Intrinsic Spin Effects and Hubbard Model in Solid State Physics Lecture
Explore the concepts of intrinsic spin effects and the Hubbard model in solid state physics through an exact analysis of a 2-site system and periodic one-dimensional systems. The lecture covers the intrinsic spin of electrons, the Hubbard Hamiltonian, possible site configurations, and the Hubbard model's representation of electron hopping and repulsion. Understand how total spin values and z-component properties contribute to the system's eigenvalues.
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PHY 752 Solid State Physics 11-11:50 AM MWF Olin 103 Plan for Lecture 34 Intrinsic spin effects -- Hubbard model (Chap. 16 & 17 in GGGPP) 1. Exact analysis of 2-site system 2. Periodic one-dimensional system 11/17/2015 PHY 752 Fall 2015 -- Lecture 34 1
11/17/2015 11/22/2013 PHY 752 Fall 2015 -- Lecture 34 PHY 711 Fall 2013 -- Lecture 34 2 2
11/17/2015 PHY 752 Fall 2015 -- Lecture 34 3
11/17/2015 PHY 752 Fall 2015 -- Lecture 34 4
11/17/2015 PHY 752 Fall 2015 -- Lecture 34 5
Intrinsic spin of an electron Magnet = ic moment of g e mc an electr o n s 0 s B = = 0.05 788 me V/ T B 2 = g 2.0023193043622 1 2 0 ( ) = + + s x y z x y z Short hand notation for eigenstates of z = + = 1 1 11/17/2015 PHY 752 Fall 2015 -- Lecture 34 6
11/17/2015 PHY 752 Fall 2015 -- Lecture 34 7
The Hubbard Hamiltonian: two particle contribution single particle contribution anti-commutator for Fermi particles = { , } 0 = c c den otes a sit e l ' ' l l c denot es sp in ( or ) { , } 0 c ' ' l l = { , } c c ' ' l l ll 11/17/2015 PHY 752 Fall 2015 -- Lecture 34 8
l = 1 2 3 4 Possible configurations of a single site 0 0 c 0 c c 11/17/2015 PHY 752 Fall 2015 -- Lecture 34 9
Hubbard model -- continued t represents electron hopping between sites, preserving spin U represents electron repulsion on a single site 11/17/2015 PHY 752 Fall 2015 -- Lecture 34 10
Two-site Hubbard model ( 1 2 H t c c = ) ( ) + + + + + c c c c c c U n n n n 2 1 1 2 2 1 1 1 2 2 where n Note that total spin = 0,1 and the z-component of the total spin z=-1,0,1 commute with H and are good eigenvalues of the states of the system. For = 1, there are 3 values of z=-1,0,1 which have the same eigenvalues. ( 1 2 For 11 z c c = = = c c l l l ) 0 = Note that: 11 11 for 0 H E E 11/17/2015 PHY 752 Fall 2015 -- Lecture 34 11
Two-site Hubbard model -- continued ( 1 2 2 H t c c c = + ) ( ) + + + + c c c c c U n n n n 1 1 2 2 1 1 1 2 2 Consider all possible 2 particle states with zero spin: 0 A c c 1 1 0 B c c 2 2 1 ( ) 0 C c c c c 1 2 1 2 2 11/17/2015 PHY 752 Fall 2015 -- Lecture 34 12
Two-site Hubbard model ( 1 H t c = ) ( ) + + + + + c c c c c c c U n n n n 2 2 1 1 2 2 1 1 1 2 2 Matrix elements of Hamiltonian for all 2 particle states with spin 0: = 0 2 U t 0 2 H U t 2 2 0 t t Eigenvectors of the Hamiltonia n: Eigenvalues of Hamiltonian: + = + 2 1 1 2 U U 2 ( ) U U = + + + 1 C A B = 2 1 E t 1 4 4 t t 2 1 4 4 t t 1 ( ) E U = A B 2 2 2 2 U U = + + 2 1 E t 2 1 1 2 U U ( ) 3 4 4 t t = + + + 1 C A B 3 4 4 t t 2 11/17/2015 PHY 752 Fall 2015 -- Lecture 34 13
Eigenvalues of the Hubbard model Eigenvalues of Hamiltonian: + = + 2 U U = 2 1 E t 1 4 4 t t E U 2 2 U U = + + 2 1 E t 3 4 4 t t 11/17/2015 PHY 752 Fall 2015 -- Lecture 34 14
Two-site Hubbard model ( 1 2 H t c c = ) ( ) + + + + + c c c c c c U n n n n 2 1 1 2 2 1 1 1 2 2 Ground state of the two-site Hubbard model 2 2 1 1 2 U U U U ( ) = + + + = + 1 C A B 2 1 E t 1 1 4 4 4 4 t t t t 2 Single particle limit (U 0) 2 E t = 1 1 2 ( ) = + + C A B 1 1 2 1 ( ) 0 0 0 A c c B c c C c c c c 1 1 2 2 1 2 1 2 2 1 2 ( )( ) = + + 0 c c c c 1 1 2 1 2 11/17/2015 PHY 752 Fall 2015 -- Lecture 34 15
In the following slides, u represents U/t: 4/20/2015 PHY 752 Spring 2015 -- Lecture 34 16
k = -2 cos(ka) k k a kF a -kF 4/20/2015 PHY 752 Spring 2015 -- Lecture 34 17
In the k-basis, the Hubbard model takes the form: 1 N ( ) = + + + k k k 2cos ( ) H ka A A u A A A A k k q q ' ' ' k q q 2 q kq k k 2 where the delta function must be satisfied modulo a reciprocal lattice vector a Simple Hartree-Fock approximation = k k 0 A A HF k k k F F = E H H F HF HF 4/20/2015 PHY 752 Spring 2015 -- Lecture 34 18
note that kF=/(2a) Hartree-Fock Exact 4/20/2015 PHY 752 Spring 2015 -- Lecture 34 19
One-dimensional Hubbard chain In our notation: ( ) ( ) w e + exa t E N J w J w = 4 dw 0 1 uw c /2 (1 ) 0 4/20/2015 PHY 752 Spring 2015 -- Lecture 34 20
Approximate solutions in terms of single particle states; broken symmetry Hartree-Fock type solutions 4/20/2015 PHY 752 Spring 2015 -- Lecture 34 21
Broken symmetry Hartree-Fock solution m=0 m=1 4/20/2015 PHY 752 Spring 2015 -- Lecture 34 22
= Here, will be determin Q ed; for exa mple / corresponds to a doubled unit cell. a Q (It can be shown that the orthogonal linear combination state does not ground state wavef unction.) contribute to the = 0 k S SDW k = = S k E H E SDW SDW S D W k ) ( 1 2 1 2 1/2 ( ) ( ) 2 = + + 2 S k where E + + k k Q k k Q = Here 2 cos( ) ka k 4/20/2015 PHY 752 Spring 2015 -- Lecture 34 23
Spin density wave solution -- continued Consistency conditions on N : 1 1 1 u = ) ( 1/2 ( ) 2 + 2 k k Q + k = tan(2 ) k + k k Q Expression for energy: 1 N 1 E u ( ) ( + ) = + + cos(2 ) 1 sin(2 ) sin(2 ) SDW N k Q + k Q + k k k k q 2 N 2 4 k kq Johannson and Berggren show that: 2 sin( ( ) Qa u Elliptic integral : = / 2) K /2 d ( ) K m ( ) 1 1/2 2 1 s in m = where 0 1/2 2 ( ) + 2 1 sin / 2 Qa 1 6 4/20/2015 PHY 752 Spring 2015 -- Lecture 34 24
Spin density wave solution -- continued Expression for energ y: ( ) 2 4 E E u = + + sin( / 2) 1 Q a SDW N 2 4 u Elliptic integral : /2 ( ) 1/2 2 ( ) 1 sin E m m d 0 = Optimal solution obtained for 2 ( ) u / 2 / 2 : Qa 1 = = where K 1/2 2 + 1 1 6 ( ) 2 4 E E u = + + 1 SDW N 2 4 u 4/20/2015 PHY 752 Spring 2015 -- Lecture 34 25
Spin density wave solution -- continued : Numerical solutions for 2 K 1 = = u where 1/2 ( ) 2 + 1 16 u 4/20/2015 PHY 752 Spring 2015 -- Lecture 34 26
Spin density wave solution -- continued Effects on single particle states = Non-interacting states: 2 cos( 1 2 ) ka k ( ) 1/2 ( ) 2 = + 2 S k Spin density wave states : 4cos( ) E ka 4/20/2015 PHY 752 Spring 2015 -- Lecture 34 27
Spin density wave solution -- continued Nature of spin density wave state: cos sin k k k S A = + = = tan(2 ) k 4cos( ) ka A + k k Q k + k a 4/20/2015 PHY 752 Spring 2015 -- Lecture 34 28
SDW form: a For spin (x cos k) For spin (x sin k) 4/20/2015 PHY 752 Spring 2015 -- Lecture 34 29
1.5 SHF=FHF (m=0) 1.0 FHF (m=1/2) 0.5 FHF (m=1) 0.0 AHF E Exact -0.5 -1.0 -1.5 0 2 4 6 8 10 4/20/2015 PHY 752 Spring 2015 -- Lecture 34 30 u
