Exact Solvability of the two-photon Rabi Hamiltonian

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by  Igor Travenec

Exact Solvability of the two-photon Rabi Hamiltonian
I.Travˇenec∗
Institute of Physics, Slovak Academy of Sciences, Du´bravska´ cesta 9, 842 28 Bratislava, Slovakia
Exactspectrumofthetwo-photonRabiHamiltonianisfound,proceedinginfullanalogywiththe
solution of standard (one-photon) Rabi Hamiltonian, published by Braak in Phys. Rev. Lett. 107,
100401(2011). TheHamiltonianisrewrittenasasetoftwodifferentialequations. Symmetriesthat
gethiddenafterfurthertreatmentarefound. Onecan plainlysee,howtheHilbertspacesplitsinto
four disjunct subspaces, categorized by four values of the symmetry parameter c = ±1,±i. There
were only two values ±1 for thestandard Rabimodel. Four analytic functions are introduced by a
recurrence scheme for the coefficients of their series expansion. All their roots yield the complete
spectrum of theHamiltonian. Eigenstates in Bargmann space are also at disposal.
2
1
PACSnumbers: 03.65.Ge, 02.30.Ik,42.50.Pq
0
2

n I. INTRODUCTION
a
J There were many trials to solve the model called (in b→ ∂ , b† →z. (2)

8 ∂z
quantum optics) Rabi Hamiltonian exactly, until Braak
1
[1]recentlysucceededtodoso. Thesamemodelisknown The eigenfunctions have to be analytic in the whole

] under several pseudonyms, e. g. single-mode spin-boson complex plane. Let us suppose that the solution of the
h system, Jaynes-Cummings model without rotating wave stationary Schr¨odinger equation is a vector composed of
p approximation and others; a brief survey can be found two analytic functions (ψ (z),ψ (z))T. We can insert
- 1 2
h in [2]. Let us introduce a more general form of Rabi eq.(2) into the Hamiltonian (1), case m = 2, and using
t Hamiltonians [3], describing the interaction between a thewell-known2×2Paulimatrices,weobtainacoupled
a
m bosonic mode with energy ω and a two-level atom with system of second-order differential equations
level spacing ω
0
[
1 Hˆ(m) = ω20σz +ωb†b+g(σ++σ−) (b†)m+bm , (1) 2gψ2′′+ωzψ1′ +2gz2ψ2+(ω20 −E)ψ1 =0
7v where m = 1,2,...; g is the interac(cid:2)tion constan(cid:3)t, σz, 2gψ1′′+ωzψ2′ +2gz2ψ1−(ω20 +E)ψ2 =0, (3)
1 σ± are the Pauli matrices, b† and b are boson creation
7 and annihilation operators, respectively. For the most where ′ means the derivative with respect to z and E is
3 studiedm=1caseBraak[1]recentlypresentedanexact the energy from the Hamiltonian’s spectrum. For better
1. algebraicsolution;wearegoingtosolvethem=2model symmetrization it is useful to go over to a linear com-
0 in full analogy. bination of the considered functions, namely φ1(z) =
2 Before Braak’s general solution, some special points, ψ1(z)+ψ2(z) and φ2(z) = ψ1(z)−ψ2(z). Thus we get
1 called also the Juddian ones, were known to be exactly the set
:
v solvable. This means that for some constraints on the
i Hamiltonian parameters, one can find both an eigen-
X 2gφ′′+ωzφ′ +(2gz2−E)φ + ω0φ =0
function and its eigenenergy. It doesn’t give the com- 1 1 1 2 2
r plete spectrum, just one excited eigenstate. This was −2gφ′′+ωzφ′ −(2gz2+E)φ + ω0φ =0. (4)
a 2 2 2 2 1
shown at first for the standard m=1 Rabi Hamiltonian
[4]. Later the same picture, of course with different con- The nextstep is to find important symmetries present
straints and eigenvalues was shown for m = 2, i. e. the in the set (4) before they become less clear after some
two-photon Rabi Hamiltonian [5]. These special points transformation followed by a series expansion. Here we
manifestthemselvesascross-sectionsofgeneralsolutions prefer a somewhat different way than Braak [1], but a
and they will serve as a check of our results. very obvious one. We will perform two transformations
Itisalsoknownthattheparametersofthetwo-photon ofthevariablez. Thefirstoneisz →−z. Onecaneasily
Rabi model are restricted by |4g| < ω, otherwise the see, that it leaves the set (4) unchanged. Simple inspec-
eigenfunctions are not normalizable [5, 6]. tion shows, that this implies common parity of both φ
1
and φ . Plainly speaking,they must be either both even
2
II. EXACT ALGEBRAIC SOLUTION
φ (−z)=φ (z)
1 1
ThefirststepofthesolutionisgoingovertoBargmann φ (−z)=φ (z), (5)
2 2
space [7], introducing complex variable z, where the
bosonic operators simplify to or alternatively, simultaneously odd
2
φ (−z)=−φ (z) ω− ω2−16g2
1 1 κ= . (11)
φ (−z)=−φ (z). (6) p8g
2 2
In fact there should be ± in front of the square root,
The second applied transformation is y = iz. The set
but the plus sign would make κ divergent in the limit
(4) becomes
g →0, which is physically not reasonable. Notice that κ
remainsrealundertheabovementionedrestriction4|g|<
d2φ dφ ω ω. ThereisalsoafurtheranalogywiththestandardRabi
−2g 1 +ωy 1 −(2gy2+E)φ + 0φ =0
dy2 dy 1 2 2 model, where the special case ω0 = 0 is exactly solved
withhelpofthecoherentstateexp(±2g/ωb†)|0i,here|0i
d2φ dφ ω
2g 2 +ωy 2 +(2gy2−E)φ + 0φ =0. (7) is the lowest bosonic state [8, 9]. In our notation, Braak
dy2 dy 2 2 1 performsthetransformationφ ∝exp(−2gz/ω)ψ¯ ,re-
1,2 1,2
call b† → z in (2). The m = 2 Rabi Hamiltonian solu-
Now the functions φ and φ evidently swapped their
1 2 tion at ω = 0 involves the squeezed vacuum [10] term
places,thusuptosomecommonmultiplicativeconstant: 0
exp(±κ b†2)|0i. We will return to this case later.
In the next step we expand the functions ψ¯ :
1,2
φ (iz)=c φ (z)
1 2
φ (iz)=c φ (z). (8) ∞
2 1
ψ¯ (z)= Q (E)zn
1 n
Possible values of the parameter c can be found by n=X−∞
inserting iz instead of z and we get φ (−z)=c φ (iz)= ∞
1 2
c2 φ (z)andanalogouslyφ (−z)=c2 φ (z). Hence c2 = ψ¯ (z)= K (E)zn. (12)
1 2 2 2 n
1 from eq.(5) or c2 = −1 from eq.(6). The symmetry n=X−∞
parameter can acquire four values, c = ±1,±i. In the
standardm=1Rabimodelsuchparameterhadonlytwo To keep the solutions analytic we expect Qn(E) =
K (E) = 0 for n < 0 [1]. Inserting these expressions
values ±1andthe transformationz →−z wassufficient. n
into eqs.(10) we get the iteration scheme
Finally let us introduce four functions G (z,E) whose
c
roots with respect to E will be used to fulfill eq.(8):
2g(n+2)(n+1)Q +[(ω−8gκ)n−4gκ−E]Q
n+2 n
ω
G+(z)=φ2(iz)−φ1(z) + 0Kn =0
2
G (z)=φ (iz)+φ (z)
− 2 1 −2g(n+2)(n+1)K +[(ω+8gκ)n+4gκ−E]K
n+2 n
Gi(z)=iφ2(iz)+φ1(z) ω0
−4ωκK + Q =0. (13)
G−i(z)=iφ2(iz)−φ1(z). (9) n−2 2 n
In the last two cases the lower of eqs.(8) was multiplied Noticethattheindexesdifferby0,2or4,thusonlythe
by i in order to keep the Gc functions real. They share coefficientsQn,Knwithcommonparitywillbenon-zero,
the common parity of their φ functions. The upper eitherforevenn=0,2,4...,eq.(5),oroddn=1,3,5...,
1,2
of eqs.(8) is then no more independent. The complete eq.(6). Theprefactore−κz2 doesn’tspoilthe parity. The
discrete spectra will be given by all roots of all G (z,E) startingpointofouriterationschemeiseitheratn=0or
c
functions, again in full analogy with the standard Rabi n=1. Letus firsthavea look onthe casen=0 andthe
model [1]. Of course, the roots are meant with respect symmetry parameterc=1. Upper eq.(8) φ (iz)=φ (z)
1 2
to E and they are independent of any chosen z. at z = 0 implies Q = K , which can itself be a func-
0 0
Having categorized the symmetries, we return to the tionofHamiltonianparameters,servingasnormalization
set(4)andperformthetransformationφ =e−κz2ψ¯ . constantfor the eigenfunctions. Butas we areinterested
1,2 1,2
We get only in the roots of G+(z,E), this constant becomes an
unimportant multiplication factor and we can choose
2gψ¯′′+(ω−8gκ)zψ¯′ −(4gκ+E)ψ¯ + ω0ψ¯ =0
1 1 1 2 2
Q =1, K =1. (14)
−2gψ¯′′+(ω+8gκ)zψ¯′ −(4ωκz2−4gκ+E)ψ¯ + 0 0
2 2 2
+ω0ψ¯ =0, (10) The second case is c = −1. We have φ (iz)= −φ (z)
2 1 1 2
and again at z =0
whereweusedtheparameterκtosimplifythefirstequa-
tion by removing the term (8gκ2−2ωκ+2g)z2ψ¯ , thus
1
specifying its value Q =1, K =−1. (15)
0 0
3
The third case with c = i is only a bit more com-
plicated. The symmetry condition from eq.(8) is now
G =−C (iz)k+C zk =0⇒ k=0,4,8,...
φ (iz) = iφ (z) and its value at z = 0 is rather trivial − 2 2
1 2 G =−C (iz)k−C zk =0⇒ k =2,6,10,...
Q = K = 0. But we will compare the first derivatives + 2 2
0 0
w. r. t. z, i. e. φ′(iz) = iφ′(z) at z = 0 and we get G =−iC (iz)k+C zk =0⇒ k =3,7,11,...
1 2 −i 2 2
iQ1 =iK1, thus now our starting point is Gi =−iC2(iz)k−C2zk =0⇒ k =1,5,9,... (19)
The second solution (ψ ,0)T with φ = φ = C zk′
1 1 2 1
Q1 =1, K1 =1. (16) and eigenenergies E =ω0/2+k′ω gives
At last for c=−i we get
G =C (iz)k′ +C zk′ =0⇒ k′ =2,6,10,...
− 1 1
G =C (iz)k′ −C zk′ =0⇒ k′ =0,4,8,...
+ 1 1
Q =1, K =−1. (17)
1 1 G =iC (iz)k′ +C zk′ =0⇒ k′ =1,5,9,...
−i 1 1
Concludingthispartwecanseethatthe Hilbertspace G =iC (iz)k′ −C zk′ =0⇒ k′ =3,7,11,... (20)
ofeigenfunctionssplitsintofourdisjunctsubspaces. The i 1 1
corresponding eigenvalues can be found separately as We will later analyze mainly the cases when ω and
0
roots of four G (E) functions, eq.(9). We substitute ω are comparable and the eigenenergies as roots of G
c c
φ = e−κz2ψ¯ , the coefficients of expanded ψ¯ func- functions reorganize as follows:
1,2 1,2 1,2
tionsarefoundapplyingtheiterationscheme(13)subse-
ω ω ω
quently with four starting points, eqs.(14-17). The coef- G : − 0, 0 +2ω, − 0 +4ω,...
−
ficients not defined by eq.(13) are zero because of parity 2 2 2
ω ω ω
demands. K−2 =K−1 =0 as well. G+ : 0, − 0 +2ω, 0 +4ω,...
Before proceeding to numerical calculations, we ought 2 2 2
ω ω ω
tomentionseveralspecialcases,wheretheexactsolution G : − 0 +ω, 0 +3ω, − 0 +5ω,...
−i
2 2 2
was already known. They will serve as a check of our
ω ω ω
0 0 0
general solution. G : +ω, − +3ω, +5ω,... (21)
i
2 2 2
We can see that real values of the symmetry parameter
III. SOME EXACTLY KNOWN CASES c are connected with even number of phonon excitations
N whereas the imaginary c is coupled to odd N. The
global ground state is always given by the lowest root of
A. Case g=0
G , which is true also for non-zero g.
−
If the interaction constant is zero, the system sepa-
rates into independent two-state atom with energy lev-
B. Case ω0 =0
els ±ω /2 and a phonon with the modes Nω, where
0
N = 0,1,2,... Thus the overall energy is ±ω /2+Nω.
0 If the gap ω between atomic levels disappears, all
It is instructive to see how these values split into four 0
eigenenergiesbecometwicedegenerate. Theyareexactly
groups as roots of four G functions. Therefore we will
c known and now the complete spectrum is given by [5]
solvethissimplecaseexplicitly. Wereturntotheoriginal
functions ψ , as the set of equations (3) decouples for
1,2
ω 1
g =0andthetwoindependentsolutionsarefoundeasily E =− +(n+ )Ωω n=0,1,2,... (22)
2 2
where another dimensionless quantity was introduced
ψ1 =C1z2E2−ωω0 =C1zk′
ψ2 =C2z2E2+ωω0 =C2zk, (18) Ω=r1− 1ω6g22 =1− 8ωgκ. (23)
wherewe denotedthe exponentsby k′ andk,asthe par- Let us reproduce this result. We return to the φ
1,2
ity conditions (5) and (6) are common for φ and ψ functions, because now the system of equations (4) de-
1,2 1,2
functions. Theyforcek′ andk tobeintegerandtheana- couples for ω =0 and the general solutions are
0
lyticity demands make them non-negative. The energies
E =−ω /2+kωandE =ω /2+k′ω shouldbecommon,
0 0 φ =exp (κ− ω)z2 .
buttheyaregenerallydifferent,thustheoverallsolutions 1 4g
h i
will have either C =0 or C =0.
1 2 C H ωΩz +C F n+1,1,ωΩz2
Let us first analyze the case (0,ψ )T with the energy 11 −n−1 4g 21 1 1 2 2 4g
2 h (cid:16)q (cid:17) (cid:16) (cid:17)i
E = −ω0/2+kω. The linear combinations φ1 = ψ1 + φ2 =exp κz2 .
ψ2 = C2zk and φ2 = ψ1 −ψ2 = −C2zk. The solutions (cid:0) (cid:1)
for the functions Gc =0 finally yield C12Hn ω4Ωgz +C22 1F1 −n2,12,ω4Ωgz2 , (24)
h (cid:16)q (cid:17) (cid:16) (cid:17)i
4
exploitingthewellknownHermitianpolynomialsH and
n
hypergeometricfunction F . Further we introduced the
1 1 ωΩ 1
quantityn=(ω+2E)/(2Ωω)−1/2. Itisnothingelsebut Q /Q ≈− →0, k →∞ (29)
2k+2 2k
eq.(22)reversed. Thusiftheparitydemandsforcentobe 2g 2k
non-negative integer again, the spectrum is reproduced.
as required. The equations for K are not so simple,
Let us show it in detail at least for the simpler case of 2k
thus we resort to the exact solution of φ and find
n even. We make use of a formula relating Hermitian 2
polynomials and hypergeometric function, valid for n =
0,2,4,... ωΩ 1
K /K ≈ →0, k →∞. (30)
2k+2 2k
2g 2k
Hn(iqz) =ine−q2z2 F n+1,1,q2z2 , (25) For n odd and especially for non-zero ω0 we performed
2n/2(n−1)!! 1 1(cid:18) 2 2 (cid:19) atleastnumericalstudyofK , Q fromthe scheme(13)
n n
and found that their ratios are also proportional to 1/n
whereas for any n, including non-integer one, the Kum- in leading term. Hence we experience no problems with
mer transformation [11] gives analyticityoftheexpandedfunctionsincludingevenvery
large z.
The next remarkconcerns the practicalnumericalcal-
n 1 n+1 1
1F1 − , ,−y =e−y1F1 , ,y . (26) culations of the roots Gc(z,E) = 0, independent of z.
(cid:18) 2 2 (cid:19) (cid:18) 2 2 (cid:19) It turns out that such calculations are numerically more
stable for large z, which is allowed by the previous no-
First of all we set C11 = 0, because the Hermitian tion. Onecanexploitthelarge-zasymptoticfortheφ1(z)
polynomial with negative index has no parity, whereas contribution to G given by [11]
±
the rest of the solutions (24) are even functions for n
even. We define q2 = ωΩ/(4g), y = q2z2 and require
G =0 using eqs. (23) - (26). We get Γ(b) 1
± F (a,b,y)= eyya−b 1+O y ≫0,
1 1
Γ(a) (cid:20) (cid:18)y(cid:19)(cid:21)
(31)
(−2)n/2(n−1)!!C +C ∓C =0. (27) which is analytic for non-negative integer power a−b=
12 22 21
n/2,i. e. n=0,2,...asexpected. φ (iz)givesthe same
2
Fornnon-integer(oroddinteger),thetermwiththeHer- after Kummer transformation (26).
mitianpolynomialwouldbecome complex,butourfunc- Furtherwereturntothecompletespectrumineq.(22).
tions φ2(iz)andG± arereal- hencewe setalsoC12 =0. For |g| → ω/4 the quantity Ω → 0 and the energy be-
We already mentioned that one parameter can be cho- comes infinitely many times degenerate. This point is
sen, say the integration constant C21 =φ1(0)=1. Thus physicallyunsound,thoughwelldefinedinthe sense ofa
C22 = φ2(0) = ±1 and the two Gc functions are iden- limit.
tical, G+ = G−. One can compare the series expansion Concluding, the eigenvalue of eq.(22) for n = 0 coin-
ofthese φ1,2(z)solutions withthose givenby the scheme cides with mutually equal lowest roots of G− and G+,
(13), starting points (14) or (15) and see that in fact we for n = 1 it is the lowest root of both G and G , for
−i i
managed to perform the complete sum of eqs. (12). n=2 the second lowest root of both G and G , etc.
− +
Havingexactformulasforsomeφ andG (z,E)func-
1,2 c
tions at disposal allows us to make several observations.
Inthe on-linesupplement toref. [1],Braakreportssome C. Special cases with non-zero ω0,ω,g
problems with the radius of convergence R of his series
in z. R seemed to be finite in some cases, though ana- Let us now recall the result of Emary and Bishop [5],
lyticity in the whole complex plane of z is required. To who found a set of isolated solutions for our model. We
keeptheseriesanalogoustooureqs.(12)convergent,asa are not going to rederive it, but the basic fact is that
necessarycondition,theratioKn+1/Knhadtogotozero under some constraint on Hamiltonian’s parameters and
for n→∞. If it was non-zero, R became finite. For the energy E, at least one of the originaleigenfunctions ψ
1,2
special case ω0 =0 we can calculate this ratio explicitly. becomes a product of some exponential function and of
The coefficients Qn with n=2k become rather simple a polynomial.
The main statement is that there exist exactly known
eigenstates with eigenenergies
1
Q = (E−ε )(E−ε )...(E−ε ),
2k (2g)k(2k)! 2k−2 2k−4 0
(28) ω 1
E =− + N + Ωω (32)
whereεn areeigenenergiesfromeq.(22). Enroutewesee 2 (cid:18) 2(cid:19)
that the energy E =ε terminates ψ¯ (z) to a (Hermite)
n 1
polynomial. The ratio if the following conditions are fulfilled
5
2.0
ωω0 == 01 8 N=4 ωω0 == 21 EEE11(((GGG-+)))
1.5 E1(G-)i
E1(G+-) 6 E1(Gi)
E(G ) 2 +
En1.0 E12(G++--)i En4 N=3 EEEE222((((GGGG-i-)i)))
3 -
E(G)
0.5 2 3 +
N=2 N=4
0.0 0
-0.50 0.05 0.1 0g.15 0.2 0.25 -20 0.1 0.2 0g.3 0.4 0.5
FIG. 1: Eigenenergies as roots of Gc(E) functions for ω0 =0
and ω=1. Full lines are exact values. FIG. 2: Eigenenergies as roots of Gc(E) functions for ω0 =1
and ω=2.
for even solutions or L=35 for the odd ones. If the val-
ω2 uesofarootwithsmallerLconvergedtothesamevalue,
2−6Ω2+ 0 =0, N =2,
4ω2 we accept it. It turns out that for smaller values of |g|
ω2 theconvergenceisexcellent,butitbecomespooreraswe
6−10Ω2+ 0 =0, N =3, approach the maximal possible value, i. e. if |g|→ω/4.
4ω2
The interval of g in our plots is limited to 0 ≤g ≤w/4.
ω2
8(3−30Ω2+35Ω4)+2(7−17Ω2) 0 It is known that although the eigenfunctions differ af-
4ω2
ter changing the sign of g, the eigenenergies remain the
+ ω04 =0, N =4, (33) same, i. e. there is a mirror symmetry E(−g)=E(g).
16ω4 Let us test our calculations at first on the exactly
solvedcaseω =0. Wechooseω =1. Partsofparabolas
etc. We will proceed so that we choose the values of 0
withcommontopformtheexactspectrumfromeq. (22),
ω and ω, than we gradually change g. Eq.(33) yields
0 see full lines in Fig. 1. It is clear, that in the vicinity of
value(s) of g and followingly the energy is got from
theinfinitelymanytimesdegeneratepointwithg =ω/4,
eq.(32). These solutions will manifest themselves as
the roots of G (E) functions become very dense and the
crossing points of appropriate roots of G and G in c
− + functions themselves are quickly oscillating. That is the
the case of N even, or crossings of G and G roots
−i i reason, why even for the orders as large as L = 34 the
in the case of N odd. This is again analogous with the
values of appropriate roots did not converge completely
standard m = 1 Rabi model, where the Juddian points
andwe haveto resortto some fitting procedure,yielding
appeared as crossings of the roots of G(1) and G(1) [1].
− + better guess of the saturation value for L → ∞. The
Recall that contrary to eq. (22), this is only one known
roots are denoted so that the lowest one is E (G ), the
1 c
excited state for chosen set of Hamiltonian parameters,
secondlowestoneE (G ),etc. Wecanseethatthecalcu-
2 c
not the complete spectrum.
lated eigenenergies fit the exact values almost perfectly,
except for some deviation at g close to ω/4 and for the
higher root, in this case E (G ).
2 ±
IV. NUMERICAL RESULTS InFig. 2wepresentthreelowestrootsofG ,fullsym-
±
bols, and two lowest roots of G , smaller empty sym-
±i
The resulting energies as roots of G functions should bols. The values ω = 1 and ω = 2 are chosen so that
c 0
not depend on z; this would be true after summing infi- the spectrum at g = 0 becomes equidistant, see eq.(21).
nite number of summands in eq.(12). In numerical cal- The large empty circles are exact solutions of eqs.(33)
culations we truncate the sum and hope that the higher and (32). The value of N is written nearby. We can see
powers are not significant. We cannot use very small z, almost perfect match with the crossings of appropriate
because of numerical instability. Somewhat surprisingly, lines. The bottom four lines with N = 0 and N = 1 do
we can use as large values as z = 1000 or even z = 104 not cross; lines with N =4 cross twice.
without real change of the roots. This fact was already Fig. 3showsthesamerootsdenotedbythesamesym-
noted for exactly solvable cases, nevertheless one even bolsasFig.2,justforω =2andω =1. Besidesthewell
0
doesn’t have to use compromise medium values despite fitted exact crossing points, there are other crossings of
ofthetruncatedexpansions. Usingthesymbolicprogram lines with different N, which are not exactly known. A
Mathematica, we could sum up to zL term with L =34 similar figure with in fact the same Hamiltonian param-
6
5 Rabi model: E(1) ≈−ω /2−4g2/(ω+ω )+... [9].
ω = 2. 0 0 0
0
4 N=4 ω = 1.
3
V. SUMMARY
2 N=3
n N=4
E
We have found the complete spectrum of the two-
1 N=2
photon Rabi Hamiltonian as roots of four analytic func-
0 tionsG± andG±i,inthewholeparametricspace. These
functions are given by the recurrence scheme (13) with
-1 fourstartingpoints(14)-(17). Theunnormalizedeigen-
functions in Bargmannspacecanbe foundas well, using
-2
0 0.05 0.1 g 0.15 0.2 0.25 ψ1(z)=[φ1(z)+φ2(z)]/2 andψ2(z)=[φ1(z)−φ2(z)]/2.
In his ”Viewpoint: The dialogue between quantum
light and matter” [12], E. Solano states that Braak [1]
FIG. 3: Eigenenergies as roots of Gc(E) functions for ω0 =2 managed to enlarge the class of exactly solvable models
and ω = 1. The description of symbols is the same as in andthathe addedthe standardRabimodelonthe short
Figure 2. listofexactlysolvablequantumsystems. Webelievethat
thispaperaddsalsothetwo-photonRabiHamiltonianon
thesamelist. Thislistcanalmostsurelybeextendedfur-
eters was already published [2], where the authors plot ther by using Braak’s approach on other related models
also numerical results from larger matrices diagonaliza- and yet another task for future is deeper understanding
tion. of the criteria of its applicability.
There is a couple of simpler analytic results, that can
begotfromourapproach. Sowecanfinde. g. thesmall-
g expansion of the (global) ground state energy E0 = Acknowledgments
E (G ):
1 −
The author is indebted L. Sˇamaj for valuable com-
ω 8g2 ments. This work was supported by the Grants VEGA
E0 ≈− 0 − +O(g4), g ≪ω,ω0 (34) No. 2/0049/12.
2 2ω+ω
0
whichcanbe comparedwithsimilarresultforthe m=1
* E-mail address: [email protected] [8] R. F. Bishop, N. J. Davidson, R. M. Quick and D. M.
[1] D.Braak, Phys.Rev. Lett 107, 100401 (2011) van der Walt, Phys.Lett. A 254, 215 (1999)
[2] V. V. Albert, G. D. Scholes and P. Brumer, Phys. Rev. [9] I.TravenecandL.Samaj,Phys.Lett.A375,4104(2011)
A 84, 042110 (2011) [10] C. Brif, Ann.Phys.251, 180 (1996)
[3] I.I. Rabi, Phys.Rev.51, 652 (1937) [11] A. D. Polyanin and V. F. Zaitsev, Exact Solutions for
[4] B. R.Judd, J. Phys. C 12, 1685 (1979) OrdinaryDifferentialEquations,Chapman&Hall/CRC,
[5] C. Emary and R. F. Bishop, J. Phys.A 35, 8231 (2002) N. Y.(2003)
[6] S.N. Dolya, J. Math. Phys.50, 033512 (2009) [12] E. Solano, Physics 4, (2011)
[7] V.Bargmann, Comm. Pure Appl.Math. 14, 197 (1961)