Department of physics, NTNU TFY4340 Mesoscopic Physics Spring 2010

Exercise 8

Assistance: Tuesday April 13.

Consider the following 4–terminal device:

We choose V_{3} = 0. According to Büttiker and Landauer, the current in terminal α is

I_{α} = \frac{2e^{2}}{h} ∑_{β≠α} \{T_{βα}V_{α} − T_{αβ}V_{β}\} .

Here, T_{βα} is the ”transmission sum” from terminal α to terminal β.

a) Let us write

I_{α} = ∑_{β} Γ_{αβ}V_{β}.

Note the difference between Γ_{αβ} and G_{αβ} = (2e^{2}/h)T_{αβ}. Determine the symmetry of the matrix Γ in the presence of a perpendicular magnetic field B. Hint: Use the Onsager symmetry relations T_{αβ}(B) = T_{βα}(−B), and the equilibrium condition

∑_{β≠α} T_{βα} = ∑_{β≠α} T_{αβ}.

b) Assume I_{1} = −I_{3} and I_{2} = −I_{4}, and write the current voltage relations in the form

I_{1} = γ_{11}V_{1} + γ_{12}(V_{2} − V_{4})
I_{2} = γ_{21}V_{1} + γ_{22}(V_{2} − V_{4})

Show that γ(B) = γ^{T} (−B) by explicitly expressing the matrix elements γ_{ij} in terms of the transmissivities T_{αβ}.

c) Finally, determine the resistances

R_{13,24} = (\frac{V_{2} − V_{4}}{I_{1}} )_{I_{2}=0}

and

R_{24,13} = (\frac{V_{1}}{I_{2}} )_{I_{1}=0}

in terms of γ_{ij}. Show that R_{13,24}(B) = R_{24,13}(−B).