Download A Semi-Classical Approach to the Jaynes

A Semi-Classical Approach to the
Jaynes-Cummings model.
Olivier Babelon, Benoı̂t Douçot, (LPTHE Paris)
Thierry Paul (DMA, ENS)
A Semi-Classical Approach to theJaynes-Cummings model. – p.1/31
Plan of the talk
1– Cold atoms Condensates.
2– The one spin 1/2 Jaynes-Cummings Model.
3– The Classical N -spin model.
4– The quantum one spin-s model.
5– Semi-classical analysis.
A Semi-Classical Approach to theJaynes-Cummings model. – p.2/31
Cold atoms Condensates.
Consider alcali atoms like Li, K, Na, etc...
~1
S
I~1
~2
S
r
I~2
~ the spin of
where I~ denotes the magnetic moment of the nucleus and S
the electron. The Hamiltonian of one atom in a magnetic field is
~ + gB S
~ ·B
~
H = g I~ · S
For two widely separated atoms the Hamiltonian is simply the sum of the
Hamiltonians of the idividual atoms.
H = H1 + H2
When they come closer however they start to interact
A Semi-Classical Approach to theJaynes-Cummings model. – p.3/31
In the Born-Openheimer we get effective potentials V (r). A Feshbach
resonnance occurs when a bound state becomes degenerate with an
atomic state. By tuning the magnetic field, one can adjust the molecule
state to be just above or below the atomic state.
V (r)
b† |0i
∝ |B|
c†k c†−k |0i
r
A particularly interesting situation is the case of a soudain perturbation. At
t = 0 the system is in an atomic state, and at t > 0 molecules start forming
in the fundamental state (at zero temperature). What is the dynamical
evolution of the system?
A Semi-Classical Approach to theJaynes-Cummings model. – p.4/31
H=
Z
dk
ǫ(k)c†kσ c−kσ
+ ωb† b + g
Z
+ ωb† b + g
X
†
†
dk b† ck↓ c−k↑ + bck↑ c−k↓
which we approximate as
H=
X
ǫj c†jσ cjσ
b† cj↓ cj↑ +
bc†j↑ c†j↓
j
j,σ
This can be rewritten in terms of pseudo spins
2szj
=
X
σ
c†jσ cjσ − 1,
s−
j = cj↓ cj↑ ,
† †
=
c
s+
j
j↑ cj↓
we finally get
H=
n−1
X
j=0
2ǫj szj
†
+ ωb b + g
n−1
X
j=0
b† s −
j
+
bs+
j
A Semi-Classical Approach to theJaynes-Cummings model. – p.5/31
The spin 1/2 Jaynes-Cummings Model.
Interaction of a two levels atom and photons.
It is described by the Jaynes-Cummings (1963) Hamiltonian.
σz
+ ~ω(b† b + 21 ) + ~Ω[σ + b + σ − b† ]
H = ~ω0
2
The resonnance condition is ∆ = ω0 − ω = 0 but we can keep ∆ 6= 0.
When b(t) = e−iωt b0 is a classical mode, we get the Rabi formula
hψ|σ z |ψi
Ω2 b†0 b0
2
sin
γt
1 − 2Pe (t) =
=
1
−
2
2
†
z
∆
2
h−1|σ | − 1i
+ Ω b0 b0
4
2
with the Rabi frequency γ =
∆2
4
+ Ω2 b†0 b0
A Semi-Classical Approach to theJaynes-Cummings model. – p.6/31
What happens if the electromagnetic field is quantum ? and in particular if
the number of photons is small (5 ≤ n̄ ≤ 40)? It turns out that the model is
Integrable. Let
N = b† b + σ + σ − ,
H0 = 12 ∆σ z + Ω(b† σ − + σ + b)
We have
H = ωN + H0 ,
[N, H0 ] = 0
The Heisenberg equations of motion are
i
d
− ω b† = Ωσ + ,
dt
i
d
− ω0 σ + = −Ωb† σ +
dt
A Semi-Classical Approach to theJaynes-Cummings model. – p.7/31
There is a beautiful formula giving the solution of these non linear
equations of motion [Ackerhalt (1975)]. The trick is to use the algebra of σ
matrices (σ z σ + = σ + , etc...) to rewrite them as
d
i − ω b† = Ωσ + ,
dt
i
d
− ω − 2H0 σ + = −Ωb†
dt
Since H0 is conserved we look for solutions of the usual form, taking care
of the ordering problem. We obtain
sin
Γt
sin
Γt
σ − (0)
b(t) = e−i(H0 +ω)t − cos Γt + iH0
b(0) + iΩ
Γ
Γ
sin Γt
sin Γt
σ − (t) = e−i(H0 +ω)t cos Γt + iH0
b(0)
σ − (0) − iΩ
Γ
Γ
with
∆2
Γ =
+ Ω2 (N − 1)
4
2
A Semi-Classical Approach to theJaynes-Cummings model. – p.8/31
Then we can compute [ Narozhny, Sanchez-Mondragon, Eberly (1981)]
∞
2
2n
X
hα, +1|σ (t)|α, +1i
|α|
sin γn t
−|α|2
2
=e
1 − 2Ω (n + 1)
2
hα, −1|σ z (0)|α, −1i
n!
γ
n
n=0
z
γn =
r
Ω2 (n
∆2
+ 1) +
4
1
0.75
0.5
0.25
20
40
60
80
100
-0.25
-0.5
-0.75
-1
A Semi-Classical Approach to theJaynes-Cummings model. – p.9/31
The Classical N -spin model.
We consider classical version of the Hamiltonian: [Bonifacio, Preparata
(1970)], [Barankov, Levitov, Spivak (2004)], [Meunier, Le Diffon, Rueff,
Degiovanni, Raimond (2006)]
H=
n−1
X
2ǫj szj
+ ω b̄b + g
n−1
X
b̄s−
j
j=0
j=0
+
bs+
j
with Poisson brackets
{b, b̄} = i,
{saj , sbj } = −ǫabc scj ,
s~j 2 = s2
The equations of motion read
ḃ = −iωb − ig
n−1
X
s−
j ,
j=0
+
z
ṡ+
j = 2iǫj sj − 2ig b̄sj ,
+
ṡzj = ig(b̄s−
−
bs
j )
j
−
z
ṡ−
j = −2iǫj sj + 2igbsj
A Semi-Classical Approach to theJaynes-Cummings model. – p.10/31
We can write these equations in the Lax form L̇ = [L, M ]. [Yurbashyan,
Kuznetsov, Altshuler (2005)]
n−1
L(λ)
M (λ)
=
ω z X sj
2
2
+
−
z
λσ + (bσ + b̄σ ) − 2 σ +
g2
g
g
λ − ǫj
j=0
= −iλσ z − ig(bσ + + b̄σ − )
Letting
L(λ) =
A(λ)
C(λ)
B(λ)
D(λ)
we have
n−1
X szj
ω
2λ
A(λ) = −D(λ) = 2 − 2 +
g
g
λ − ǫj
j=0
n−1
−
2b X sj
+
B(λ) =
,
g
λ − ǫj
j=0
n−1
+
2b̄ X sj
+
C(λ) =
g
λ − ǫj
j=0
[Gaudin (1972)], [Sklyanin (1979)]
A Semi-Classical Approach to theJaynes-Cummings model. – p.11/31
The consequence of this equation is that the spectral curve
Γ(λ, µ) ≡ det(L(λ) − µ) = 0
is independent of time. Since L(λ) is traceless, it reads
2
Q2n+2 (λ)
2
µ = A (λ) + B(λ)C(λ) ≡ Qn−1
j=0 (λ
− ǫ j )2
2
X
X Hj
s
1
4
2
= 4 (2λ − ω)2 +
+
H + 2
Qn−1
2 n
2
2
g
ωg
g
λ
−
ǫ
(λ
−
ǫ
)
j
j
(λ
−
ǫ
)
j
j=0
j
j
Q2n+2 (λ)
The (n + 1) Hamiltonians can be chosen as
X
Hn = ωbb̄ + ω
szj
j
Hj = (2ǫj −
ω)szj
+
g(bs+
j
+
b̄s−
j )
X sj · sk
+g
,
ǫj − ǫk
2
k6=j
j = 0, · · · , n − 1
The Hamiltonian of the system is
H = Hn +
X
j
Hj
A Semi-Classical Approach to theJaynes-Cummings model. – p.12/31
We are interested in the initial condition where all spins are up and no
molecule
szj = s
b = b̄ = 0, s±
j = 0,
The values of the conserved quantities are
Hn = nsω,
X
ω
1
2 2
Hj = 2s(ǫj − ) + g s
2
ǫj − ǫk
k6=j
The associated energy is H = 2s
B(λ) = 0, C(λ) = 0, so that
Pn−1
j=0
ǫj . For these solutions we have
2

ω X g2 s 1 
Q2n+2 (λ)
4 
2
Q
= A0 (λ) = 4 λ − +
2
g
2
2 λ − ǫj
(λ
−
ǫ
)
j
j
j
is a perfect square: Q2n+2 (λ) =
4
g4
Qn+1
i=1
(λ − Ei )2
A Semi-Classical Approach to theJaynes-Cummings model. – p.13/31
The zeroes of Q2n+2 (λ) are located at the zeroes of A0 (λ) and satisfy the
condition
n−1
ω g2 s X 1
,
(∗)
E= +
2
2 j=0 ǫj − E
These equations have another interesting interpretation. Let us perform
an analysis of the small fluctuations around the solution b = 0, szj = s. The
linearized equations of motion are
˙ = −iωδb − ig
δb
X
δs−
j ,
˙ − = −2iǫj δs− + 2igsδb
δs
j
j
−
−2iEt
, ∀j. The
=
δs
We look for eigenmodes δb(t) = δb(0)e−2iEt , δs−
j (0)e
j
second equation gives
gs
−
δsj = −
δb
E − ǫj
Inserting into the first equation, we get the self-consistency equation for E
wich is exactly eq.(*). The zeroes of Q2n+2 (λ) are precisely the eigen
frequencies of the small perturbations around the static solution
b = 0, szj = s.
A Semi-Classical Approach to theJaynes-Cummings model. – p.14/31
Let us assume that ǫj < 0. The ground state corresponds to all the spins
up: szj = s, j = 0, · · · , n − 1. The graph looks like this:
we see that we have n + 1 real roots if ω > ωsup or ω < ωinf . In between
we have n − 1 real roots and a pair of complex conjugate roots.
A Semi-Classical Approach to theJaynes-Cummings model. – p.15/31
Separated variables λj are the zeroes B(λj ) = 0. The conjugate variables
are µj = −A(λj ). One can reconstruct the Lax matrix itself once we know
the coordinates of these n points. For B(λ), we simply have
n−1
X
s−
j
Qn
2b
2b j=1 (λ − λj )
Qn
B(λ) =
+
=
g
λ − ǫj
g
j=1 (λ − ǫj )
j=0
The equations of motion of the separated variables are
Q
k (λi − Ek )
Q
λ̇i = 2i
j6=i (λi − λj )
One can solve explicitely these equations, hence finding the solution of
the reduced model on the critical surface. The solution is through the Abel
map. Here we have to adapt it to this degenerate situation.
A Semi-Classical Approach to theJaynes-Cummings model. – p.16/31
Introducing the polynomial
P (λ) =
b
−1
(t) =
X
X
2iEl t
Al e
Y
k6=l
Al e2iEl t ,
(λ − Ek )
B(λ) = b(t)2 P (λ)
In the one spin case
2s − κ2
√
b̄b =
2
cosh g 2s − κ2 (t − t0 )
X
λ1
X
λ2
X
X
λ3
X
λn
The motion of the n separated variables
X
A Semi-Classical Approach to theJaynes-Cummings model. – p.17/31
The quantum one spin-s model.
Let b, b† and sj be now quantum operators
[b, b† ] = ~,
−
z
]
=
2~s
,
s
[s+
j,
j
j
±
]
=
±~s
[szj , s±
j
j
The quantum Hamiltonian is
H=
n−1
X
2ǫj szj + ωb† b + g
j=0
n−1
X
+
+
bs
(b† s−
j )
j
j=0
We will consider the case n = 1, spin s. Then
H = H0 + H1
with
H1 = ω(b† b + sz ),
H0 = (2ǫ − ω)sz + g(b† s− + bs+ )
Notice that
[H0 , H1 ] = 0
A Semi-Classical Approach to theJaynes-Cummings model. – p.18/31
We will work with the Bargman spaces. For the oscillator b, b† this is the
space
Z
2
|z|
Bb = f (z), entire function of z |f (z)|2 e− ~ dzdz̄ < ∞
On this space we have
b=~
d
,
dz
b† = z
Similarly, for the spin s, we define
Bs = g(w), entire function of w
Z
|g(w)|2
dwdw̄
(1 + |w|2 )2s+2
<∞
This space is of dimension 2s + 1. We have
s+ = ~
d
,
dw
s− = −w2 ~
d
+ 2~sw,
dw
sz = −w~
d
+ s~
dw
A Semi-Classical Approach to theJaynes-Cummings model. – p.19/31
Since [H0 , H1 ] = 0, we can fix the total number of particules, to its critical
value s, i.e. the value of H1 = ω(b† b + sz ) on this subspace is ~ωs.
The Hilbert space is the span of
n
Hs = (zw)n ,
n = 0, · · · , 2s
o
Hence H0 acts on polynomials in x = zw of degree ≤ 2s.
h
i
H0 f (x) = g x~2 ∂ 2 + [−2κx − x2 + ~]~∂ + s~[2κ + 2x] f (x)
ω
ǫ
−
where we have defined
2
κ=
g
P2s
Eigenstates are polynomials f (x) = n=0 fn xn . The Schroedinger
equation becomes
~
3/2
√
√
3/2
(n+1) 2s − n fn+1 +~κ(2s−2n)fn +~ n 2s + 1 − n fn−1 = g −1 Efn
In this form, the system is represented by a symmetric Jacobi matrix.
A Semi-Classical Approach to theJaynes-Cummings model. – p.20/31
An example of spectrum is shown below.
Moreover the dependence in κ is rather smooth.
A Semi-Classical Approach to theJaynes-Cummings model. – p.21/31
If |Ei is the state with spin up and zero molecules at time t = 0, the
average number of molecules at time t is
i tH
~
n̄(t) = hE|e
−i tH
~
x∂x e
|Ei
In the regime κ2 > 2, classically there is no molecule formation. Quantum
mechanically we get the behavior shown here (s = 100).
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
-20
20
40
60
80
A Semi-Classical Approach to theJaynes-Cummings model. – p.22/31
In sharp contrast, in the regime κ2 < 2, we expect molecule formation. We
get the curve (for s = 100). The green curve corresponds to the
preliminary semi classical calculation we now explain.
120
100
80
60
40
20
-20
20
40
60
80
A Semi-Classical Approach to theJaynes-Cummings model. – p.23/31
Semi-classical analysis.
The initial state ψ|t=0 is a coherent state that one should interpret as
ψ|t=0 (z, w) = e
z̄ ′ z
~
(1 + w̄′ w)2s |z̄ ′ =0,w̄′ =0 = 1
It is localized around zero because it is multiplied by the measure
1
e− ~ z̄z−2s log(1+w̄w) ,
s=
1
~
H
We want to compute ψ|t = e−i ~ t ψ|t=0 . At the beginning the state remains
localized around x = 0. We look for a wave function of the form
ψ=e
2iQ(x)
~
,
x = zw,
Q(x) = l(t) + α(t)x
This is linear in x but quadratic in zw. Writing the Schroedinger equation
i~∂t ψ = H0 ψ, we obtain
′′
′
2
′
′ 2
iQ̇ = g ~(xQ + Q ) − (x + 2κx)Q + 2ix(Q ) − is~(x + κ)
A Semi-Classical Approach to theJaynes-Cummings model. – p.24/31
we find the equations
α̇
2
= g 2α + 2iκα − s~
il˙ = g [~α − is~κ]
and the rest is −gx2 α. Imposing α(0) = 0, we get
iE (E −
iE
+
α(t) =
g
g Ē −
√
−2g 2−κ2 t
Ē)e
√
−2g
2−κ2 t
Ee
The "phase" l(t) plays no role in what follows and we will not write it.
Hence α(t) is of the form
α(t) = A(1 + ρ(t)),
−Γt
ρ(t) = O(e
),
p
Γ = 2g 2 − κ2
Putting this into the coherent state, we get
2i
1
2i
e ~ l(t) e− ~ (z z̄+2ww̄−2iAzw) e ~ (Aρ(t)zw)
Since ρ(t) goes to zero rapidly, we consider the first exponential.
A Semi-Classical Approach to theJaynes-Cummings model. – p.25/31
Diagonalizing this quadratic form introduces two orthogonal directions VT
and VL
z
w̄
= T VT + LVL ,
r 2 1
VL ≃
,
iA
3
with corresponding eigenvalues ΛT = 3,
r 1
1
VT ≃
3 −2iA
ΛL = 0, so that
2
3
ψ ≃ e− ~ T̄ T e− 3~ ρ(t)L̄L
We see that the state is localized in the VT direction but spreads in the VL
2
direction since the localizing factor is e− 3~ ρ(t)L̄L and ρ(t) becomes very
small as time becomes large.
w̄
VL
q
VT
h̄
ρ(t)
√
h̄
z
A Semi-Classical Approach to theJaynes-Cummings model. – p.26/31
How long is this a good approximation ? One has to make sure that the
rest R = gx2 αψ remains small. Let ψexact (t) be the exact solution of the
Schroedinger equation, and ψapprox (t) be the approximate solution. We
have
i~∂t ψexact
i~∂t ψapprox
= Hψexact
= Hψapprox + R
Taking the norm of this expresion we see that
|ψapprox − ψexact | = O(~ǫ ) ⇐⇒ |R| ≃ O(~1+ǫ )
2
2
Hence |gz w α| = |g|
integral, therefore
2
2
3 L̄L
≃ ~1+ǫ . But L is localized by the gaussian
2
~
L̄L ≃ ≃ ~eΓt
3
ρ
hence, we get the bound
1
1−ǫ
log
t≃
2Γ
~
A Semi-Classical Approach to theJaynes-Cummings model. – p.27/31
The extension at later time is of the form
ψ(x, t) = e
2igκ
~ t
2i
e− ~ S(x) a(x, t)
We have gone from a coherent state to a WKB state. Writing the
Schroedinger equation, we get
i∂t a =
~
−1
′ 2
′
gx (2iS ) + (x + 2κ)(2iS ) + 2 a
−~0 g [(x(2(2iS ′ ) + x + 2κ)a′ + (x(2iS ′′ ) + (2iS ′ ))a]
+~g [xa′′ + a′ ]
Setting
px = −2iS ′
we see that the coefficient of a(x, t) in the first term vanishes if
p2x − (x + 2κ)px + 2 = 0
A Semi-Classical Approach to theJaynes-Cummings model. – p.28/31
when x is small the two solutions of the above equation are
px = −2iA,
px = 2iĀ
Choosing the first solution, the reduced action reads
S(x) = −Ax + O(x2 )
Hence we have z z̄ + 2ww̄ − 2iAzw = z z̄ + 2ww̄ + 2iS(zw)
px
a(x, 0)
Ax
p2x − (x + κ)px + 2 = 0
x
Āx
a(x, t)
A Semi-Classical Approach to theJaynes-Cummings model. – p.29/31
Next we kill the ~ term by solving
i∂t a = g [(x(2px − x − 2κ)a′ + (xp′x + px )a]
Now, the equation for a is of the form
ȧ = ẋ a′ + β(x) a
This is a typical transport equation. Its solution is of the form
a(x0 , t) = px (x(t, x0 ))
∂x(t, x0 )
∂x0
−1/2
a0 (x(t, x0 )),
x(t = 0, x0 ) = x0
The solution of the equation of motion for the separated variable λ1 is
E0 (λ0 − Ē0 ) − Ē0 (λ0 − E0 )e−Γt
,
λ1 (t) =
(λ0 − Ē0 ) − (λ0 − E0 )e−Γt
λ1 (0) = λ0
we see that λ(−∞) = Ē0 and λ(+∞) = E0 .
A Semi-Classical Approach to theJaynes-Cummings model. – p.30/31
Our Lagrangian variety is
p2x − (x + 2κ) px + 2 = 0
It is parametrized in terms of λ1 as folllows
ω
2
λ1 (t) −
px (t) =
g
2
x(t) =
2
g
(λ1 (t) − ǫ) +
g
(λ1 (t) − ω/2)
The Jacobian is easy to calculate:
∂x(t, x0 ) ∂λ1 (t) ∂λ0
∂x(t, x0 )
=
∂x0
∂λ1 (t)
∂λ0 ∂x0
Separated variables uniformize the Liouville torus.
A Semi-Classical Approach to theJaynes-Cummings model. – p.31/31