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### Transcript of 1 Fermions in one · PDF file 2g( x) (x) (x) 1. To simplify matters even further we consider...

• 1 Fermions in one dimension

We will be using the effective continuum Hamiltonian:

H = ∑ σ=±

∫ dx ψ†σ(x)

( − ivf

∂x

) σ3ψσ(x) +

∫ dx

[ Π2(x)

8Ma2o +

1

2 ∆2(x)

] + ∑ σ=±

∫ dx √

2g∆(x)ψ̄σ(x)ψσ(x)

1. To simplify matters even further we consider the case of a heavy Carbon atom: M →∞. When we do this, we can drop the kinetic term, which amounts to disregarding the quantum fluctuations. Therefore, we can work with a constant classical field, so we replace ∆(x) with ∆o. Our Hamiltonian reduces to:

H = ∑ σ=±

∫ dx ψ†σ(x)

( − ivf

∂x

) σ3ψσ(x) +

∫ dx

1

2 ∆2o +

∑ σ=±

∫ dx √

2g∆oψ̄σ(x)ψσ(x) (1)

= ∑ σ=±

∫ dx ψ†σ(x)

( − ivf

∂x

) σ3ψσ(x) +

∫ dx

1

2 ∆2o +

∑ σ=±

∫ dx ψ†σ(x)

(√ 2g∆o

) σ2ψσ(x) (2)

Let’s group some terms together and write this out in two separate ways:

H =

∫ dx

(∑ σ=±

iψ†σ(x)

[ −vf ∂∂x −

√ 2g∆o√

2g∆o vf ∂ ∂x

] ψσ(x) +

1

2 ∆2o

) (3)

= ∑ σ=±

∫ dxψ̄σ

( − iγ1∂x +

√ 2g∆o

) ψσ +

∫ dx

1

2 ∆2o (4)

Notice that the second term looks like a Dirac Hamiltonian with a mass term: m = √

2g∆o. We could solve this as was done in lecture, but we will opt for a straight forward Fourier expansion instead,

ψσ(x) =

∫ dp

m

ω(p) ψ±σ (p)e

±ipx (5)

Here I used superscripts for the modes, so that we do not confuse them with the ”spin” subscripts. ω(p) will turn out to be the absolute value of our eigen energies. These factors are chosen to make the integration measure Lorentz invariant.

In order for the anticommutation relations to be preserved, we need:

{ ψσα(p), ψ

† σ′α′(p

′) }

= 2π ω(p)

m δσσ′δαα′δ(p− p′)

With all other anti-commutation relations zero.

Throwing in our Fourier transforms, and performing the necessary manipulations, we see the following matrix equation is what concerns us:

( ψ±σ (p)

)† [ ±pvf −i√2g∆o i √

2g∆o ∓pvf

] ψ±σ (p) (6)

In either case the egienvlaue equation is the same:

Det

[ ±pvf − E −i

√ 2g∆o

i √

2g∆o ∓pvf − E

] = −

[ (pvf + E)(pvf − E) + (

√ 2g∆o)

2 ]

= − [ (pvf )

2 − E2 + ( √

2g∆o) 2 ]

= 0 (7)

Which means:

E(p) = ± √

(pvf )2 + 2g∆2o := ±ω(p)

1

• Pushing the analogy between the above relation and the Pythagorean identity further we can define:

1 =

√ (pvf/ω(p))2 + (

√ 2g∆o/ω(p))2 (8)

:=

√ sin2(θ) + cos2(θ) (9)

Let’s expand the field about the positive and negative eigenvalues (energies):

ψσ =

∫ dp

m

ω(p)

( âσ(p)uσ(p)e

−ipx + b̂σ(p)vσ(p)e ipx )

I choose to set:

{âσ(p), â†σ′(p ′)} = 2πω(p)

m δσσ′δ(p− p′) (10)

{b̂σ(p), b̂†σ′(p ′)} = 2πω(p)

m δσσ′δ(p− p′) (11)

etc. The factor of ω/m in the mode expansion as well as the above anti-commutation relations will set our normalization of our eigenvectors. Indeed:{ ψσ(x), ψ

† σ′(x

′) }

=

∫ dpdq

(2π)2 m2

ω(p)ω(q)

({ âσ(p), â

† σ′(q)

} u(p)u†(q)e−i(px−qx

′) + { b̂σ(p), b̂

† σ′(q)

} v(p)v†(q)ei(px−qx

′) )

(12)

= δσσ′

∫ dp

m

ω(p)

( u(p)u†(p)e−ip(x−x

′) + v(p)v†(p)eip(x−x ′) )

(13)

= δσσ′δ(x− x′) (14)

⇒ u(p)u†(p) = v(p)v†(p) = ω(p) 2m

(15)

We will remember this for later.

We solve the following matrix equation:

ω(p)u(p) =

[ pvf −i

√ 2g∆o

i √

2g∆o −pvf

] u(p) (16)

−ω(p)v(p) = [

pvf −i √

2g∆o i √

2g∆o −pvf

] v(p) (17)

Re-expressing in terms of sin and cos:

u(p) =

[ sin(θ) −icos(θ) icos(θ) −sin(θ)

] u(p) (18)

−v(p) = [

sin(θ) −icos(θ) icos(θ) −sin(θ)

] v(p) (19)

This suggests (do not confuse a±, b± with our creation and annihilation operators!):

± [ a± b±

] =

[ sin(θ) −icos(θ) icos(θ) −sin(θ)

] [ a± b±

] (20)

⇒ ±a± = a±sin(θ)− ib±cos(θ) (21) ±b± = ia±cos(θ)− b±sin(θ) (22)

⇒ b± = a± icos(θ)

sin(θ)± 1 (23)

˙. .

[ a± b±

] = c±

[ sin(θ)± 1 icos(θ)

] (24)

Here c± is a normalization constant, which can be determined by recalling the normalization we want:

2

• ω(p)

2m =

1

2cos(θ) = |c±|2

[ (sin(θ)± 1)2 + cos2(θ)

] (25)

= 2|c±|2(1± sin(θ)) (26)

⇒ |c±| = 1

2

1√ cos(θ)

( 1± sin(θ)

) (27) =

1

2

ω(p)√ m(ω(p)± pvf )

(28)

Which means:

u(p) = 1

2 √ m(ω(p) + pvf )

[ pvf + ω(p)

im

] and

v(p) = 1

2 √ m(ω(p)− pvf )

[ pvf − ω(p)

im

] Re-expressing our Hamiltonian in terms of the mode expansion we arrive at:

H = ∑ σ=±

∫ dp

m

ω(p) ω(p)

[ â†σâσ − b̂†σ b̂σ

] + L

2 ∆2o (29)

Notice I replaced the integral over x with the length or our polymer chain L = Nao. We can now find the ground state with the usual prescription for a fermionic system: we define hole creation and annihilation operators:

d̂σ = b̂ † σ, d̂

† σ = b̂σ

So we have: b̂†σ b̂σ = d̂σd̂

† σ

using

{d̂σ(p), d̂†σ′(p ′)} = 2πω(p)

m δσσ′δ(p− p′)

We normal order our Hamiltonian:

H = ∑ σ=±

∫ dp

m

ω(p) ω(p)

[ â†σ(p)âσ(p) + d̂

† σ(p)d̂σ(p)− 2π

ω(p)

m δσσδ(p− p)

] +

1

2

∫ dx∆2o(x) (30)

We now define our ground state in the usual manner:

âσ(p)|gnd〉 = d̂σ(p)|gnd〉 = 0

Thus, for the ground state we have:

Egnd = − ∑ σ=±

∫ Λ −Λ

dpδσσδ(p− p) √

2g∆2o + (pvf ) 2 +

L

2 ∆2o (31)

= −L π

∫ vfΛ −vfΛ

dp̄

vf

√ 2g∆2o + p̄

2 + L

2 ∆2o (32)

Note: the Dirac delta function in momentum space gives us the factor of L/2π, and we summed over sigma. Λ is our momentum cutoff.

3

• We can finally find the distortion of the lattice in the ground state by minimizing the above equation in respect to the classical coordinate ∆o:

1

L

d ( Egnd)

) d∆o

= − 1 π

∫ vfΛ −vfΛ

dp̄

vf

2g∆o√ 2g∆2o + p̄

2 + ∆o = 0 (33)

⇒ πvf 2g

=

∫ vfΛ −vfΛ

dp̄ 1√

2g∆2o + p̄ 2

(34)

Of course ∆o = 0 satisfies this equation, but let’s find a non-trivial solution. The first integral is easy if we realize: ∫ vfΛ

−vfΛ dp̄

1√ 2g∆2o + p̄

2 =

∫ vfΛ −vfΛ

dp̄√ 2g∆o

1√ 1−

( i p̄√

2g∆o

)2 (35) =

1

i sin−1

( i

p̄√ 2g∆o

)∣∣∣∣∣ vfΛ

−vfΛ

(36)

= sinh−1 ( p̄√

2g∆o

)∣∣∣∣∣ vfΛ

−vfΛ

(37)

= 2sinh−1 ( vfΛ√

2g∆o

) (38)

Plugging in we can solve for the ∆o which minimizes 1 our ground state energy:

πvf 2g

= 2sinh−1 ( vfΛ√

2g∆o

) (39)

⇒ ∆o =

[√ 2g

Λvf sinh

( πvf 4g

)]−1 (40)

∼ √

2

g Λvfe

− πvf 4g (41)

This is the spontaneous mass generation which was referred to as dimerization in the original SSH model!

We could plug ∆o back into the ground state energy to find a more explicitly expression, but the integral is nasty, so we only note this in passing.

2. We build our single particle states out of the ground state defined above in the usual manner:

â†σ(p)|gnd〉, b̂†σ(p)|gnd〉

Recall:

Ĥ = ∑ σ′=±

∫ dp′

2π E(p′)

[ â†σ′(p

′)âσ′(p ′) + d̂†σ′(p

′)d̂σ′(p ′) ]

+ Egnd (42)

= : Ĥ : +Egnd (43)

Since:

â†σ(p ′)âσ′(p

′)

[ â†σ′(p)|gnd〉

] = â†σ′(p

′) [ â†σ(p)âσ′(p

′) + δσσ′δ(p− p′) ] |gnd〉

1We do need to take a second derivative of our expression to show that this ∆o is ind