Alexandre R. Rocha200.145.112.249/webcast/files/Journeys17-Class03.pptx.pdf · 2017-07-12 · T=8...

37
Alexandre R. Rocha [email protected] Instituto de Física Teórica - Brazil

Transcript of Alexandre R. Rocha200.145.112.249/webcast/files/Journeys17-Class03.pptx.pdf · 2017-07-12 · T=8...

Page 1: Alexandre R. Rocha200.145.112.249/webcast/files/Journeys17-Class03.pptx.pdf · 2017-07-12 · T=8 βΓ2 β 1 sink 1 asink 2 a 1 Γ2eik 1a+i2βsink 2 a 2 T≈2 Γ2 β 1 β sink 1 a

Alexandre R. Rocha

[email protected]

Instituto de Física Teórica - Brazil

Page 2: Alexandre R. Rocha200.145.112.249/webcast/files/Journeys17-Class03.pptx.pdf · 2017-07-12 · T=8 βΓ2 β 1 sink 1 asink 2 a 1 Γ2eik 1a+i2βsink 2 a 2 T≈2 Γ2 β 1 β sink 1 a

AlexandreReilyRocha 2

IL=IR

Page 3: Alexandre R. Rocha200.145.112.249/webcast/files/Journeys17-Class03.pptx.pdf · 2017-07-12 · T=8 βΓ2 β 1 sink 1 asink 2 a 1 Γ2eik 1a+i2βsink 2 a 2 T≈2 Γ2 β 1 β sink 1 a

ΓL

ΓR εV µR

µL

V ≠ 0 2|EF-ε|

V = 0

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AlexandreReilyRocha 5

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AlexandreReilyRocha 6

( ) ( ) - 0 0 f N U - + = µ ε ε ε

Page 7: Alexandre R. Rocha200.145.112.249/webcast/files/Journeys17-Class03.pptx.pdf · 2017-07-12 · T=8 βΓ2 β 1 sink 1 asink 2 a 1 Γ2eik 1a+i2βsink 2 a 2 T≈2 Γ2 β 1 β sink 1 a

( ) ( ) - 0 0 f N U - + = µ ε ε ε

Canyoudoabitofcoding?

Mathema4ca/Matlab/Fortran/C/C++/Python/Punchcards/byhand?

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( ) ( ) - 0 0 f N U - + = µ ε ε ε

L L R

R

ΓL

ΓR εV µR

µL

V ≠ 0 V = 0

Page 9: Alexandre R. Rocha200.145.112.249/webcast/files/Journeys17-Class03.pptx.pdf · 2017-07-12 · T=8 βΓ2 β 1 sink 1 asink 2 a 1 Γ2eik 1a+i2βsink 2 a 2 T≈2 Γ2 β 1 β sink 1 a
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@⇢ (~r, t)

@t= �r ~J

⇢ (~r, t) �! ⇢ = | lih l|

@⇢

@t=

1

i~

hH, ⇢

i=

1

i~

⇣H | lih l|� | lih l| H

Con4nuityequa4on

Classical DensityMatrix

Page 11: Alexandre R. Rocha200.145.112.249/webcast/files/Journeys17-Class03.pptx.pdf · 2017-07-12 · T=8 βΓ2 β 1 sink 1 asink 2 a 1 Γ2eik 1a+i2βsink 2 a 2 T≈2 Γ2 β 1 β sink 1 a

I = j ∂ρ∂t

j

Ψ =e−iEkt

Neilka

l∑ l = cl

k t( )l∑ l

Page 12: Alexandre R. Rocha200.145.112.249/webcast/files/Journeys17-Class03.pptx.pdf · 2017-07-12 · T=8 βΓ2 β 1 sink 1 asink 2 a 1 Γ2eik 1a+i2βsink 2 a 2 T≈2 Γ2 β 1 β sink 1 a

j ∂ρ∂t

j =Yj+1→ j +Yj−1→ j

Yj+1,j = � i

~

hhj

���H��� j + 1ickj+1 (t) c

k⇤j (t)� hj + 1

���H��� jickj+1 (t) c

k⇤j (t)

i

Yj�1,j = � i

~

hhj

���H��� j � 1ickj�1 (t) c

k⇤j (t)� hj � 1

���H��� jickj�1 (t) c

k⇤j (t)

i

Yj�1,j = � 2�

~N sin (k) Yj+1,j =2�

~N sin (k)

Page 13: Alexandre R. Rocha200.145.112.249/webcast/files/Journeys17-Class03.pptx.pdf · 2017-07-12 · T=8 βΓ2 β 1 sink 1 asink 2 a 1 Γ2eik 1a+i2βsink 2 a 2 T≈2 Γ2 β 1 β sink 1 a

Ψ =1N

eilkal∑ ψl

Ψ =1N

e−ilkal∑ ψl

Page 14: Alexandre R. Rocha200.145.112.249/webcast/files/Journeys17-Class03.pptx.pdf · 2017-07-12 · T=8 βΓ2 β 1 sink 1 asink 2 a 1 Γ2eik 1a+i2βsink 2 a 2 T≈2 Γ2 β 1 β sink 1 a

Yj�1,j = � 2�

~N sin (k) Yj+1,j =2�

~N sin (k)

v E( ) = 1∂E∂k

= −2βasin ak( )

Yj�1,j =vkNa

=vkL

Page 15: Alexandre R. Rocha200.145.112.249/webcast/files/Journeys17-Class03.pptx.pdf · 2017-07-12 · T=8 βΓ2 β 1 sink 1 asink 2 a 1 Γ2eik 1a+i2βsink 2 a 2 T≈2 Γ2 β 1 β sink 1 a

E = ε0 + 2β cos ka( ) v E( ) = 1!∂E∂k

= −2βasin ak( )

!

∂n(E)∂E

= DOS E( ) = 22π

La

12β sinka

DOS = 2h

Lv E( )

I = n(E)ev(E)E=µR

µL

∑ = eµR

µR

∫ ∂n(E)∂E

v E( )dE = 2ehe µ1 −µ2( ) = 2e

2

hΔV

Page 16: Alexandre R. Rocha200.145.112.249/webcast/files/Journeys17-Class03.pptx.pdf · 2017-07-12 · T=8 βΓ2 β 1 sink 1 asink 2 a 1 Γ2eik 1a+i2βsink 2 a 2 T≈2 Γ2 β 1 β sink 1 a

I =2ehe µ1 −µ2( ) =

2e2

hΔV

dIdV

=G =2e2

h

Page 17: Alexandre R. Rocha200.145.112.249/webcast/files/Journeys17-Class03.pptx.pdf · 2017-07-12 · T=8 βΓ2 β 1 sink 1 asink 2 a 1 Γ2eik 1a+i2βsink 2 a 2 T≈2 Γ2 β 1 β sink 1 a
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Page 20: Alexandre R. Rocha200.145.112.249/webcast/files/Journeys17-Class03.pptx.pdf · 2017-07-12 · T=8 βΓ2 β 1 sink 1 asink 2 a 1 Γ2eik 1a+i2βsink 2 a 2 T≈2 Γ2 β 1 β sink 1 a

Daniel Ugarte e Varlei Rodrigues (Unicamp)

Page 21: Alexandre R. Rocha200.145.112.249/webcast/files/Journeys17-Class03.pptx.pdf · 2017-07-12 · T=8 βΓ2 β 1 sink 1 asink 2 a 1 Γ2eik 1a+i2βsink 2 a 2 T≈2 Γ2 β 1 β sink 1 a
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AlexandreReilyRocha 26€

ψ0

ψ1

ψ2

ψ−1

β

ε0

ε0

Γ

ε0

ε0

ε0

ε0

Ψ =1N

eilkal∑ ψl

Page 25: Alexandre R. Rocha200.145.112.249/webcast/files/Journeys17-Class03.pptx.pdf · 2017-07-12 · T=8 βΓ2 β 1 sink 1 asink 2 a 1 Γ2eik 1a+i2βsink 2 a 2 T≈2 Γ2 β 1 β sink 1 a

AlexandreReilyRocha 27

β

ε0

ε0

ε0

ε0

ε0

ε0

β

Page 26: Alexandre R. Rocha200.145.112.249/webcast/files/Journeys17-Class03.pptx.pdf · 2017-07-12 · T=8 βΓ2 β 1 sink 1 asink 2 a 1 Γ2eik 1a+i2βsink 2 a 2 T≈2 Γ2 β 1 β sink 1 a

AlexandreReilyRocha 28

β

ε0

ε0

Γ

ε0

ε0

ε0

ε0

Page 27: Alexandre R. Rocha200.145.112.249/webcast/files/Journeys17-Class03.pptx.pdf · 2017-07-12 · T=8 βΓ2 β 1 sink 1 asink 2 a 1 Γ2eik 1a+i2βsink 2 a 2 T≈2 Γ2 β 1 β sink 1 a

AlexandreReilyRocha 29

E k( ) = ε0 + 2β cos ka( )

Page 28: Alexandre R. Rocha200.145.112.249/webcast/files/Journeys17-Class03.pptx.pdf · 2017-07-12 · T=8 βΓ2 β 1 sink 1 asink 2 a 1 Γ2eik 1a+i2βsink 2 a 2 T≈2 Γ2 β 1 β sink 1 a

T = 4β 2Γ2 sin2 kaβ 2e−ika −Γ2eika

2 ≈ 4Γβ

$

%&

'

()

2

sin2 ka

Γ << β

β

ε0

ε0

Γ

ε0

ε0

ε0

ε0

Page 29: Alexandre R. Rocha200.145.112.249/webcast/files/Journeys17-Class03.pptx.pdf · 2017-07-12 · T=8 βΓ2 β 1 sink 1 asink 2 a 1 Γ2eik 1a+i2βsink 2 a 2 T≈2 Γ2 β 1 β sink 1 a

T = 4β 2Γ2 sin2 kaβ 2e−ika − Γ2eika

2€

β

ε0

ε0

Γ

ε0

ε0

ε0

ε0

Page 30: Alexandre R. Rocha200.145.112.249/webcast/files/Journeys17-Class03.pptx.pdf · 2017-07-12 · T=8 βΓ2 β 1 sink 1 asink 2 a 1 Γ2eik 1a+i2βsink 2 a 2 T≈2 Γ2 β 1 β sink 1 a

AlexandreReilyRocha 32

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β

ε0

ε0

Γ

ε0

ε0

ε0

ε0

ε1

β1

Page 32: Alexandre R. Rocha200.145.112.249/webcast/files/Journeys17-Class03.pptx.pdf · 2017-07-12 · T=8 βΓ2 β 1 sink 1 asink 2 a 1 Γ2eik 1a+i2βsink 2 a 2 T≈2 Γ2 β 1 β sink 1 a

| i =

8>><

>>:

| k, 1i + r | �k, 1i if j �1 andchain 1c |0, 0i if j = �1 andchain 0

t1 | �k, 0i if j �1 andchain 0t2 | k, 0i if j � 1 andchain 0

T =

����v1kv0k

���� |t1|2 +

����v1kv0k

���� |t1|2

|vk| =2a�

~ sin ka

Page 33: Alexandre R. Rocha200.145.112.249/webcast/files/Journeys17-Class03.pptx.pdf · 2017-07-12 · T=8 βΓ2 β 1 sink 1 asink 2 a 1 Γ2eik 1a+i2βsink 2 a 2 T≈2 Γ2 β 1 β sink 1 a

T = 8 βΓ2

β1sink1asink2a

1Γ2eik1a + i2β sink2a

2

T ≈ 2 Γ2

β1βsink1asink2a

DOSsurf E( ) = 2 12π

12β sinka

Page 34: Alexandre R. Rocha200.145.112.249/webcast/files/Journeys17-Class03.pptx.pdf · 2017-07-12 · T=8 βΓ2 β 1 sink 1 asink 2 a 1 Γ2eik 1a+i2βsink 2 a 2 T≈2 Γ2 β 1 β sink 1 a

DOSsurf E( ) = 2 12π

12β sink2a

DOStip E( ) = 2 12πsink1aβ1

T ≈16π 2Γ2DOSsurf (E)DOStip(E)

Page 35: Alexandre R. Rocha200.145.112.249/webcast/files/Journeys17-Class03.pptx.pdf · 2017-07-12 · T=8 βΓ2 β 1 sink 1 asink 2 a 1 Γ2eik 1a+i2βsink 2 a 2 T≈2 Γ2 β 1 β sink 1 a
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T ≈16π 2Γ2DOSsurf (E)DOStip(E)

Page 37: Alexandre R. Rocha200.145.112.249/webcast/files/Journeys17-Class03.pptx.pdf · 2017-07-12 · T=8 βΓ2 β 1 sink 1 asink 2 a 1 Γ2eik 1a+i2βsink 2 a 2 T≈2 Γ2 β 1 β sink 1 a

β

ε0ε1

ε0

ε0

ε0

ε0