Polaron reference

Holstein T, Annals of physics, V8,p 325 (1959)
Li ZZ, Theory of Solid State Physics, p 294, (1985)
Yuri M. Galperin, Introduction to Modern Solid State Physics,

Patterson J.,Bailey B., Solid-State Physics, Introduction to the theory, Springer Verlag (2007) p213.

Electron phonon coupling

coupling strength

α/2=Δ/ħω_L

where Δ is the deformation energy.

Added name

name added 06/27/2008

Confusing matrix elements

When we discuss about the matrix element of optical transition, we often can use either of the following ways:

1) H₁=exE, where e is the electronic charge, E is the electric field
2) H1=epA/m, where p is the momentum of the electron and A is the vector potential.

Following approach 1):

transition rate:
W=|<φ_i|exE₀|φ_f>|² δ(E_i-E_f-ħω)=e²E₀²|<φ_i|x|φ_f>|²δ(E_i-E_f-ħω)

Power:
P=ħωe²E₀²|<φ_i|x|φ_f>|²δ(E_i-E_f-ħω)

Dielectric constant (imaginary part):
ε₂=P/(ε₀ωE₀²)=ħe²/ε₀|<φ_i|x|φ_f>|²δ(E_i-E_f-ħω)

However, if we use the approach 2)

transition rate:
W=|<φ_i|epA/m|φ_f>|² δ(E_i-E_f-ħω)=(e/m)²A₀²|<φ_i|p|φ_f>|²δ(E_i-E_f-ħω)

Power:
P=ħω(e/m)²A₀²|<φ_i|p|φ_f>|²δ(E_i-E_f-ħω)

Dielectric constant (imaginary part):
using the relation that E₀=iωA₀
ε₂=P/(ε₀ωE₀²)=ħ(e/m)²/(ε₀ω²)|<φ_i|p|φ_f>|²δ(E_i-E_f-ħω)

We can easily see that the function ε₂(ω) in the two cases are very different, which one is correct and why?

In all the books I read, approach 2) seems to be used, why?

Quantization of electromagnetic field:

Vector potential:

A_n(r)=sqrt(ħ/2ε₀ω)(a_n+a_n^†)u_n(r)

Useful thermoconductivities

Material	Temperature	Thermoconductivity (W/mK)	Working temperature (K)
Apiezon N	293	0.194	4 - 300
	4	0.095
Varnish	300	0.44	423
	100	0.24
	77	0.22
	4.2	0.062
	1	0.034
Crycon grease	4.2	0.1
	2	0.03
Copper	300	400
Silver	300	429
Gold	300	318
Stainless steel	300	11-- 45
air	300	0.025

evaporation

423Kcryoncongrease

What I understand for the Mossbauer s...

What I understand for the Mossbauer spectroscopy

1) Recoil-free absorption spectra, similar to the optical absorption spectra.
Here the key is that in a solid, the atoms can not move freely. In fact, the motion of atoms are described by the vibrational spectra as collective phenomena. In this case, the energy involved is quantized, and there is a lower cut off.
We can estimate the lower cut off using

E_{min}~k_BT_{Debye}/N^{1/3}

For a 300K Debye temperature and N~10^23, we get E_{min}~1e-9 J.

At the same time, we can calculate the recoil energy of a certain nucleus

E_{recoil}=\frac{p^2}{2m}=\frac{1}{2m}(\frac{h}{\lambda})^2

Using Fe⁵⁷ as an example, if it is excited by 100 keV gamma ray, the recoil energy is 3e-20, which means that no phonon can be created. Therefore, the whole crystal will recoil, leaving the recoil speed 1e-21 m/s. In contrast the single nuclear recoil speed can be as high as 500 m/s.

This is the way to get recoil free absorption.

2) discrete peaks due to separate nuclear energy levels

3) nuclear energy levels may change due to .

    A electric interaction

    a) isomer shift, the monopole effect of the environment to the nucleus
     This comes from the interaction between charge of electrons and the nucleus. Because the nucleus is so small, only the s electron that has some density at the center of the atom can have significant interaction. For example, Fe3+ has more isomer shift then Fe2+ because 3d electrons can screen 4s electron, putting them slightly out of the center of the atom.

    b) quadropole splitting, the interaction between the electric quadrapole of nucleus, i.e. the field gradient from the electrons surrounding the nucleus.
   
    B magnetic interaction
    a) Zeeman effect: E=\muB, where \mu is the magnetic moment of the nucleus.


Other remarks,

   Normally, Co⁵⁷ in Rh is used as source, because Co57 can decay into excited state of Fe57 (I=5/2) state, it will later decay  I=5/2->I=3/2 state and I=3/2->I=1/2. Because the I=1/2 is the ground state of Co57. The real useful radiation source is I=3/2->I=1/2.

Polaron activation energy

In the simplest scenario, we can assume two parabolic functions as the potential wells for the two lattice cites.

V₁=(x-x₀)²
V₂=(x+x₀)²

then the activation energy is where the two potential cross: V1=V2.

Here we get E=0 and E_a=V₁=V₂=x₀².

In the case of optical process, we are talking about vertical excitation:
then the excitation is from the bottom of a well to the edge of the other well.

Let x=x₀, V₂=4*x₀²=4E_a, this is where the E_opt=4*E_a comes from.

Operation of Bruker 113V

Turn on

Bolometer

Change diamond filter to position 2 before pumping

Connect pump station to the bolometer

Pump overnight using roughing pump according to instruction

Switch to turbo to get better vacuum

Precool bolometer using liquid nitrogen (fill twice to make sure it is full in both chamber) (see manual)

Precool for 15 minutes

Close the valve that connects the pumping station to the bolometer

Shutdown the pumping station according to the instruction and move it away

Blow out the liquid nitrogen from the Helium reservoir using He gas

Transport liquid He to the reservoir

Turn the diamond filter to position 1 for measurement

Turn the 3 switches on the preamplifier on

Spectrometer

Open the nitrogen gas valve (main, scanner and vent) on the wall

Close the valve that connects spectrometer to the pump

Vent the system (by ordering vent optics from the control panel)

Fire the Hg lamp

Turn on the scanner

Put in the sample assembly

Do a few test runs on the mirror and the sample to check if the level is consistent with previous runs

Evacuate the system by ordering so from the control panel

Close the vent valve

Open the valve that connects the spectrometer to the pump slowly

Do a few test runs to check if the system is stabilized to run the real experiement

Masurement…

Shut down

Turn off the three switches of the bolometer

Open the vent valve

Close the valve that connects spectrometer to the pump

Vent the system

Take the sample assembly out and put back the original lid

Close nitrogen gas valves (main, scanner and vent) on the wall

Evacuate the system by ordering so from the control panel

Open the valve that connects the spectrometer to the pump slowly

Magnet Lab manual is now posted to the Xpy.Daov space.
The link is:
http://xpydaov.spaces.live.com/

vibrational modes for WS2

We can solve the vibrational modes from the Hamiltonian of classical mechanics.

Assuming:

1) There is one spring constant k, and two kinds of masses: m for S and M for W.

2) The coordinates are (x1,y1) and (x2,y2) for S and (x3,y3) for W.

Using {x1,y1,x2,y2,x3,y3} as basis, we can get the matrix:

,

Mode set one:

Mode set two (translation):

Mode set three:

DC conductivity of metal

σ = nq²τ/m

where n is the density of carrier, q is the charge of the carriers, tau is the τ time, m is the effective mass.

The temperature dependence of σ comes from the temperature dependence of τ.

τ can be found from scattering theory. It turns out that

1/τ ~ A², where A is the amplitude of the vibration.

A² ~ n, where n is the number of phonons.

As we know that <n>=1/(exp(hω/k_BT)-1).

At hight T, <n> ~ T, which is the linear relation found for 1/τ or σ.

At low T, we have to consider not only the number of vibrational modes excited, but also the scattering effect of these mode.

The former gives T^-3 relation and the latter gives T². So the total relation will be T^-5.

sum rule in SI

n_eff=∫₀^ωσ₁dω/(π/2*ω_p²)

where ω_p²=e²/(V₀*m*ε₀), V₀ is the volume for the unit cell.

Making a pellet with correct density and thickness: As we know the mean free path of a pellet is: lambda=4*R/3/rou, where R is the particle radius, rou is volume density. We want to make a pellet because our sing crystal sample is too thick to transmit light. Then we want to grind sample to small particles that can transmit light and put them into pellets. The keys are: 1) pellet thickness d should be slightly larger than the mean free path: d > lambda =4*R/3/rou 2) effective thickness (deff) should be small enough so that light can penetrate: deff =d*rou < 1/alpha, where alpha is the absorption coefficient. Thus the correct thickness is: 4R/3/rou < d < 1/alpha/rou is the real condition that we want to meet. There is a implicit condition here: R < 3/alpha/4, which is very important. In conclusion, to make a good pellet: 1) particle size has to be R < 3/alpha/4 2) sample thickness 4R/3/rou < d < 1/alpha/rou, which can be tuned by the volume concentration rou. Example: If we have alpha=1e7 m^{-1}, then the condition of a good pellet is : 1) R < 75 nm. 2) d < 1e7/rou This condition can not be realized because the particle size needs to be very small.

where is the most color of the world coming from?

All the colors of insulators are decided by transmittance instead of reflectance. This has been a huge confusion that I just make it clear myself not so long ago. In most insulators, the refractive index n is between 1 and 3, therefore the reflectance R=(1-n)²/(1+n)² is between 10% and 20%. This is not a small amount of light, however, this part of light is normally not so colorful either. That's because n is not rapidly frequency dependent. We can see this from our everyday experience. Any flat surface will reflect white light, no matter what color the surface is.

If everything reflects white light, how come we see a colorful world? The reason is that surface roughness disperses reflection, making this part of light only important when you have a flat surface and you are looking at a appropriate angle. So, the colorful light we see on normal stuff is actually from double transmittance, because it does not care about the surface roughness. Remember, this part of light can be as much as 80-90% of the total light illuminated on the material. It acts like reflectance because it is transmitted twice. It is colorful because the color is filtered by the media. (As we know, absorption is truly a strong frequency dependent quantity.)

The question immediately follows is that if what we see is transmittance, why most of the stuff is not transparent at all. This is another confusing phenomenon that took me a long time to understand. Actually this has to do with scattering again. If you have a big piece of NaCl single crystal, it should be transparent. But if you grind it into powder, it will keep the color, but become not transparent due to too much scattering at the powder surface. In everyday life, the stuff we see all contain so called pigments, which are colorful transparent small particles. They act like powders, which keeps the color, but not the transparency. If you look everything under optical microscopy, you can see that. Another easy way to verify this is that there is no opaque and colorful (not black) single crystal.

Back to the transition metal oxides. Their band gap is often large enough to leave all the visible range in the gap. In that case, in principle, all the visible lights should go through, making these oxides white or colorless. However this is only true for Sc₂O₃ and TiO₂ when there is not 3d electrons left on the metal ion. Other oxides are normally black and their powder shows colors. The reason is there is the d-d transition. For example, NiO, the electronic configuration is 3d_⁸, (5t_2g,3e_g), which can be excited to 3d⁸* (4t_2g, 3e_g). The energy is in the visible range. The oscillator strength (the absolute value of the absorption) is however much smaller (normally 2 to 3 order of magnitude) than the O-p to Ni-d charge transfer excitation, simply because these kind of excitations are symmetry forbidden.

If it is so small, should it be so important? The answer is yes, because the absorption does not need to be large to make a 1 mm thick sample colorful or even black. We can easily estimate this: to make a 1mm thick sample black, the absorption coeffcient α> 1/1mm = 1000 1/m. (definition of alpha is that transmitted light intensity I=I₀*exp(-α*d), where d is the thickness). If we take one step further, we can estimate extinction coefficient: k=α*c/(ω*2), where c is the speed of light, ω is the angular frequency. For a light with wave length 500 nm, α = 100 /m corresponds to k= 4e-3. This is so small that you can not it in a figure that is scaled to a charge transfer excitations. In fact, you have to use logarithmic scale to bring this detail up. This also explains why the powder is colorful. Because the transmittance depends on the absorption coefficient alpha and sample thickness. As long as the sample is small enough, it is not black anymore. Instead it will show the intrinsic color which corresponds to the small excitation in the visible range. After all, green color for NiO, red color for Fe₂O₃ (the rust) are their intrinsic properties.

bandgap_and _colors

Color and band gap of 3d transition metal oxides

mateials	Band gap (eV)	IBulk	Film (100 nm)	Powder (10 micron)
alpha Fe2O3 (Hametite)	2.34	black
Fe3O4	0.05	black
BiFeO3	2.7	black	yellow
NiO	3.4	black
TiO2	3.03	colorless

People often use interference between the reflected light from front and back surface to measure the thickness of a film. However, there is this a very important formula:

d=nu/(2*n)*), where nu is the frequency difference (difference of 1/lambda ), n is the refractive index, and d is the film thickness.

The caveat is that, we should watch out for the absorption (or the extinction coefficient k).

The formula comes from interference between reflection of from and back surface of a material. However, it does not consider the phase shift of the reflectance, and besides, it does not consider the phaseshift of the transmission of the back-surface reflection.

The problem only goes away when k is very small.

Because the phaseshift dependce on

tan(theta)=epsilon2/epsilon1=2nk/(n^2-k^2),

where theta is the phase shift.

If k<<n, theta~0. Then there is no problem, fortunately, it is the case most of the time.

Interesting color

If the material is has a single band (band gap Eg), the color depends on that only.

Eg	Tranmission/ color	reflection/ color	color of powder
>3.2 eV	large transmission / no color	small reflection / no color	white!
1.6eV< <3.2 eV	depends on thickness	depends on thickness	change color with powder size
<1.6 eV	none	metallic lusters may have different color (e.g. copper: red)	never change color with size

The color is very interesting and confusing. Actually, all the color we see is the color of transmittance color!!

Some may say, there are a lot of non-transparent colorful stuff. That's illusion of color. Let's use paint as an example. The paint contains many small transparent small particles (which is also called pigments). It is the reflection between interfaces makes the light not able to go all the way through, and in turn make the paint not transparent. The real color we normally see is the double transmitted light from the back surface.

Relation between optical functions

We measure reflection R in the expreiment.

Using R we can calculation the phase shift:

θ(ω) = ω/π ∫ log(√(R/R₀))/(ω₀²-ω²)dω₀

Given R and θ as functions of ω, one can calculate all the optical functions.

1) n and k (refractive index and extinction factor):

because R=((n-1)²+k²)/((n+1)²+k²)

k²=4n/(1-R)-(n+1)²

therefore:

n=(1-R)/(1+R+2*√(R)cos(θ))

hence:

k=√(R(n+1)²-(n-1)²)/√(1-R)

2) other functions:

ε₁=n²-k²;

ε₂=2nk

σ₁=ε₂*ω*ε₀

α=k*ω/c, where c is the speed of light