Physics 12c, Problem Set 8 Solutions - Caltech Theoretical Particle ...

Physics 12c, Problem Set 8 Solutions
June 13, 2013
[1] Hyperscaling
For λ = 0, the scaling hypothesis becomes
G(Ω p ɛ) = ΩG(ɛ);
hence choosing Ω p ɛ = constant, or Ω ∼ |ɛ| −1/p , we find
G(ɛ) ∼ Ω −1 ∼ |ɛ| 1/p .
The hyperscaling relation, together with the definition of ν implies
G(ɛ) ∼ (r corr ) −d ∼ |ɛ| dν .
Comparing the two expressions for G(ɛ) we find
dν = 1/p.
From Problem 2c, we have 1/p = 2 − α; therefore
the Josephson identity.
[2] Evaporative cooling
dν = 2 − α,
(a) The Maxwell distribution is of the form
√
2
( m
) 3/2
P (v) =
v 2 e − 1 2 mv2 /τ . (2.1)
π τ
Check the normalization,
∫ ∞
√ ∫ 2 ∞
dvP (v) =
π 23/2
where
0
0
= √ 4 ∫ ∞
dx π
Probability that 1 2 mv2 > x 0 is
And
0
x =
P (x 0 ) = √ 4 ∫ ∞
π
which is the fraction of particle that escapes.
( m
) 1/2 1
dv
2τ 2τ mv2 e − 1 2 mv2 /τ
(
x 2 e −x2) = 1, (2.2)
( 1
2τ mv2 )
. (2.3)
x 0
(
dx x 2 e −x2) . (2.4)
P (x 0 = 1) = 0.572407, (2.5)
1

(b) Number of particles with speed in (v, v + dv) is NP (v)dv, carrying energy
1
2 mv2 NP (v)dv. Total energy is
U = N
∫ ∞
0
dvP (v) 1 2 mv2 = Nτ √ 4 ∫ ∞
π
Eenrgy of particles with 1
2τ mv2 > x 0 is
U(x 0 ) = Nτ 4 √ π
∫ ∞
where g(x 0 ) is fraction of energy lost.
(c) The remaining particles have energy
0
dxx 4 e −x2 = 3 Nτ. (2.6)
2
x 0
dxx 4 e −x2 = 3 2 Nτg(x 0), (2.7)
g(1) = 0.849145 (2.8)
Ū = (1 − g)U = (1 − g) 3 Nτ. (2.9)
2
After evaporation,
¯N = (1 − f)N (2.10)
Ū = (1 − g) 3 2 Nτ = 3 2 ¯N ¯τ = 3 (1 − f)N ¯τ. (2.11)
2
So,
for x 0 = 1.
¯τ = 1 − g τ = 0.3528τ, (2.12)
1 − f
[3] Einstein relation on a lattice
(a)
So,
Γ(s → s + 1)
Γ(s + 1 → s) = eF ∆/τ . (3.13)
And,
P R =
P R
= e F ∆/τ →
P L
∆/τ
eF
1 + e , P 1
F ∆/τ L = . (3.14)
1 + eF ∆/τ
P R − P L = eF ∆/τ ( )
− 1 F ∆
e F ∆/τ + 1 = tanh . (3.15)
2τ
(b) For F ∆
τ

[4] Einstein relation from linear response theory
(a)
Z 0 = ∑ a
e −Ea/τ (4.18)
and
Z λ = ∑ a
e −(ɛa−λAa)/τ ∼ Z 0 + λ τ Z 0 < A > . (4.19)
And we know that
Therefore,
∑
B a e −(ɛa−λAa)/τ ∼ Z 0 < B > + λ τ Z 0 < AB > . (4.20)
a
< B > λ
Z 0 < B > + λ τ Z 0 < AB >
Z 0 + λ τ Z 0 < A >
∼< B > 0 + λ τ (< AB > 0 − < A > 0 < B > 0 ).
(4.21)
(b) Next consider H → H−F x. Find drift velocity < v > F , where < v > 0 = 0.
< v > F = F τ < vx > 0 (4.22)
[5] The cost of erasure
(a) According to Boltzmann distribution, the probability is proportional to
e −E/τ . Thus, we have
1
P (0) =
(5.23)
1 + e −λ/τ
(b)
(c)
W =
∫ ∞
0
P (1) =
e−λ/τ
1 + e −λ/τ (5.24)
dW = P (0)dE 0 + P (1)dE 1 (5.25)
dW =
dW =
e−λ/τ
dλ (5.26)
1 + e−λ/τ ∫ ∞
The result agrees with the Landauer’s Principle.
0
e −λ/τ
dλ = τ ln 2 (5.27)
1 + e−λ/τ 3