Construction of free energy using Landau theory

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Discussion Overview

The discussion revolves around the construction of free energy in the context of the Ising model, particularly focusing on antiferromagnetic systems using Landau theory. Participants explore the implications of temperature on the order parameter, the symmetry properties of the system, and the effects of external magnetic fields.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • Some participants describe the total energy of an Ising model and the implications of negative coupling constant J, leading to an antiferromagnetic ground state at T=0.
  • There is a proposal that the Landau free energy can be expressed as a polynomial in the order parameter, with specific forms for the coefficients A(T) and B(T) based on symmetry considerations.
  • One participant expresses confusion about how the temperature dependence of A(T) is determined and why B(T) is considered constant.
  • Another participant confirms that at the critical temperature, the equilibrium state corresponds to zero magnetization, and discusses the implications of this for temperatures above and below the critical temperature.
  • Questions arise regarding the effect of an external magnetic field on the Landau free energy expression, particularly concerning the symmetry of the order parameter.
  • Some participants argue that even with a preferred direction introduced by an external magnetic field, the symmetry of the order parameter for antiferromagnets remains intact.
  • Clarifications are made regarding the distinction between ferromagnetic and antiferromagnetic systems in the context of external magnetic fields and their impact on the order parameter.

Areas of Agreement / Disagreement

Participants generally agree on the basic framework of Landau theory and the behavior of the order parameter at critical temperatures. However, there are multiple competing views regarding the effects of external magnetic fields on the symmetry of the order parameter and the implications for the free energy expression.

Contextual Notes

Participants express uncertainty regarding the definitions of terms and the implications of symmetry in the context of different magnetic states. There are unresolved questions about the temperature dependence of coefficients in the free energy expression and the effects of external fields on antiferromagnetic systems.

CAF123
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Consider an Ising model system where the total energy is ##E = −J \sum_{<ij>} S_iS_j ##, ##S_i = \pm 1## and ##< ij >## implies sum over nearest neighbours. For ##J < 0## the ground state of this system at ##T = 0## is antiferromagnetic. (All adjacent spins misaligned so net magnetisation zero and thus antiferromagnetic).

Let ##m_{1,2}## be the magnetisations in the two sublattices of an antiferromagnetic Ising model. Define the order parameter ##\psi ≡ m_1−m_2##. The Landau free energy for this system should have the form, $$F = at\psi^2 + b\psi^4 + ...,$$ where ##t \equiv (T − T_c)/T_c ## and ##a, b > 0## are constants. Why is this the case?

I know that ##F## is some polynomial expansion in order parameter ##\psi## and as far as I understand from the Landau theory, it is constructed based solely on the symmetries of the system, particularly that obeyed by the order parameter. ##m_i \rightarrow -m_i## can be viewed as a rotation of the system by ##\pi## so F should beinvariant under ##m_i \rightarrow -m_i## which is to say ##\psi \rightarrow -\psi## is a symmetry. So we construct $$F = C + A(T)\psi^2 + B(T)\psi^4 + ...,$$ where ##C## is a constant, can be set to 0. I am just a bit confused as to how they obtained the expressions for ##A(T)## and ##B(T)##? Imposing that at equilibrium, ##\partial F/\partial \psi = 0## then this means $$2\psi(A(T) + 2B(T) \psi^2) = 0$$ ie ##\psi = 0## or ##\psi^2 = -A(T)/2B(T)##. The case ##\psi = 0## corresponds to the case when ##m_1 = m_2## so that this could be realized in the ground state. I just would like the argument as to why we infer the dependence of ##A## on ##T## and that ##B## is a constant.

Or by using the given expression, at the critical temperature (or temperature at which we reach criticality) the first term vanishes which implies the equilibrium situation is one in which we have ##\psi=0.## Similarly for the case T>T_c. (We must choose ##\psi=0## otherwise we get an imaginary solution for ##\psi## which is unphysical. For T<Tc we get two minima. Is this correct understanding? Thanks :)
 
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CAF123 said:
Or by using the given expression, at the critical temperature (or temperature at which we reach criticality) the first term vanishes which implies the equilibrium situation is one in which we have ##\psi=0.## Similarly for the case T>T_c. (We must choose ##\psi=0## otherwise we get an imaginary solution for ##\psi## which is unphysical. For T<Tc we get two minima. Is this correct understanding? Thanks :)
That's right. At equilibrium state, the magnitisation will be such that we have the derivative of the Free energy (with respect to magnetisation) be equal to zero. Below the critical temperature, we want to have antiferromagnetic states possible. But above the critical temperature, we want to have only the ground state of random spins. Therefore, we choose A's dependence on temperature so that this is true.
 
Hi BruceW, :)
BruceW said:
That's right. At equilibrium state, the magnitisation will be such that we have the derivative of the Free energy (with respect to magnetisation) be equal to zero. Below the critical temperature, we want to have antiferromagnetic states possible. But above the critical temperature, we want to have only the ground state of random spins. Therefore, we choose A's dependence on temperature so that this is true.
What do you mean by above ##T_c## we only have the ground state of random spins?

Another question I had is if we apply an external magnetic field, ##h## to the system , how will this affect the Landau free energy expression? I thought that if this was the case then we would no longer have the symmetry ##\psi = -\psi## since there is now a preferred direction in the system so the lowest order term that should be added is ##\psi h## but apparently this is not correct. Any thoughts here? thanks!
 
CAF123 said:
Hi BruceW, :)

What do you mean by above ##T_c## we only have the ground state of random spins?
Hi :) yes, I was just confirming what you said "we must choose psi=0 or we get an imaginary solution". In other words, above ##T_c## we must choose psi=0 or we get an imaginary solution. and psi=0 is the state where spins are just random. I guess by 'ground state' I just mean a state which is minimum of free energy. But maybe this is not correct terminology.

CAF123 said:
Another question I had is if we apply an external magnetic field, ##h## to the system , how will this affect the Landau free energy expression? I thought that if this was the case then we would no longer have the symmetry ##\psi = -\psi## since there is now a preferred direction in the system so the lowest order term that should be added is ##\psi h## but apparently this is not correct. Any thoughts here? thanks!
Even if there is a preferred direction to the system, we still have the symmetry transform ##\psi## to ## - \psi## Remember how ##\psi## is defined, as the difference between magnetisation of the two sublattices. In other words, if we paint a checkerboard pattern on to our lattice, it doesn't matter which squares we call black and which squares we call white. And this is still true even if the system does not have a directional symmetry. The two sublattices are not related to the directional symmetry of the lattice, but to the translational symmetry.
 
BruceW said:
Hi :) yes, I was just confirming what you said "we must choose psi=0 or we get an imaginary solution". In other words, above ##T_c## we must choose psi=0 or we get an imaginary solution. and psi=0 is the state where spins are just random. I guess by 'ground state' I just mean a state which is minimum of free energy. But maybe this is not correct terminology.
Ok thanks

Even if there is a preferred direction to the system, we still have the symmetry transform ##\psi## to ## - \psi## Remember how ##\psi## is defined, as the difference between magnetisation of the two sublattices. In other words, if we paint a checkerboard pattern on to our lattice, it doesn't matter which squares we call black and which squares we call white. And this is still true even if the system does not have a directional symmetry. The two sublattices are not related to the directional symmetry of the lattice, but to the translational symmetry.
If we have a system where the order parameter is just ##m## say, the magnetisation of the whole system, then the correct addition to the free energy upon imposing a magnetic field is the ##\psi h##. Why is this? Sorry maybe I didn't quite understand what you said.
 
If the order parameter was magnetisation of the whole system, then we would be trying to describe a ferromagnet, not an antiferromagnet. For a ferromagnet, yes the symmetry of ##m## to ##-m## is broken by the external magnetic field. But for an antiferromagnet, the symmetry of ##\psi## to ##- \psi## is not broken by the external magnetic field. To see why this is true, draw an 8by8 grid of squares, and color them black and white like a chessboard. The order parameter for the antiferromagnet is the spins on the white squares minus the spins on the black squares. So for ##T<T_c##, the state which minimises free energy is the state where all spins on white squares are 'up' and all spins on black squares are 'down'. So overall, the total magnetisation is zero. And for this reason, the state is not affected by an external magnetic field. The symmetry related to our definitions of black squares and white squares is not affected by an external magnetic field.
 
Hi BruceW, sorry I am still not understanding a few things.
BruceW said:
If the order parameter was magnetisation of the whole system, then we would be trying to describe a ferromagnet, not an antiferromagnet. For a ferromagnet, yes the symmetry of ##m## to ##-m## is broken by the external magnetic field.
Why?

So for ##T<T_c##, the state which minimises free energy is the state where all spins on white squares are 'up' and all spins on black squares are 'down'. So overall, the total magnetisation is zero.
This makes sense since in this configuration, the Ising energy is minimised so constitutes the ground state which is antiferromagnetic because the magnetisation vanishes. But did I not show in the above that for ##T < T_c## we have two solutions? Are these solutions one where all spins on white are up and the all spins on black down and vice versa for the other solution? I am still not sure why the application of the field doesn't change the symmetry, is it because whatever the field does to the spins on the white squares it does the opposite to the ones on the black, so overall the system is unchanged? Thanks!
 
##at## is the general first order Taylor expansion of A(T). For B(T), only the lowest order, i.e. a constant, is taken into account as B has to stay positive to guarantee the stability of the system ( at least as long as there are no higher order terms in the field).
 
CAF123 said:
Hi BruceW, sorry I am still not understanding a few things.
Why?
that's ok! For the ferromagnet, when ##T<T_c## we have a nonzero net magnetisation (up or down). Therefore, if we add an external magnetic field (up or down), there would be some energy preference of one state over the other. i.e. the symmetry between the states is lost.

CAF123 said:
This makes sense since in this configuration, the Ising energy is minimised so constitutes the ground state which is antiferromagnetic because the magnetisation vanishes. But did I not show in the above that for ##T < T_c## we have two solutions? Are these solutions one where all spins on white are up and the all spins on black down and vice versa for the other solution? I am still not sure why the application of the field doesn't change the symmetry, is it because whatever the field does to the spins on the white squares it does the opposite to the ones on the black, so overall the system is unchanged? Thanks!
yes. those are the two solutions. yeah, well the system has changed, because of the external magnetic field. But we still have the symmetry operation of turning black squares into white squares and vice versa. Yeah, you're right. the external field would affect the system in the same way if the (white squares were 'up' and black were 'down') or (black square were 'up' and white squares were 'down'). Therefore the symmetry remains.
 
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