Department of physics, NTNU TFY4340 Mesoscopic Physics Spring 2010

Solution to Exercise 8

a) From the Büttiker–Landauer formula (first equation in the exercise), we have diagonal elements

Γ_{αα} = \frac{2e^{2}}{h} ∑_{β≠α} T_{βα}

and off–diagonal elements

Γ_{αβ} = −\frac{2e^{2}}{h} T_{αβ}.

From the symmetry relation T_{αβ}(B) = T_{βα}(−B), we have directly for the off–diagonal matrix elements

Γ_{αβ}(B) = Γ_{βα}(−B).

Combination of the equilibrium condition and Onsager symmetry yields

∑_{β≠α} T_{βα}(B) = ∑_{β≠α} T_{αβ}(B) = ∑_{β≠α} T_{βα}(−B),

and hence

Γ_{αα}(B) = Γ_{αα}(−B).

b) We write out the equations (with V_{3} = 0):

I_{1} = \frac{2e^{2}}{h} \{∑_{β≠1} T_{β1}V_{1} − T_{12}V_{2} − T_{14}V_{4}\}

I_{2} = \frac{2e^{2}}{h} \{∑_{β≠2} T_{β2}V_{2} − T_{21}V_{1} − T_{24}V_{4}\}

I_{3} = −I_{1} = \frac{2e^{2}}{h} \{−T_{31}V_{1} − T_{32}V_{2} − T_{34}V_{4}\}
I_{4} = −I_{2} = \frac{2e^{2}}{h} \{∑_{β≠4} T_{β4}V_{4} − T_{41}V_{1} − T_{42}V_{2}\}

Pairwise summation of eqn 1 and 3, respectively eqn 2 and 4, yields

0 = (T_{21} + T_{41})V_{1} − (T_{12} + T_{32})V_{2} − (T_{14} + T_{34})V_{4}
0 = −(T_{21} + T_{41})V_{1} + (T_{12} + T_{32})V_{2} + (T_{14} + T_{34})V_{4}

which are identical equations. This shows that we had only 3, and not 4, independent equations for the 3 voltages V_{1}, V_{2}, and V_{4} (for given currents I_{1} and I_{2}).

Here, we need the currents in terms of the difference d = V_{2} − V_{4}, whereas we are not interested in the sum s = V_{2} + V_{4}. Therefore, we rewrite the last equation,

0 = −(T_{21} + T_{41})V_{1} + (T_{12} + T_{32})(s + d)/2 + (T_{14} + T_{34})(s − d)/2
= −(T_{21} + T_{41})V_{1} + (T_{12} + T_{32} + T_{14} + T_{34})s/2 + (T_{12} + T_{32} − T_{14} − T_{34})d/2

where we have used V_{2} = (s + d)/2 and V_{4} = (s − d)/2. From this equation, we can express s in terms of V_{1}, d, and T_{αβ}:

s = \frac{1}{τ} \{2(T_{21} + T_{41})V_{1} + (T_{14} + T_{34} − T_{12} − T_{32})d\} ,

with

τ = T_{12} + T_{32} + T_{14} + T_{34}.

Clearly, τ will enter the elements of the matrix γ, so let us study the symmetry properties of τ first. We may write (using ∑_{β≠α} T_{βα} = ∑_{β≠α} T_{αβ})

τ = ∑_{β≠2} T_{β2} − T_{42} + ∑_{β≠4} T_{β4} − T_{24}
= ∑_{β≠2} T_{2β} − T_{42} + ∑_{β≠4} T_{4β} − T_{24}
= T_{21} + T_{23} + T_{41} + T_{43}.

These two expressions for τ, together with Onsager symmetry (T_{αβ}(B) = T_{βα}(−B)), show that

τ(B) = τ(−B).

What remains now is to write the equations for I_{1} and I_{2} in terms of V_{1}, s, and d, and eliminate s with the expression found above. We start with I_{1}:

I_{1} = \frac{2e^{2}}{h} \{∑_{β≠1} T_{β1}V_{1} − \frac{s}{2} (T_{12} + T_{14}) − \frac{d}{2} (T_{12} − T_{14})\}

= \frac{2e^{2}}{h} \{∑_{β≠1} T_{β1} − \frac{(T_{21} + T_{41})(T_{12} + T_{14})}{τ}\} V_{1}

+ \frac{2e^{2}}{h} \{−\frac{T_{14} + T_{34} − T_{12} − T_{32}}{2τ} · (T_{12} + T_{14}) − \frac{1}{2}(T_{12} − T_{14})\} d

Hence, the (1, 1)–element is

γ_{11} = \frac{2e^{2}}{hτ} \{τ ∑_{β≠1} T_{β1} − (T_{21} + T_{41})(T_{12} + T_{14})\} ,

and it is clear that

γ_{11}(B) = γ_{11}(−B).

The first sum is symmetric in B, and the second term is a product of two sums, where a change of sign in B turns one sum into the other, and vice versa. The (1, 2)–element is

γ_{12} = −\frac{e^{2}}{hτ} \{(T_{14} + T_{34} − T_{12} − T_{32}) (T_{12} + T_{14}) + (T_{14} + T_{34} + T_{12} + T_{32}) (T_{12} − T_{14})\}
= −\frac{e^{2}}{hτ} \{2 (T_{14} + T_{34}) T_{12} − 2 (T_{12} + T_{32}) T_{14}\}
= \frac{2e^{2}}{hτ} \{T_{32}T_{14} − T_{34}T_{12}\}

Similarly, for I_{2}:

I_{2} = \frac{2e^{2}}{h} \{∑_{β≠2} T_{β2}V_{2} − T_{21}V_{1} − T_{24}V_{4}\}

= \frac{2e^{2}}{h} \{(T_{12} + T_{32} + T_{42}) · \frac{1}{2} (s + d) − T_{21}V_{1} − T_{24} · \frac{1}{2} (s − d)\}
= \frac{2e^{2}}{h} \{−T_{21} + \frac{1}{τ} (T_{21} + T_{41}) (T_{12} + T_{32} + T_{42} − T_{24})\} V_{1}
+ \frac{2e^{2}}{h} \{\frac{1}{2τ} (T_{12} + T_{32} + T_{42} − T_{24}) (T_{14} + T_{34} − T_{12} − T_{32}) + \frac{1}{2} (T_{12} + T_{32} + T_{42} + T_{24})\} d

Hence, the (2, 2)–element is (where we use the expression for τ found earlier)

γ_{22} = \frac{e^{2}}{hτ} \{(T_{12} + T_{32} + T_{42} − T_{24}) (T_{14} + T_{34} − T_{12} − T_{32})\}
+ \frac{e^{2}}{hτ} \{(T_{12} + T_{32} + T_{42} + T_{24}) (T_{14} + T_{34} + T_{12} + T_{32})\}
= \frac{e^{2}}{hτ} \{2 (T_{12} + T_{32} + T_{42}) (T_{14} + T_{34}) + 2T_{24} (T_{12} + T_{32})\}
= \frac{2e^{2}}{hτ} \{(T_{14} + T_{24} + T_{34}) (T_{12} + T_{32}) + T_{42} (T_{14} + T_{34})\}
= \frac{2e^{2}}{hτ} \{τ ∑_{β≠4} T_{β4} + T_{42} (T_{14} + T_{34}) − (T_{14} + T_{34}) (T_{14} + T_{24} + T_{34})\}

= \frac{2e^{2}}{hτ} \{τ ∑_{β≠4} T_{β4} − (T_{41} + T_{43}) (T_{14} + T_{34})\}

This expression will be unchanged if we invert the direction of the magnetic field, hence γ_{22}(B) = γ_{22}(−B). Finally, the (2, 1)–element is

γ_{21} = \frac{2e^{2}}{hτ} \{−T_{21} (T_{12} + T_{32} + T_{14} + T_{34}) + (T_{21} + T_{41}) (T_{12} + T_{32} + T_{42} − T_{24})\}
= \frac{2e^{2}}{hτ} \{−T_{21}T_{14} − T_{21}T_{34} + T_{21}T_{42} − T_{21}T_{24} + T_{41}T_{12} + T_{41}T_{32} + T_{41}T_{42} − T_{41}T_{24}\}
= … = \frac{2e^{2}}{hτ} \{T_{23}T_{41} − T_{43}T_{21}\}

Now, since T_{αβ}(B) = T_{βα}(−B), and τ(B) = τ(−B), we see that

γ_{12}(B) = γ_{21}(−B).

In conclusion, the 2 × 2 matrix γ has the property

γ(B) = γ^{T} (−B).

c) The condition I_{2} = 0 yields

V_{1} = −\frac{γ_{22}}{γ_{21}} (V_{2} − V_{4}) ,

and

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

i.e.,

R_{13,24} = −\frac{γ_{21}}{γ_{11}γ_{22} − γ_{12}γ_{21}} .

Similarly, the condition I_{1} = 0 yields

V_{2} − V_{4} = −\frac{γ_{11}}{γ_{12}} V_{1},

and

I_{2} = (γ_{21} − \frac{γ_{11}}{γ_{12}} γ_{22}) V_{1},

i.e.,

R_{24,13} = −\frac{γ_{12}}{γ_{11}γ_{22} − γ_{12}γ_{21}} .

Since γ_{12}(−B) = γ_{21}(B), we see immediately that

R_{13,24}(B) = R_{24,13}(−B).

Note that these 4–point resistances may have both signs. This is nothing more strange than the fact that Hall voltages may have both signs. Of course, the dissipation is always positive.