1. for each odd i

(1)

Statistical Physics 3 December 8th, 2010

Solution 10

1. for each odd i

e

^−β^Hⁱ⁰

= ∑

si+1=±1,t_i+1=±1

e

^K¹^(sⁱ^sⁱ⁺¹^+tⁱ^tⁱ⁺¹^+sⁱ⁺¹^sⁱ⁺¹^+tⁱ⁺¹^tⁱ⁺²^)+K²^(sⁱ^tⁱ^+2sⁱ⁺¹^tⁱ⁺¹^+sⁱ⁺²^tⁱ⁺²^)+4C

2. for each combination of {s

_i

= ±1, s

_i+2

= ±1, t

_i

= ±1,t

_i+2

= ±1} we have to solve the equation log[e

^2K²

2 cosh(K

₁

(s

_i

+ t

_i

+ s

_i+2

+ t

_i+2

)) + e

^−2K²

2 cosh(K

₁

(s

_i

− t

_i

+ s

_i+2

− t

_i+2

))]

= D + B(s

_i

+ s

_i+2

)(t

_i

+ t

_i+2

) + A(s

_i

+ s

_i+2

)

²

+ A(t

_i

+ t

_i+2

)

²

+ E(s

_i

+ s

_i+2

)

²

(t

_i

+ t

_i+2

)

²

The work is made easier by recognizing that only the following configurations give independent equations:

s

_i

+ s

_i+2

| t

_i

+ t

_i+2

2 , 2 2 , 0 2 , −2 0 , 0 for a total of 4 equations with 4 unknowns: good news!

3. the decimation procedure gives, for odd i, once the sum over s

_i+1

,t

_i+1

was performed:

e

^−β^Hⁱ⁰

= e

^K²^(sⁱ^tⁱ^+sⁱ⁺²^tⁱ⁺²^)+4C

· h

e

^2K²^+K⁴^(sⁱ^tⁱ^+sⁱ⁺²^tⁱ⁺²⁾

2 cosh((K

₁

+ K

₃

)(s

_i

+ t

_i

+ s

_i+2

+t

_i+2

)) + e

^−2K²^−K⁴^(sⁱ^tⁱ^+sⁱ⁺²^tⁱ⁺²⁾

2 cosh((K

₁

− K

₃

)(s

_i

− t

_i

+ s

_i+2

− t

_i+2

)) i 4. by substituting

−β H

⁰

=

N

∑

i=1

[K

₁⁰

(s

_i

s

_i+1

+t

_i

t

_i+1

) + K

₂⁰

(s

_i

t

_i

+ s

_i+1

t

_i+1

) + K

₃⁰

(s

_i

t

_i+1

+ s

_i+1

t

_i

) + K

₄⁰

(s

_i

t

_i+1

s

_i+1

t

_i

) + 2C

⁰

]

and over all the combinations for {s

_i

= ±1, s

_i+2

= ±1,t

_i

= ±1, t

_i+2

= ±1} we obtain only 5 independent equations:

• evidently 8 configurations are equivalent to the corresponding configurations by (s,t) → (−s, −t);

• you can explicity verify that the configurations (+++-), (++-+) and (+-++) also give one unique equation.