Experimental Thermodynamics II: Open System

Experimental Thermodynamics I: Closed System

Vapor pressure

Chemical thermodynamics, chemical equilibrium, and chemical reaction rate

Band gap of periodic potential

Puzzle

Puzzle

Thermaldynamic functions

As we know, combining the first and second law of thermaldynamics, one gets

dU=TdS-PdV, (1)

where T,S,P and V are temperature, entropy, pressure and volume.

In addition, we also know that S and V are extended functions, meaning:

U(nS,nV)=nU(S,V),

which leads to Euler’s equation (the deviation can be found on the Internet):

U=TS-PV.

On the other hand, we know there are other three energy related functions, say H (enthalpy), F(free energy) and G(Gibbs free energy). Their relation to internal energy U is the following:

H=U+PV
F=U-TS
G=U-TS+PV.

Therefore, with those functions, one can describe four different process, adiabatic+isometric, adiabatic+isobaric, isothermal+isometric, isothermal+isobaric.

However, if we consider Euler’s equation, we get
H=TS
F=-PV
G=0
what’s wrong here?

Periodic potential problem

When we study solid state physics, we often deal with periodic potential, which may make you think that it is trivial. Normally, we don’t really know the function form of the potential. So approximation is made and perturbation theory is invoked, e.g. in free electron approximation and tight binding approximation.

I was surprised when I tried to solve this Schodinger equation:

d2Ψ/dx2+cos(Kx)Ψ=EΨ, here hbar, m are omitted for simplicity.

We know from Bloch theorem that the wave function takes the form

Ψ=exp(-ikx)u(x),

where u(x)=u(x+2n/K).

But to it takes me only so far. I can not find the exact solution.

Does the analytical solution exist?

RHEED simulation

Periodic function and Fourier expansion

Periodic function and Fourier expansion

The theorem is that if a function f is periodic with frequency w (period a), the function can be expanded as Fourier polynomials. i.e.

If f(x+na)=f(x)

Then f(x)=\sum_{k=nK}f(k)e^{-ikx}

where K=\frac{2\pi}{a}

.

The proof actually relies on common sense.

define:

 f(k)=\int f(x)e^{-ikx}dx

One can see that both f(x) and exp(-ikx) are perodic, where average of exp(-ikx) is zero. So if the two periods are not commensurate, f(k) will be zero.

The only possible nozero f(k) occurs when kx=2npi, where k=nK.

pydao0.979 released

Pydao, a new software for data organize and analysis is released as a trial version.

Data organization is based on hierachical data file (HDF) structure.

Data analysis is based on the plugins designed for special usage.

Data visualization is based on matplotlib and mayavi pakage.

Now I have lattice dynamics as a useful plugin.

There are some build in analysis and visualization tools, not as much as the xpy1.xx. But we will make pydao more and more complete in the future.

Right now, only source code is provided. Win32 compiled will come soon.

Surface preparation of MgO

Surface preparation of MgO

What's the use of MgO

1) Good material for hight temperature processing 2800 C melting temperature

2) low dielectric constant ~10, low loss

Problem of MgO

1) surface reacts with H2O and CO2 after exposing to air

2) impurities like Ca

Preparation steps

1) Cleaving

Normally with charged surface, difficult for scanning

2)Polishing

sub-micron polishing is normally used.

3) Acid etching

This is to remove the additional material left from the polishing. So basically, the sample as received is already etched.

agent: phosphoric and nitric-acid

4) Annealing

Most important part. This is to solve problem 1) and to make atomically flat surface.

Surface orientation	Duration	Temperature	Step height	Terrace width	Comments	Reference
(100)	2 hours	Tanneal>1000 C, the surface changes remarkably The annealing is better for higher temperature until  Tanneal>1350 C		200-300 nm	Step comes from the Ca atoms diffuse to the surface	Ahmed1996
(100)	12 hours	> 700 C can get rid of Mg(OH)2 and MgCO3 >1100C, one can get atomically smooth surface	4 nm	700 nm		Aswal2002
(100) 2 degree miscut	360 min	1000	3-7 nm	120 nm		Benedetti2007
(110)	10 min	1000 in 10-7 torr			facet created because of liquid solid interface in the etching	Giese2000
(111)	30 min	1700 to heal the facet			The surface may reorganize into (332) 1700 C to heal the facet

Reference

[Ahmed1996] Ahmed F. et al J. of Low Temperature v105, p1343 (1996)

[Plass1998] Plass R. Surface Science, v414, p 268 (1998)

[Giese2000] Giese D.R. Surface Science, v 457, p 326 (2000)

[Aswal2002] Aswal D.K et al. Journal of Crystal Growth, v236, p.661. (2002)

[Benedetti2007] Benedetti S et al. Surface Science v601, p 2636. (2007) 

Lattice constant

Lattice constant of some triangular lattice materials

Material	Lattice Constant	Crystal Structure
LuFe2O4	3.441 (LuO) / 2.086 (FeO)
SiC	a=3.086; c=15.117	Wurtzite
Si	5.43095	Diamond
C	3.56683	Diamond
GaN	a=3.189; c=5.185	Wurtzite
ZnO	3.249	P 63 m c
SiO2	4.916	Quantz

Effective number of oscillators and B...

Effective number of oscillators and Born effective charge

The operational definition of effective number of oscillators is:

n_{eff}=\frac{2}{\pi\omega_p^2}\int_{\omega_1}^{\omega_2}{\epsilon_2\omega d\omega

where 

\omega_p^2=\frac{e^2}{V_0m\epsilon_0}

here e is electronic charge, V_0 is the unit cell volume, m is the reduced mass of the oscillator and \epsilon_0 is the vacuum dielectric constant.

If we introduce (Born) effective charge q_{eff}, we also get

n_{eff} =\frac{Nq_{eff}^2V_0\mu\epsilon_0}{e^2V_0m\epsilon_0} = N(\frac{q_{eff}}{e})^2 \frac{m}{\mu}

LFO

LuFe2O4

BFO

LFO

Introduction to multiferroics

Introduction to multiferroics

Landau theory of charge order: symmet...

Landau theory of charge order: symmetry breaking

Landau theory is a good way to describe the symmetry breaking of of charge order using group theory.

Suppose the charge pattern can be described as

\rho(\vec{r})=\rho_0(\vec{r})+\delta\rho(\vec{r})

where ρ is the total charge density, ρ₀ is the charge density of the high temperature phase corresponding to symmetry group G₀ and δρ correspond to one or more irreducible representations (except the identity) of G₀.

At high temperature δρ=0, giving the high symmetry phase. While at low temperature, δρ≠0, giving the low symmetry phase.

A good example of cubic to tetragonal structural phase transition can be very revealling.

In this case:

\rho(\vec{r})=\sum_i\rho_0(\vec{r}-\vec{R}_i)+\rho_1(\vec{r}-\vec{R}_i+z_0)

where R_i represents the cubic lattice and z₀ is the distortion along z direction.

One can look at the character table (below) and recognize that the z₀ distortion corresponds T_1u irreducible representation (IR).



A closer look at the character table shows that only C₄, σ_h and σ_d leave z₀ invariant. Therefore, in the low temperature (low symmetry phase), the group has E, C₄, σ_d and σ_h, which gives a new group C_4v, corresponding to the symmetry of the low T phase.

Useful special symbols

Useful special symbols

Greek letters	α	β	γ	δ	ε	ζ	η	θ	ι	κ
	λ	μ	ν	ξ	ο	π	ρ	ς	σ	τ
	υ	φ	χ	ψ	ω
	Α	Β	Γ	Δ	Ε	Ζ	Η	Θ	Ι	Κ
	Λ	Μ	Ν	Ξ	Ο	Π	Ρ		Σ	Τ
	Υ	Φ	Χ	Ψ	Ω
Roman letters	ℂ	℃	ℇ	℉	ℐ	ℑ	ℒ	ℏ	℞	Ω
	ℋ	ℛ	ℱ	Å	ℰ
	𝓖
Additional Greek letters	ħ	ϴ	ϵ
Arrows	↔	↕	￫	￩	￬	￪	→	←	↓	↑
	↙	↘	↗	￨
Operators	±	∓	×	÷	∙	†	≈	≣	≡	≠
	∫	∏	∑	∆	∇	∂	∅
	√	∛	∜
	∝	∞
Geometry	‖	⊥
Fractions	¼	½	¾
Units	Å

Conductivity of conductor

Conductivity of conductor

How to define a conductor? The very natural way to say that a conductor is a material that conducts electric current. But for a physicist, that's not enough, most materials do have a measurable conductivity, just the numbers vary by 20 order of magnitude. It is hard to draw the lines between conductors, semiconductors and insulators. However, some good examples (300 K) should be able to at least give us some idea.

Material	Conductivity Sm-1 or 1/(Ωm)	Ref
Silver	63.0 × 106
Copper	59.6 × 106
Gold	45.2 × 106
Mercury	1.0× 106
Carbon	2.8 × 104
Fe3O4	103
LuFe2O4	~1	[1] After breakdown 60V/cm-1
Germanium	2.2
LuFe2O4	10-2	[1] Before breakdown 10V/cm-1
Silicon	1.5× 10-3
BiFeO3	~1× 10-3	[2] 1kV/cm-1
Deionized water	5.5 × 10-6
Glass	10-10 -10-14
Paraffin	10-17
Teflon	10-22 - 10-24

Reference:
[1] Title: Nonlinear current-voltage behavior and electrically driven phase transition in charge-frustrated LuFe2O4
Author(s): Zeng LJ, Yang HX, Zhang Y, et al.
Source: EPL   Volume: 84   Issue: 5 Article Number: 57011   Published: DEC 2008

[2]Title: Switchable Ferroelectric Diode and Photovoltaic Effect in BiFeO3
Author(s): Choi T, Lee S, Choi YJ, et al.
Source: SCIENCE   Volume: 324   Issue: 5923   Pages: 63-66   Published: APR 3 2009

Frustrated spin triangles

Frustrated Ising spin triangles

When we talk about spin liquid, one often invoke the frustrated spin system, for which the classical example is the Ising spin triangle with antiferromagnetic interaction.

Let's then look at such a model system see how it behaves, which will be very revealing for understanding more complicated system.

1. Hamiltonian and basis

Suppose there is a (localized) spin triangle of sites A,B,C, in the language of second quantization, there are 8 possible states, which we take as the basis:

$\phi_1 =|\uparrow_A\uparrow_B\uparrow_C>$;
$\phi_2 =|\uparrow_A\uparrow_B\downarrow_C>$;
$\phi_3 =|\uparrow_A\downarrow_B\uparrow_C>$;
$\phi_4 =|\uparrow_A\downarrow_B\downarrow_C>$;
$\phi_5 =|\downarrow_A\uparrow_B\uparrow_C>$;
$\phi_6 =|\downarrow_A\uparrow_B\downarrow_C>$;
$\phi_7 =|\downarrow_A\downarrow_B\uparrow_C>$;
$\phi_8 =|\downarrow_A\downarrow_B\downarrow_C>$;

The spin Hamiltonian will be:
$H=\frac{1}{2}[\sum_{i \ne j}^{}J_{ij} S^z_iS^z_j+(S^+_iS^-j+S^-_iS^+_j)/2]$

2. Eigenstates

We can diagonalize the Hamitonian and get the eigenstates and eigenenergies.

$\xi _i=\sum_{j}^{}c_{ij}\phi_{j}$


Table:

	E=-3/4J				E=3/4J
	ξ1	ξ2	ξ3	ξ4	ξ5	ξ6	ξ7	ξ8
ci1	0.000	0.000	0.000	0.000	1.000	0.000	0.000	0.000
ci2	-0.816	0.000	0.000	0.000	0.000	0.577	0.000	0.000
ci3	0.408	0.707	0.000	0.000	0.000	0.577	0.000	0.000
ci4	0.000	0.000	0.000	0.816	0.000	0.000	-0.577	0.000
ci5	0.408	-0.707	0.000	0.000	0.000	0.577	0.000	0.000
ci6	0.000	0.000	0.707	-0.408	0.000	0.000	-0.577	0.000
ci7	0.000	0.000	-0.707	-0.408	0.000	0.000	-0.577	0.000
ci8	0.000	0.000	0.000	0.000	0.000	0.000	0.000	1.000

It turns out that the system become two energy subspaces.

The first subspace corresponds to degenerate ground states with energy -3/4J.
One can see that |S_z|= 0 for those states.

The second subspace corresponds to a spin quartet with S=3/2.

3. Field and temperature dependence:  magnetization plateau

By varying temperature and magnetic field, one can study the magnetization change.
Above is the result we found for different J value. Starting from left are J=0,1,2,3,4,5,6,7,9,10,11

1) J=0 corresponds to isolated spins, which is actually Brolluvin function
2) J>=9: there is a clear plateau at low magnetization which corresponds to the saturation magnetization of the first subspace of the eigenstates.
3) J=1-7: intermediate cases

4. Field and temperature dependence.

In real experiment, one can not vary J unfortunately. Instead, we only have control over B and T. Next, we show the B,T dependence of the magnetization with fixed J.

In this picture, we can see that at very low temperature, the two plateau is obvious.

One can show that a system with 6 Ising spin which form triangular lattice also has similar behavior, in the sense that there is a low magnetization plateau of 1/3 of the magnitude. The 10 Ising spin system is not calculable for me however due to the computational difficulty but one can imaging the similarity. The key is that for the frustrate the spin system, there is a ground state with huge degeneracy which can behave like a paramagnetic system with reduced spin magnitude.

5. specific heat

As shown in the above figure (J=1,2,3 from the left), the specific heat has a peak at a energy scale proportional to the exchange interaction.

This is actually very typical behavior of two-level system.

6. specific field in magnetic fields

From left to right, the curves correspond to different magnetic fields. One can see that at intermediate fields, there are two peaks in the specific heat.