### Goldstone modes are massless

```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:
~k
So, the energy dispersion relation E or
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 natural
and ~ k needs special condition.
jkj =
p
k2
Rotation symmetry SO(3) of spin & spin-wave
H = ¡ J
P
ij
Si ¢Sj
Ferromagnetism with J >0
Ground state with E0=-J NS2
Low energy Excitations.
What is the Eexc ?
This is classical picture.
H j0 > = E 0j0 >
j #j > = Sj¡ j0 >
Lowest excitation, but not Eigenstate
H j #j > 6
= ¸ j #j >
Coherent superposition of |j >
H jk > =
p1
N
P
j
ei k r j [¡
¡
1
4 º J (N ¡
P
1
± (j
2J
2)j #j > + 12 º J j #j >
#j + ± > + j #j ¡ ± > ]
f :::g = 1 ¡ coska » k2
H = ¡ J
P
ij
Si ¢Sj = ¡ J
P
ij
[Six ¢Sjx + Siy ¢Sjy + Siz ¢Sjz ]
E(k) » k2
So, the Goldstone boson in Ferromagnetism has
We are familiar with E(k) ~k dispersion of the massless modes in rel. field theory.
What is the difference and origin for this ?
The same method has a difficulty to deal with
Quantization of Ferromagnetic spin-wave (Holstein-Primakoff method)
j0 > =
j #j > = Sj¡ j0 >
j #j > =
y
Consider this is a creation of a boson ! ai
j #j > = ayj j0 >
In each site, Si has a definite Sz value
nS¡ j " > = nay j " > = jn >
and Sz jn > = (S ¡ n)jn >
This mapping looks good.

jk > =
ayk j0 >
with
ayk
=
p1
N
P
i
ei k r i ayi
Sz counts S - (boson number)
Spin commutation rules & Boson commutation rule slightly mismatch.
= 2(S-n)
Cure
1
Linear approximation !!
= 2(S-n)
~ 2S
Just the same result as before,
E(k) » f 1 ¡ ° (k)g = 1 ¡ coska » k2
Now let us consider Antiferromagnetism.
H = ¡ J
P
ij
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
P
ij
Si ¢Sj
Simple trick: Rotate all spins on B sites by 180 by Sx then,
Sj§ ! + Sj¨ ; Sjz ! ¡ Sjz then
!
H = ¡ J
H = ¡ J=2
!
H = ¡
P
P
P
i< A
+ +
1
z z
[
(S
S
+
h:c)
¡
S
i Sj ]
i
j
j<B 2
y y
y
y
2
[S(a
a
+
h:c:)
+
S(a
a
+
a
i
i j
i
i ai ) ¡ S ]
i (j )
1
2
2 N zS
+ zS
P
y
a
k k ak
+ zS
P
y y
°
(k)(a
k
k a¡ k + ak a¡ k )
Ferromagnetic case
H = ¡
1
2
N
zS
2
+ zS
P
y
a
k k ak
+ zS
P
y y
°
(k)(a
k
k a¡ k + ak a¡ k )
Just like a SC (but with Boson) and A-B site symmetry  2x2 mtx. diagonalization
µ
H = (ayk a¡ k )
1
° (k)
° (k)
¡ 1
¶µ
ak
ay¡ k
¶
E 2(k) = [° 2 (k) ¡ 1] » k2 , so E(k) » k
“Doubling” makes k-linear dispersion naturally appear !!
Dirac Eq. has doubling : E2 = p2+ m2
Ferromagnetic Goldstone boson:
E(k) » k2
Anti-Ferromagnetic Goldstone boson:
E(k) » k
Spinwave dispersion
-

-

CF (T) » T 3=2 and
CA F (T) » T 3 like phot on and phonon
Superconductivity also has doubling in p-h states with fermions .
µ
H = (cyk ;" c¡
2
2
k ;# )
²(k)
¢
¢
¡ ²(k)
¶µ
2
E (k) = [² (k) + ¢ ] , so E(k) = §
p
ck ;"
cy¡ k ;#
¶
² 2 (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 =
1
2
P
2
2
[p
+
(q
¡
q
)
]
i
+
1
i
i i
Quantize :
qi =
p1
N
H =
1
2
P
P
[qi ; pj ] = i ±i j
i kr
k e Qk and pi =
i [Pk P¡ k
1=2
[! k Q¡
P
P
k
ei k r Pk
+ (1 ¡ cos(k))Qk Q¡ k ]
Harmonic oscillator
ayk = (2! k ) ¡
p1
N
2
2 2
H ho = [p + ! x ] and E »
k
¡ i Pk ] ; ak = (2! k ) ¡
H = k !pk ayk ak
wit h ! (k) = [2(1 ¡ cos(k))] » k
1=2
p
!
[! k Qk + i P¡ k ]
2
» !
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
and
determined.
~ k2 is more natural
~ k needs
special condition  “Doubling”.
jkj =
p
k2
```