• Non ci sono risultati.

The stochastic limit in the analysis of the open BCS model 1

N/A
N/A
Protected

Academic year: 2021

Condividi "The stochastic limit in the analysis of the open BCS model 1"

Copied!
26
0
0

Testo completo

(1)

The stochastic limit in the analysis of the open BCS model

1

Fabio Bagarello

Bjalowieza, 2004

1J. Phys. A, 2004 and IDAQP, submitted

(2)

General problem

Suppose that

System

S

interaction

←→ Reservoir

R

How R affects the time evolution of S?

Description nr.1: S is not a conservative system (it changes energy with R) ⇒ its time evolution αtS is not a group but only a semigroup (of completely positive maps,...., Lindblad, Gorini,...);

Description nr.2: S + R is a conservative system, so that its time evolution αtS+R is a group whose generator is a self-adjoint operator, the hamiltonian H

H = H| S {z+ HR}

=:H0

+λHI Strategy nr.1: Master equation approach

Main reference: P.A. Martin, Mod`eles en M´ecanique Statis- tique des Processus Irr´eversibles, Lecture Notes in Physics, 103,

(3)

Springer-Verlag, Berlin, (1979)

Strategy nr.2: Stochastic limit approach

Main reference: L. Accardi, Y.G. Lu, I. Volovich, Quantum Theory and its Stochastic Limit, Springer (2002)

F.B., Many-body applications of the stochastic limit: a re- view, Reports in Math. Phys., submitted

(4)

the stochastic limit approach

Starting point: the interaction picture Let

H = H0 + λHI

Ψ(t) = e−i(H0+λHI)tΨ0 (Schr¨odinger picture) and

ΨI(t) = eitH0Ψ(t) = eitH0e−i(H0+λHI)tΨ0 (interaction picture)

tΨI(t) = −iλHI(t)ΨI(t), where HI(t) = eiH0tHIe−iH0t and ΨI(t)|t=t0 = ΨI(t0). Its solution is

ΨI(t) = U(t, t0I(t0) where the wave operator U(t, t0) is such that:

tU(t, t0) = −iλHI(t)U(t, t0) with U(t0, t0) = I;

U(t, t0) = U(t, t1)U(t1, t0);

U(t, t0) = U(t0, t).

(5)

We can also write

U(t, t0) = I − iλ Z t

0

HI(t1)U(t1, t0) dt1 =

= I − iλ Z t

0

HI(t1) dt1+ (−iλ)2 Z t

0

HI(t1) dt1 Z t1

0

HI(t2) dt2+ ....

Let now < . > be a state of S + R. We assume that:

< HI(t) >= 0, for all t. This is automatic, e.g., for all the dipole hamiltonians;

< HI(t1+s)HI(t2+s)...HI(tn+s) >=< HI(t1)HI(t2)...HI(tn) >, which is a typical requirement for the n-points functions (Wight- man);

R

R| < HI(0)HI(t) > | dt < ∞ (the two point functions are in L1(R)).

Under these assumptions the first non trivial contribution in the van Hove limit (λ → 0, t → ∞ with λ2t = τ finite) of the iterated

(6)

series for < U (t, t0) > is

< −λ2 Z t

0

HI(t1) dt1 Z t

1

0

HI(t2) dt2 >−→ −τ Z 0

−∞

< HI(0)HI(s) > ds

How to implement the van Hove limit:

we rewrite ∂tU(t, t0) = −iλHI(t)U(t, t0) as

tUλ(t) = −iλHI(t)Uλ(t);

rescaling t → t/λ2 this becomes

tUλ(t/λ2) = −i

λHI(t/λ2)Uλ(t/λ2) or

Uλ(t/λ2) = I − i λ

Z t

0

HI(t12)Uλ(t12) dt1 =

= I − i λ

Z t

0

HI(t12) dt1+ +

µ

−i λ

2Z t

0

HI(t12) dt1 Z t

1

0

HI(t22) dt2 + ...

(7)

if ϕ0 is the ground vector for R (T=0) and ξ is an arbitrary vector for S we consider vector states like

< . >=< ϕ0 ⊗ ξ, . ϕ0 ⊗ ξ > .

The first non-trivial term in the expansion of Uλ(t/λ2) is there- fore

Iλ(t) :=

µ

−i λ

2Z t

0

dt1 Z t

1

0

< HI(t12)HI(t22) > dt2. The first aim of the SLA consists in getting I(t) = limλ,0Iλ(t) and finding a new time-dependent operator HI(sl)(t), acting on a new Hilbert space H(sl) with cyclic vector η0 6= ϕ0 such that

I(t) = − Z t

0

dt1 Z t

1

0

dt2< HI(sl)(t1)HI(sl)(t2) >η0⊗ξ

| {z }

nothing depends onλ

HI(sl)(t) is now used to deduce the expression of the generator, looking for the time evolution of operators of the algebra A = B(HS) ⊗ IR.

(8)

The Physicals Model

Our model consists of two main ingredient, the system, which is described by spin variables, and the reservoir, which is given in terms of bosonic operators. It lives in a box of volume V = L3, with N lattice sites.

The system

We define, following Martin and Buffet, [MB] (JSP, 1978 and P.A.Martin, Lecture Notes in Physics, 103),

HN(sys) =

int. with a mag. f ield

˜² XN

j=1

σj0

mean f ield interaction

g N

XN i,j=1

σi+σj ,

i, j −→ discrete values of the momentum that an electron in a fixed volume can have (lattice sites);

σj+ creates a Cooper pair with given momentum while σj an- nihilates the same pair;

˜² is the energy of a single electron and −g < 0 is the interaction close to the Fermi surface.

(9)

The algebra of the Pauli matrices is given by

i+, σj] = δijσi0, ±i , σj0] = ∓2δijσi±. We will use the following realization of these matrices:

σ0 ≡ σz =

 1 0 0 −1

 , σ+ =

 0 1 0 0

 , σ =

 0 0 1 0

 .

Introducing

SNα = 1 N

XN i=1

σiα, RN = SN+SN = RN,

HN(sys) can be simply written as HN(sys) = N(˜²SN0 − gRN) and we have:

[SN0 , RN] = [HN(sys), RN] = [HN(sys), SN0 ] = 0, for any given N > 0.

The intensive operators SNα are all bounded by 1 in the operator norm, and the commutators [SNα, σjβ] go to zero in norm when N → ∞ as N1, for all j, α and β.

(10)

The reservoir

We consider here a realistic bosonic reservoir (in contrast with [MB]), so that some of the following formulas must be under- stood to be formal (....but they can be made rigorous using, e.g., [B,JMP1998]).

We introduce here as many bosonic modes a~p,j as lattice sites are contained in V : j = 1, 2, ..., N . ~p is the value of the momentum of the j-th boson which, because of the periodic boundary condition, is ~p = L~n, where ~n = (n1, n2, n3), nj ∈ Z. These operators satisfy the following CCR,

[a~p,i, a~q,j] = [a~p,i, a~q,j ] = 0, [a~p,i, a~q,j] = δijδ~p~q and their free dynamics is given by

HN(res) = XN

j=1

X

~p∈ΛN

²~pa~p,ja~p,j.

Here ΛN = {~p = L~n, ~n ∈ Z3} and ²~p = 2m~p2 = 2(n2mL21+n222+n23).

(11)

The interaction

The form of the interaction between reservoir and system is assumed to be of the following form:

HN(I) = XN

j=1

j+aj(f ) + h.c.), where aj(f ) = P

~p∈ΛN a~p,jf (~p), f being a given test function.

Thefinite volumeopen system is now described by the following hamiltonian,

HN = HN0 + λHN(I), where HN0 = HN(sys) + HN(res), and λ is the coupling constant.

The free evolution of the interaction hamiltonian is:

HN(I)(t) = eiHN0 tHN(I)e−iHN0 t =

= XN

j=1

(eiHN(sys)tσj+e−iHN(sys)teiHN(res)taj(f )e−iHN(res)t + h.c.).

The computation of the part of the reservoir is trivial and produces eiHN(res)taj(f )e−iHN(res)t = aj(f e−it²),

(12)

where aj(f e−it²) = P

~p∈ΛN a~p,jf (~p)e−it²p.

Let us now compute the free evolution of the spin operators.

The differential equations of motion for the spin variables are



+j (t)

dt = 2i˜²σj+(t) + igSN+(t)σj0(t)

0j(t)

dt = 2ig(σj+(t)SN(t) − σj(t)SN+(t)).

(1)

where we have called, with a little abuse of language useful to maintain the notation simple, σjα(t) = eiHN(sys)tσjαe−iHN(sys)t (instead of σj,Nα,f ree(t)).

Let us now call Sα = F − strong limN→∞SNα. The existence of this limit (and of all its powers) has been discussed, e.g., in [BM, JSP1993] and references therein. With this in mind we can take the sum over j = 1, 2, ..., N of (both sides of) the system (1), divide the result by N, and consider the F −strong limN→∞ of the equations obtained in this way. We find that ˙S0(t) = 0 and ˙S+(t) = i(2˜² + gS0(t))S+(t), which can be easily solved: S0(t) = S0 = (S0) and S+(t) = S+ei(2˜²+gS0)t. Of course S(t) = (S+(t)). System (1) is

(13)

now replaced by its semiclassical approximation,



j+(t)

dt = 2i˜²σj+(t) + igS+(t)σj0(t)

j0(t)

dt = 2ig(σj+(t)S(t) − σj(t)S+(t)),

which can be explicitly solved using, for instance, the Laplace transform technique, [MB]. It gives

σj+(t) = eiνtρj0 + ei(ν+ω)tρj+ + ei(ν−ω)tρj, where we have defined the following operators











ρj0 = g2ωS2+ ¡

2Sσj+ + S0σj0 + 2S+σj¢ ρj+ = gSω2+

³

gS− ω−gSω+gS00σj+ + ω−gS2 0σj0 − gS+σj

´ ρj = gSω2+

³

gS− ω+gSω−gS00σj+ ω+gS2 0σj0 − gS+σj

´ , and the following quantities

ω = gp

(S0)2 + 4S+S, ν = 2˜² + gS0.

Defining further να(~p) = ν − ²~p + αω, α = 0, ±, we obtain HN(I)(t) =

XN j=1

X

α=0,±

¡ρjαaj(f eitνα) + h.c¢ .

(14)

Remark:– This same free time evolution can be found also for a fermionic reservoir, since CCR and CAR produce the same free time evolution for both the annihilation and the creation operators.

The next step in the SLA consists in computing the following quantity

Iλ(t) = µ

−i λ

2Z t

0

dt1 Z t1

0

dt2ωtot µ

HN(I)(t1

λ2)HN(I)(t2 λ2)

, and its limit for λ going to zero. Here ωtot = ωsysωβ, ωsys being a state of the system, and ωβ a state of the reservoir, which we will take to be a KMS state corresponding to an inverse temperature β = kT1 . It is convenient here to use the so-called canonical rep- resentation of thermal states, [ALV,2002]. For that we introduce two sets of mutually commuting bosonic operators {c(γ)~p,j}, γ = a, b, as follows:

a~p,j = p

m(~p)c(a)~p,j +p

n(~p)c(b),†~p,j ,

(15)

where

m(~p) = ωβ(a~p,ja~p,j) = 1

1 − e−β²~p, n(~p) = ωβ(a~p,ja~p,j) = e−β²~p 1 − e−β²~p, and

[c(α)~p,j, c~q,k(γ)] = δjkδ~p ~qδαγ,

while all the other commutators are trivial. Then, if we define fm(~p) = p

m(~p)f (~p) and fn(~p) = p

n(~p)f (~p), we get aj(f eitνα) = c(a)j (fmeitνα) + c(b)j (fneitνα),

(both c(γ)j (f ) and c(γ)j (f ) are linear in their argument f). We conclude that

HN(I)(t) = XN

j=1

X

α=0,±

½ ρjα

µ

c(a)j (fmeitνα) + c(b)j (fneitνα)

+ h.c

¾ . Let us now introduce the vacuum of the operators c(α)~p,j, Φ0:

c(α)~p,jΦ0 = 0, ∀~p ∈ ΛN, j = 1, ..N, α = a, b.

The KMS state ωβ can be represented as the following vector state, as in a GNS-like representation:

ωβ(Xr) =< Φ0, XrΦ0 >,

(16)

for Xr = a]~p,ja]~p,j, where a~p,j] = a~p,j or a]~p,j = a~p,j. Therefore ωtot

µ

HN(I)(t1

λ2)HN(I)(t2 λ2)

= XN

j=1

X

α,β=0,±

X

~p∈ΛN

sysjαρjβ)|fm(~p)|2·

·eiλ2t1να(~p)e−iλ2t2νβ(~p) + ωsysjαρjβ)|fn(~p)|2e−iλ2t1να(~p)e+iλ2t2νβ(~p)}.

Since we are interested to limλ→0Iλ(t), we must require that the following integral exists finite:

Z 0

−∞

X

~p∈ΛN

|fr(p)|2e±iτ να(~p) < ∞,

where fr(p) is fm(p) or fn(p). Under this assumption we find:

I(t) = lim

λ→0Iλ(t) = −t XN

j=1

X

α=0,±

n

ωsysjαρjα(a)α + ωsysjαρjα(b)α o

, where

Γ(a)α = Z 0

−∞

X

~p∈ΛN

|fm(p)|2e−iτ να(~p), Γ(b)α = Z 0

−∞

X

~p∈ΛN

|fn(p)|2eiτ να(~p). To this same result we could also arrive starting with the follow- ing stochastic limit hamiltonian

HN(sl)(t) = XN

j=1

X

α=0,±

½ ρjα

µ

c(a)αj(t) + c(b)αj(t)

+ h.c

¾ ,

(17)

where the operators c(γ)αj(t) are assumed to satisfy the following commutation rule,

[c(γ)αj(t), c(µ)βk(t0)] = δjk δαβ δγµδ(t − t0(γ)α , for t > t0. It is now a simple exercise to check that

J(t) = (−i)2 Z t

0

dt1 Z t

1

0

dt2tot(HN(sl)(t1)HN(sl)(t2))

coincides with I(t), where Ωtot = ωsysΩ = ωsys < Ψ0, Ψ0 >.

Here Ψ0 is the vacuum of the operators c(γ)αj(t): c(γ)αj (t)Ψ0 = 0 for all α, j, γ and t.

Following the SLA we now use HN(sl)(t) to compute thegenerator of the theory. Let X be an observable of the system and 11r the identity of the reservoir. Its time evolution (after limit λ → 0 is taken) is jt(X ⊗11r) = Ut(X ⊗11r)Ut, where Ut is the wave operator satisfying the following differential equation ∂tUt = −iHN(sl)(t)Ut (then ∂tUt = iUtHN(sl)(t)).

(18)

Then we find

tjt(X ⊗ 11r) = iUt[HN(sl)(t), X ⊗ 11r]Ut =

= iUt XN

j=1

X

α=0,±

½

jα, X](c(a)αj(t) + c(b)αj(t)) + [ρjα, X](c(a)αj(t) + c(b)αj(t))

¾ Ut To order normally this formula above, i.e. to move to the right all

the annihilation operators c(γ)αj and to the left the creation operators c(γ)αj , we need the following commutators

[c(a)αj(t), Ut] = −iρjαΓ(a)α Ut, [c(b)αj(t), Ut] = −iρjαΓ(b)α Ut, and, taking their adjoint,

[Ut, c(a)αj(t)] = iUtρjαΓ(a)α and [Ut, c(b)αj(t)] = iUtρjαΓ(b)α . (These results are easily found using the time consecutive prin- ciple, [ALV,2002]).

Going back to ∂tjt(X ⊗ 11r) and recalling that Ωtot(∂tjt(X ⊗ 11r)) = Ωtot(jt(L(X))), we get

L(X) = XN

j=1

X

α=0±

{[ρjα, X]ρjαΓ(a)α + [ρjα, X]ρjαΓ(b)α

(19)

−ρjαjα, X]Γ(a)α − ρjαjα, X]Γ(b)α } If X is self-adjoint (X = X) we have

L(X) = L1(X) + L2(X), where

L1(X) = XN

j=1

X

α=0±

n

jα, X]ρjαΓ(a)α + h.c.

o ,

L2(X) = XN

j=1

X

α=0±

n

jα, X]ρjαΓ(b)α + h.c.

o

The phase transition

As discussed in [MB], the phase structure of the model is given by the dynamical behavior of SN0 and RN := SN+SN. These inten- sive operators are both self-adjoint, so that we can use equations above.

As a first step we compute L(SN0 ) = L1(SN0 )+L2(SN0 ). We have L1(SN0 ) = 1

N XN

j=1

L1j0) = 1 N

XN j=1

X

α=0,±

n

jα, σj0jαΓ(a)α + h.c.

o

(20)

which can be written as L1(SN0 ) = X

α=0,±

n

(bα+SN+ + bα−SN + bα0SN0 + bα111N(a)α + h.c.

o , where 11N = N1 PN

j=111j and the various coefficients {bα γ} have been introduced here only to stress the fact that L1(SN0 ) is linear in the intensive operators. We have already discussed the fact that the limit of the right hand side of this formula exists in the strong topology restricted to a certain family F of states, since all the operators SNα (and their powers) converge in this topology, [BM, JSP1993]. Therefore, also the limit of the left hand side does exist in the same topology. After some non trivial algebra we find

L1(S0) := F − strong lim

N→∞L1(SN0 ) =

= −8g4S0(S+S)2 ω3

½

(a)+ ω − g

(ω + gS0)2 + <Γ(a) ω + g (ω − gS0)2

¾ , where <Γ(a)± indicates the real part of Γ(a)± .

The computation of L2(S0) := F − strong limN→∞L2(SN0 )

(21)

follows essentially the same steps and produces L2(S0) = −8g4S0(S+S)2

ω3

½

(b)+ ω + g

(ω + gS0)2 + <Γ(b) ω − g (ω − gS0)2

¾ , so that

L(S0) = −8g4S0(S+S)2

ω3 h(S0, S+S), where

h(S0, S+S) = {<Γ(a)+ ω − g

(ω + gS0)2 + <Γ(a) ω + g (ω − gS0)2+ +<Γ(b)+ ω + g

(ω + gS0)2 + <Γ(b) ω − g

(ω − gS0)2},

and we have written explicitly the dependence of h on S+S = F −strong limN→∞SN+SN (via the pulsation ω). It is interesting to observe that the same function h(S0, S+S) appear also in the computation of L(S+S) := F − strong limN→∞L(SN+SN). Here it is convenient to use the fact that limN→∞[SNα, σjβ] = 0, for all α, β and j. We get

L(S+S) = −16g4(S+S)3

ω3 h(S0, S+S).

(22)

The phase structure of the model is now given by the right-hand sides of equations for L(S+S) and L(S0), see [MB], that is, from the zeros of the functions

f1(x, y) = −8g4xy2

ω3 h(x, y), f2(x, y) = −16g4y3

ω3 h(x, y), where x = S0, y = S+S ( ⇒ ω = gp

x2 + 4y and ν = 2˜²+ gx).

In particular, the existence of a super-conducting phase corre- sponds to the existence of a non trivial zero of f1 and f2, [MB], or, equivalently, to the existence of a non trivial zero of h.

To find these zeros, we need the explicit expression for the coef- ficients <Γ(γ)± :

(a)± = Z

−∞

X

~p∈ΛN

|fm(~p)|2e−iτ ν±(~p) dτ = 2π X

~p∈ΛN

|fm(~p)|2δ(ν±(~p)), and

(b)± = 2π X

~p∈ΛN

|fn(~p)|2δ(ν±(~p)).

It is now almost straightforward to recover the results of [MB]. We look for solutions corresponding to ν = 0. Then x = S0 = −2˜²/g

(23)

and ν+(~p) = ω − ²~p, which is zero if and only if ω = ²~p. Also, ν(~p) = −ω − ²~p, which is never zero. Therefore <Γ(γ) = 0, γ = a, b, while the sums for <Γ(γ)+ are restricted to the smaller set, EN ⊂ ΛN, of values of ~p such that, if ~q ∈ EN then ²~q = ω. Finally, recalling the expression of m(~p) and n(~p), we get

(a)+ = 2π eβω eβω − 1

X

~p∈EN

|f (~p)|2, <Γ(b)+ = 2π 1 eβω − 1

X

~p∈EN

|f (~p)|2.

Equation h(x, y) = 0 becomes now eβω

eβω − 1 X

~p∈EN

|f (~p)|2 ω − g

(ω + gx)2+2π 1 eβω − 1

X

~p∈EN

|f (~p)|2 ω + g

(ω + gx)2 = 0,

=⇒ eβω = g + ω g − ω.

This equation replaces the equation obtained in [MB],g tanh

³βω 2

´

= ω. The conclusion is now straightforward:

1) let ξ = ωg. Then equation above has a non-trivial solution if and only if the function g(ξ) = eβgξ 1+ξ1−ξ has a zero ξ 6= 0. This solution does exist only if the first derivative of g(ξ), computed in

(24)

ξ = 0 is positive, i.e. when βg − 2 > 0, that is if T < Tc = 2kg . We recover therefore the first result of [MB].

2) it is also possible to find numerically the value of y = S+S from our equation. However, it is more convenient to observe that:

g tanh

µβω 2

= geβω2 − eβω2

eβω2 + eβω2 = geβω − 1 eβω + 1 = g

g+ω g−ω − 1

g+ω

g−ω + 1 = ω, which is the same result found in [MB].

Remarks:– (1) These results are independent of the cutoff N.

This suggests that, at least for what concerns the critical temper- ature and the related value of S+S, nothing changes after the thermodynamical limit is considered (i.e., when N → ∞).

(2) Concerning a comparison with [MB], the SLA allows to avoid the use of a perturbative expansion on the master equation, which is very hard to control and which is also based on the assumption that the observables whose time evolution we are interested in com- mute with the free hamiltonian. This makes the SLA more flexible and powerful wrt the standard techniques.

(25)

Further results and future projects

(1) In a recent paper (IDAQP, submitted) we have analyzed the role of a second reservoir interacting with R in the attempt of increasing the value of the critical temperature. This approach has produced surprising (to me!!) results: superconducting phases exist also for T > Tc under very special assumptions on the pa- rameters of the model, also for the original [BM] model.

(2) We are now considering, again in the same perspective, a double stochastic limit.

Indeed: suppose that the free evolution of the annihilation op- erator of the reservoir, a~p,i(t) = a~p,ie−i²~pt, is replaced, for some reason, by a~p,i(t) = a~p,ie−iγ²~pt, with γ < 1. Then the function να(~p) will be replaced by να(~p) = ν − γ²~p + αω. All the other formulas are left unchanged, at least formally. It is easy to check now that equation h(x, y) = 0 produces now, taking ν = 0 as

(26)

before, the following equation:

eβω/γ = g + ω g − ω,

which admits a non trivial solution in ]0, g[ if gβ/γ − 2 > 0, that is under a new critical temperature Tc(γ) = 2kγg = Tγc, which is larger than Tc since γ is smaller than 1 by construction. There- fore, the critical temperature increases leaving invariant all the parameters! It is worth stressing that a similar conclusion was by no means evident in Martin’s papers.

How to get this change in the free evolution of the bosons?

No-go results:

1) reservoir with different statistic;

2) exact solution of two reservoirs mutually interacting.

The working idea: a nested stochastic limit.

We introduce a second reservoir R2 interacting only with R with an O(λ) term and the R interacts with S with an O(λ2) hamiltonian.

Riferimenti

Documenti correlati

Nei primi anni Cinquanta, la ricerca di Fo e di Lecoq segue strade parallele, in entram- bi i casi volta a recuperare le condizioni ‘pragmatiche’ del genere teatrale, dalla «scrittura

Table S2 Number of rice reads per pool deconvoluted to one, two, or three BACs; the percentage column reports the fraction of the total number of reads that were deconvoluted to

Le scelte degli autori dei testi analizzati sono diverse: alcuni (per esempio Eyal Sivan in Uno specialista) usano materiale d’archivio co- me fotografie e filmati, altri invece

Here, to make up for the relative sparseness of weather and hydrological data, or malfunctioning at the highest altitudes, we complemented ground data using series of remote sensing

Inizia una fase in cui diventa gradualmente più accessibile per i genitori la riflessione sullo stato mentale della figlia e, pur non riuscendo ad attivare questi pensieri quando

Since, as said, most observed fragments belong to the QP region of the phase-space (see Fig. 1, left panel), the observed difference clearly demonstrates the action of an

This result strongly suggests that we are observing discrete shifts from part-time to full-time work, as conjectured by Zabalza et al 1980 and Baker and Benjamin 1999, rather