This is Eric's logbook for Nano Japan project from 2012.6.3-2012.7.27.

updated 2012.5.18

Daily schedule†

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Goal of the project.†

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Schedule for discussion†

Time and DayMonTueWedThuFri
09:00-10:00SatoNugrahaTapsanitTatsumiNugraha
13:30-14:30SaitoSimonHasdeoSimonTapsanit
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Basic Questions and Answers†

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Q: About Brillouin Zone†

Hey guys! So I believe this is where I should post questions. I don't have any specific questions today, but was just trying to understand Brillouin zones and Bloch's theorem, which appear a lot in Saito-sensei's book on Carbon Nanotubes.

From google I have an ok idea of what Brillouin zones are, but am still trying to figure out how they relate to the energy gap of the cell such as on page 28 of the CN book. I was also wondering how on page 47 the Brillouin zone can be a long line segment instead of a polygon.

If anybody has any quick words of advice for how to understand Bloch's theorem that would be great as well! It's a little bit intimidating and I'm not sure if there's a simple way to begin to understand it.

Thanks so much and see you guys in a couple of weeks!

Eric

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Answer (by Tapsanit)†

Please consider my answers one by one in order to understand the Brillouin zone.

1. What is the Bloch'theorem written in page. 17 of Saito's book?
A: The electron in solid can behave as wave and its wave function which is 
 denoted by Psi(r) should satisfy the Bloch's theorem due to the periodicity 
 of the atoms in solid. The Psi is sometimes called the Bloch wave function. 
 It can be written as the phase factor exp(ikr) times the periodic function in 
 real space u(r), u(r+T) = u(r), : Psi(r) = exp(ikr)u(r). It is recommended to 
 check that the Psi(r)=exp(ikr)u(r) satisfies the Bloch's theorem.
2. What is the Brillouin zone? (not solved)
A: We must understand the reciprocal lattice before answering this question,  
 because the Brillouin zone is defined from the reciprocal lattice.
3. What is the reciprocal lattice that define the Brillouin zone?
A: As mentioned, the electron in solid is a kind of wave whose wave
 function satisfies the Bloch's theorem. This means that it has the  
 wavevector with amplitude and direction just like normal wave. The dimension 
 of wavevector is 1/[length] which is a reciprocal of length in real  
 lattice. Then, reciprocal lattice is constructed to obtain the wavevector   
 of the electron. Please note that there are many wavevectors possibly 
 obtained from the reciprocal space. The unit vectors of the reciprocal 
 lattice (b1,b2,b3) are obtained from the unit vectors of the real lattice 
 (a1,a2,a3) by following relations 
 (* denotes  dot product of two vectors) :
 b1*a1 = 2*Pi, b2*a2 = 2*Pi, b3*a3 = 2*Pi and bi*aj = 0 if i /= j. 
 It is recommended to try to use this relations to derive the unit vectors of 
 reciprocal lattice, b1 and b2, of graphene in Eq. (2.23) of Saito's book 
 using a1 and a2 in Eq. (2.22).
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Q: About copy machine†

Can I use a copy machine or printer?

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A: (not solved)†

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Q: About library†

I would like to know where the library is and how to use it.

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A: (not solved)†

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Report†

This part is basically written by Eric-san. Any other people can add this. Here the information should be from new to old so that we do not need to scroll.

Note: Format didn't carry over from powerpoint. Check that the &tex(): Error! The expression contains invalid characters.; satisfies the Bloch's theorem: Bloch?s theorem is: T_a ?=e^(ik?a) ? T ?_(a ?_i )is a translation operator ? T ?_(a ?_i ) ?(r ?)= ?(r ?+a ?_i) r ?+(a_i ) ?=r ? because (a_i ) ? is a lattice vector. u(r) is a periodic function. T ?_(a ?_i ) ?(r ?)= ?(r ?+a ?_i )= e^(ik?(r ?+(a_i ) ? ) ) u(r ?+(a_i ) ?)= e^(ik?((a_i ) ? ) ) [e^(ik?(r ? ) ) u(r ?+(a_i ) ?)]=e^(ik?((a_i ) ? ) ) [e^(ik?(r ? ) ) u(r ?)] Theorem is satisfied.

Derive the unit vectors of reciprocal lattice, b1 and b2, of graphene in Eq. (2.23) of Saito's book using a1 and a2 in Eq. (2.22).

	a ?_i=unit vector of real lattice
	 b ?_i=unit vector or reciprocal lattice
	a ?_i?b ?_i=2? 
	a ?_i?b ?_j=0 if i?j
	from eqn 2.22 in CN book:

\begin{eqnarray}

a_1=\left(\frac{\sqrt{3}}{2}a,\frac{a}{2}\right), a_2=\left(\frac{\sqrt{3}}{2}a,-\frac{a}{2}\right)

\end{eqnarray}

	a ?_1=(?3/2 a,a/2), a ?_2=(?3/2 a,(-a)/2)
	a ?_1?b ?_1=a_1x b_1x+a_1y b_1y=b_1x  (a?3)/2+b_1y   a/2=2?
	a ?_2?b ?_1=	a_2x b_1x+a_2y b_1y=b_1x  (a?3)/2-b_1y   a/2=0
	?b_1y   a/2  = b_1x  (a?3)/2
	?b_1x  (a?3)/2+b_1x  (a?3)/2=2?
	?b_1x=2?/(a?3)  ? ?-b_1y   a/2=0 ? b_1y=2?/a 
                   b_1=(2?/(a?3),2?/a)

Same process gives: b_2=(2?/(a?3),(-2?)/a)

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June 5 (Hasdeo teaches XX and YY. from 9:00-9:30) etc†

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June 4†

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