HΦの演習問題 (問題)三澤 貴宏
東京大学物性研究所 特任研究員（PCoMS PI）

例題1: spin 1/2 dimer (fulldiag)
例題2: spin 1/2 chain (Lanczos)
例題3: J1-J2 Heisenberg model(Lanczos,TPQ)
例題4: Kitaev model (Lanczos,TPQ)
例題5: Hubbard chain (Lanczos,TPQ)
好きなものからやって下さい
ほとんどlaptop PCでできるはずです(TPQはちょっと重いですが…)
もちろん、自分のやりたい別の課題もやっても OKです例題1: Heisenberg dimer, Hubbard dimer~0 S~1
H = JS
H=t(c†0 c1 + h.c.) + U (n0" n0# + n1" n1# )1. 全対角化でエネルギー固有値を求めましょう。
Emin=-3/4(1重), Emax=1/4(３重縮退) となるはず
2. S=1,2/3,2 …として同じことをやりましょう
Emin=-S(S+1), Emax=S2 となるはず
3.Hubbard 模型でも同じことをやってみましょう(half filling , Sz=0)pU
E = 0, U, ⇥ (1 ± 1 + (4t/U )2 )2

例題2: Heisenberg chain
H=JX

Si Sjhi,ji

1. Lanczosでエネルギーを計算 (サイズL= 20位まで)→基底状態と第一励起状態のエネルギー差(ギャップ)を計算→ギャップの大きさを1/Lでプロットしてみましょう
2. XY模型でも同じことをやってみましょう
3. S=1のハイゼンベルク模型でも同じことをやってみましょう(Haldane gap)

例題3: J1-J2ハイゼンベルク模型
H = J1X

Si Sj + J 2hi,ji

G, HONG YAO, AND LEON BALENTS

XS i Sjhhi,jii

PHYSICAL REVIEW B 86, 024424 (201

最近接 J1,次近接J2

lattice.gpで描画可能

0ms(k0)0.3∆Tms(kx)

0.3

∆S

120.2

4012ms

80.2

0.1

0.0
0.0

Neel AFM

Stripe AFM

0.1

0.8

0.0
1.0

QSL

0.2

0.4

J2/J1

0.6

∆T, ∆S

ture on the 3intermediatephase
of2 the J13-J2
01
on the possibility of symmetry breaking
f these priorstudiedhavesuggestedthat
1513141512ate has VBS order. We note, however,results for the J1 -J2 model are based
891011 orechniques 11(such asseries
expansion
thods, or fixed node or related versions
apted to avoid
thewhich
4 sign 5 problem,
677
ased Monte Carlo in this system), or on
n of very smallsystems.
Some theoretical
01
2
3
3
possibility of VBS order comes from theed quantum criticality,28 which predicts
1512131415
quantum phasetransition—adeconfinedoint (DQCP)—should occur between an
and a plaquette or columnar VBS state,owever, the existence of such a transition
imply that it occurs for the J1 -J2 model inis particular model even harbors a VBS
etical motivation for VBS order comes024424(2012)
FIG. 1. PRB(Color 86,online)
The ground-state phase diagram for t
1
spin- 2 AFM Heisenberg J1 -J2 model on the square lattice,determined by accurate DMRG calculations on long cylinders wi
Ly up to 14. Changing the coupling parameter J2 /J1 , three differe
phases are found: Néel antiferromagnet (AFM), topological quantu
spin liquid (QSL), and stripe AFM phase. ms (k0 = (π,π )) [ms (kx(π,0))] denotes the staggered magnetization in the Néel AFM pha

J2/J1~0.5で非磁性の基底状態(スピン液体?)例題3: J1-J2ハイゼンベルク模型
H = J1

Xhi,jiSi Sj + J 2X

S i Sjhhi,jii

1. Lanczosでエネルギーを計算 (サイズ4 4位)
2. TPQで比熱を計算(J2/J1~0.5でどうなるか？)
3. (発展)余裕があればスピン相関も計算してみましょう
L = 4
W = 4
model = "Spin"method = "Lanczos"
lattice = "square lattice"J = 2.0
J’ = 1.02Sz = 0例題4: Kitaev model
H=Jx

XSix SjxX

Jyx bond

Siz Sjzz bond

20

10X

相図Jx,Jy,Jzで相互作用
4Jzy bond3方向のそれぞれが5Siy Sjy

32A. Kitaev / Annals of Physics 321 (2006) 2–111

5Jz=1,4

Jx= Jy=0

gapless

1713151716121416121179117106810651351412130240131517131416

12

lattice.gpで描画可能

Az

gappedB
Jx=1,
Jy= Jz=0

Ax

Ay

Jy=1,Jx= Jz=0

Fig. 5. Phase diagram of the model. The triangle is the section of the positive octant (Jx, Jy, Jz P 0) byJx + Jy + Jz = 1. The diagrams for the other octants are similar.

Annals of Physics 321, 2-111 (2016)is impossible, even if we introduce new terms in the Hamiltonian. On the other height copies of each phase (corresponding to diﬀerent sign combinations of Jx, Jy,
the same translational properties. It is unknown whether the eight copies of thephase are algebraically diﬀerent.
We now consider the zeros of the spectrum that exist in the gapless phase. The
tum q is defined modulo the reciprocal lattice, i.e., it belongs to a torus. We repre
momentum space by the parallelogram spanned by (q1, q2)—the basis dual to (nthe symmetric case (Jx = Jy = Jz) the zeros of the spectrum are given by

可解模型→スピン液体

例題4: Kitaev model
H=Jx

Xx bondSix Sjx

JyXSiy Sjy

y bondJz

XSiz Sjzz bond

1. Lanczosでエネルギーを計算 (サイズ18サイト位)
2. TPQで比熱を計算: マヨラナ粒子の兆候がみえるか？
3. (発展)次近接のスピン相関が厳密に0を確認
4. (発展)ハイゼンベルク項をたすとどうなるか？
5. (発展)磁場をかけて磁化の温度依存性から帯磁率が計算可能
W = 3
L = 3model = "SpinGC"
method = "Lanczos"
lattice = "Honeycomb"
J0x = -1.0J0y = 0.0
J0z = 0.0
J1x = 0.0
J1y = -1.0
J1z = 0.0
J2x = 0.0
J2y = 0.0J2z = -1.0

例題5: Hubbard chain
H=tXhi,ji†(ci cj+ h.c.) + U

Xni" ni#i

1. Lanczosでエネルギー・二重占有度を計算 (サイズ8サイト位)
2. TPQで比熱・二重占有度を計算:
3. (発展) 全対角化でアンサンブル平均を計算してTPQと比較

L = 8model = "FermionHubbard"
method = "Lanczos"
lattice = "chain"t = 1.0
U = 8.0
nelec = 82Sz = 0