Goldstone Bosons in Condensed Matter System

Outline of lectures:

1. Examples (Spin waves, Phonon, Superconductor)

2. Meissner effect, Gauge invariance and Phase mode (GS boson)

3. Higgs mode (amplitude mode) in Superconductors

Condensed matter systems exist because of broken symmetries: Translation, Rotation, U(1) gauge, etc

Goldstone theorem (’62): “ breaking a continuous symmetry should cause massless excitations.This is natural response of the system to restore the broken symmetry.”

Goldstone modes in CM:Phonons (??), Spin-wave (‘40), phase mode in SC (‘57-60), CDW, etc

Goldstone mode in SC Higgs mechanism (Meissner effect)Higgs mode (amp. Mode) in SC always exists but its observations are rare.

Goldstone modes are massless:

So, the energy dispersion relation E or ~ k

In CM or in Nature, we have ~ k and ~ k2

Question (very important ) : what and/or how is the power determined.w ~ k may be more familiar, but ~ k2 is more naturaland ~ k needs special condition.

jkj =p

k2

Rotation symmetry SO(3) of spin & spin-wave

H = ¡ J Pi j Si ¢Sj Ferromagnetism with J >0

Ground state with E0=-J NS2

Low energy Excitations.What is the Eexc ?

This is classical picture.

Lowest excitation, but not Eigenstate

H = ¡ J Pi j Si ¢Sj = ¡ J P

i j [Sxi ¢Sx

j + Syi ¢Sy

j + Szi ¢Sz

j ]

H j0>= E0j0>

j #j >= S¡j j0>

Coherent superposition of |j >

H jk >= 1pN

Pj ei kr j [¡ 1

4ºJ (N ¡ 2)j #j > +12ºJ j #j >

¡ 12J P

±(j #j +±> +j #j ¡ ±>]

f :::g= 1¡ coska » k2

H j #j >6= ¸j #j >

So, the Goldstone boson in Ferromagnetism hasWe are familiar with E(k) ~k dispersion of the massless modes in rel. field theory.What is the difference and origin for this ?

E (k) » k2

The same method has a difficulty to deal with

Quantization of Ferromagnetic spin-wave (Holstein-Primakoff method)

j #j >= S¡j j0>

Consider this is a creation of a boson ! ayi j #j >= ay

j j0>

j0>=j #j >=

In each site, Si has a definite Sz value

nS¡ j ">= nayj ">= jn >

and Sz jn >= (S ¡ n)jn >

jk >= ayk j0> with ay

k = 1pN

Pi ei kr i ay

i

Sz counts S - (boson number)

This mapping looks good.

Spin commutation rules & Boson commutation rule slightly mis-match.

Cure

= 2(S-n) 1

Just the same result as before, E (k) » f1¡ °(k)g= 1¡ coska » k2

Linear approximation !!

= 2(S-n)~ 2S

Now let us consider Antiferromagnetism.

H = ¡ J Pi j Si ¢Sj Same Hamiltonian but J <0 gs is different.

Not every site is the same.There are A site & B site

Unit cell increases, BZ decreases. “Doubling”

Creation and annihilation change its role in A and B sites.

H = ¡ J Pi j Si ¢Sj

Simple trick: Rotate all spins on B sites by 180 by Sx then,

S§j ! +S¨

j ;Szj ! ¡ Sz

j then

H = ¡ J =2Pi (j )[S(ay

i ayj + h:c:) + S(ay

i ai + ayi ai ) ¡ S2]

! H = ¡ 12N zS2 + zS P

k aykak + zS P

k °(k)(aykay

¡ k + aka¡ k)

! H = ¡ J Pi<A

Pj <B [1

2(S+i S+

j + h:c) ¡ Szi Sz

j ]

Ferromagnetic case

H = ¡ 12N zS2 + zS P

k aykak + zS P

k °(k)(aykay

¡ k + aka¡ k)

Just like a SC (but with Boson) and A-B site symmetry 2x2 mtx. diagonalization

H = (ayka¡ k)

µ 1 °(k)°(k) ¡ 1

¶ µ akay

¡ k

¶

E 2(k) = [°2(k) ¡ 1] » k2, so E (k) » k

“Doubling” makes k-linear dispersion naturally ap-pear !!Dirac Eq. has doubling : E2 = p2+ m2

Ferromagnetic Goldstone boson: E (k) » k2

Anti-Ferromagnetic Goldstone boson: E (k) » k

- -

CF (T) » T3=2 and

CA F (T) » T3 like photon and phonon

Spinwave dispersion

H = (cyk;" c¡ k;#)

µ ²(k) ¢¢ ¡ ²(k)

¶ µ ck;"cy¡ k;#

¶Superconductivity also has doubling in p-h states with fermions .

E 2(k) = [²2(k) + ¢ 2] , so E (k) = §p

²2(k) + ¢ 2 ! not E (k) » k

This is a quasi particle dispersion not a dispersion of Goldstone boson.

Goldstone boson in SC has indeed E(k) ~ k

We will come back to this question.

Acoustic phonon (Goldstone boson) dispersion : E(k) ~ k

H = 12

Pi [p2

i + (qi+1 ¡ qi )2]

Quantize : [qi ;pj ] = i±i j

qi = 1pN

Pk eikr Qk and pi = 1p

NP

k ei kr Pk

H = 12

Pi [PkP¡ k + (1¡ cos(k))QkQ¡ k]

Hho = [p2 + ! 2x2] and E »p

! 2 » !Harmonic oscillator

ayk = (2! k)¡ 1=2[! kQ¡ k ¡ iPk] ; ak = (2! k)¡ 1=2[! kQk + iP¡ k]

H = Pk ! kay

kakwith ! (k) =

p[2(1¡ cos(k))] » k

Goldstone modes in CM are abundant .They are massless.

In CM or in Nature, we have ~ k and ~ k2

Question (very important ) : what and/or how is the power determined. ~ k2 is more natural and ~ k needs special condition “Doubling”.

jkj =p

k2