assignment 3 solutions - university of calgary in...

Assignment 3 SolutionsQuestion 1

1(a) First let’s expand the constitutive equation for the magnetization, m = Tanh[b(Jz*m+B)] around small magnetiza-tion and small (but finite) magnetic fields. Note that I am using B for magnetic field because I define H to be some-thing different by default in Mathematica. Clear@B, b, Jz, m, T, cDrhs = Tanh@b HJz * m + BLD;ans1 = Normal@Series@rhs, 8m, 0, 1<DDans2 = Normal@Series@ans1, 8B, 0, 2<DD

Tanh@B bD + m IJz b - Jz b Tanh@B bD2M

B b + Jz m b - B2 Jz m b3

The most important thing to notice immediately is that there is no solution corresponding to m=0, unless B=0 or b=0.Neither of these is the case, which can only mean one thing: the magnetization is NEVER zero in the presence of amagnetic field except at infinitely high temperature. In fact, you already knew this from looking at the Pauli paramag-net. So the magnetic field always provides a small magnetization. This means that there isn’t a real phase transition asthere was for the field-free case, but things still look quite similar. Under this assumption, let’s define the transitiontemperature as the temperature at which the slope of the magnetization as a function of the temperature is largest. Inthe field-free case, this happens at the usual Tc=Jz/kB. ans3 = Solve@m ã ans2, mDm = m ê. ans3@@1DD;b = 1 ê k ê T;Jz = 1; k = 1; B = 0.5; Plot@m, 8T, 0, 2<DClear@B, Jz, kDDm = FullSimplify@D@m, TDD

::m ØB b

1 - Jz b + B2 Jz b3>>

0.5 1.0 1.5 2.0

0.5

1.0

1.5

2.0

2.5

2 B3 Jz k2 T - B k5 T4

IB2 Jz + k2 T2 H-Jz + k TLM2

The slope is largest when the denominator approaches zero:

ans4 = SimplifyASolveAIB2 Jz + k2 T2 H-Jz + k TLM2 ã 0, TEE

::T Ø1

3

Jz

k+ IJz2 kM ì -

27

2B2 Jz k6 + Jz3 k6 +

3

23 Jz2 I27 B4 - 4 B2 Jz2M k12

1ê3

+

1

k3-27

2B2 Jz k6 + Jz3 k6 +

3

23 Jz2 I27 B4 - 4 B2 Jz2M k12

1ê3

>,

:T Ø1

3

Jz

k+ IJz2 kM ì -

27

2B2 Jz k6 + Jz3 k6 +

3

23 Jz2 I27 B4 - 4 B2 Jz2M k12

1ê3

+

1

k3-27

2B2 Jz k6 + Jz3 k6 +

3

23 Jz2 I27 B4 - 4 B2 Jz2M k12

1ê3

>, :T Ø1

12 k3

4 Jz k2 - J2 Â J-Â + 3 N Jz2 k4N ì -27

2B2 Jz k6 + Jz3 k6 +

3

23 Jz2 I27 B4 - 4 B2 Jz2M k12

1ê3

+

Â 22ê3 JÂ + 3 N K-27 B2 Jz k6 + 2 Jz3 k6 + 3 3 Jz2 I27 B4 - 4 B2 Jz2M k12 O1ê3

>, :T Ø1

12 k3

4 Jz k2 - J2 Â J-Â + 3 N Jz2 k4N ì -27

2B2 Jz k6 + Jz3 k6 +

3

23 Jz2 I27 B4 - 4 B2 Jz2M k12

1ê3

+

Â 22ê3 JÂ + 3 N K-27 B2 Jz k6 + 2 Jz3 k6 + 3 3 Jz2 I27 B4 - 4 B2 Jz2M k12 O1ê3

>, :T Ø1

12 k3

4 Jz k2 + J2 Â JÂ + 3 N Jz2 k4N ì -27

2B2 Jz k6 + Jz3 k6 +

3

23 Jz2 I27 B4 - 4 B2 Jz2M k12

1ê3

-

Â 22ê3 J-Â + 3 N K-27 B2 Jz k6 + 2 Jz3 k6 + 3 3 Jz2 I27 B4 - 4 B2 Jz2M k12 O1ê3

>, :T Ø1

12 k3

4 Jz k2 + J2 Â JÂ + 3 N Jz2 k4N ì -27

2B2 Jz k6 + Jz3 k6 +

3

23 Jz2 I27 B4 - 4 B2 Jz2M k12

1ê3

-

Â 22ê3 J-Â + 3 N K-27 B2 Jz k6 + 2 Jz3 k6 + 3 3 Jz2 I27 B4 - 4 B2 Jz2M k12 O1ê3

>>

Let’s expand these solutions around small magnetic fields:T = T ê. ans4@@1DD;Tc = Normal@FullSimplify@Series@T, 8B, 0, 2<D, 8k œ Reals, k > 0, Jz œ Reals, Jz > 0<DD

-B2

Jz k+Jz

k

So, the critical temperature is shifted lower by a factor of B2

Jz kB.

1(b) To plot the magnetization at various magnetic fields, I will calculate it exactly. I’ll also plot the susceptibility soyou can see the lack of a true divergence at the (new) transition temperature.Clear@b, m, BD; func = Tanh@b HJz * m@BD + BLD

Tanh@b HB + Jz m@BDLD

D@func, BD

b Sech@b HB + Jz m@BDLD2 H1 + Jz m£@BDL

func2 = b Sech@b HB + Jz mLD2 H1 + Jz cL

b H1 + Jz cL Sech@HB + Jz mL bD2

2 ass3sol.nb

ans = Solve@func2 ã c, cD

::c Ø -b Sech@HB + Jz mL bD2

-1 + Jz b Sech@HB + Jz mL bD2>>

Here is the general expression for the susceptibility:c = c ê. ans@@1DD

-b Sech@HB + Jz mL bD2

-1 + Jz b Sech@HB + Jz mL bD2

Clear@TD; b = 1 ê k ê T;c

-SechA B+Jz m

k TE2

k T -1 +Jz SechB

B+Jz m

k TF2

k T

Now let’s do a loop over temperature for various choices of the magnetic field

ass3sol.nb 3

B = 0.00001; Jz = 1; k = 1; grain = 1000; range = 2;For@i = 1, i § grain,

Clear@mD;T = range * i ê grain;TT@iD = T;ans = FindRoot@m - rhs, 8m, 1<D;m = m ê. ans@@1DD;mm@iD = m;chi@iD = c;H* Print@T," ",mm@iDD; *Li++;

D;func = Table@8TT@iD, mm@iD<, 8i, 1, grain<D;ListPlot@funcDfunc2 = Table@8TT@iD, chi@iD<, 8i, 1, grain<D;ListPlot@func2D

0.5 1.0 1.5 2.0

0.2

0.4

0.6

0.8

1.0

0.5 1.0 1.5 2.0

2

4

6

8

10

For very small external fields, the transition is indistinguishable from the field-free case.

4 ass3sol.nb

B = 0.001; Jz = 1; grain = 1000; range = 2;For@i = 1, i § grain,



0.5 1.0 1.5 2.0

0.2

0.4

0.6

0.8

1.0

0.5 1.0 1.5 2.0

2

4

6

8

10

For larger fields, the magnetization change is no longer discontinuous.

ass3sol.nb 5




0.5 1.0 1.5 2.0

0.2

0.4

0.6

0.8

1.0

0.5 1.0 1.5 2.0

2

4

6

8

10

12

For yet larger fields, the transition is very smooth and the susceptibility no longer goes to infinity.

6 ass3sol.nb




0.5 1.0 1.5 2.0

0.2

0.4

0.6

0.8

1.0

0.5 1.0 1.5 2.0

0.5

1.0

1.5

2.0

2.5

For very large fields the susceptibility is strongly reminiscent of the heat capacity in the 1D case!

1(c) The actual transition temperature is related to the original one via:Clear@Jz, B, T, kD; Tc

-B2

Jz k+Jz

k

First solve for B in terms of Tc:

ass3sol.nb 7

Clear@TcD; ans = FullSimplifyBSolveBTc ã Tc0 -B2

Jz k, BF, 8k œ Reals, k > 0, Jz œ Reals, Jz > 0<F

::B Ø - Jz k H-Tc + Tc0L >, :B Ø Jz k H-Tc + Tc0L >>

B = B ê. ans@@2DD

Jz k H-Tc + Tc0L

Define Jz in terms of the unperturbed transition temperature:Jz = k * Tc0;

Now express m solely in terms of T, Tc0, and Tc

m2 = FullSimplifyBB b

1 - Jz b + B2 Jz b3,

8k œ Reals, k > 0, Tc œ Reals, Tc > 0, T œ Reals, T > 0, Tc0 œ Reals, Tc0 > 0<F

T2 Tc0 H-Tc + Tc0L

T3 - T2 Tc0 + Tc02 H-Tc + Tc0L

T = Tc * H1 - tL; m3 = FullSimplify@m2,8k œ Reals, k > 0, Tc œ Reals, Tc > 0, T œ Reals, T > 0, Tc0 œ Reals, Tc0 > 0, t œ Reals<D

-JH-1 + tL2 Tc2 Tc0 H-Tc + Tc0L N í IH-1 + tL3 Tc3 + H-1 + tL2 Tc2 Tc0 + Tc Tc02 - Tc03M

Now expand the magnetization around small t:Series@m3, 8t, 0, 0<D

Tc2 -HTc - Tc0L Tc0

HTc - Tc0L2 HTc + Tc0L+ O@tD1

It’s clear that there is no critical behaviour of the magnetization at the new critical temperature, because the leadingorder behaviour is simply the value of the magnetization, which is non-zero. So the critical exponent has gone from 1/2to 0. This is not surprising: there is no real phase transition in this system!1(d) Now let’s calculate the susceptibility. Assume that the magnetic field is small. Then:

Clear@B, T, mD; m = 0; T = Tc H1 - tL; c = NormalBSeriesB-SechA B+Jz m

k TE2

k T -1 +Jz SechB

B+Jz m

k TF2

k T

, 8B, 0, 3<FF

B2

k3 H-1 + tL Tc H-Tc + t Tc + Tc0L2-

1

k H-Tc + t Tc + Tc0L

B = Jz k H-Tc + Tc0L ; k = 1; c2 = FullSimplify@c,8k œ Reals, k > 0, Tc œ Reals, Tc > 0, T œ Reals, T > 0, Tc0 œ Reals, Tc0 > 0, t œ Reals<D

Series@c2, 8t, 0, 0<D

-H-1 + tL2 Tc2 - t Tc Tc0 + Tc02

H-1 + tL Tc HH-1 + tL Tc + Tc0L2

Tc + Tc0

Tc HTc - Tc0L+ O@tD1

There is no divergence in the susceptibility in this model for T>Tc (m=0), everything is nice and smooth. Let’s look atT<Tc:

8 ass3sol.nb

Clear@BD; m =B b

1 - Jz b + B2 Jz b3; c = NormalBSeriesB-

SechA B+Jz m

k TE2

k T -1 +Jz SechB

B+Jz m

k TF2

k T

, 8B, 0, 3<FF

1

Tc - t Tc - Tc0+

B2 H-Tc + t TcL

H-Tc + t Tc + Tc0L4

B = Jz k H-Tc + Tc0L ; k = 1; c2 = FullSimplify@c,8k œ Reals, k > 0, Tc œ Reals, Tc > 0, T œ Reals, T > 0, Tc0 œ Reals, Tc0 > 0, t œ Reals<D

Series@c2, 8t, 0, 0<D

1

Tc - t Tc - Tc0-

H-1 + tL Tc HTc - Tc0L Tc0

HH-1 + tL Tc + Tc0L4

1

Tc - Tc0+

Tc Tc0

HTc - Tc0L3+ O@tD1

Again, there is no divergence in the susceptibility in this model, everything is nice and smooth.

1(e) The free energy per particle near the critical temperature is:

B = Jz k H-Tc + Tc0L ; f = FullSimplify@Jz * m^2 ê 2 - k * T * Log@2 * Cosh@b * HJz * m + BLDD,8Tc œ Reals, Tc > 0, T œ Reals, T > 0, Tc0 œ Reals, Tc0 > 0, t œ Reals<D

-IH-1 + tL4 Tc4 HTc - Tc0L Tc02M ë I2 IH-1 + tL3 Tc3 + H-1 + tL2 Tc2 Tc0 + Tc Tc02 - Tc03M2M +

H-1 + tL Tc LogB2 CoshBJ Tc0 H-Tc + Tc0L I-H-1 + tL3 Tc3 - Tc Tc02 + Tc03MN í

IH-1 + tL Tc IH-1 + tL3 Tc3 + H-1 + tL2 Tc2 Tc0 + Tc Tc02 - Tc03MMFF

fapprox = Normal@Series@f, 8t, 0, 1<DD

-Tc4 Tc02

2 HTc - Tc0L3 HTc + Tc0L2-

Tc LogB2 CoshBJ -HTc - Tc0L Tc0 ITc3 - Tc Tc02 + Tc03MN í ITc HTc - Tc0L2 HTc + Tc0LMFF +

t -Tc4 Tc02 ITc3 + 2 Tc Tc02 - 2 Tc03M

HTc - Tc0L5 HTc + Tc0L3+

Tc JLogB2 CoshBJ -HTc - Tc0L Tc0 ITc3 - Tc Tc02 + Tc03MN í ITc HTc - Tc0L2 HTc + Tc0LMFF -

J -HTc - Tc0L Tc0 ITc6 - 2 Tc4 Tc02 + 5 Tc3 Tc03 - 2 Tc2 Tc04 - 2 Tc Tc05 + Tc06M

TanhBJ -HTc - Tc0L Tc0 ITc3 - Tc Tc02 + Tc03MN í ITc HTc - Tc0L2 HTc + Tc0LMFN í ITc

HTc - Tc0L4 HTc + Tc0L2MN

To obtain the mean energy, we use U/N = k T^2*D[Log[Z],T]. Because T=Tc(1-t) then the derivative with respect to Tis -(1/Tc)*(d/dt). The full derivative is too complicated to carry out, but we can perform it instead near the transitiontemperature:

ass3sol.nb 9

U = FullSimplify@k * T^2 * H-1 ê TcL D@fapprox, tD,8Tc œ Reals, Tc > 0, T œ Reals, T > 0, Tc0 œ Reals, Tc0 > 0, t œ Reals<D

-H-1 + tL2 Tc2 -Tc3 Tc02 ITc3 + 2 Tc Tc02 - 2 Tc03M


LogB2 CoshBJ Tc0 H-Tc + Tc0L ITc3 - Tc Tc02 + Tc03MN í ITc HTc - Tc0L2 HTc + Tc0LMFF -

J Tc0 H-Tc + Tc0L ITc6 - 2 Tc4 Tc02 + 5 Tc3 Tc03 - 2 Tc2 Tc04 - 2 Tc Tc05 + Tc06M

TanhBJ Tc0 H-Tc + Tc0L ITc3 - Tc Tc02 + Tc03MN í ITc HTc - Tc0L2 HTc + Tc0LMFN í ITc

HTc - Tc0L4 HTc + Tc0L2M

Series@U, 8t, 0, 0<D

-Tc2 -Tc3 Tc02 ITc3 + 2 Tc Tc02 - 2 Tc03M


LogB2 CoshBJ Tc0 H-Tc + Tc0L ITc3 - Tc Tc02 + Tc03MN í ITc HTc - Tc0L2 HTc + Tc0LMFF -

J Tc0 H-Tc + Tc0L ITc6 - 2 Tc4 Tc02 + 5 Tc3 Tc03 - 2 Tc2 Tc04 - 2 Tc Tc05 + Tc06M

TanhBJ Tc0 H-Tc + Tc0L ITc3 - Tc Tc02 + Tc03MN í ITc HTc - Tc0L2 HTc + Tc0LMFN í

ITc HTc - Tc0L4 HTc + Tc0L2M + O@tD1

Clearly the mean energy is well-behaved at the transition temperature. Now to the heat capacity, which is C_V/N =-T*(d^2f/dT^2). To do this we need to expand the free energy to second order in t. But we already know that every-thing is well-behaved, and we will simply get the coefficient proportional to t^2! So I will stop this question here.

Question 2In the spin-1 Ising model, we can make use of the general results for the mean-field Hamiltonian, Hi = Jzm^2/2 -(Jzm+B)SiSi. So the partition function is:Clear@b, J, B, T, Jz, m, cD;Z = FullSimplify@HExp@-b * Jz * m^2 ê 2D * Sum@Exp@b * HJz * m + BL * sD, 8s, -1, 1<DL^Num,

8b œ Reals, b > 0, Jz œ Reals, Jz > 0, m œ Reals, m ¥ 0, B œ Reals, B ¥ 0<D

K‰-1

2Jz m2 b

H1 + 2 Cosh@HB + Jz mL bDLONum

The free energy is:F = FullSimplify@-Log@ZD ê b,

8b œ Reals, b > 0, Jz œ Reals, Jz > 0, m œ Reals, m ¥ 0, B œ Reals, B ¥ 0, Num œ Reals, Num > 0<D

-1

bNum LogB‰-

1

2Jz m2 b

H1 + 2 Cosh@HB + Jz mL bDLF

Now let’s impose that the magnetization is an explicit function of the magnetic field:

F = -1

bNum LogB‰-

1

2Jz m@BD2 b

H1 + 2 Cosh@HB + Jz m@BDL bDLF

-1

bNum LogB‰-

1

2Jz b m@BD2

H1 + 2 Cosh@b HB + Jz m@BDLDLF

The magnetization is:

10 ass3sol.nb

M = FullSimplify@-D@F, BD,8b œ Reals, b > 0, Jz œ Reals, Jz > 0, m œ Reals, m ¥ 0, B œ Reals, B ¥ 0, Num œ Reals, Num > 0<D

-Jz Num m@BD m£@BD +2 Num Sinh@b HB + Jz m@BDLD H1 + Jz m£@BDL

1 + 2 Cosh@b HB + Jz m@BDLD

Substituting in the expression for the susceptibility gives:

M = -Jz Num m@BD c +2 Num Sinh@b HB + Jz m@BDLD H1 + Jz cL


-Jz Num c m@BD +2 Num H1 + Jz cL Sinh@b HB + Jz m@BDLD


This expression must be zero:FullSimplify@m@BD * Num - M,8b œ Reals, b > 0, Jz œ Reals, Jz > 0, m œ Reals, m ¥ 0, B œ Reals, B ¥ 0, Num œ Reals, Num > 0<D

Num m@BD + Jz c m@BD -2 H1 + Jz cL Sinh@b HB + Jz m@BDLD


There is a common factor of 1+Jz c, just as in the spin-1/2 case. So we must have m = 2 Sinh@b HB+Jz m@BDLD1+2 Cosh@b HB+Jz m@BDLD

as the

equation for the magnetization. In the absence of an external magnetic field, we can determine the critical temperatureby expanding the right hand side:

SeriesB2 Sinh@b HJz m@BDLD

1 + 2 Cosh@b HJz m@BDLD, 8m@BD, 0, 2<F

2

3Jz b m@BD + O@m@BDD3

So the critical temperature is defined by 1 = 23Jz b, giving Tc = 2Jz/3k. This is lower than the critical temperature for

the spin-1/2 case. I suppose that it is harder to magnetize a magnet when there are higher degrees of freedom. In onedimension, the critical temperature is predicted to be Tc = 4J/kB, which is completely ridiculous because the transitionis at zero temperature. To obtain the critical behaviour of the magnetization, let’s expand the right hand side to higherorder. First though let’s substitute Jz = 3kTc/2:Clear@TcD; Jz = 3 * k * Tc ê 2; b = 1 ê k ê T;

ans = NormalBSeriesB2 Sinh@b HJz m@BDLD

1 + 2 Cosh@b HJz m@BDLD, 8m@BD, 0, 3<FF

FullSimplify@Solve@ans ã m@BD, m@BDD, 8T œ Reals, T > 0, Tc œ Reals, Tc > 0<D

Tc m@BD

T-3 Tc3 m@BD3

8 T3

:8m@BD Ø 0<, :m@BD Ø -

2 2

3T -T + Tc

Tc3ê2>, :m@BD Ø

2 2

3T -T + Tc

Tc3ê2>>

The only interesting root is the last one. Let’s substitute T=Tc*(1-t):

T = Tc * H1 - tL; ans = FullSimplifyB

2 2

3T -T + Tc

Tc3ê2F

-

2 2

3H-1 + tL t Tc

Tc

ass3sol.nb 11

The behaviour near Tc is given by the Taylor expansion around t=0:Series@ans, 8t, 0, 1<D

22

3t + O@tD3ê2

So the critical exponent is the same as for the spin-1/2 case, namely b=1/2. Now let’s consider the susceptibility. Weneed to bring the field back into the expansion:

ans = NormalBFullSimplifyB

SeriesB2 Sinh@b HB + Jz mLD

1 + 2 Cosh@b HB + Jz mLD, 8m, 0, 3<F, 8T œ Reals, T > 0, Tc œ Reals, Tc > 0<FF

ans2 = Normal@FullSimplify@Series@ans, 8B, 0, 1<D, 8T œ Reals, T > 0, Tc œ Reals, Tc > 0<DD

-3 m I2 + CoshA B

Tc-t TcEM

H-1 + tL I1 + 2 CoshA B

Tc-t TcEM

2-

9 m3 -28 - 12 CoshBB

Tc - t TcF + 12 CoshB

2 B

Tc - t TcF + CoshB

3 B

Tc - t TcF ì

8 H-1 + tL3 1 + 2 CoshBB

Tc - t TcF

4

+2 SinhA B

Tc-t TcE

1 + 2 CoshA B

Tc-t TcE

-9 m2 I7 + 2 CoshA B

Tc-t TcEM SinhA B

Tc-t TcE

4 H-1 + tL2 I1 + 2 CoshA B

Tc-t TcEM

3

3 m3

8 H-1 + tL3-

m

-1 + t+B I9 m2 - 8 H-1 + tL2M

12 H-1 + tL3 Tc

rhs =3 m@BD3

8 H-1 + tL3-

m@BD

-1 + t+B I9 m@BD2 - 8 H-1 + tL2M

12 H-1 + tL3 TcD@rhs, BD

-m@BD

-1 + t+

3 m@BD3

8 H-1 + tL3+B I-8 H-1 + tL2 + 9 m@BD2M

12 H-1 + tL3 Tc

-8 H-1 + tL2 + 9 m@BD2

12 H-1 + tL3 Tc-m£@BD

-1 + t+3 B m@BD m£@BD

2 H-1 + tL3 Tc+9 m@BD2 m£@BD

8 H-1 + tL3

Re-expressing the derivative of the magnetization with respect to the field as the susceptibility gives (B=0 again now):

rhs =-8 H-1 + tL2 + 9 m@BD2

12 H-1 + tL3 Tc-

c

-1 + t+

9 m@BD2 c

8 H-1 + tL3

-c

-1 + t+

9 c m@BD2

8 H-1 + tL3+

-8 H-1 + tL2 + 9 m@BD2

12 H-1 + tL3 Tc

Finally we arrive at an expression for the susceptibility:ans = FullSimplify@Solve@c ã rhs, cDDc = c ê. ans@@1DD

::c Ø -2 I8 H-1 + tL2 - 9 m@BD2M

3 Tc I8 H-1 + tL2 t - 9 m@BD2M>>

-2 I8 H-1 + tL2 - 9 m@BD2M

3 Tc I8 H-1 + tL2 t - 9 m@BD2M

12 ass3sol.nb

First consider the case for T>Tc:m@BD = 0; c

-2

3 t Tc

The critical exponent is g=1 as before (note that t<0 which explains the negative sign). For T<Tc the magnetization is

-2

2

3H-1+tL t Tc

Tc so:

m@BD = -

2 2

3H-1 + tL t Tc

Tc; FullSimplify@cD

1 - 3 t

3 t Tc

So for T<Tc, c again goes like 1/t, but with a factor of 1/3 instead of 2/3. So these critical exponents don’t depend onthe spin dimension. Last let’s consider the heat capacity at zero field:m@0D =.; B = 0; k =.; ans = Series@FullSimplify@-k * T * D@F, 8t, 2<DD ê Tc^2, 8t, 0, 2<D

Unset::norep : Assignment on m for m@0D not found. à

9 k Num I2 + CoshA 3 m@0D

2EM m@0D2

2 I1 + 2 CoshA 3 m@0D

2EM

2-

1

4 I1 + 2 CoshA 3 m@0D

2EM

39 k Num m@0D2 -8 - 20 CoshB

3 m@0D

2F - 8 CoshB

3 m@0D

2F2+

21 m@0D SinhB3 m@0D

2F + 6 CoshB

3 m@0D

2F m@0D SinhB

3 m@0D

2F t -

1

16 I1 + 2 CoshA 3 m@0D

2EM

4

27 k Num m@0D2 -16 - 72 CoshB3 m@0D

2F - 96 CoshB

3 m@0D

2F2- 32 CoshB

3 m@0D

2F3+ 21 CoshB

3 m@0D

2F

m@0D2 + 48 CoshB3 m@0D

2F2m@0D2 + 12 CoshB

3 m@0D

2F3m@0D2 + 84 m@0D SinhB

3 m@0D

2F +

192 CoshB3 m@0D

2F m@0D SinhB

3 m@0D

2F + 48 CoshB

3 m@0D

2F2m@0D SinhB

3 m@0D

2F -

120 m@0D2 SinhB3 m@0D

2F2- 24 CoshB

3 m@0D

2F m@0D2 SinhB

3 m@0D

2F2

t2 + O@tD3

For T>Tc the heat capacity is simply zero:m@0D = 0; ans

O@tD3

For T<Tc we find:

m@0D = -

2 2

3H-1 + tL t Tc

Tc; ans

4 k Num + O@tD3

So for T<Tc the heat capacity is a constant 4NkB; because neither of these have any t-dependence, a=0, as we foundin the spin-1/2 case. Of course all of these results are totally different from the exact solution in 1D, because there is nophase transition there!

ass3sol.nb 13

So for T<Tc the heat capacity is a constant 4NkB; because neither of these have any t-dependence, a=0, as we foundin the spin-1/2 case. Of course all of these results are totally different from the exact solution in 1D, because there is nophase transition there!

Question 3In the absence of an external field, the left and right parts of the equation for the auxiliary field are:Clear@b, Hp, J, bcD; lhs = HCosh@b * HJ + HpLD ê Cosh@b * HJ - HpLDL^Hz - 1Lrhs = Exp@2 * b * HpD

HCosh@HHp + JL bD Sech@H-Hp + JL bDL-1+z

‰2 Hp b

Let’s expand these around small values of H’:lhs2 = Normal@Series@lhs, 8Hp, 0, 1<DDrhs2 = Normal@Series@rhs, 8Hp, 0, 1<DD

1 + Hp H-2 b Tanh@J bD + 2 z b Tanh@J bDL

1 + 2 Hp b

When they are equal this defines the transition temperature:ans = Solve@lhs2 ã rhs2, bD

Solve::ifun : Inverse functions are being used by Solve, sosome solutions may not be found; use Reduce for complete solution information. à

:8b Ø 0<, :b ØArcTanhA 1

-1+zE

J>>

temp = b ê. ans@@2DD; ans = Solve@temp ã bc, JD

::J ØArcTanhA 1

-1+zE

bc>>

Define the coupling constant in terms of the transition temperature:J = J ê. ans@@1DD

ArcTanhA 1

-1+zE

bc

Define the b’s in terms of temperature:bc = 1 ê k ê Tc; b = 1 ê k ê T;

Now to go beyond the transition temperature it is necessary to expand the equation governing the H’ to higher order. Itturns out that expanding to order H'2 is too low (the only solutions are either H’=0, which we already know, orH’=const > 0, which is not applicable to the transition). So we expand to order H'3. We also expand the result aroundt=0:

14 ass3sol.nb

t =.; lhs2 = FullSimplify@Normal@Series@lhs, 8Hp, 0, 3<DD, 8z œ Reals, z > 0, t œ Reals, t ¥ 0<Dlhs3 = FullSimplify@Normal@Series@lhs2, 8t, 0, 1<DD, 8z œ Reals, z > 0, t œ Reals, t ¥ 0<Drhs2 = Normal@Series@rhs, 8Hp, 0, 3<DDrhs3 = FullSimplify@Normal@Series@rhs2, 8t, 0, 1<DD, 8z œ Reals, z > 0, t œ Reals, t ¥ 0<D

1

3 k3 H-1 + tL3 Tc33 k3 H-1 + tL3 Tc3 +

2 Hp TanhBArcCoth@1 - zD

-1 + tF IHp2 - 3 k2 H-1 + tL2 Tc2M H-1 + zL + Hp TanhB

ArcCoth@1 - zD

-1 + tF

3 k H-1 + tL Tc H-1 + zL2 + Hp H3 + z H-7 - 2 H-3 + zL zLL TanhBArcCoth@1 - zD

-1 + tF

1

3 k3 Tc3 H-1 + zL3

IH-1 + zL I6 Hp2 k H1 + 2 tL Tc H-1 + zL2 + 6 Hp k2 H1 + tL Tc2 H-1 + zL2 + 3 k3 Tc3 H-1 + zL2 +

2 Hp3 H1 + 3 tL H2 + H-2 + zL zLM -

2 Hp t H-2 + zL z I6 Hp k Tc H-1 + zL2 + 3 k2 Tc2 H-1 + zL2 + Hp2 H8 + 5 H-2 + zL zLM ArcCoth@1 - zDM

1 -4 Hp3

3 k3 H-1 + tL3 Tc3+

2 Hp2

k2 H-1 + tL2 Tc2-

2 Hp

k H-1 + tL Tc

1 +1

3 k3 Tc3I4 Hp3 H1 + 3 tL + 6 Hp2 k H1 + 2 tL Tc + 6 Hp k2 H1 + tL Tc2M

Now equate the two sides and solve for H’:ans = FullSimplify@rhs3 - lhs3, 8z œ Reals, z > 0, t œ Reals, t ¥ 0<Dans2 = Solve@ans ã 0, HpD

1

3 k3 Tc3 H-1 + zL32 Hp H-2 + zL z IHp2 H1 + 3 tL H-1 + zL +

t I6 Hp k Tc H-1 + zL2 + 3 k2 Tc2 H-1 + zL2 + Hp2 H8 + 5 H-2 + zL zLM ArcCoth@1 - zDM

:8Hp Ø 0<, :Hp Ø J6 k t Tc ArcCoth@1 - zD - 12 k t Tc z ArcCoth@1 - zD + 6 k t Tc z2 ArcCoth@1 - zD -

-II-6 k t Tc ArcCoth@1 - zD + 12 k t Tc z ArcCoth@1 - zD - 6 k t Tc z2 ArcCoth@1 - zDM2 -

4 I1 + 3 t - z - 3 t z - 8 t ArcCoth@1 - zD + 10 t z ArcCoth@1 - zD - 5 t z2 ArcCoth@1 - zDM

I-3 k2 t Tc2 ArcCoth@1 - zD + 6 k2 t Tc2 z ArcCoth@1 - zD - 3 k2 t Tc2 z2 ArcCoth@1 - zDMMN í

I2 I1 + 3 t - z - 3 t z - 8 t ArcCoth@1 - zD + 10 t z ArcCoth@1 - zD - 5 t z2 ArcCoth@1 - zDMM>,

:Hp Ø J6 k t Tc ArcCoth@1 - zD - 12 k t Tc z ArcCoth@1 - zD + 6 k t Tc z2 ArcCoth@1 - zD +

-II-6 k t Tc ArcCoth@1 - zD + 12 k t Tc z ArcCoth@1 - zD - 6 k t Tc z2 ArcCoth@1 - zDM2 -

4 I1 + 3 t - z - 3 t z - 8 t ArcCoth@1 - zD + 10 t z ArcCoth@1 - zD - 5 t z2 ArcCoth@1 - zDM

I-3 k2 t Tc2 ArcCoth@1 - zD + 6 k2 t Tc2 z ArcCoth@1 - zD - 3 k2 t Tc2 z2 ArcCoth@1 - zDMMN í

I2 I1 + 3 t - z - 3 t z - 8 t ArcCoth@1 - zD + 10 t z ArcCoth@1 - zD - 5 t z2 ArcCoth@1 - zDMM>>

Yikes! These are pretty ugly. Let’s make sure though that they all go to zero when t->0:t = 0; ans2

88Hp Ø 0<, 8Hp Ø 0<, 8Hp Ø 0<<

Great! The first one is the metastable solution, so the interesting result must be the second or third solution:

ass3sol.nb 15

t =.; Hp = Hp ê. ans2@@2DD;Series@Hp, 8t, 0, 1<D

3 -k2 Tc2 H-1 + zL3 ArcCoth@1 - zD t

-1 + z- 3 Hk Tc H-1 + zL ArcCoth@1 - zDL t + O@tD3ê2

Yay! The leading order term goes like t , as was found for the magnetization in the usual MFT model. Let’struncate the solution to the leading-order term:

Hp =1

-1 + z3 -k2 Tc2 H-1 + zL3 ArcCoth@1 - zD t ;

Now we can obtain the magnetization, using the first equation (the central site):m0 = HExp@b * BD * H2 * Cosh@b * HJ + HpLDL^z - Exp@-b * BD * H2 * Cosh@b * HJ - HpLDL^zL ê

HExp@b * BD * H2 * Cosh@b * HJ + HpLDL^z + Exp@-b * BD * H2 * Cosh@b * HJ - HpLDL^zL

-2z CoshB -1 ê H-1 + zL 3 t ,I-k2 Tc2 H-1 + zL3 ArcCoth@1 - zDM + k Tc ArcTanhB1

-1 + zF ì

Hk H1 - tL TcLFz+

2z CoshB 1 ê H-1 + zL 3 t ,I-k2 Tc2 H-1 + zL3 ArcCoth@1 - zDM + k Tc ArcTanhB1

-1 + zF ì

Hk H1 - tL TcLFz

ì

2z CoshB -1 ê H-1 + zL 3 t ,I-k2 Tc2 H-1 + zL3 ArcCoth@1 - zDM + k Tc ArcTanhB1

-1 + zF ì

Hk H1 - tL TcLFz+

2z CoshB 1 ê H-1 + zL 3 t ,I-k2 Tc2 H-1 + zL3 ArcCoth@1 - zDM + k Tc ArcTanhB1

-1 + zF ì

Hk H1 - tL TcLFz

Now expand this around t=0, assuming B=0:B = 0; m0fin = FullSimplify@Normal@Series@m0, 8t, 0, 1<DD, 8z œ Reals, z > 0, t œ Reals, t ¥ 0<D

3 z -k2 t Tc2 H-1 + zL3 ArcCoth@1 - zD

k Tc H-1 + zL2

So, the magnetization scales as t , giving a critical exponent of 1/2, just like in the regular MFT model. Note thatwhile the auxiliary field H’ and the magnetization are closely related, they actually differ by the factor z/[(z-1)k Tc]:FullSimplify@m0fin ê Hp, 8z œ Reals, z > 0, t œ Reals, t ¥ 0<D

-z

k Tc - k Tc z

Now to the calculation of the susceptibility. We need to go back to the original equations for the central spin and theperipheral spins, including the magnetic field. To make life simpler, let’s make use of the fact that H’ = m k Tc (z-1)/z,so that the left and right hand sides of the equation both only involve m:

16 ass3sol.nb

Clear@B, Hp, mD; Hp = m * k * Tc * Hz - 1L ê z;m0 = HExp@b * BD * H2 * Cosh@b * HJ + HpLDL^z - Exp@-b * BD * H2 * Cosh@b * HJ - HpLDL^zL ê

HExp@b * BD * H2 * Cosh@b * HJ + HpLDL^z + Exp@-b * BD * H2 * Cosh@b * HJ - HpLDL^zL

-2z ‰-

B

k H1-tL Tc CoshB-

k m Tc H-1+zL

z+ k Tc ArcTanhA 1

-1+zE

k H1 - tL TcFz+

2z ‰B

k H1-tL Tc CoshB

k m Tc H-1+zL

z+ k Tc ArcTanhA 1

-1+zE

k H1 - tL TcFz

ì

2z ‰-

B

k H1-tL Tc CoshB-

k m Tc H-1+zL

z+ k Tc ArcTanhA 1

-1+zE

k H1 - tL TcFz+

2z ‰B

k H1-tL Tc CoshB

k m Tc H-1+zL

z+ k Tc ArcTanhA 1

-1+zE

k H1 - tL TcFz

Now let’s expand the right hand side around small m, t, and B:ans = FullSimplify@Normal@Series@m0, 8m, 0, 1<DD, 8z œ Reals, z > 0, t œ Reals, t ¥ 0<Dans2 = FullSimplify@Normal@Series@ans, 8t, 0, 1<DD, 8z œ Reals, z > 0, t œ Reals, t ¥ 0<Dans3 = FullSimplify@Normal@Series@ans2, 8B, 0, 1<DD, 8z œ Reals, z > 0, t œ Reals, t ¥ 0<D

K-1 + ‰4 B

k Tc-k t TcO H-1 + tL - 4 ‰2 B

k Tc-k t Tc m H-1 + zL TanhBArcCoth@1 - zD

-1 + tF ì K1 + ‰

2 B

k Tc-k t TcO2H-1 + tL

4 ‰2 B

k Tc -1 ê H-1 + zLK1 + ‰2 B

k TcO m t H-2 + zL z ArcCoth@1 - zD + 1 ê Hk TcL

B K1 + ‰2 B

k Tc H1 - 2 mL + 2 mO t + ‰B

k Tc k Tc CoshBB

k TcF 2 m H1 + tL + SinhB

2 B

k TcF ì K1 + ‰

2 B

k TcO3

H1 + tL HB + k m TcL

k Tc-m t H-2 + zL z ArcCoth@1 - zD

-1 + z

If we re-define this result with a B-dependent magnetization,

m1 =H1 + tL HB + k m@BD TcL

k Tc-m@BD t H-2 + zL z ArcCoth@1 - zD

-1 + z

-t H-2 + zL z ArcCoth@1 - zD m@BD

-1 + z+

H1 + tL HB + k Tc m@BDL

k Tc

then we can take the derivative with respect to the external field:c1 = D@m1, BD

-t H-2 + zL z ArcCoth@1 - zD m£@BD

-1 + z+

H1 + tL H1 + k Tc m£@BDL

k Tc

The left hand side is simply the susceptibility, so:

Clear@cD; ans = SolveB-t H-2 + zL z ArcCoth@1 - zD c

-1 + z+

H1 + tL H1 + k Tc * cL

k Tcã c, cF

99c Ø -HH-1 - tL H-1 + zLL ë Ik t Tc I1 - z - 2 z ArcCoth@1 - zD + z2 ArcCoth@1 - zDMM==

c = c ê. ans@@1DD; Series@c, 8t, 0, 0<D

H-1 + zL ë Ik Tc I1 - z - 2 z ArcCoth@1 - zD + z2 ArcCoth@1 - zDM tM +

H-1 + zL ë Ik Tc I1 - z - 2 z ArcCoth@1 - zD + z2 ArcCoth@1 - zDMM + O@tD1

Evidently, even with this lowest-order expansion, the critical exponent of the susceptibility is the same as in theregular mean-field approximation, namely g=1. We can do better, however. Let’s expand to higher-order in themagnetization, and go through the same procedure:

ass3sol.nb 17

Evidently, even with this lowest-order expansion, the critical exponent of the susceptibility is the same as in theregular mean-field approximation, namely g=1. We can do better, however. Let’s expand to higher-order in themagnetization, and go through the same procedure:Clear@c, mD;ans = FullSimplify@Normal@Series@m0, 8m, 0, 2<DD, 8z œ Reals, z > 0, t œ Reals, t ¥ 0<Dans2 = FullSimplify@Normal@Series@ans, 8t, 0, 1<DD, 8z œ Reals, z > 0, t œ Reals, t ¥ 0<Dans3 = FullSimplify@Normal@Series@ans2, 8B, 0, 1<DD, 8z œ Reals, z > 0, t œ Reals, t ¥ 0<D

1

K1 + ‰2 B

k Tc-k t TcO3

8 ‰3 B

k Tc-k t Tc CoshBB

k Tc - k t TcF2SinhB

B

k Tc - k t TcF -

1

-1 + tm H-1 + zL CoshB

B

k Tc - k t TcF

TanhBArcCoth@1 - zD

-1 + tF -

1

H-1 + tL2m2 H-1 + zL2 SinhB

B

k Tc - k t TcF TanhB

ArcCoth@1 - zD

-1 + tF2

1

K1 + ‰2 B

k TcO42 ‰

3 B

k Tc

1

-1 + z4 K1 + ‰

2 B

k TcO m t H-2 + zL z ArcCoth@1 - zD -CoshBB

k TcF + 2 m SinhB

B

k TcF +

1 ê Hk TcL 2 ‰B

k Tc 2 B I1 - 4 m2M t + 2 k m H1 + tL Tc + 2 IB It + 2 m2 tM + k m H1 + tL TcM CoshB2 B

k TcF +

k Tc - 2 m H2 B t + k m H1 + 2 tL TcL + k Tc CoshB2 B

k TcF SinhB

2 B

k TcF

I-IB I-1 + m2 - t + 3 m2 tM - k m H1 + tL TcM H-1 + zL + m t H2 B m - k TcL H-2 + zL z ArcCoth@1 - zDM ë

Hk Tc H-1 + zLL

m1 =1

k Tc H-1 + zLI-IB I-1 + m@BD2 - t + 3 m@BD2 tM - k m@BD H1 + tL TcM H-1 + zL +

m@BD t H2 B m@BD - k TcL H-2 + zL z ArcCoth@1 - zDM

1

k Tc H-1 + zLIt H-2 + zL z ArcCoth@1 - zD m@BD H-k Tc + 2 B m@BDL +

H-1 + zL Ik H1 + tL Tc m@BD - B I-1 - t + m@BD2 + 3 t m@BD2MMM

FullSimplify@D@m1, BDD

1

k Tc H-1 + zLIH1 + tL H-1 + zL + H-H1 + 3 tL H-1 + zL + 2 t H-2 + zL z ArcCoth@1 - zDL m@BD2 +

k Tc HH1 + tL H-1 + zL - t H-2 + zL z ArcCoth@1 - zDL m£@BD +

2 B H-H1 + 3 tL H-1 + zL + 2 t H-2 + zL z ArcCoth@1 - zDL m@BD m£@BDM

c1 =1

k Tc H-1 + zLIH1 + tL H-1 + zL + H-H1 + 3 tL H-1 + zL + 2 t H-2 + zL z ArcCoth@1 - zDL m2 +

k Tc HH1 + tL H-1 + zL - t H-2 + zL z ArcCoth@1 - zDL c +2 B H-H1 + 3 tL H-1 + zL + 2 t H-2 + zL z ArcCoth@1 - zDL m cM

1

k Tc H-1 + zLIH1 + tL H-1 + zL + k Tc c HH1 + tL H-1 + zL - t H-2 + zL z ArcCoth@1 - zDL +

m2 HH-1 - 3 tL H-1 + zL + 2 t H-2 + zL z ArcCoth@1 - zDL +

2 B m c HH-1 - 3 tL H-1 + zL + 2 t H-2 + zL z ArcCoth@1 - zDLM

Now the susceptibility is an explicit function of the magnetization:

18 ass3sol.nb

B = 0; Clear@m, cD; ans = Solve@c ã c1, cDc = c ê. ans@@1DD;

99c Ø I-1 + m2 - t + 3 m2 t + z - m2 z + t z - 3 m2 t z - 4 m2 t z ArcCoth@1 - zD + 2 m2 t z2 ArcCoth@1 - zDM ë

Ik t Tc I1 - z - 2 z ArcCoth@1 - zD + z2 ArcCoth@1 - zDMM==

m = 0; Series@c, 8t, 0, 0<D


H-1 + zL ë Ik Tc I1 - z - 2 z ArcCoth@1 - zD + z2 ArcCoth@1 - zDMM + O@tD1

m =3 z -k2 t Tc2 H-1 + zL3 ArcCoth@1 - zD

k Tc H-1 + zL2; Series@c, 8t, 0, 0<D


I-1 + z + 3 z2 ArcCoth@1 - zDM ë Ik Tc I1 - z - 2 z ArcCoth@1 - zD + z2 ArcCoth@1 - zDMM + O@tD1

So the results for the scaling are the same as before.

ass3sol.nb 19

assignment 3 solutions - university of calgary in...

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