1. Resonant Conduction through Mesoscopic Systems
Naomichi Hatano
Institute of Industrial Science, University of Tokyo
Collaborators:
Akinori Nishino (U. Tokyo)
Takashi Imamura (U. Tokyo)
Keita Sasada (U. Tokyo)
Hiroaki Nakamura (Natl. Inst. Fusion Science)
Tomio Petrosky (U. Texas at Austin)
Sterling Garmon (U. Texas at Austin)
I’m going to talk on the effect of resonance on electronic conduction of mesoscopic systems. These are my collaborators.

2. S. Datta “Electronic Transport in Mesoscopic Systems”
Landauer Formula
S. Datta “Electronic Transport in Mesoscopic Systems” 
Perfect conductor  L
T = 0 k
In the first of my talk, I review the Landauer formula which gives the conductance at zero temperature. I basically follow Datta’s excellent textbook. Let me begin with computing the conductance of a perfect conductor. Assume a perfect conductor has this dispersion relation. K is the wave number along the conductor. Suppose we attach two baths to the conductor, one with a chemical potential higher than the other. Then, the right-going electrons have the chemical potential mu one, so these states are filled up to here, while the right-going electrons have the chemical potential mu two, so these states are filled up to here. The currents carried by these electrons are cancelled by the currents carried by these electrons. So, the total current is therefore carried by these electrons.  m2

3. S. Datta “Electronic Transport in Mesoscopic Systems”
Landauer Formula
S. Datta “Electronic Transport in Mesoscopic Systems”
for a perfect conductor
voltage
difference
spin k m2 
density
n =1/L
This is the total current. This factor is due to the spin degree of freedom. Each state carries the current of this amount. Now, the density is given by this, where L is the length of the conductor. This is the velocity. We then move from the wave number summation over to the energy integration. This is then the voltage difference. So, this coefficient is the conductance. That is, this is the conductance for a perfect conductor. Now, the inverse of the conductance is the resistance. Why does a perfect conductor have a resistance?

4. S. Datta “Electronic Transport in Mesoscopic Systems”
Landauer Formula
S. Datta “Electronic Transport in Mesoscopic Systems” 
Perfect conductor 
contact resistance
Actually, this is a contact resistance. At the contact, the electrons coming from the bath into the conductor have to go through rearrangement of conduction channels. This causes the resistance. Once the electrons are inside the conductor, it runs without any resistance.

5. Landauer Formula   L scatterer transmission probability T
S. Datta “Electronic Transport in Mesoscopic Systems”   L
scatterer
transmission probability T
Linear response
Now, if we have a scatterer with the transmission probability T in the conductor, the current is simply decreased by the factor T. Here the linear response comes in, because the transmission probability is generally energy dependent. So, this is the linear response conductance for a conductor with a scatterer. This is called the Landauer formula.
for a conductor with a scatterer

6. Landauer Formula   L scatterer transmission probability T T1 T2
S. Datta “Electronic Transport in Mesoscopic Systems”   L
scatterer
transmission probability T T1 T2
We can rewrite the conductance in the following way. The resistance is given by the summation of the contact resistance and the rest. So, we can interpret this term as a bare resistance of the scatterer. In fact, if you have two scatterers, then this is the resistance.
contact resistance
“bare” resistance
of the scatterer
(incoherent case)

7. Fisher-Lee Relation Complex Potential flux normalization Kubo formula:
S. Datta “Electronic Transport in Mesoscopic Systems”
Effective Hamiltonian: H. Feshbach, Ann. Phys. 5 (1958) 357
Complex Potential
Once the conductance is given by the transmission probability, the next problem is the calculation of the transmission probability. The Fisher-Lee relation gives the general formula. It basically claims that the transmission probability is proportional to the square modulus of the Green’s function. Of course, it is difficult to calculate the Green’s function of the whole Hamiltonian, which is an infinite system, but it is possible to contract the semi-infinite leads to a complex effective potential.I don’t have time to go into details, but this effective potential goes back to Feshbach in 58. Remember that this is a complex potential. I’m coming back to this point in a minute. Anyway, now the Green’s function is the one of the Hamiltonian of a finite system and is reasonably tractable. The conductance is finally given by this. These are the proportional constants. They appear because of the normalization. In general, these states are normalized by the density, but for the calculation of the transmission probability, we actually need flux normalization. This difference causes these factors. Now, this final formula is almost equivalent to the Kubo formula for the conductivity. You have a square modulus of the Green’s function and you have velocities which correspond to these factors.
flux normalization
Kubo formula:

8. Non-Hermiticity of open quantum systems
N. Hatano, K. Sasada, H. Nakamura and T. Petrosky, arXiv:
semi-infinite leads →
complex eigenvalues
solutions of E
resonant states
bound states
Now I move to the second part of my talk. In this part, I would like to extract contribution of resonant states to the conductance. Before that, I discuss the definition of the resonant state for a while. First, I’m going to argue that the total Hamiltonian with semi-infinite leads is in fact non-Hermitian and have complex eigenvalues. That’s why we had complex potentials when we contracted the leads in the previous slide. We then define resonant states as eigenstates of the Hamiltonian with complex eigenvalues. Then we claim that the conductance is approximately given by the contributions from discrete eigenvalues including bound states and resonant states.

9. Non-Hermiticity of open quantum systems
N. Hatano, K. Sasada, H. Nakamura and T. Petrosky, arXiv: 
The non-Hermiticity of the total Hamiltonian basically comes from the kinetic term. Let me calculate the difference between the expectation value of the squared momentum operator and its complex conjugate. Note here that I restrict the integration to a finite volume. The reason for that will be self-evident in a minute. For each term we carry out the partial integration. These terms cancel each other but the boundary terms do not. The difference, or the imaginary part of the Hamiltonian expectation is given in this form of the momentum flux at the boundaries. In general dimensions, the imaginary part of the Hamiltonian expectation over a domain is equal to the momentum flux out of the domain boundary. This suggests that if a state has a constant outgoing wave, it has a complex eigenvalue. For this double delta potential, for example, assume this form of the wave function. On the left of the scattering potential the state has a left-going wave. On the right of the potential the state has a right-going wave. When we plug this into the Schroedinger equation, we find that the wave number satisfying this equation is a solution. The solutions is generally complex and hence the eigenenergy is also complex.
ex.)

10. Resonant State anti-resonant states K E bound states bound states
continuum
continuum
In the complex wave number plane, we bave bound states on the positive imaginary axis. You see, the pure imaginary gives a bound state. Then there are resonant states here and so-called anti-resonant states here, which are time-reversal to the resonant states. Finally, we have continuum of the scattering states. We can map this onto the complex energy plane through this dispersion relation. The bound states are on the negative real axis. The resonant states are here but note that they are on the second Riemann sheet. You see, because of this dispersion relation, the complex energy plane has two Riemann sheets. The anti-resonant states are also on the second Riemann sheets. Finally, the continuum of the scattering states are on the positive real axis. An important point is, we have a branch point here and the branch cut here.
branch point
anti-resonant
states
resonant states
branch cut
resonant states

11. K. Sasada and N. Hatano, Physica E 29 (2005) 609
Resonant State
K. Sasada and N. Hatano, Physica E 29 (2005) 609 E
resonant states
bound states
Now I’m coming back to the conductance. The conductance was proportional to the square modulus of the Green’s function. We can transform this by inserting a complete set which consists of the integration over the continuum and the bound states. Let me swing the integration contour in this way. We move the upper arm into the second Riemann sheet and the lower arm to here. We can thus extract resonant state contributions. There is still this branch point effects, but it is generally small except for the low energy region. By neglecting it, we have the conductance as the summation over the poles of bound states and resonant states.
neglecting the branch point effect

12. Resonance and Conductance
System
Eigenstates
Conductance
One adatom on a chain
One resonant state
Lorentzian peak
Two adatoms on a chain
Two resonant states
Fano peak
One adatom on a ladder
Resonant states with very long lifetimes
Very narrow peak
A dot with Coulomb interaction
Many-body scattering states
Resonance peaks due to Coulomb interaction
In the rest of my talk, I’m going to show three examples of the resonant state contributions to the conductance. The first example is one adatom on a tight-binding chain. This system has only one resonant state. The resulting conductance has a Lorentzian peak. The second example is two adatoms on a tight-binding chain. We have then two resonant states. We claim that the interference between the two resonances produce a Fano asymmetric peak. In the third example, we have an adatom on a tight-binding ladder. Then we find resonant states with very very long lifetimes. The resulting conductance has a very narrow peak. So far, all what I’ve talked about has been one-electron problems. I finally report our recent result on a many-body scattering state before conclusion.

13. S. Tanaka, S Garmon, and T. Petrosky, PRB 73, 115340 (2006)
One Resonance
S. Tanaka, S Garmon, and T. Petrosky, PRB 73, (2006)
th
th′ Ed
quantum wire
adatom
This is the first example, one adatom on a tight-binding chain. We have tight-binding hopping on the chain, the hopping between the chain and the adatom, and the energy level of the adatom. As before, we can assume this form of the eigenfunctions and thereby obtain the resonant states.

14. One Resonance symmetric Lorentzian
These are the solutions on the complex energy plane. These are bound states. Because this is a lattice model, we have an energy band, that is we have the energy maximum as well as the energy minimum. This bound states is just below the energy minimum and this bound state is just above the energy maximum. Actually, this is a bound state of a hole. Then we have one resonant state here and the corresponding anti-resonant state here on the second Riemann sheet. On the other hand, this is the conductance calculated from the standard Landauer formula. Since we have one resonance here, the conductance is given lilke this, which produces a Lorentzian peak here.

15. Fano Peak th th′ Ed th′ Ed quantum wire
K. Sasada and N. Hatano, Physica E 29 (2005) 609
th
th′ Ed
quantum wire
adatom
th′ Ed
Now we are interested in the interference between two resonances. For that purpose, we add another adatom here. Then the conductance is given by the square modulus of the summation of two resonances. We claim that the cross term produces a Fano asymmetric peak.
Cross term of two resonances
Fano asymmetric peak

16. K. Sasada and N. Hatano, Physica E 29 (2005) 609
Fano Peak
K. Sasada and N. Hatano, Physica E 29 (2005) 609
asymmetric Fano
These are poles. The bound states scarecely move, but the resonant states split into two. These open circles are the poles for the case of one adatom. The filled circles are the poles for the case of two adatoms. The cross term between the two resonances actually produce this asymmetry. We revealed in this paper that the asymmetry becomes greater if the difference in the imaginary parts gets greater.

17. Quasi-Bound State in Continuum
H. Nakamura, N. Hatano, S Garmon, and T. Petrosky, PRL, to appear.
th
th′ Ed
quantum wire
(two channel)
th″
adatom
The third example is an adatom on a ladder. We found in this paper that there are resonant states with very long lifetime close to the energy band. There is a classic problem of quantum mechanics called bound state in continuum. The question is can we have a bound state inside the continuum of scattering state? It may be quite strange because we have a bound state and an extended state with the same energy. The answer to the question is yes, but at a limited point of the parameter space of the system. In contrast, our state, which we call quasi-bound state, can exist in the continuum of scattering state in a wide region of the parameter space. So, we suggest that it may be easy to detect it experimentally.
Quasi-Bound State in Continuum
A resonant state with a very long lifetime close to the energy band

18. Quasi-Bound State in Continuum
H. Nakamura, N. Hatano, S Garmon, and T. Petrosky, PRL, to appear.
lower energy band
upper energy band
These are poles on the complex energy plane. These are bound states and these are resonant states which produce this peak. These are actually resonant states very close to the real axis. We have an upper energy band here and lower energy band here. The upper energy band would have a bound state just below its energy minimum but since it’s in the middle of the lower energy band, it is slightly destabilized and become a quasi-bound state. So, the we have quasi-bound state when we have two dispersions that overlap. Now, because of this quasi-bound state, we have actually an almost invisible dip here. Expanding the scale, it is actually a very narrow Lorentzian.
tiny imaginary part

19. Quantum Dot Two-lead interacting resonant-level model dot lead 1
A. Nishino and N. Hatano, J. Phys. Soc. Jpn. 76 (2007)
Two-lead interacting resonant-level model
dot
lead 1
lead 2
Finnally, let me introduce our recent result on the interacting electron system. Here we have two leads with linear dispersion attached to a quantum dot. On the dot, we have Coulomb repulsion. Then we want to calculate the current defined by this. We found an exact solution of the scattering state by the Bethe ansatz method.
Exact solution by the Bethe ansatz method

20. A. Nishino and N. Hatano, J. Phys. Soc. Jpn. 76 (2007) 063002
Quantum Dot
A. Nishino and N. Hatano, J. Phys. Soc. Jpn. 76 (2007)
Input energy
One-particle resonance k1
We found a general solution, but let me show you an example of a scattering state with three particles. What we would expect are three resonance peaks in the current when the dot energy is equal to each particle energy. k2 d k3

21. Resonances due to the interaction
Quantum Dot
We have a line shape as expected for the non-interacting case. When we introduce the Coulomb repulsion, additional peaks emerge like this.
Resonances due to the interaction

22. A. Nishino and N. Hatano, J. Phys. Soc. Jpn. 76 (2007) 063002
Quantum Dot
A. Nishino and N. Hatano, J. Phys. Soc. Jpn. 76 (2007)
Input energy
Many-body resonance k1
In fact, the exact solution of the current has many-body resonances at these points. These resonances occur exactly at the mid point of a pair of particle levels. This is actually the first to find an exact solution with many-body resonances for an open quantum system. k2 Ed k3

23. Summary Landauer Formula and Fisher-Lee Relation
Resonant State of Quantum Open System
One Resonance and Lorentzian Peak
Two Resonances and Fano Peak
Two Channels and Quasi-Bound State in Continuum
Exact Solution of Quantum Dot with Coulomb Interaction
Let me summarize my talk. I first review the Landauer formula and the Fisher-Lee relation. Then I extract the contributions of the resonant states out of the conductance. I showed one resonance produces a Lorentzian peak, but the cross term between two resonances can produce an asymmetric Fano peak. I also reported the existence of a quasi-bound states in continuum, and finally an exact solution of a case with Coulomb interaction. Thank you for your attention.