Magnons contribution to spontaneous magnetization

Question

In Statistical Physics part II of Landau's course in theoretical physics it is stated that the magnon part of the spontaneous magnetization can be calculated as $$ M_m \equiv M(T) - M(0) = -\frac{1}{V} \left(\frac{\partial \Omega_m}{\partial \mathfrak{H}}\right)_{\mathfrak{H}\to0} $$ where $M_m$ is the spontaneous magnetization, $M$ the magnetization, $\Omega_m$ the grand-potential and $\mathfrak{H}$ the external magnetic field applied to the body.

Landau gives as reference his own book on electrodynamics of continuous media, and in particular to the formula $$ dF= -SdT+ \zeta d\rho + \mathbf{H}\cdot \frac{d\mathbf{B}}{4\pi}, $$ where $\zeta$ is the chemical potential.

Now, I can see from this that reasonably $$ M=-\frac{1}{V} \left(\frac{\partial \Omega_m}{\partial \mathfrak{H}}\right) $$ but where does the expression for $M_m$ above come from?

I tried $\Omega = F - \zeta \rho$ and $$ d\Omega= -SdT- \rho d\zeta + \mathbf{H}\cdot \frac{d\mathbf{B}}{4\pi} $$ setting $\Omega_m=\Omega-\mathbf{H}\cdot\mathbf{B}/4\pi$ we get $$ d\tilde{\Omega}=-SdT- \rho d\zeta + \mathbf{B}\cdot \frac{d\mathbf{H}}{4\pi}. $$ Using $\mathbf{B}=\mathbf{H}+4\pi \mathbf{M}$ we have $$ d\Omega= -SdT- \rho d\zeta + \mathbf{B}\cdot \frac{d\mathbf{B}}{4\pi} - \mathbf{M}\cdot d\mathbf{B} = -SdT- \rho d\zeta + \frac{d\mathbf{B}^2}{8\pi} - \mathbf{M}\cdot d\mathbf{B} $$ $$ d\tilde{\Omega}=-SdT- \rho d\zeta + \frac{d\mathbf{H^2}}{8\pi} + \mathbf{M}\cdot d\mathbf{H} $$ hence $$ \mathbf{M}=-\frac{\partial \Omega}{\partial \mathbf{B}} =+\frac{\partial \tilde{\Omega}}{\partial \mathbf{H}}. $$ This still does not explain why $M(T)-M(0)$ and why $\mathfrak{H}\to0$, even though this fact is clear from the intuitive notion of spontaneous magnetization.

Answer 1

You should not use the free energy $$ dF= -SdT+\ldots + \mathbf{H}\cdot \frac{d\mathbf{B}}{4\pi}. $$ Instead, as you said in your post, it is the following thermodynamic potential that matters $$ d\Omega_m= -SdT+\ldots - (\mathbf{M}- \mathbf{M(0)})\cdot d\mathbf{H} . $$ This is because in a magnetic system, what you can control is $\mathbf{H}$ not $\mathbf{B}$. Please read more in Electrodynamics of Continuous Media. Notice that I have subtracted $\mathbf{M(0)}$ because he wants to calculate the magnon contribution to the spontaneous magnetization.
Then keeping $T$ constant, we have $$ M_m \equiv M(T) - M(0) = -\frac{1}{V} \left(\frac{\partial \Omega_m}{\partial \mathfrak{H}}\right)_{\mathfrak{H}\to0}. $$

Answer 2

In spontanious symmetry breaking we minimize lagrangian of zero temperature. These lagrangians are phenomenological lagrangians like Landau or Gizburg-landau or something like mean field. After we found a minimum we expand the previous lagrangian around one of the minima (optional(spontanious symmetry breaking)) we get to "mass term" for one field and we see that some field lost their mass. These massless particles play a great role in conveying long range order or collective mode.

Some of the well-known massless particles are phonons in superfluid He-4 and Magnons in magnetism or 3-D ising model. People call them Goldstone modes ( collective behavior(primary excitations) caused by waving of massless particles).

All in all, we have two mechanisms of symmetry breaking in physics. One is Goldstone mechanism and the second is Higgs mechanism. First one for short range forces like between two spin : $\sum_{ij}J S_i\cdot S_j$, and second kind longe range forces : $$\sum_{ij}\frac{e^2}{(r_i-r_j)^2}.$$