Friday, December 11, 2009

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
5.43095Diamond

C
3.56683Diamond

GaNa=3.189;
c=5.185
Wurtzite

ZnO3.249P 63 m c

SiO2
4.916Quantz









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


eq=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}

eq=\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}


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






Tuesday, September 29, 2009

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})


eq=\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)

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

where Ri 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 T1u irreducible representation (IR).



A closer look at the character table shows that only C4, σh and σd leave z0 invariant. Therefore, in the low temperature (low symmetry phase), the group has E, C4, σd and σh, which gives a new group C4v, 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






























Wednesday, September 16, 2009

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

LuFe2O410-2[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

Tuesday, September 1, 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:

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



eq=
\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



ξ1ξ2ξ3ξ4ξ5ξ6ξ7ξ8
ci10.0000.0000.0000.0001.0000.0000.0000.000
ci2-0.8160.0000.0000.0000.0000.5770.0000.000
ci30.4080.7070.0000.0000.0000.5770.0000.000
ci40.0000.0000.0000.8160.0000.000-0.5770.000
ci50.408-0.7070.0000.0000.0000.5770.0000.000
ci60.0000.0000.707-0.4080.0000.000-0.5770.000
ci70.0000.000-0.707-0.4080.0000.000-0.5770.000
ci80.0000.0000.0000.0000.0000.0000.0001.000






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

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 |Sz|= 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.



Friday, August 28, 2009

Python real time class methods

Python real time class methods

Python is a great programming language with countless merits. However, one short coming (compared with Matlab e.g.) is that one often has to restart the program in order to refresh the modified the code, which is kind of a hassle. Below, I will show what I did to circumvent this problem. Basically, I define the class method in another file, say methods.py, which is different from the file that contains the definition of the class, say myclass.py. Then I found a way to reload (or redefine) all the instance (or class) methods.

class MyClass():
    def __init__(self):
        import methods as methods;
        self.add_extended_methods(methods);
        pass;

    def add_extended_methods(self, methods):
        import inspect;      
        if not methods.lazy:
            reload(methods);
        self.methods=methods;
        
        modulefile=inspect.getfile(methods);
        modulefile=modulefile.replace('.pyc','.py');
        
        for k in methods.__dict__.keys():
            expr="methods."+k;
            v=eval(expr);
            if inspect.isfunction(v) and inspect.getfile(v)==modulefile:
                cmd="def "+k+"(self,*args, **kwargs):return self.methods."+k+"(self,*args, **kwargs);"
                exec(cmd);
                import new;
                cmd="self."+k+"=new.instancemethod("+k+", self, MyClass)";
                exec(cmd);

One can also call the "add_extended_methods" anytime when he wants to use updated methods.

This saved me a lot of time from restarting the whole Python/Program

Wednesday, August 12, 2009

Parameters of multiferroics:

Parameters of multiferroics:

Type
Name
TC
(K)
TN
(K)
Polarization
(uC/cm^2)
Critical field
Other critical temperatures



I.1 Lone pair FEBiFeO3
1100
640
100
20 T quenching spiral,
10 T spiral rotation [Xu2009]
140 K spin reorientation

































I.2 Geometric FEYMnO3
1200
42
5

























II.1 Valence order of magnetic ionsLuFe2O4
320
240
25
see phase diagram of [Xu2008]


































III.1 Spiral MagnetsTbMnO341
28
6e-2

























III.2 frustrated collinear MCa3CoMnO616.5
16.5
9e-2



























[Xu2008] Charge Order, Dynamics, and Magnetostructural Transition in Multiferroic LuFe2O4

Author(s): Xu XS, Angst M, Brinzari TV, et al.
Source: PHYSICAL REVIEW LETTERS   Volume: 101   Issue: 22 Article Number: 227602   Published: NOV 28 2008


[Xu2009] Optical properties and magnetochromism in multiferroic BiFeO3

Author(s): Xu XS, Brinzari TV, Lee S, et al.
Source: PHYSICAL REVIEW B   Volume: 79   Issue: 13 Article Number: 134425   Published: APR 2009

classes of multiferroics

category of multiferroics


Type-I
FE, M order from different subsystem
Tc>>TN
Type-II
FE, M order from same subsystem, but independent origin

Tc>TN
Type-III
FE caused by M order
TC<=TN
SubType:I.1 Lone pair FE
II.1 Valence order of magnetic ions
III.1 Spiral Magnets
Example:
BiFeO3
TC: 1100
TN: 640 K
P: 100uC/cm^2 [Xu2009Ref]

Intro: perovskite, A disp along (111)
LuFe2O4
TCO: 320 K
TN: 240 K  [Xu2008]
P: 25 uC/cm2
TbMnO3
TN=41K
TC=28K
P: 6e-2 [Kimura2007]

Intro: inverse Dzyaloshinskii–Moriya effect. P~Qxe, where three symbols are ploarization, spiral propagation direction and normal vector of the spiral plan
Other Examples:
{BiMnO3, PbVO3}

Ni3V2O6, MnWO4, CuO
orthorhombic RMnO3 (R=Tb,Dy)
SubType:I.2 Geometric FE
III.2 frustrated collinear M
ExampleYMnO3
TC: 1200 K
TN: 42 K
P: 5 μC/cm^2 [VanAken2004]
Intro: perovskite, BO6 title, then A goes up and down asymmetrically

Ca3CoMnO6
TN, TC=16.5 K,
P: 9e-2 uC/cm2

Intro: magnetostriction
Other Examples:hexagonal RMnO3 (R=Ho-Lu)
RMn2O5 (R=Pr-Lu, Bi,Y)



[Kimura2007] Spiral magnets as magnetoelectrics
Author(s): Kimura T
Source: ANNUAL REVIEW OF MATERIALS RESEARCH   Volume: 37   Pages: 387-413   Published: 2007


[VanAken2004] The origin of ferroelectricity in magnetoelectric YMnO3

Author(s): Van Aken BB, Palstra TTM, Filippetti A, et al.
Source: NATURE MATERIALS   Volume: 3   Issue: 3   Pages: 164-170   Published: MAR 2004


[Xu2008] Charge Order, Dynamics, and Magnetostructural Transition in Multiferroic LuFe2O4

Author(s): Xu XS, Angst M, Brinzari TV, et al.
Source: PHYSICAL REVIEW LETTERS   Volume: 101   Issue: 22 Article Number: 227602   Published: NOV 28 2008


[Xu2009] Optical properties and magnetochromism in multiferroic BiFeO3

Author(s): Xu XS, Brinzari TV, Lee S, et al.
Source: PHYSICAL REVIEW B   Volume: 79   Issue: 13 Article Number: 134425   Published: APR 2009

Tuesday, June 23, 2009

Python, global function

Python, global function

Problem

1) One want to organize the code using modules (files)

2) One wan packages to see other packages (at least a global function that is aware of all the definition of classes and can be used by all module)

3) One can not using from xx import * in every package

The solution is dynamic importing (which is one of the beauty of Python)

We can

1) define the global function, say, isa() in __ini__.py in the package's root directory

2) import in the METHOD (not __init__ method or as a module import) where you want to use.

Monday, June 8, 2009

Xpy version 1.47 is released

See http://xpydaov.spaces.live.com/

Electric dipole-dipole interaction

Electric dipole-dipole interaction between f electron atoms


Estimate of electric dipole moment:

According to Yen et al.[1], the electric dipole moment can be estimated as:
 
p~e<r>Vc/D,

where e is the elctronic charge, <r>~10-10 m is the mean radius of the 4f electronic orbit, Vc~500 cm-1 is the characteristic crystal field splitting, and D~60000 cm-1 [2] is the energy separation between (4f)n and (4f)n-1(5d)' configuration.

Then p~1.3e-31 C.m

Then the dipole-dipole interaction is

W=\frac{ p_1 p_2}{4\pi \epsilon_0 R^3}


eq=W=\frac{ p_1 p_2}{4\pi \epsilon_0 R^3}

For example, if R=0.42 nm, we get

W=1.2 e-5 eV, or 0.01 meV.



[1]    W. M. Yen et al., Physical Review 140, 1188 (1965).
[2]    G. H. Dieke, and H. M. Crosswhite, Applied Optics 2, 675 (1963).

Friday, June 5, 2009

Transfer helium from (100 Liter) Dewa...

Transfer helium from (100 Liter) Dewar to bolometer

In order to transfer helium smoothly, we need to position the bolometer at a good altitude, which depends on the dimensions of the Dewars and the trasfer line. Here is an illustration of how to determine how high the bottom of the bolometer should be.


Thursday, May 21, 2009

From oscillator strength to spontaneo...

From oscillator strength to spontaneous decay rate:

Approach I


Using
quantum mechanics, one can derive the relation between the oscillator
strength calculated from the absorption coefficient and the transition
matrix element |x|2. And the spontaneous decay rate is also decided by the matrix element |x|2, therefore, one can find the decay rate from the absorption coefficient.





f=\frac{2m\omega}{\hbar}|x|^2



eq=<br />f=\frac{2m\omega}{\hbar}|x|^2



while spontaneous decay rate:
A=\frac{e\omega^3}{3\pi\hbar c^3\epsilon_0 n^2}|x|^2



eq=A=\frac{e\omega^3}{3\pi\hbar c^3\epsilon_0 n^2}|x|^2






Therefore,




\frac{A}{f}=\frac{\omega^2e^2}{6m\pi c^3\epsilon_0n^2}







eq=\frac{A}{f}=\frac{\omega^2e^2}{6m\pi c^3\epsilon_0n^2}




Approach II: Judd-Ofelt theory,


f=\frac{2m\omega}{\hbar}S



eq=f=\frac{2m\omega}{\hbar}S



A=\frac{\omega^3e^2n^2}{\hbar c^3\pi\epsilon_0}S



eq=A=\frac{\omega^3e^2n^2}{\hbar c^3\pi\epsilon_0}S



Therefore,

\frac{A}{f}=\frac{\omega^2e^2n^2}{2mc^3\pi\epsilon_0}






eq=\frac{A}{f}=\frac{\omega^2e^2n^2}{2mc^3\pi\epsilon_0}

One can see that, in two approaches, there is an factor of 3n^4.

I am not so sure why two are not the same, but at this moment I tend to believe the second approach because it is specialized just for this case.

Usage

In a real experiment, oscillator strength can be found using partial sum rule:
The dimensionless oscillator strength is:
f=\frac{2c}{N_e\hbar \pi \omega_p^2}\int_{E_2}^{E_1}n\alpha dE

eq=f=\frac{2c}{N_e\hbar \pi \omega_p^2}\int_{E_2}^{E_1}n\alpha dE

Here
\omega_p^2=\frac{e^2 \rho}{m\epsilon_0}





eq=\omega_p^2=\frac{e^2 \rho}{m\epsilon_0}

[Reference:]

Kumar G.A. et al, Journal of Luminescence vol. 99, p. 141-148 (2002)







Conservation of

Conservation of Σλi

Description:

For a Helbert space (wave function subspace) {φi} defined by the operator O whose eigenvalues are λi, corresponding to wavefunction  φi.

Simply from linear algebra,
Σλi is actually the trace of the operator matrix O, which is of course conserved, no matter what unitary transformation of the {φi} is taken to form new set of basis, say {ψi}.

To generalize this, Σλin, is also conserved, because On also defines the space.

Example:

For a 4f electron multiplet, say, Nd3+, 4I9/2, ΣJz,in is conserved and of course Σμz,in is conserved too because they differ only by a factor of gμB

Just to give a few numbers,

sqrt(3Σμz,i2/5)=

3.6 μB(theoretical value)
3.4, 3.5 and 3.8 from measurement on crystal field split 4I9/2 multiplet.

Tuesday, April 21, 2009

Polishing sample

Polishing sample

We have successfully polished a sample down to 45 micron thickness using template method. Following is a summary.

As shown in the figure, the idea is to create a template using much harder materials (such as glass, rasor blade) than the sample. One advantage of this way is that you can measure the sample thickness using a micrometer (basically sample+glass -glass). Everything is glued using paraffin, which is safe and surprisingly strong for polishing and easy for removing.

The steps are:

1) we can start with as thick as 500 micron

2) polish the template down to 75 micron (150 grit, 400 grit)

3) put on sample, using 800 grit to polish easily down to the same thickness of the template

4) continue polishing with 800 grit until thickness is less than 50 micron

5) continue with 1500 grit, at this time one of the templates may be gone, which is ok.

6) polish until the satisfactory thickness, the lower limit may be 30 micron.