Why Is Quantum Contact Conductance Maximized at 2*q^2/h Even Without Scattering?

[chenhon5]

My question is how the maximum quantum contact conductance 2*q^2/h comes from. The maximum contact conductance can only be obtained when there is no scattering in the contacts between nano-channel and two big pads. (To my understanding, resistance comes from scattering, but if there is no scattering, how come we got the conductance with only 2*q^2/h, but not infinity.)

How my question comes from:When a nano-channel (quantum contact point, nanowire or nanotube) (only one energy level with two spins) contact with two big pads, there is a maximum conductance 2*q^2/h. This maximum conductance can be obtained only if the transport in the nano-channel is ballistic. In other words, the conductance comes from the contacts only. (It was described "The origin of the resistance: redistribution of the current among the current-carrying modes at the interfaces").

However, people said, the maximum conductance can be got only if there is no scattering at the contacts. But what my understanding is the resistances come from scattering, if there is no scattering, why the conductance is not infinity, but only 2*q^2/h.

Therefore, my understanding is that there are some scatterings at the contact between big pads with the nano-channel and that limit the conductance to be 2*q^2/h (but there is no scattering in the channel due to the ballistic transport). Am I wrong? I appreciate any comment.

 

[Galileo]

There conductance is quantized because the wavefunction is 'constricted' in the transverse directions across the contact. The contact acts like waveguide for the electrons and so they can exhibit only a discrete number of modes.
The conductance of one mode is independent of the mode and equal to 2e^2/h.

There are some simple models for the contact you can use to show this. I think a rectangular wave-guide would work, and it's the easiest to work with.

 

[chenhon5]

There conductance is quantized because the wavefunction is 'constricted' in the transverse directions across the contact. The contact acts like waveguide for the electrons and so they can exhibit only a discrete number of modes.
The conductance of one mode is independent of the mode and equal to 2e^2/h.

There are some simple models for the contact you can use to show this. I think a rectangular wave-guide would work, and it's the easiest to work with.

Thank you so much for your replay. You answer is very helpful. But I do not know that much about waveguide. Could you give me more explanation about the waveguide?

1. Whether the contacts or the channel is the waveguide?

2. What is the "mode" you are talking about in this case? Relate to the eigenvalues, or the TE, TM mode...

3. Do you know how to derive the value 2e^2/h?

 

[chenhon5]

There conductance is quantized because the wavefunction is 'constricted' in the transverse directions across the contact. The contact acts like waveguide for the electrons and so they can exhibit only a discrete number of modes.
The conductance of one mode is independent of the mode and equal to 2e^2/h.

There are some simple models for the contact you can use to show this. I think a rectangular wave-guide would work, and it's the easiest to work with.

Just one more question. Whether the quantum contact conductance comes from two contacts or only one contact?

As you mentioned, the wavefunction is "constricted" in the transverse direction when electron flow from big pad to the nano-channel (say from the left pad to the channel). Therefore when the electron flow out from the nano-channel to the other pad (on the right), there should be no constriction on the electron. Therefore I am wondering can we say the 2e^2/h only origin from one contact (the left contact in this case).

Of course, if the electron flows from the other direction (from right to the left), the contact conductance will only attribute to the contact on the right handside.

 

[Galileo]

Thank you so much for your replay. You answer is very helpful. But I do not know that much about waveguide. Could you give me more explanation about the waveguide?

1. Whether the contacts or the channel is the waveguide?

2. What is the "mode" you are talking about in this case? Relate to the eigenvalues, or the TE, TM mode...

3. Do you know how to derive the value 2e^2/h?

Not that I`m very knowledgeable about this, but it's like this:
If you consider an electron constrained to travel through a rectangular tube with sides of length L, i.e. with potential:
V(x,y,z)=0 if (0<x,y<L) and V=infinity otherwise. There is free propagation in the z-direction. So in the transverse direction the particle is simply in a 2D potential well. For each eigenstates (labeled by 2 quantum numbers) you have a mode of conductance (actually, you have to account for spin degeneracy too, which doubles the the number of modes/density of states).
An electron in a particular mode with momentum $\hbar k$ in the z-direction contributes a current $-e\hbar k/m$. The relation between k and the energy is ofcourse: k^2=2mE.
To get the total current, you have to add the contributions for different k, from where k corresponds to the energy at the left side of the contact (Fermi energy EF) to the right side: (EF-eV), since the energy difference is eV, where V is the potential across the contact.
This is most easily done by imposing periodic boundary conditions (so k=2pi/L times some integer) you can let L go to infinity at the end, but it won't make a difference.
If you calculate the total current from the mode this way you'll find:
$$I=2e^2/h V$$
So the conductance G=I/V =2e^2/h.

 

[chenhon5]

Not that I`m very knowledgeable about this, but it's like this:
If you consider an electron constrained to travel through a rectangular tube with sides of length L, i.e. with potential:
V(x,y,z)=0 if (0<x,y<L) and V=infinity otherwise. There is free propagation in the z-direction. So in the transverse direction the particle is simply in a 2D potential well. For each eigenstates (labeled by 2 quantum numbers) you have a mode of conductance (actually, you have to account for spin degeneracy too, which doubles the the number of modes/density of states).
An electron in a particular mode with momentum $\hbar k$ in the z-direction contributes a current $-e\hbar k/m$. The relation between k and the energy is ofcourse: k^2=2mE.
To get the total current, you have to add the contributions for different k, from where k corresponds to the energy at the left side of the contact (Fermi energy EF) to the right side: (EF-eV), since the energy difference is eV, where V is the potential across the contact.
This is most easily done by imposing periodic boundary conditions (so k=2pi/L times some integer) you can let L go to infinity at the end, but it won't make a difference.
If you calculate the total current from the mode this way you'll find:
$$I=2e^2/h V$$
So the conductance G=I/V =2e^2/h.

Thank you very much. Your explanation is excellent!