The Kondo Effect

7/30/2019 The Kondo Effect

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The Kondo Effect

Marc Alomar, Yuxuan Chen and Rex Lundgren

Department of Physics,

The University of Texas at Austin,

Austin, TX 78712,

USA

In this report, we study the Kondo effect problem. First, an effective Hamiltonianis derived from an Anderson impurity model. Then, by considering the effect of themagnetic impurities perturbatively, we show that the resistivity diverges at second orderat low tempertures. Finally, we study the non-perturbative effects of the magneticimpurity, using the renormalization group method.

The resistivity of pure metals usually drops as its tem-perature is lowered. However, when magnetic impuritiesare added, the low-temperature behavior changes dra-matically: the resistivity reaches a minimum, and thenincreases as the temperature is lowered further. Thisphenomenon was first observed in the 1930s1 , but it was

not until 1964 when a theoretical explanation was givenby J. Kondo.2 The effect was described in terms of theexchange interaction between a localized magnetic mo-ment and the conduction electrons, and Kondo used aperturbative approach to find the resistance minimum.However, his theory predicted a divergence at T 0,inconsistent with the experimental results. The correctexplanation at low temperature was given by Andersonusing the method of the renormalization group.3

In this report we study the Kondo effect using the tech-niques of many-body theory. In section I, we study howthe sd-Hamiltonian is obtained from the Anderson im-purity model. In section II, we use second-order pertur-

bation theory to obtain the temperature dependence ofthe resistivity. We find that this analysis is only valid upto the characteristic Kondo temperature, TK. Finally, insection III we use the Poor mans scaling approach tounderstand the behavior below TK.

I. MODEL HAMILTONIAN

In this section we derive the sd-Hamiltonian from theAnderson impurity model. The Anderson Hamiltonian isgiven by

H =k

kckck +

Vkdck + h.c.

+

dnd + U ndnd (1)

1 W. J. de Haas et al., Physica 1, 1115 (1934)2 J. Kondo, Prog. Theor. Phys. 32, 37 (1964)3 P. W. Anderson, J. Phys. C 3, 2436 (1970)

where the operators d create electrons of spin at the

localized impurity, and ck creates an itinerant electronof spin and energy k in the metallic host. The totalwavefunction of the Hamiltonian, |, can be written asthe sum of three terms, |0, |1 and |2, where thesubscript refers to the occupancy of the impurity site.

Using this decomposition, we can write the Schrodingerequation in matrix form,

2n=0

Hmn |n = E|m (2)

where Hmn = PmHPn. The operators Pn are the projec-tors onto the subspace with n electrons on the impuritysite. They satisfy the relation P0 + P1 + P2 = 1, and aregiven by

P0 =

(1 nd)

P1 = 1 P0 P2

P2 = ndnd

(3)

Now, lets find the matrix elements Hmn. The diagonalelements leave the occupation number fixed, and they canbe identified with the diagonal terms of the AndersonHamiltonian,

H00 = k kc

kck

H11 =

k kc

kck + dH22 =

k kc

kck + 2d + U

(4)

The off-diagonal terms are due to the hybridizationterm of the Hamiltonian, Hhyb =

k(Vkd

ck + h.c.).

This term only contains one d operator, and therefore itcan only connect states that differ by one impurity elec-tron. For this reason, H20 = H02 = 0. The other matrixelements can be found using the explicit expressions ofP0 and P2,


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H10 = P1HhybP0 = (1 P2 P0)HhybP0 = HhybP0

=k

Vkdck(1 nd)(1 nd) =

k

Vkdck(1 nd)

(5)

H21 = P2HhybP1 = P2Hhyb(1 P0 P2) = P2Hhyb

=k

Vkndnddck =

k

Vkdndck

(6)

where we have used ndndd = ndd

nd = d

nd and

ndndd = d

nd. The other matrix elements follow

from Hmn = Hnm,

H01 = k Vk ck(1 nd)d

H12 =

k Vk ckndd(7)

We are interested in the effect of virtual excitationsfrom the |1 subspace. Thus, we have to write (2) onlyin terms of |1. Writing out (2),

H00 |0 + H01 |1 = E|0

H21 |1 + H22 |2 = E|2 (8)

we can formally define |0 and |2 in terms of |1,

(E H00)1H01 |1 = |0

(E H22)1H21 |1 = |2 (9)

Now we have

E|1 = H10 |0 + H11 |1 + H12 |2

= [H101

E H00H01 + H11 + H12

1

E H22H21] |1

(10)

In order to simplify the above expression we have to

commute the terms (EH)1 and H. First, we commuteonly the matrix elements,

H00H01 =

kk

Vkknkck(1 nd) d

=

kk

Vkkck(kk c

kck)(1 nd)d

(11)

Cleaning up this expression gives

H00H01 =k

Vkkck(1 nd)d

kk

Vkkckc

kck(1 nd)d = H01(k + H00)

(12)

One can do the same thing for H00H21.

H00H21 =

kk

Vkknkdndck

=

kk

Vkkdnd(kk ckc

k)ck

(13)

Cleaning up this expression as done with H00H01 gives

H00H21 =

kk

Vkdndck(k + H00) =

H21(k + H00) (14)

Using the above expressions its straightforward to ob-tain

(E H00)1H01 =

kk

Vkck(1 nd)d(E k H

= kk

Vkck(1 nd)d

k d(1

E d H0k d

kk

Vkck(1 nd

k d

(15)

and

(E H22)1H21 = H21(E+ k H22)

1

=

kk

Vkdndck(E U 2d H00 + k)

1

=

kkVk d

nd

ck

U + d k

1 E d H00U + d k1

kk

Vkdndck

U + d k

(16)

where we have expanded the expressions in parenthesesto leading order in U and d. Plugging these expressionsin (10) we get


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E H11

kk

[VkVk

d(1 nd)ckck(1 nd)d

k d+

+VkVk

ckndddndck

U + d k]

(17)

The one-particle terms nd can be dropped, becausethey act on a single-particle subspace. If we swap k andk in the first term, we obtain

EH11 =

kk

VkVk

dckc

kd

k d+

ckddck

U + d k

(18)

Finally, using the fermonic commutation relations weobtain

E H11 =

kk

VkVk [

( ckck)dd

k d+

ckck( dd)

U + d k] =

kk

VkVk(

1

U + d k+

+1

k d)ckckd

d

kk

(ckck

U + d k

ndk d

)

(19)

The last term in the second parenthesis is a constant,and can be absorbed into a shift of the single-particle

energy of the itinerant electron. This expression can berelated to the spin operators by using the Pauli identity,

= 2 (20)

In particular, we can rewrite

ckckdd =

ckckdd

=1

2 ckckd

d ( + )

=1

2

ckckdd +

1

2

ckcknd

= 2skkSd +1

2

ckcknd (21)

where Sd =12

d

d is the impurity spin and

skk =

ckck is the conduction electron spin.

Using this result, we can finally write

E H11 =kk

Jkk

2skkSd +

1

2

ckcknd

kk

VkVk

ckckU + d k

=kk

2JkkskkSd + Kkk

ckck

(22)

where

Jkk = VkVk

1

kd+ 1U+dk

Kkk =

VkV

k

2

1

kd 1U+dk

(23)The total effective Hamiltonian acting on the subspace

|1 is

Heff =k

kckck+

kk

2JkkskkSd + Kkk

ckck

(24)

The energy scale of the typical excitations is muchsmaller than U + d and d. Thus, we can neglect theenergy dependence of Jkk and Kkk . In that case, theinteraction Jkk can be treated as local, and the scatter-ing term Kkk can be absorbed into a shift of the single-particle energy of the itinerant electron. These approxi-mations lead to the sd-Hamiltonian,

Hsd =k

kckck + 2JSd s(r = 0) (25)

II. IMPURITY SCATTERING

The goal of the this section is to calculate the effectof magnetic impurties on transport properties. To do so,we calculate the scattering rate. This is given by

1

2= Nimp

k,

< | < k, |T+()|k, > |2 >s (kk)

(26)where T is the T-matrix. This formula is derived in

problem 5.5.3 and the T is given by

Tn < Himp + HimpG0,nHimp + ... > (27)


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where n is the Matsubara frequency. This equation issimply Fermis Golden Rule. Once we obtain the scatter-ing time, we can estimate the resistivity via the Drudeformula = m/e2n, where n is the electron density.

A. First Order Correction

To first order T = Himp. Now we work out the Matrixelement. First we use the fact that

< k, |Himp|k, >= JLd < |S | > (28)

Plugging this into the formula for T gives

1

2= NimpJ

2L2d

(< |S | >< |S | >)k

(k k) =

cimpJ2 >s=

J2S(S+ 1)cimp(29)

Here we have used the fact

k (k k) = Ld and

>s= S(S+ 1) 1. We have also usedthe fact

>s=>s (30)

In this model the average occupation number is 1, sothe spin is 12 , thus the linear term is zero in thise case

since the S = 12 terms cancel each other out.Thus the first order correction is

1

2= J2S(S+ 1)cimp (31)

The key result here is that it does not explain the ex-istence of a resistance minimum

B. Second Order Perturbation Theory

We now turn to second order perturbation theory. Inthis case the scattering matrix is given by

T(2) = Himp( H0)1Himp (32)

In order to get the matrix element we first calculateH|k, >= Himpc

k| >= 2JS sc

k| >. s is the spin

of the conduction electron and | > is the Fermi Sea.Plugging in what s is in term of operators give

Jk1,k2

1,2

S 1,2ck11

ck22ck| >=

Jk1,1

S 1,ck1,1

| >

J k1,k2

1,2

S 1,2ck11

ckck22 | > (33)

Because | > is the Fermi sea, ck11 is non-zero whenk1 is greater then kf. We also have k2 < kf and k1, k >kf in the second term. This term is a combination ofone-particle and two-particle-one-hole excitations.

Looking at the matrix element now, we can see thatthe cross terms between one and two particles will bezero. Lets look at the one-particle term first. This gives

J2 < |

k11 k11(S ,

1)ck

11

1

+ H0S (,1)c

k11

| >

(34)

Focusing on the denominator we see

1

+ H0ck1,1 | >=

1

+ k1ck1,1 | > (35)

The previous term becomes

J2 < |

k11 k11(S ,

1)(S ,1)

ck11

ck11+ H0

| >

(36)

Using the commutation relations

ck1,

1ck11 = k1,k1;1, c

k11

ck1,

1(37)

and the fact that the second term vanishes because itcounts the number of particles above the fermi sea, whichis zero, this becomes

J2(S ,1

)(S ,1)k1

(k1)

k k1(38)

One can follow similar steps for the two particle term.This is done in the appendix. It is given by

J2(S )(S )p

(p)

k p(39)

Thus the total matrix element is


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FIG. 1 Two possible diagrams for the matrix element.

J2((S )(S )k1

(k1)

k k1+

(S )(S )p

(p)

k p) (40)

The next step is to evaluate some of these spin terms.

(S )(S ) = SjjSi

i =

SiSj(ij) = SiSj(ij ijkk) =

S2 S (41)

One can follow similar steps for the other spin term forthe two particle part of the matrix element. Thereforethe matrix element can be written as

J2((S2 S)p

(p)

k p+

(S2 + S)

p(p)

k p) (42)

One can simplify and combine these two equations toget

J2((S(S+ 1)p

(p) + (p)

k p+

S)p

(p) (p)

k p) (43)

At finite temperature (E) = nf(E) and (E) =1 nf(E) and we can even further simpilfy the matrixelement to

< k , |T(2)|k, >= J2p

1

k p(S(S+ 1)+

+(2nf(p) 1) S)(44)

The first term is dropped because it is a non-singularconstant with no dependance on temperature. Rewriting2nf(p) 1 = tanh(p/2) and squaring gives

| < k , |T(2)|k, > |2 =

J2(p

tanh(p2 )

k p)2(< (S )2 >s=

J2(p

tanh(p2 )

k p)2S(S+ 1)

Now we approximate this integral. Using p andp

gives

p

tanh(p2 )

k p=

DD

dtanh(2 )

k 2ln(

D

kbT) (45)

We notice that we can now write an effective J as

Jeff = J(1 + 2ln(D

kbT)) (46)

Finally combining the first and second order pertuba-

tions give

1

2= cimpJ

2effS(S+ 1) =

1

20(1 4Jln(

T

D))

(47)Thus, we see when T=0 this diverges.

III. RENORMALIZATION GROUP

The model we consider is s-d Hamiltonian. One can de-rive it from the Anderson Impurity Hamiltonian as seen

in an eariler section. This is the subject of Problem 2.4in Atland and Simons. The Hamiltonian is given by

H =k

kckck +

kk

JzSz(ckck c

kck)+

J+S+ckck + JS

ckck (48)

where S represents the spin of the impurity electron.This model allows for possible anisotropy. Lets consider


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removing degrees of freedom at the top of the conduc-tion band in a region D/b < |k| < D where D is theconduction bandwidth and b > 1. Following problem 2.4we can use projection operators and can come up withan effective Hamiltonian for the case when no conductionelectrons are in the high energy portion the conducitonband. This allows us to see virtual flucations to the highenergy part of the electron band. The Schodinger equa-

tion for the wavefunction with no conduction electrons,|1 > is

(H101

E H00H01 + H11 + H12

1

E H22H21)|1 >= E|1 >

(49)

Where Hmn = PmHPn, where Pm is the projectionoperator. Let us focus on the term that scatters a con-duction electron in the upper region of the band.

H121

E H22 H21|1 > (50)

There are eight possible terms that arise from thisterm. One can see these eight terms from Feynman di-agrams.The allowed terms must take return to the sub-space where the high-energy part of the conduction bandremains unoccupied. Some of these processes involvespin-flips. An example of one term is

J+J

kskskf

Scksckf1

E H22S+ckfcks|1 > (51)

where kf lies withn the band edge and ks lies outsidethe band edge.

A. Deriving RG Equations

Lets clean up this term. Lets look at

kf

cksckf1

E H22ckfcks|1 > (52)

Next we use that fact that the wavefunction has no

electrons in the upper conduction band. Thus ckfc

kf =1 and we are left with a sum over kf. We also assumekf D. We can evaluate this sum using the density ofstates.

kf

=

DD/b

d() 0D(1 1/b) (53)

Thus we arrive at

kf

cksckf

1

E H22ckfcks|1 >

0D(1 1/b)cks

cks1

E H22|1 >

0D(1 1/b)cks

cks1

E D + ks H0

|1 > (54)

where H0 is the energy of the electrons in the bandwithout the impurity. Thus we have

J+Jksks

0D(1 1/b)cks

cksSS+

E D + ks H0|1 >

(55)

Next we approximate H0 as zero and we use the spincommutation relations to obtain

J+Jksks

0D(1 1/b)(1/2 Sz)cks

cks1

E D + ks|1 >

(56)

Doing this for the other eight equations give J as afunction of b. The two equations ares

J(b) = J JzJ0D(1 1/b)(1

E D + k+

1

E D + k) (57)

Jz(b) = Jz J+J0D(1 1/b)( 1E D + k

+

1

E D + k) (58)

Next we ignore E compared to D and since typicalinternal excitations small compared to D we can approx-imate these as

J(b) = J + 2JzJ0(1 1/b) (59)

Jz(b) = Jz + 2J+J0(1 1/b) (60)

These give the renormalization group equations

dJdl

= 20JzJ (61)

and

dJzdl

= 20J+J (62)


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FIG. 2 RG Flow.

B. Analysis of RG equations

One notes we can divide the difference equations toobtain

dJdJz

=JzJ

(63)

One can integrate these equations to obtain

J2z J2 = constant. (64)

It is important to note that Jz always increases underrenormalization. Thus when |Jz| < J, J will scaleto zero. If |Jz| > J, both will flow to strong couplingwhere the perturbation theory breaks down. A figure ofthe RG flow is shown in Fig 2.

For Jz = J = J the scaling equation becomesdJ/dl = 20J

2. Integrating this equation gives

1

J

1

J(l)= 20l = 20ln(

D(l)

D) (65)

Therefore, we have

Dexp(1

0J) = D(l)exp(

1

0J(l)) = kbTk (66)

where Tk

is the Kondo Temperature. Thus when T =Tk we have scale invariance. We can write J as

J =J0

1 + J0ln(DD0

)(67)

and we see it will diverege when you are at an energyscale of Tk. However one must keep in mind this ap-poarch is only valid for small J and keeping correctionswill change the function.

FIG. 3 Screening by a magnetic impurity.

C. Large Coupling Limit

Finally we conclude with a few words about the largecoupling limit, which is predicted by RG flow. In thelarge J limit, the Hamiltonian of the system is just

H =kk

JzSz(ckck c

kck)+

J+S+ckck + JS

ckck (68)

In this case the impurity spin and some of the conduc-tion electrons form a spin singlet state, with each con-duction electron contrubting a fraction of its total spin.A physical picture of this effect is imagining the spinimpurity surrounded by a sea of conduction electrons ofopposite, but smaller spin. This is seen in Fig 3.

Thus the spin impurity is screened and has a screeninglength, which one can determine from the Kondo Tem-perature. This length scale or radius of the cloud is givenby Rk vf/kbTk.

D. Exact Solutions

We close this report by mentioning some exact meth-ods to investigate the large J limit exactly. Thereare two common approachs to the Large J limit,Numerical Renormalization Group pioneered by Wil-son and the Bethe Ansatz. Both of these, alongwith other methods, are discussed in Hewsons book,The Kondo Problem to Heavy Fermions.


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IV. WORK DONE BY GROUP MEMBERS

Marc Alomar worked on the introduction and Section1. Yuxuan Chen worked on Section 2. Rex Lundgrenworked on Section 3.

V. APPENDIX

In the Appendix, we derive the two particle overlap,that is needed to calculate the scattering time. It is givenby

k1,k2,k1,k

2

1,2,1,

2

J2 < |(S 2,

1)(S 1,2)c

k22

ckck11

1+ H0

ck11ckck22 | >

Now we look at how the first three states act on theFermi Sea.

1

+ H0ck11c

kck22 | >=

ck11ckck22

k (k1 + k k2)| >

(69)

Plugging this in gives

k1,k2,k1,k

2

1,2,1,

2

J2(S 2,

1)(S 1,2)

< |ck22

ckck11

ck11ckck22 | >

k (k1 + k k2)

Now the goal is to evaluate the six operator piece. Todo this, first look at

ckck11

ck11ck = ck11ckc

k11

ck =

ck11

(k1,k1;1 ck11

ck)ck (70)

one can continue moving particles past and arrive at

ckck11

ck11ck = k1,k1;1,k1,k1;1,+k1,k1;1,1k,k;

(71)

One can then ignore the second term because it doesntnot contribute to scattering (it is possible however thatit will renormalize single-particle energy but it will notdepend on temperature), and thus we see the matrix el-ement is

J2(S )(S )p

(p)

k p(72)

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The Kondo Effect