Karlsruher Institut f¨ur Technologie www.tkm.kit.edu/lehre/
Moderne Theoretische Physik III SS 2015
Prof. Dr. A. Mirlin Blatt 09, 100 Punkte
Dr. U. Karahasanovic, Dr. I. Protopopov Besprechung, 26.06.2015
1. Domain walls in Ising model
(5 + 5 + 5 + 10 + 5 + 10 + 20 = 60 Punkte, schriftlich) (a) The energy of a state with a domain wall is just
−(N −1)J + 2J. (1)
There areN−1 positions where a domain wall can sit. The partition function with the account of states with a single domain wall reads
Z=Z0+δZ =e(N−1)βJh
1 + (N−1)e−2βJi
. (2)
(b) Due to fixing of spin σ1 to 1 the positions of domain walls immediately determine the spins . The energy of a state with m domain walls is
−N J+ 2Jm. (3)
The number of states withm domain walls is just the number of ways to choosem places where we put a domain wall among available N−1 places.
(c)
Z =e(N−1)βJ
N−1
X
m=0
e−2JβΓ(m) =e(N−1)βJ(1 +e−2Jβ)N−1 (4) In thermodynamic limit we reproduce the results of exercise sheet 8.
Z = [2 coshβJ]N (5)
(d) Suppose we have a domain wall between spin k and k+ 1. It means that we have exactlyN −k spins down in the system. The average number of spins down is
N−= exp((N −1)βJ −2Jβ)PN−1
k=1(N −k)
[exp(N Jβ) + (N −1) exp(N βJ−2βJ)] = exp(−2Jβ)N(N −1)
2[1 + (N−1) exp(−2βJ)] (6) Obviously,N−→1/2N in thermodynamic limit.
(e) The spinσ1 is fixed in our system. On the other hand in state withmdomain walls σN = (−1)m. Thus
hσ1σNi= 1 Z
N−1
X
m=0
(−1)mexp[(N−1)βJ−2Jmβ]Γ[m] = 1
Z exp((N−1)βJ) [1−exp(−2Jβ)]N−1 (7)
In thermodynamic limit
hσ1σNi= [tanhJ/kBT]N =e−N/ξ, ξ=− 1
ln tanhJ/kBT. (8)
(f) The excitation with smallest energy is just a single flipped spin. The energy of such a state (we concentrate on the thermodynamic limit and omit the boundary terms in the energy) is
−2N J + 8J (9)
Average number of spins down reads N− = 1
ZNexp(2N Jβ−8Jβ) (10)
Z = exp(2N Jβ) [1 +Nexp(−8Jβ)] (11)
IfN e−8Jβ ≪1 we get
N−≈N e−8Jβ. (12)
In the opposite limit N− ≈ 1. Thus, in this approximations N− is always smaller then 1.
Obviously,N−is an extensive quantity and should be proportional toN. Thus, our answer is actually valid only in the limitN e−8Jβ ≪1. We can improve it slightly by considering the configurations where arbitrary number of single-spin domains are present and neglecting interaction between them. This analogous to the ideal gas approximation. We then get
Z =e2N Jβ
∞
X
m=0
1
m!Nme−8mβJ = exph
2N Jβ+N e−8βJi (13) N− = 1
Z
∞
X
m=0
m 1
m!Nme−8mβJ = N e−8βJ Z exph
2N Jβ+N e−8βJi
=N e−8βJ (14) We see once again that N−/N ≪1 at low temperatures.
(g) So far we neglected completely the possibility of having domains of spin up with more than one spin (or equivalently the possibility of domain walls with length larger than 4). Let us show that the contribution of such configurations is negligible at low temperatures. The energy of a domain wall of length m is just 2mJ. How many different domain walls exist in the system? We can estimate it as follows. Lets start a domain wall somewhere and build it step by step. At each step we will have 3 options how to continue the domain wall (forward, right or left). We have alsoN points to start the domain wall. Thus, we estimate the number of different domain walls as 3mN. This is an estimate from above. It does not take into account the fact that a domain wall is always closed and aftermsteps we should return to the point we started from. Thus, we have an estimate for the contribution of configurations with a single domain wall of lengthm into the partition function
δZm<3me−2mJβN (15) We see that at low temperaturesZmdecays monotonously as a function ofm. Thus, the simplest single-spin domains are indeed most important.
2. Infinite-range interaction and mean-field theory.
(10 + 10 + 10 + 10 = 40 Punkte, m¨undlich)
(a) We have for the partition function
Z =X
σ
exp
"
βHX
i
σi
# exp
Jβ 2N
X
i,j
σiσj
=X
σ
exp
"
βHX
i
σi
# exp
Jβ 2N
X
i
σi
!2
= rN β
2πJ Z
dhX
σ
exp
"
βHX
i
σi
# exp
"
−βN h2
2J +βhX
i
σi
#
. (16) (b) After the Hubbard-Stratonovich transformation we can sum over the spins inde-
pendently Z =
rN β 2πJ
Z
dhexp
−βN h2 2J
X
σ
exp
"
β(h+H)X
i
σi
#
= rN β
2πJ Z
dhexp
−βN h2 2J
[2 coshβ(h+H)β]N (17) The partition function can be written as
Z = rN β
2πJ Z
dhexp[−βG(H, T, h)] (18)
G(H, T, h) = N h2
2J −kBT Nln [2 coshβ(h+H)β] (19) The function G is proportional to the number of particles. At large N only small vicinity of minimum of G is important and the integration can be done by the saddle-point method. The saddle-point equation reads
h
J = tanhβ(h+H) (20)
If we identify nowh=Heff − Hwe get Heff− H
J = tanhβHeff. (21)
The filedHeff is an effective field felt by a spin due to the combined effect of external magnetic filed and all the other spins.
(c) Leth0(T,H) be the solution of the saddle-point equation h
J = tanhβ(h+H). (22)
To compute the integral we writeh =h0+δh and expand the function G(h, T,H) to the second order in δh.
G= h20N
2J +kBT N
2 lnJ2−h20
4J2 +N(J+h20β−J2β)
2J2 δh2. (23)
We see now explicitly that the fluctuations δh are suppressed by N ≫ 1. We eva- luate the remaining gaussian integral over δhand find the Gibbs free energy (only extensive part is interesting)
G(T,H) =G(h0(T,H), T,H) = h20N
2J +kBT N
2 lnJ2−h20
4J2 . (24)
-3 -2 -1 1 2 3
-1.0 -0.5 0.5 1.0
-3 -2 -1 1 2 3
-1.0 -0.5 0.5 1.0 1.5
Abbildung 1: Graphical solution of the saddle-point equation and function G in non-zero fieldH at high temperaturesJβ <1
Here h0 =h0(T, H) is defined by the equation h0
J = tanhβ(h0+H). (25)
Let us now discuss in more detail the saddle-point solution h0. We start with the consideration of H= 0. We have the equation.
h
J = tanhβh (26)
We can solve this equation graphically. Plotting hJ and tanhβh on the same plot as functions ofh we observe the following simple facts. At high temperatures,J/kBT <1.
The only solution of the equation is h = 0. This corresponds to the fact that at high temperatures the spins are fluctuating strongly and their joined action on the given spin vanishes due to averaging.
At J/kBT > 1 we get two solutions ±h0 and, of course a trivial solution h = 0.
Correspondingly, the function G(h) has two (degenerate) minima at h = ±h0 and a maximum at h= 0. Computing the integral we should limit our the integration to the vicinity of one of the two minima of G. This is the spontaneous symmetry breaking. To understand this point formally we can imaging that we have an infinitesimally small by finite external field H in the system. This field will break the symmetry between the minima and make one of the minima slightly deeper then the other (which one depends on the sign of H). SinceG is proportional to the number of particles in the system and stays in the exponent. The contribution of the minimum with higher value of Gwill be strongly suppressed even for extremely small H.
Let us now include finite field Hinto consideration. We assume for definiteness thatH is positive. At high temperatures,J/kBT <1, we get again only one minimum inG.
At low temperatures and small magnetic field we get two minima, one of which is deeper and should be take into account.
At low temperatures and sufficiently large H we are again in the situation with one minimum.
Let us now find the magnetization in the system. We have m
N =−1
N∂HG(H, T) =−∂h0
h20
2J +kBT
2 lnJ2−h20 4J2
∂Hh0 =− h0
J + h0 (h20−J2)β
∂Hh0 (27)
-3 -2 -1 1 2 3
-1.0 -0.5 0.5 1.0
-3 -2 -1 1 2 3
-0.5 0.5 1.0 1.5
Abbildung 2: Graphical solution of the saddle-point equation and function G in non-zero but small fieldHand low temperaturesJβ >1.
-3 -2 -1 1 2 3
-1.0 -0.5 0.5 1.0
-3 -2 -1 1 2 3
-0.5 0.5 1.0 1.5
Abbildung 3: Graphical solution of the saddle-point equation and functionGrelatively large fieldH and low temperaturesJβ >1.
To find ∂Hh0 we use the saddle point equation (differentiate it with respect to H) We
find ∂Hh0
J − β(1 +∂Hh0)
cosh2β(h0+H) = 0 (28)
We take into account that 1
cosh2β(h0+H) = 1−tanh2β(h0+H) = 1− h20
J2 (29)
and find
∂Hh0=− (h2−J2)β
J+ (h2−J2)β. (30)
Substituting this now into the expression for the magnetization we get m= h0
J (31)
