059 - pr 02c - particle and wave regimes of Hubbard model: The Hubbard Hamiltonian for quadratic h ss
and spin-interaction U s , with s the site-index and
[2.29] Hˆ Hˆ 0 Hˆ int
the spin-index, is,
h
ss
ss
dˆs† dˆs
U nˆ
ˆ ; nˆ ; nˆs dˆs† d s
† s s s
(1.1)
s
ˆ 0 and as “particles” if H ˆ 0 (how they behave when both Show that electrons behave as “waves” if H 0 int
ˆ are nonzero is a fascinating problem intimately related to wave – particle dualism). Hˆ 0 , H int ˆ 0 the Hubbard Hamiltonian describes noninteracting electrons on a lattice. Introduce the – For H the – ket ket of the int one-particle eigenkets with energies k ,
k cˆk† 0 [one[one-pa partic rticle le eigen eigenke kets ts with with ener energie giess k ]; The state
k 1;
k
(1.2)
1 of lowest energy with N / electrons of spin up/down ( , ) and momenta cˆk† cˆk is,
[2.30]
N
N
k 1 k 1
We can say that in the state
eigenket of quadratic ˆ ; H 0 E/ ; ˆ (unperturbed) H 0
cˆk†cˆk † 0
(1.3)
defined by (1.3) the electrons behave as “waves” since they have probability have probability 0
or 1 of being in some delocalized state state k cˆk † 0 (depending on
, and not depending depending on k ). To see
2
this, effect normal-ordering upon the expression of the amplitude k | 0 cˆk
N
N
cˆ k 1 k 1
N
†
cˆ
†
k k
0
N
N
0 cˆ k 1 k 1
† † k k k
N
cˆ cˆ
0 N
N
(1.4)
0 kk cˆk† k k cˆk† cˆk†cˆk†cˆk 0 kk k 0 kk k0 ; k 1 k 1
k 1 k 1
The amplitude (1.4) four different cases depending on whether k is a quantum number of an occupied or unoccupied momentum state, k [1, N ], k [1, N ] k | ; k [1, N ], k [1, N ] k | ; k [1, N ], k [1, N ] k | 0 0 0 ;
Correspondingly, the probability, k |
2
k [1, N ], k [1, N ] k | ;
, is either zero (0) or one (1) (further depending on
(1.5)
, but in a
more obvious manner).
1
†
The operators cˆk
† ˆ are cˆk are for momentum ( momentum ( k ) being a good quantum number (a.k.a., delocalized state), while dˆ s d s
for position for position ( ( s ) being a good quantum number (a.k.a., localized state). 2
†
Specifically: move all annihilation operators to the right using the anticommutation relation [cˆk , cˆk ]
kk :
ˆ† ˆ† cˆ†kcˆ†kcˆ k cˆk cˆk†cˆk† ( kk cˆk†cˆk )cˆk† kk k k ck kk c k
ˆ 0 . The Hamiltonian H ˆ is already in a diagonal form and the generic eigenket Next we discuss the case H 0 int
XY can be written as, XY dˆs† dˆs† 0 ; X , Y two arbitrary collections of atomic sites ;
(1.6)
sX sY
We can say that in XY the electrons behave like “particles” since they have probability 0 or 1 of being on a
ˆ † 0 . In analogy with (1.4), given atomic site, s d s
dˆ
s | XY 0 dˆs
sX sY
sX sY
† s
† dˆs 0
sX sY
† ˆ† 0 dˆs dˆs d 0 s
† † † † 0 ss dˆ s s s dˆs dˆs dˆs dˆs 0
sX sY
s 0 s s s 0
ss
(1.7)
The cases in (1.7) divide up exactly as they do in (1.5), illustrating particle behaviour.
ˆ , and the energy-eigenvalue of the state Compute the energy-eigenvalue of the state under the operator H 0 ˆ . XY under the operator H int The energy eigenvalue of
Hˆ 0
ˆ , is brought out by H 0
ss
† † h ss (dˆ s dˆs dˆs dˆs )
N
N
k 1
k
k 1
k
E / ;
(1.8)
ˆ The energy eigenvalue of XY is brought out by H int Hˆ int XY
U nˆ
† s s
nˆs XY
s
U dˆ s
† s
dˆs dˆs† dˆs XY
s
U
s
XY E XY XY ;
(1.9)
sX Y
ˆ , , and contrast it to Obtain an expression for the – ket which is the ground state of Hˆ H 0 0
of (1.3).
ˆ , , is obtained by minimizing (1.3)’s The state of lowest energy is (1.3). The N-particle ground state of H 0 0 3 eigenenergy (1.8) in with respect to N and N under the constraint N N N ,
( N* ) ( N * ) ( * ) N N ); ( N , N ; ) k k ( N 0; N N k 1 k 1 N
N
The system of equations resulting from (1.10) is, N* N * ( N* ) ( N * ) ( * ) 0 ; 0 ; 0 N* N* N ; N N N N
3
In two dimensions: to maximize a function f ( x, y) subject to the constraint
multipliers: namely, solve
(1.10)
(1.11)
g ( x, y) c , one must use the met hod of Lagrange
x y 0 , in which ( x, y; ) f ( x, y) ( g ( x, y) c) .
st
nd
4
The 1 and 2 equations in (1.11) are identical, implying N* N * , so the solution is, N
*
1 2
N
N /2 N /2
cˆ
N N N 0 N * N * *
† † k k
k 1 k 1
cˆ
0 cˆ cˆ
0 *
k 1 k 1
*
N N
† † k k
0
N /2
cˆ
† k
k1
cˆk† 0 ; (1.12)
Show that 0 of (1.12) is also an eigenstate of the total spin operators, defined as, ˆ (Sˆ x , Sˆ y , Sˆ z ) S
Sˆ (Sˆ , Sˆ s
s
x s
s
y s
, Sˆsz ) [total spin operator];
(1.13)
5 th j Recall the definition of the spin-operator, and the algebra they obey (in which is the th entry of the j 6 ˆ j Pauli matrix (i.e., )). We rewrite S in terms of the Fourier transform of S s j as, s ˆ
1 1 j ˆ Sˆ s j dˆs† d s 2 2V
e
iks †
cˆk
kk
j iks k
e cˆ
e i ( k k ) s 1
kk
V
2 cˆ cˆ † k
j k
e i( k k ) s ˆ j V S s ; (1.14) kk
Act upon the -ket (1.12) with the operator S s j of (1.14) in which k k , as, ˆ
j † j N /2 cˆk† cˆk cˆk cˆk † † † † 1 † j † † 1 Sˆ 0 2 cˆk cˆk cˆk cˆk 0 † j cˆ1cˆ1cˆ2cˆ2 ...cˆ † cˆ † 0 ; (1.15) † j k 1 2 cˆk cˆk cˆk cˆk j k
N 2
N 2
From (1.15) we have four terms to consider which each act upon the vacuum – ket. Let us assume that k [1, N 2 ] ˆ j 0 ), (for the case k [1, N 2 ] , we have S k 0 j j cˆk† cˆk 0 cˆk† cˆk cˆ1† cˆ1† ...cˆk† cˆk† ...cˆ N† cˆ N† 0 cˆk † j (cˆ1† )(cˆ1† )...(1 cˆk† cˆk )cˆk† ...cˆ N† cˆ N † 0 2
2
2
2
(1) 2( k 1) cˆk † jcˆ1†cˆ1† ... cˆk† ...cˆ† cˆ † cˆk † (cˆk † )...(cˆ† )(cˆ† )cˆk 0 N 2
N 2
N 2
N 2
(1.16)
† j j † † † † † † 2( k 1) j j cˆk cˆk 0 (cˆ1 )(cˆ1 )...cˆk cˆk ...cˆ N cˆ N 0 (1) 0 0 ; 2
2
Having done (1.16) in full bloody detail, we can use some of the reasoning presented there to skip some steps. 7 We see that the null -ket is produced ,
4
In the last step, we used the property:
N
even
and
N /2 i 1
N /2
ˆ Bˆ A i j j 1
N /2
i1
ˆ Bˆ (1)2 N /2 A i i
N /2 i1
ˆ Bˆ for the special case of A i i
ˆ , Bˆ ] 0 (i.e., an even number of fermions obeying the appropriate anticommutation-relations). [ A i j
5
From 034 - pr 08 - the 2nd quantized spin operators, we have,
[1.92] Sˆ s j
1 2
dˆ
† s
j ˆ ˆi ˆ j d s ; j x, y, z; [ S s , S s ] i ss
ijk Sˆ sk ;
k x , y, z
However,
ˆ d n
U k R.C .
cˆ
nk k
1 Vb
We see that (1.14) is in position-space ( dˆ s
6
†
e
ik n
kR.C .
D.C . cˆ k dˆ n†
U kR.C .
* † nk k
cˆ
1 V b
e
ik n †
k R.C .
cˆ k ;
d ˆs , atomic-site index s ), while (1.12) is in momentum space ( cˆk† cˆk ,
momentum-index k ); we must effect a canonical transform of (1.14) to momentum space. 7 Justification for the braced-claim in (1.17) (below) is found in 033 - pr 04 - proof of pauli exclusion principle.
j j j cˆk† cˆk 0 cˆk† cˆk cˆ1†cˆ1† ...cˆk†cˆk† ...cˆ N† cˆ N† 0 cˆk † cˆ1†cˆ1† ...cˆk † ...cˆ N† cˆ N † 0 2
2
2
2
zero in acting upon any -ket j cˆ1†cˆ1† ...( (1) 2( k 1)
cˆk †cˆk †
(1.17)
)...cˆ N† cˆ †N 0 0 ; 2
2
rd
nd
The evaluation of the 3 term of (1.16) proceeds in a manner identical to the 2 term evaluation done in (1.17), j j j cˆk† cˆk 0 cˆk † cˆk cˆ1† cˆ1† ...cˆk† cˆk † ... cˆ N† cˆN† 0 cˆk † cˆ1† cˆ1† ...cˆk † ... cˆ†N cˆ†N 0 ( 1)2(k 1) 0 0 ; (1.18) 2
2
2
2
th
st
Unsurprisingly, the evaluation of the 4 term of (1.16) proceeds in a manner identical to the 1 term evaluation j j done in (1.16), so we needn’t provide the gory detail. In fact, one just replaces cˆk † cˆk with cˆk † cˆk , j j cˆk† cˆk 0 cˆk † cˆk cˆ1† cˆ1† ...cˆk† cˆk † ...cˆ†N cˆN † 0 j 0 ; 2
(1.19)
2
th Hence, (1.15) becomes the trace of the j Pauli matrix; all three Pauli matrices are traceless, so we get
eigenvalues of zero as promised (thus proving tha t 0 is a magnetic-singlet state), j j ) 0 12 Tr j 0 12 0 0 0 0 S ˆk j 0 0 0 0 ; Sˆk j 0 12 (
(1.20)
ˆ 0 (wave) and H ˆ 0 (particle) exhibit Pauli paramagnetism (that is a Show that both regimes H 0 int system whose total spin is zero for B 0 and grows parallel to B for B 0 ) 8
If we perturb the system with a weak external magnetic field B along, say, the z-axis ( B Bz ˆ ) and discard the 9 coupling to the orbital motion , the noninteracting part of the Hamiltonian (1.1) changes to, ˆ B ˆ ); (1.21) Hˆ h dˆ † dˆ Hˆ Hˆ g S h dˆ† dˆ 1 g B( Nˆ N 0
ss
ss
s
s
0
0
B
ss
ss
s
s
2
B
In (1.21) we introduced,
g electron gyromagnetic ratio 2; B Bohr magneton ; Nˆ
nˆ s
s
ˆ ; ; nˆs dˆs† d s
(1.22)
† It will be handy to have transformed the Hamiltonian (1.21) to the cˆk cˆk basis as done in (1.14).
ˆ (the latter appearing in N ˆ and dˆ † d ˆ of (1.22), and the former Transforming the individual terms dˆ s† d s s s 10
being diagonal ), kinetic energy diagonal
1 dˆ s† dˆs Vb
kk
ei ( k k ) s cˆk† cˆk
1 V b
e kk
i ( k k ) s
dˆ s† dˆ s ; dˆ s† dˆ s dˆ s† dˆ s ;
(1.23)
The Hamiltonian (1.21) is rewritten using (1.23), and it appears as, 8
The Pauli paramagnetic behavior is due to the spin degrees of freedom and it is therefore distinct from the paramagnetic behavior due to the orbital degrees of freedom, see 089 - continuity and paramagnetic vs diamagnetic currents. 9
The magnetic moment due to an orbiting species of mass m , charge
q , and angular momentum L is L
presuming the nonrelativistic regime which, since p mv (q / c) A (for 10
q 2 mc
L , so we are
B A ), is defined by p (q / c)A i .
Not so if one wishes to consider a tight-binding (rather than free) model: then, one would replace
1 Hˆ 0 V b 1
Vb
e
i ( k k ) s
kk s
h s
k
† † † 1 ˆ ˆ ˆ ˆ ˆ ˆ h c c g B ( c c c c ) s s k k B 2 k k k k s
1
V b
cˆ cˆk 12 g B B(cˆk†cˆk cˆk†cˆk )
† ss k
e kk
i ( k k ) s
(1.24)
Hˆ 0 ;
The eigenkets (1.3) are also eigenkets of this new Hamiltonian (1.21) but with a different eigenvalue. Using the new representation (1.23), and the resulting Hamiltonian (1.24), we have, N N † † ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ h c c g B ( c c c c ) c c 0 k B k k k k k k s k 1 k 1
Hˆ 0
† ss k
†
1 2
†
(1.25)
nd
By the same steps in (1.16) through (1.19), the 2 term in (1.25) becomes, k[1, N ]; or k[1, N ]
cˆ cˆk cˆ cˆk † k
† k
N
N
cˆ k 1 k 1
† † k k
cˆ
0 1
N
N
cˆ k 1 k 1
† † k k
cˆ
0 1 ; ( );
(1.26)
Using (1.26) in (1.25), we have,
Hˆ 0
h s
† † cˆ cˆk 12 g B B(cˆk cˆk cˆk cˆk )
† ss k
s
s
12 g B B((1) ( 1))
N N ˆ s g B B H 0 k k ( N N ) g B B ; k 1 s k 1
(1.27)
Thus, in the presence of an external magnetic field the state (1.3) is no longer the lowest in energy since for, e.g., B 0 , it becomes energetically convenient to have more electrons of spin up than of spin down. This is the typical behavior of a Pauli paramagnet, that is a system whose total spin is zero for B 0 and grows parallel to B for B 0 .
ˆ 0 . The Hamiltonian H ˆ of (1.1) is already in a diagonal form (as illustrated in Next we discuss the case H 0 int (1.9)) and the generic eigenket was found to be (1.6) (an eigenket of position, as shown in (1.7)), and the eigenenergy was computed in (1.9) to be E XY
U s . The ground state(s) for a given number N N N
sX Y
was computed in (1.10) through (1.12). Let N be the total number of atomic sites. For N N all states with V
X
V
Y 0 are ground states (with zero energy) and, again, the system behaves like a Pauli paramagnet . 11
ˆ nor H ˆ In conclusion, neither H favors any kind of magnetic order. However, their sum (1.1) sometimes int 0 does. In 064 - Heisenberg model we illustrate an example of this phenomenon.
11
I don’t see how this line of reasoning is sensitive to
B 0 vs. B 0 .
