}}} } \\ {e^{{ - \frac{J}{T}}} } & {e^{{\frac{{J - \mu_{B} H}}{T}}} } \\ \end{array} } \right). \hfill \\ \end{gathered}$$

(A15)

A diagonalization of the operator $\hat{V}_{kk + 1}$ gives the following:

$$\begin{gathered} \left| {\begin{array}{*{20}c} {e^{{\frac{{J + \mu_{B} H}}{T}}} - \lambda } & {e^{{ - \frac{J}{T}}} } \\ {e^{{ - \frac{J}{T}}} } & {e^{{\frac{{J - \mu_{B} H}}{T}}} - \lambda } \\ \end{array} } \right| = 0, \Rightarrow \hfill \\ \lambda^{2} - \lambda e^{\frac{J}{T}} 2\cosh \left( {\frac{{\mu_{B} H}}{T}} \right) + 2\sinh \left( \frac{2J}{T} \right) = 0, \hfill \\ \lambda_{1,2} = e^{\frac{J}{T}} \cosh \left( {\frac{{\mu_{B} H}}{T}} \right) \pm \sqrt {e^{\frac{2J}{T}} \left( {\cosh \left( {\frac{{\mu_{B} H}}{T}} \right)} \right)^{2} - 2\sinh \left( \frac{2J}{T} \right)} , \hfill \\ \hat{V}_{kk + 1} = \left( {\begin{array}{*{20}c} {e^{{\frac{{J + \mu_{B} H}}{T}}} } & {e^{{ - \frac{J}{T}}} } \\ {e^{{ - \frac{J}{T}}} } & {e^{{\frac{{J - \mu_{B} H}}{T}}} } \\ \end{array} } \right) = \left( {\begin{array}{*{20}c} {\lambda_{1} } & 0 \\ 0 & {\lambda_{2} } \\ \end{array} } \right). \hfill \\ \end{gathered}$$

(A16)

Therefore, that statistical sum z will have a form

$$z = Sp\hat{V}_{12} \hat{V}_{23} ...\hat{V}_{N1} = Sp\left( {\begin{array}{*{20}c} {\lambda_{1}^{N} } & 0 \\ 0 & {\lambda_{2}^{N} } \\ \end{array} } \right) = \lambda_{1}^{N} + \lambda_{2}^{N} .$$

(A17)

λ₁ is the highest eigenvalue of V, while λ₂ is the other eigenvalue and |λ₂| < λ₁. This gives the formula of the free energy.

$$\begin{gathered} F = - TN\ln \lambda_{1} = \hfill \\ = - TN\ln \left\{ {e^{\frac{J}{T}} \cosh \left( {\frac{{\mu_{B} \mathscr{H}}}{T}} \right) + \sqrt {e^{\frac{2J}{T}} \left( {\cosh \left( {\frac{{\mu_{B} \mathscr{H}}}{T}} \right)} \right)^{2} - 2\sinh \left( \frac{2J}{T} \right)} } \right\} = \hfill \\ = - TN\ln \left\{ {e^{\frac{J}{T}} \cosh \left( {\frac{{\mu_{B} \mathscr{H}}}{T}} \right) + e^{\frac{J}{T}} \sqrt {\left( {\cosh \left( {\frac{{\mu_{B} \mathscr{H}}}{T}} \right)} \right)^{2} - 2\left( {1 - e^{{ - \frac{4J}{T}}} } \right)} } \right\} = \hfill \\ = - NT\left\{ {\frac{J}{T} + \ln \left\{ {\cosh \left( {\frac{{\mu_{B} \mathscr{H}}}{T}} \right) + \sqrt {\left( {\cosh \left( {\frac{{\mu_{B} \mathscr{H}}}{T}} \right)} \right)^{2} - 2\left( {1 - e^{{ - \frac{4J}{T}}} } \right)} } \right\}} \right\}, \hfill \\ \end{gathered}$$

(A18)

$$\begin{gathered} - NT\left\{ {\frac{J}{T} + \ln \left\{ {\cosh \left( {\frac{{\mu_{B} H}}{T}} \right) + \sqrt {\left( {\cosh \left( {\frac{{\mu_{B} H}}{T}} \right)} \right)^{2} - 2\left( {1 - e^{{ - \frac{4J}{T}}} } \right)} } \right\}} \right\}, \hfill \\ M = - \frac{\partial F}{{\partial H}} = NT\frac{{\mu_{B} }}{T}\frac{{\sinh \left( {\frac{{\mu_{B} H}}{T}} \right) + \frac{{2\cosh \left( {\frac{{\mu_{B} H}}{T}} \right)\sinh \left( {\frac{{\mu_{B} H}}{T}} \right)}}{{2\sqrt {\left( {\cosh \left( {\frac{{\mu_{B} H}}{T}} \right)} \right)^{2} - 2\left( {1 - e^{{ - \frac{4J}{T}}} } \right)} }}}}{{\left\{ {\cosh \left( {\frac{{\mu_{B} H}}{T}} \right) + \sqrt {\left( {\cosh \left( {\frac{{\mu_{B} H}}{T}} \right)} \right)^{2} - 2\left( {1 - e^{{ - \frac{4J}{T}}} } \right)} } \right\}}}, \hfill \\ \mathop {Lim}\limits_{H \to 0} M = N\mu_{B} \sinh \left( 0 \right)\frac{{1 + \frac{1}{{\sqrt {1^{2} - 2\left( {1 - e^{{ - \frac{4J}{T}}} } \right)} }}}}{{\left\{ {1 + \sqrt {1^{2} - 2\left( {1 - e^{{ - \frac{4J}{T}}} } \right)} } \right\}}} \to 0. \hfill \\ \end{gathered}$$

(A19)

In the one-dimensional Ising model, there is no remnant magnetization. Spontaneous spin polarization does not occur.