Eigenstates and partition function of the Ising model at mean field level


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6 months ago by

According to the H. Bruus's textbook (electronic version), page 73:

The effective model to study crystals of magnetic ions is the Heisenberg model. In this model, the source of interaction originates from exchange spin-spin interaction between the magnetic ions.   It reads

  $H=-2\sum_{i,j=1}^NJ_{ij}\vec{S_i}\cdot\vec{S_j}$H=2i,j=1NJijSi·Sj   ,

where  $\vec{S_i}$Si  is the spin operator of the ion at site  $i$i and  $J_{ij}$Jij is the overlap integral or exchange integral, measuring the strength of the interaction. Since the spin-spin interaction is very short, we can simplify it to a constant value  $J_0$J0 for the nearest neighbors and zero for any site beyond.
This model can easily explain a ferromagnetic chain for  $J_{ij}>0$Jij>0  and antiferromagnet for  $J_{ij}<0$Jij<0, depending on the operator  $\vec{S_i}\cdot\vec{S_j}$Si·Sj taking positive or negative values, respectively. By writing the spin operators  $\vec{S_i}$Si and $\vec{S_j}$Sj in terms of fermionic operators and the Pauli matrices,

$\vec{S_i}=\sum_{\alpha,\beta}c_{\alpha}^{\dagger}\vec{\sigma_{\alpha\beta}}c_{\beta}$Si=α,βcασαβcβ ,

one immediately see that the Heisenberg model, which is an effective model in essence, have quartic operators. This prevents one to study the system in general, and further treatments would be necessary. Therefore, as a first step can study the system at mean field.
By applying the mean field decomposition scheme

$\vec{S_i}=<\vec{S_i}>+\delta\vec{S_i}$Si=<Si>+δSi ,

to the Heisenberg model Hamiltonian, one can find

$H_{MF}=-2J_0\sum_{ij}<\vec{S_i}>\cdot\vec{S_j}-2J_0\sum_{ij}\vec{S_i}\cdot<\vec{S_j}>+2J_0\sum_{ij}<\vec{S_i}>\cdot<\vec{S_j}>$HMF=2J0ij<Si>·Sj2J0ijSi·<Sj>+2J0ij<Si>·<Sj>.

Again, we encounter the concept of symmetry breaking. We can easily see that if the system is completely disordered, then the Heisenberg mean field parameter$<\vec{S_i}>$<Si> is vanishing. This is because, in the disordered phase, the spin of the magnetic ion can take equal value in each direction of the spin space, or more explicitly

  $<\vec{S_x}>=<\vec{S_{-x}}>$<Sx >=<Sx> ,   $<\vec{S_y}>=<\vec{S_{-y}}>$<Sy>=<Sy>  and  $<\vec{S_z}>=<\vec{S_{-z}}>$<Sz>=<Sz> .

(such configuration is a symmetry in essence and the word disordered phase seems misleading to me!)

To allow the mean field parameter \( <\vec{S}_i> \) to take nonzero value one would need to break the aforementioned symmetry in the spin space, and let the spin of the magnetic ion take a preferred direction, say \( z \)- axis. This explicit assumption, symmetry breaking, that survive the mean field parameter, is called mean field assumption! (according to the text, page 74).

\( <\vec{S}_i > = < \vec{S}_z > e_z \),

where due to the translational symmetry argument, the expectation value of the spin operator, mean field parameter, is taken to be the same on all lattice sites.  One can further simplify the mean field Hamiltonian by defining the local magnetic moment  \( \vec{m} \) that spin \( \vec{S}_i \) feels at site i, due to the other spin at site j as follows

\( \vec{m} = 2 \sum_j \, J_{ij} <\vec{S}_j> =
2 \sum_j \, J_0 < \vec{S}_z > e_z =
2 n J_0 < \vec{S}_z > e_z \)

where sum  j , runs over the nearest neighbor spins around site i, resulting in n sites.
Thus one can write the mean field Hamiltonian in terms of the magnetic moment as follows

\( H_{MF} = -\sum_i \vec{m} \cdot \vec{S}_i - \sum_j \vec{m} \cdot \vec{S_j}
+ \sum_i \, m <\vec{S_z}> \)
\( \rightarrow H_{MF} = -2 \sum_i \vec{m}\cdot \vec{S}_i
+ m N <\vec{S}_z> \)

where N is the total number of sites and \( m= \left| \vec{m} \right| \) is size of the magnetic moment that every spin feels.

Now we come to questions:
1) What is the eigenstate of the mean field Hamiltonian before and after symmetry breaking?
2) After obtaining the wave function, how to write the mean field partition function?

#The Art of Mean-Field Theory

This is a very important question; yet, before providing an answer, there are two points:

-- Your first equation for the Hamiltonian needs a minus sign, according to your conventions for anti-/ferromagnetism.

-- "We can easily see that if the system is completely asymmetrical or disordered then the Heisenberg mean field parameter is vanishing."

You call a disordered phase "completely asymmetrical", but you argue for its symmetry just a few lines later. There is some inconsistency (?).
written 6 months ago by AlQuemist  
I removed the world "asymmetry", which I supposed it is a synonym of disorder.
written 6 months ago by Claude Shannon  

2 Answers


2
6 months ago by
Spontaneous symmetry breaking (SSB)
[This is the first part of my answer, since your question is multi-faceted.]

“Whenever the ground state of a given physical system fails to exhibit a symmetry that is present in the fundamental equations of that system, it is said that this symmetry has spontaneously been broken.” [Muñoz-Vega, et al. “Spontaneous Symmetry Breakdown in non-relativistic Quantum Mechanics”, Am. J. Phys. 80.10 (2012) 891--897, arXiv:1205.4773].

Consider the Ising model with the Hamiltonian \( H = -J \sum_{\langle i , j \rangle} S_i^z \, S_j^z - h \sum_i S_i^z \) for the ferromagnetic (FM) case with \( J > 0 \). The Hamiltonian has a discrete \( \mathbb{Z}_2 \) symmetry; ie., it is invariant under a transformation
\[
S_i^z \mapsto -S_i^z ~,~ \forall i ~.
\]

Possible ground-states in FM case

* A paramagnetic (PM) state is defined as
\[
\langle S_i^z \rangle = 0 ~,~ \forall i
\]
* A ferromagnetic (FM) state is defined as
\[
\langle S_i^z \rangle = \langle S^z \rangle \neq 0 ~,~ \forall i ~,
\]
where
\[
\langle S_i^z \rangle = \sum_{n: \text{configurations} } S_{n,i}^z \, e^{-\beta \, E_n[\{S_{n,i}^z\}]}
\]

To see how spontaneous symmetry breaking occurs, we should study the analyticity of \( m(h) := \langle S_i^z (h) \rangle \) at \( h = 0 \), where \( h \) is an external magnetic field and we have assumed translational invariance on the lattice.

Considering
\[
m_+ := \lim_{h \rightarrow 0^+} m(h) \equiv \lim_{h \rightarrow 0^+} \langle S^z (h) \rangle \\
m_- := \lim_{h \rightarrow 0^-} m(h) \equiv \lim_{h \rightarrow 0^-} \langle S^z (h) \rangle ~,
\]
we readily notice that, due to the \( \mathbb{Z}_2 \) symmetry of the Hamiltonian (and \( E_n \)),
\[
m_+ = - m_- ~. \tag{1}
\]
Provided \( m(h) \) is analytic at \( h = 0 \), then by definition, we should have
\[
m_+ = m_- ~. \tag{2}
\]
Putting (1) and (2) together, we see that
\[
m(h=0) = \langle S^z (h) \rangle = 0 ~;
\]
therefore, the system should be always in a PM phase.

Nonetheless, we observe FM phases, therefore the analyticity is a wrong assumption. In fact, when the system size goes to infinity (at the “thermodynamic limit”), the partition function, the free energy, and their derivatives can become non-analytic in a certain parametre range (This implies a phase transition according to the Ehrenfest classification). That means in turn, that the ground-state (or low-energy states) might have a lower symmetry compared to the Hamiltonian. For instance, in the case of Ising model, the ground-state lacks the \( \mathbb{Z}_2 \) symmetry: a spontaneous breaking of symmetry has happened, so to say.

The good thing is that we can quantify all of these, most simply on the mean-field level -- but we can indeed use higher approximations to show this.
SSB is a modern way of understanding continuous phase transitions à la Landau. Yet, not all phase transition fit into this paradigm. Personally, I think it is much better to think of phase transitions in terms of entropy, correlations and entanglement (essentially, microscopics), rather than Landau's SSB.
1
6 months ago by
2) Microscopic view of phase transitions
[This is the 2nd part of my answer.]

To have a more quantitative idea of how phase transitions happen _microscopically_, we can consider the mean-field approximation to the Ising 1D chain for \( S= \frac{1}{2} \),

\(
H_{MF} = - h_{eff} \sum_{i \in \{1, \cdots, N\} } S_i^z - J N m^2 ~,
\)

with \( h_{eff} := J m + h \), where \( N \) is the number of lattice sites, \( J > 0 \) is the coupling, \( h \) is the external magnetic field, \( S_i^z = \pm 1 \) and \( m = \langle S \rangle \) is the mean-field. The derivation of the mean-field can be found in any modern statistical physics textbook and will not be repeated here.

\( H_{MF} \) is an _effective_ Hamiltonian for the system at different temperatures; that means, within this approximation, at each temperature, the original Ising model,

\( H_{Ising} = -J \sum_{\langle i, j \rangle} S_i^z \, S_j^z - h \sum_{i} S_i^z \)

behaves like a _non_-interacting system with an effective applied field \( h_{eff} \). Hence, let's consider the eigenstates and eigenenergies of \( H_ {MF} \).

The eigenstates are simply spin configurations of size \( N \) = \( N_\uparrow \) + \( N_\downarrow \). The corresponding eigenenergies will be

\(
\begin{align}
E[N_\uparrow, N_\downarrow] &= - (N_\uparrow - N_\downarrow) \, h_{eff} \notag \\
&= - (N - 2N_\downarrow) \, h_{eff} ~. \notag
\end{align}
\tag{1}
\)

More explicitly, the eigenstates are

\(
\begin{align}
\uparrow \uparrow \uparrow \uparrow \uparrow \quad&:
E_0 = -N \, h_{eff} ~,~
\text{degeneracy} = 1 \notag \\
\uparrow \uparrow \downarrow \uparrow \uparrow \quad&:
E_1 = -(N-2) \, h_{eff} ~,~
\text{degeneracy} = {}^NC_1 \notag \\
\downarrow \uparrow \downarrow \uparrow \uparrow \quad&:
E_2 = -(N-4) \, h_{eff} ~,~
\text{degeneracy} = {}^NC_2 \notag \\
\cdots & \notag \\
\downarrow \downarrow \downarrow \downarrow \downarrow \quad&:
E_F = N \, h_{eff} ~,~
\text{degeneracy} = 1 ~, \notag
\end{align}
\tag{2}
\)

where a representative of each set of degenerate eigenstates, their energy and degeneracy are given, from the ground-state up to the highest excited state. Notice that the energies are separated by a certain _mean-field gap_,

\(
E_{gap} = 2 h_{eff} ~,
\tag{3}
\)

where \( h_{eff} \) is **proportional to the mean-field** \( m \), in absence of external magnetic fields:

\(
E_{gap}\vert_{h = 0} \propto m
\tag{4}
\)

Now the picture is clear:
Deep inside the ordered phase, \( m \neq 0 \), there will be a _finite_ energy gap between the ground-state and the first excited state. This gap protects the mean-field state or configuration from variations – since such variations are energetically costly. When we approach the phase transition, \( m \rightarrow 0 \), and the gap vanishes. That implies the ground-state becomes degenerate (so to say, the 1st excited state collapses onto the ground-state, \( E_1 \rightarrow E_0 \) ), and thus, the **nature of the ground-state changes radically**. This is the general meaning of a phase transition.

Furthermore, near the phase transition, the gap is small, therefore, the energy cost of excitations is very low; this is precisely the reason for the increase in “fluctuations” which are merely transitions to the excited states above the mean-field ground-state. This also leads to an increase in entropy and hence, disorder.

All together, this provides a generic microscopic picture for phase transitions of any kind – a view which I prefer to Landau's symmetry-based view.
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