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1. Important length scales
2. Potential well
3. 2-dimensional electron gas
4. Landauer formula
5. Landauer-Büttiker formalism
6. S-Matrix
Contents
1. Important length scales
2. Potential well
3. 2-dimensional electron gas
4. Landauer formula
5. Landauer-Büttiker formalism
6. S-Matrix
Contents
Important length scales
• Wavelength: at low temperatures current transport occurs for electrons with energies near the Fermi-energy. The Fermi-wavelength is given as:
𝜆𝐹 =2𝜋
𝑘𝐹 𝐸𝐹 =
ℏ2𝑘𝐹2
2𝑚
For low temperatures, meaning 𝑘𝑇 ≪ 𝐸𝐹 , the Fermi distribution is almost a step function
E
f(E)
GaAs: 𝜆𝐹 ≈ 40 nm Si: 𝜆𝐹≈ 35 - 112 nm
𝑘𝑥
𝑘𝑦
𝑘𝐹
−𝑘𝐹
−𝑘𝐹
𝑘𝐹
Important length scales
• Wavelength: at low temperatures current transport occurs for electrons with energies near the Fermi-energy. The Fermi-wavelength is given as:
𝜆𝐹 =2𝜋
𝑘𝐹 𝐸𝐹 =
ℏ2𝑘𝐹2
2𝑚
For low temperatures, meaning 𝑘𝑇 ≪ 𝐸𝐹 , the Fermi distribution is almost a step function
E
f(E)
GaAs: 𝜆𝐹 ≈ 40 nm Si: 𝜆𝐹≈ 35 - 112 nm
Important length scales
• Mean free path 𝑳𝒎: distance an electron travels until its initial momentum is destroyed
• Phase-relaxation length 𝑳𝝋: distance an electron travels until its initial
phase is randomized
Ballistic regime: 𝜆𝐹 < 𝐿 < 𝐿𝑚 Diffusive regime 𝐿 > 𝐿𝑚 Coherent transport: 𝐿 < 𝐿𝜑 Incoherent transport: 𝐿 > 𝐿𝜑
GaAs: 𝐿𝑚 ≈ 100 − 10000 nm 𝐿𝜑 ≈ 200 nm
Si : 𝐿𝑚 ≈ 37 − 118 nm 𝐿𝜑 ≈ 40 − 400 nm
𝑳𝒎
𝑳𝝋
1. Important length scales
2. Potential well
3. 2-dimensional electron gas
4. Landauer formula
5. Landauer-Büttiker formalism
6. S-Matrix
Contents
Solve
−ℏ2
2𝑚∆ + 𝑉 𝑥 Ψ 𝑥 = 𝐸Ψ 𝑥
with boundary conditions
Ψ −𝐿
2= Ψ +
𝐿
2= 0
⇒ 𝐸𝑛 =ℏ2𝜋2𝑛2
2𝑚𝐿2
𝑘𝑛 =𝑛𝜋
𝐿
with 𝑛 ∈ ℕ
Number of occupied states
𝐸𝐹 =ℏ2𝑘𝐹2
2𝑚= 𝐸𝑀 =
ℏ2𝜋2𝑀2
2𝑚𝐿2
⇒ 𝑀 = 𝐼𝑛𝑡𝑘𝐹𝐿
π
Potential well
E
L
1. Important length scales
2. Potential well
3. 2-dimensional electron gas
4. Landauer formula
5. Landauer-Büttiker formalism
6. S-Matrix
Contents
Two semiconductor layers:
n-AlGaAs and i-GaAs layer
electrons flow from n- to i-layer
electrons are „caught“ in z-direction
2-dimensional electron gas
z
z
𝐸𝐶
𝐸𝐶
𝐸𝐹
𝐸𝐹
𝐸𝑉
𝐸𝑉
E
E
2-deg
Wavefunction of a free particle
Ψ 𝑥, 𝑦, 𝑧 = 𝜙𝑚 𝑧 𝑒𝑖𝑘𝑥𝑥𝑒𝑖𝑘𝑦𝑦
Its energy is given as:
𝐸(𝑘) = 𝐸𝐶 + 𝜀𝑚(𝑧) +ℏ2
2𝑚𝑘𝑥2 + 𝑘𝑦
2
𝜀𝑚(𝑧): transverse energy in z-direction
2-deg: only 𝑚 = 1 is occupied
𝜀1(𝑧) < 𝐸𝐹 𝜀𝑚>1
(𝑧) > 𝐸𝐹
2-dimensional electron gas
𝑘 = 𝑘𝑥2 + 𝑘𝑦
2
E(k)
𝐸𝐹
𝜀1(𝑧)
𝜀2(𝑧)
𝜀3(𝑧)
2-dimensional electron gas
contact 1 contact 2 L
W
Ohm‘s law for the resistance 𝑅 =𝐿
𝜎𝑊
We can also use the conductance 𝐺 = 𝑅−1 =σ𝑊
𝐿
For 𝐿 → 0 we would expect 𝐺 → ∞
But: experiments show that G is quantized
𝑊 ≪ 𝐿 𝑘𝐹 < 𝐿 < 𝐿𝑚
x
y
2-dimensional electron gas
First experimental results were obtained by B.J. van Wees in 1988
2-dimensional electron gas at an AlGaAs-GaAs interface
The width is controlled with the gate voltage
Width
2-dimensional electron gas
𝐸𝐹
Because W is so small, only a few modes are occupied. We can rewrite the wavefunction:
Ψ 𝑥, 𝑦, 𝑧 = 𝜙1 𝑧 𝜉𝑛(𝑦)𝑒𝑖𝑘𝑥
Its energy is given as:
𝐸 𝑘 = 𝐸𝐶 + ε1(𝑧)+ 𝜀𝑛(𝑦)+ℏ2𝑘2
2𝑚
𝜀𝑛(𝑦): transverse energy in y-direction
(e.g. energies of the potential well)
𝐸𝐶 + ε1(𝑧)+ 𝜀𝑛(𝑦)< 𝐸𝐹 : open transport channel
𝐸𝐶 + ε1(𝑧)+ 𝜀𝑛(𝑦)> 𝐸𝐹 : closed transport channel
𝑘 = 𝑘𝑥
E(k)
ε1(𝑦)
ε2(𝑦)
ε3(𝑦)
ε4(𝑦)
1. Important length scales
2. Potential well
3. 2-dimensional electron gas
4. Landauer formula
5. Landauer-Büttiker formalism
6. S-Matrix
Contents
Landauer formula
Assumptions:
• Reflectionless contacts: the current flowing from the conductor to the contacts is not reflected
• Ballistic conductor: no reflection within the conductor
• Low temperatures
contact 1 contact 2 Ballistic
conductor
µ1 µ2
-k -k
+k +k
Landauer formula
Result: finite contact resistance that is quantized for a ballistic
conductor 𝐺𝐶 =2𝑒2
ℎ𝑀
How do we calculate M?
Number of modes can be estimated to be (for zero magnetic field)
𝑀 = 𝐼𝑛𝑡𝑘𝐹𝑊
π
because of 𝐸𝑀 = 𝐸𝐹
Landauer formula
A very large number of modes has to be carried by a few modes.
k k k
E E E
resistance = contact resistance
𝜇1
𝜇2 𝜇2
𝜇1
Landauer formula
Now consider a conductor with two ballistic leads. There is a
transmission probability T that an electron crosses the conductor.
𝐼1+ =2𝑒
ℎ𝑀[μ1 − 𝜇2] and 𝐼2
+ =2𝑒
ℎ𝑀𝑇 μ1 − 𝜇2
𝐼1− =2𝑒
ℎ𝑀(1 − 𝑇)[μ1 − 𝜇2]
Total current: 𝐼 = 𝐼1+ − 𝐼1
− = 𝐼2+ =2𝑒
ℎ𝑀𝑇 𝜇1 − 𝜇2 ⇒ 𝐺 =
2𝑒2
ℎ𝑀𝑇
µ1 µ2 Lead 1 Lead 2 conductor
𝐼1+
𝐼1−
𝐼2+
transmission
reflection
Landauer formula
𝐺 =2𝑒2
ℎ𝑀𝑇 Landauer formula
Generalization: 𝐺 =2𝑒2
ℎ 𝑀𝑇𝑛 𝑛
Can we obtain Ohm‘s law from the Landauer formula? Yes, we can see that:
𝐺−1 =𝐿
σ𝑊+𝐿0σ𝑊
1. Important length scales
2. Potential well
3. 2-dimensional electron gas
4. Landauer formula
5. Landauer-Büttiker formalism
6. S-Matrix
Contents
Landauer-Büttiker-formalism
The formula has to be modified
𝐼𝑝 =2𝑒
ℎ 𝑇𝑞←𝑝𝜇𝑝 − 𝑇𝑝←𝑞μ𝑞𝑞 with 𝑇 = 𝑀𝑇
We can introduce 𝐺𝑝𝑞 =2𝑒2
ℎ𝑇𝑝←𝑞
and obtain
𝐼𝑝= 𝐺𝑝𝑞𝑉𝑝 − 𝐺𝑞𝑝𝑉𝑞𝑞
Sum rule (Kirchhoff-laws):
𝐺𝑞𝑝𝑞 = 𝐺𝑝𝑞𝑞
1. Important length scales
2. Potential well
3. 2-dimensional electron gas
4. Landauer formula
5. Landauer-Büttiker formalism
6. S-Matrix
Contents
S-Matrix
For a coherent conductor the transmission function can be expressed with the scattering matrix. The scattering matrix relates the incoming amplitudes for each state with the outgoing amplitudes after the scattering process.
𝑏 = 𝑆 𝑎 For the transmission probabilities:
𝑇𝑚←𝑛 = 𝑠𝑚←𝑛
2
𝑆 = 𝑟 𝑡′𝑡 𝑟′
𝑟, 𝑡, 𝑟‘, 𝑡‘ are matrices
Incoming states {a}
µ1 µ2
{b} states after scattering
Scattering
S
S-Matrix
Properties of the S-matrix: • Calculating the S-matrix is equivalent to solving the problem. • dim𝑆 = 𝑀𝑇 ×𝑀𝑇 with 𝑀𝑇(𝐸) = 𝑀𝑝(𝐸)𝑝
• S has to be unitary 𝑆†S = 𝑆𝑆† = 𝐼 • Reversing the magnetic field transposes the S-Matrix
𝑆 𝐵 = 𝑆𝑡(−𝐵)
S-Matrix
Combining S-matrices: Instead of solving the problem rightaway, one can divide it into smaller problems that have already been solved. Example:
𝑎1
𝑎3 𝑏1
𝑏3 𝑏2
𝑎2
𝐬(𝟏) 𝐬(𝟐)
S-Matrix
𝑏1𝑏2= 𝑟
(1) 𝑡′(1)
𝑡(1) 𝑟′(1)𝑎1𝑎2 and
𝑏3𝑎2= 𝑟
(2) 𝑡′(2)
𝑡(2) 𝑟′(2)𝑎3𝑏2
Eliminate 𝑎2 and 𝑏2: 𝑏1𝑏3=𝑟 𝑡′
𝑡 𝑟′𝑎1𝑎3
Where 𝑡 = 𝑡(2)𝑡(1)
1−𝑟′(1)𝑟(2)
𝑡′ =𝑡′(1)𝑡′(2)
1−𝑟(2)𝑟′(1)
𝑟 = 𝑟(1) +𝑡′(1)𝑟(2)𝑡(1)
1−𝑟′(1)𝑟(2)
𝑟′ = 𝑟′ 2 +𝑡(2)𝑟′(1)𝑡′(2) 1−𝑟′(1)𝑟(2)
𝑠(1)
𝑠(2)
𝑠(1+2)
Summary
• For mesoscopic ballistic conductors the conductance is 𝐺𝐶 =2𝑒2
ℎ𝑀
• Conductor with 2 ballistic leads 𝐺 =2𝑒2
ℎ𝑀𝑇
• Generalization for many conductors 𝐺𝑝𝑞 =2𝑒2
ℎ𝑇𝑝←𝑞
• We can use the S-matrix to calculate the conductance