Download - Goldstone Bosons in Condensed Matter System
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