a r X i v :h e p -p h /0111158v 1 13 N o v 2001DCPT/01/108
IPPP/01/54
QCD Finite Energy Sum Rules
and the Isoscalar Scalar Mesons
S.N.Cherry a and M.R.Pennington b a Groupe de Physique Th′e orique,IPN Universit′e de Paris-Sud,F-91406Orsay C′e dex,France.b Institute for Particle Physics Phenomenology,University of Durham,Durham,DH13LE,U.K.Abstract We apply QCD Finite Energy Sum Rules to the scalar-isoscalar current to deter-mine the lightest u d meson in this channel.We use ‘pinch-weights’to improve the reliability of the QCD predictions and reduce the sensitivity to the cut-o?s 0.A decaying exponential is included in the weight function to allow us to focus on the contribution from low mass states to the phenomenological integral.On the theoretical side we include OPE contributions up to dimension six and a contribu-tion due to instantons taken from the Instanton Liquid Model.Phenomenologically,
we incorporate experimental data by using a coupling scheme for the scalar current which links the vacuum polarisation to the ππscattering amplitude via the scalar form factor.We ?nd that the sum rules are well saturated for certain instanton
parameters.We conclude that the f 0(400?1200)de?nitely contains a large u
d component,whereas th
e
f 0(980)most likely does not.We are able to estimate the average light quark mass and ?nd m q (1GeV 2)=5.2±0.6MeV.
1
1Introduction
Currently,experiment suggests that there are more scalar mesons than can be accommo-dated in a single quark-model nonet.Many QCD motivated models predict the existence of non-q qq[3]and K
q states,which are the extra states?In the work that follows we hope to shed some light on this question by attempting to determine which of the5scalar-isoscalar mesons currently listed in the Particle Data Group tables[5]is the lightest u d meson in this channel.To address this question we will use the QCD sum rule technique.The answer has importance for chiral symmetry breaking in QCD, see for instance[6,7].
QCD sum rules are integral expressions that relate the hadronic and partonic regimes, i.e.the low energy world of resonances with the high energy world of(tractable)QCD. Since their conception over twenty years ago[8],QCD sum rules have become an estab-lished technique both for calculating hadronic properties in channels,where the QCD expressions are under control,and,conversely,estimating QCD parameters(such as the masses of the quarks)in channels where there is good experimental information.
Using QCD sum rules in the scalar channel is not a new idea.They were?rst applied to the scalar mesons in[9]where Laplace sum rules were used(for a recent review of Laplace Sum Rules see[10]).Here the phenomenological side was represented using the,now standard,resonance+continuum ansatz,with resonances being represented as δ-functions and the continuum being calculated entirely within perturbative QCD.As only the Operator Product Expansion(OPE)terms,discussed in Sect.3,were taken into account on the theory side,exact mass degeneracy between the lightest isoscalar
and isovector was found,with m f
0=m a
=1.00±0.03GeV.The s
q interpretation of these two mesons could not be ruled out with the data then available.
The e?ects of going beyond the OPE were studied in[12],where instantons(see Sect.4)were included on the theoretical side,but not in the QCD continuum on the phenomenological side.Once again the resonances were treated asδ-functions and only the isospin-1channel was considered.It was concluded that the lightest q
For the light scalar mesons the zero-width approximation is not a good one and the?rst attempt to go beyond it in a sum rule investigation was by Elias et al[13].Here resonances were represented by Breit-Wigner formulae,which,whilst better than aδ-function,still do not adequately describe the complex structure in the scalar sector.A further abstraction was introduced by replacing these Breit-Wigner shapes with a Riemann sum of rectangular https://www.doczj.com/doc/dd5304559.html,place sum rules were used,but in an integral form which requires the calculation of perturbative expressions at rather low energy scales.Again instantons were included, but not in the continuum.In the isovector channel,the a0(980)was found to decouple from the sum rule and the mass of the lightest quarkonium isovector was consistent with the a0(1450).In the isoscalar channel the conclusions were not so clear cut,it was predicted that the lightest quarkonium here should have a width less than half of its mass, and although the f0(980)could not be ruled out as the main contributor a lighter state was preferred.Similar conclusions were reached in[14].This study used both Laplace and Finite Energy sum rules,again requiring perturbative expressions to be evaluated at low energies.The authors noted,that for a consistent treatment of the Borel sum rules,the instanton e?ects should also be included in the QCD continuum contribution. The sum rules were dominated by an isoscalar with mass around1GeV and an isovector with mass around1.5GeV,thus suggesting the f0(980)and a0(1450)as the lightest q
q,K?0(1430).Instantons were included and all QCD integrals were carried out in the complex plane,which improves the convergence of perturbative expres-sions.The isovector decay constants were found to be of comparable size,suggesting a similar structure for both.The author concluded that this favoured a Unitarised Quark Model[16]scenario,where two physical hadrons can arise from one‘seed’state.
In the work that follows we will apply QCD Finite Energy Sum Rules introduced in Sect.2to the scalar-isoscalar channel.On the theoretical side we will include in Sects.3 and4instanton e?ects as well as the OPE terms up to dimension six and use the contour improvement prescription to increase the convergence of perturbative expressions.On the the phenomenological side in Sect.5we will incorporate experimental data directly by relating the hadronic vacuum polarisation to theππscattering amplitude via the pion’s scalar form-factor.The sum rules are evaluated in Sect.6and we present our conclusions in Sect.7.
3
2The Sum Rules
The connection between quarks and hadrons is through the vacuum polarisation,Π(s). This is de?ned by the correlation function of two currents,i.e.
Π(s)=i d4x e iqx 0|T{J(x)J(0)}|0 ,(1) where s=q2=?Q2and the currents are chosen so as to select the desired channel.In this work we are considering the scalar-isoscalar mesons and so we choose the Renormalisation Group invariant current
J(x)=m q
u(x)u(x)+√n(x)n(x),(2)
where m q≡1
π s00w(s)ImΠ(s)d s=?1
a phenomenological parameterisation).On the right-hand side we have an integral in the non-physical region(complex s),and here we use?eld theory(i.e.QCD)to calculate the integrand.The assumption that Eq.(3)holds for a range of values of s0is a statement of global quark-hadron duality,i.e.the integral of an hadronic expression is equal to the integral of an expression written in terms of partonic degrees of freedom.
Early sum rule work used positive integer powers of s as weight functions.In this work we wish to investigate the lightest meson in the I=J=0channel,and so we choose a decaying exponential,w(s)=exp(?s/M2).On the phenomenological side,this weight function will suppress the region of the integral s>M2,and so by varying M2we can control which regions of the integrand contribute most signi?cantly to the sum rule.If we also include integer powers of s in the weight function then we can build up a family of sum rules.
Almost by de?nition,at energies where resonances dominate the spectral density,the implications of QCD are not straightforwardly calculable.If QCD were tractable then the technology of QCD sum rules would be redundant.This means that,for moderate values of s0,our straightforward QCD description ofΠ(s)must fail,at the very least, in the region of the positive real axis(see Fig.1).However,it is believed[17]that it is only in the region of the positive,real axis that the straightforward QCD description is particularly unreliable and that in the rest of the s-plane calculable QCD provides a good description of the correlator.Consequently we introduce a zero into the weight function at the point where the circle joins the real axis,i.e.s0.Thus our sum rules have the weight function w(s)=s k(1?s/s0)e?s/M2and we denote them by the symbol T k.Elsewhere in the literature they are often referred to as’pinched weight’sum rules.Then the left hand side of Eq.(3)becomes
T k(s0,M2)≡1
s0 exp ?s
Although the current Eq.(2)is RG-invariant,the correlator it gives rise to is not.As well as making the calculation of the correlator within QCD more di?cult,this would also lead to an unwanted scale dependence in our?nal results.The second derivative of this correlator,Π′′(s),is RG-invariant.AsΠ′′(s)is also analytic within the contour shown in Fig.1,we could choose to write the sum rules completely in terms of this quantity,but ImΠ′′(s)is not directly related to any physical quantity and so we would lose our link to experiment.Instead we use partial integration to re-express the theoretical side of the sum rules in terms ofΠ′′(s).To do this we write the integral on the theoretical side as
i1= v′′(s)Π(s)d s.(5) We now integrate this expression by parts twice to give,
i1=v′(s)Π(s)?v(s)Π′(s)+ v(s)Π′′(s)d s,(6) and arrange the constants of integration which appear in v(s)and v′(s)so that these functions disappear at s=s0.Then our sum rule is
1
2πi |s|=s0v(s)[Π′′(s)]QCD d s.(7) 3The Operator Product Expansion
In order to calculate the vacuum polarisation within QCD we make use of the Operator Product Expansion(OPE)[20].This replaces the matrix element of the product of currents in Eq.(1)by a linear combination of local operators ordered in terms of increasing dimensionality,
Π(s=?Q2)= n C n(Q2) O n .(8)
The?rst term in this expansion is the purely perturbative contribution,with higher dimension terms giving the contribution due to various condensates.We will consider terms up to dimension six,higher order terms are expected to be negligible.Expressions are available in the literature for the dimension zero term to four-loop order,the dimension four term to two loops and the dimension six term at leading order.We quote these expressions with refernces to the original works for the full information.All expressions are given in the
The dimension zero expression can be found in [21]
and
is
given
by Π
′′0(Q 2)=6m 2q
3
?2log Q 2144
?356
log Q 24log 2Q 29
+b 14ζ(3) + ?27204ζ(3) log Q 28log 2Q 224
log 3Q 2
864?
915192
ζ(5),and ζ(n )is the Riemann Zeta function.As Π′′(s )is renormalisation group invariant,we are free to make any convenient choice of the renormalisation scale.We use the ‘contour improvement’prescription,whereby the logs in Eq.(9)are ‘re-summed’by making the choice μ2=Q 2.This improves the convergence of the perturbative series by removing the dependence on large powers of possibly large logs which would appear in the (unknown)higher order terms.The contour improved perturbative contribution to our correlator is then,Π′′0(Q 2)=6m 2q (Q 2)3+a 2s (Q 2) 50712ζ(3) +a 3s (Q 2) ?47816+4752Q 6 6 m q
3
?I g 18 +I s 4a s (Q 2)28π2
,(11)
7
where
I G=?9 a s G2
9 +4a s m s24
+3m4s3 ,(12)
and
I s= m s
7π2 124 .(13)
The leading order dimension six contribution can be taken from[25]and we follow this reference in taking this term at the?xed scale of1GeV.The dimension six contribution is then
Π′′6(Q2)=3m2q(1GeV2)
qσμνGμνaλa q
+π2a s qσμνλa q +2qγμλa q
2{γμ,γν}andλa are the usual Gell-Mann matrices
normalised such that Tr[λaλb]=2δab.Currently there is no reliable method to calculate the vacuum expectation values appearing in Eq.(14),so we follow standard practice and relate these dimension six condensates to the(RG variant)dimension three light quark condensate qσμνGμνaλa q ,this is done via the parameterisation
g qq ,(15) and for the four-quark condensates, qσμνλa q and qγμλa q ,via the vac-uum saturation hypothesis[8].To take into account the possible violation of the vacuum saturation hypothesis we include a multiplicative factor V vs.Thus our?nal expression for the dimension six contribution is
Π′′6(Q2)=6m2q
qq ?
176π2a s V vs
qq 2 .(16)
In order to evaluate all these expressions we need to calculate the running coupling and quark masses at complex values.This is done by solving the four-loop Renormalisation Group Equations that de?ne the QCDβandγfunctions(our conventions for these functions are given in Appendix A).
For the coupling,we choose to solve the implicit equation
a s(μ2)a s(m2τ)d a′m2τ =0,(17)
8
iteratively using Newton’s method.As input,we take the experimental valueαs(m2τ)=
0.334±0.022,as measured by the ALEPH collaboration[28].For reference,had we
used the standard expansion forαs,then at four-loops and with three active?avours,
this central input value would correspond toΛQCD≈365MeV.For higher values of αs(m2τ),or correspondinglyΛQCD,the radiative corrections to Eq.(10)become more
important and start to swamp the higher dimension terms,i.e it becomes more important
to include higher loop corrections to the perturbative contribution than to include the
condensates.It has been argued[29]that the sum rule methodology could become invalid
forΛQCD 330MeV.This is especially problematic for attempts to determine condensate
values via the sum rule technique.As this is not our aim and considering the success of
previous sum rule studies with similar and higher values ofαs(m2τ)(andΛQCD),we feel
justi?ed in applying sum rules to this problem.
The momentum dependence of the quark mass is given by
m q(μ2)=m q(1GeV2)exp a s(μ2)a s(1GeV2)γ(a′)
2π2 K1(ρc
Taking two derivatives we obtain
Π′′i(s)=3ρ2c m2q
Q2)+K21(ρc
qq|ππ .(21)
Inserting a complete set of hadronic states into the correlator Eq.(1)and keeping only the lowest state,i.e.two pions,gives the spectral density in the elastic region,
ImΠ(s)
32π2|d(s)|2,(22) whereβ(s)=
where the coupling function,α(s),is real for real s>4m2π.Substituting Eq.(23) into Eq.(22)and making use of the elastic unitarity relationship Im T=β|T|2,we obtain
ImΠ(s)
32π2
Im T
π=f
Im T(s)
K.Even after
this has opened theππ?nal state will continue to pick out the non-strange contribution to the sum over states.Thus we might expect Eq.(26)to give a reasonable approximation to the true spectral density even above1GeV.We call this Coupling Scheme I.
The multi-pion?nal states mentioned above will increase the n
π=f
(Im T(s))2
K threshold,whereas Coupling Scheme I ignores all
inelastic channels and so will underestimate the spectral density,Coupling Scheme II takes into account all inelastic channels,including those with hidden strangeness,and so may be an overestimate.
As input we take the parameterisation of the I=J=0ππscattering data carried out by Bugg et al[39].However,when continued down to threshold this fails to reproduce the scattering length predicted by Chiral Perturbation Theory(χPT)and so in the region
11
Im T
Energy (GeV)
Figure2:A sketch showing the absorptive part of the I=J=0ππscattering
amplitude as calculated from the data set described in the text(line).Superim-
posed are experimental points from[42](diamonds),[43](boxes),[44](triangles)
and[45](circles).
below450MeV we use the Roy equation extrapolation of the classic Ochs-Wagner phase-shifts[40],as carried out by Pennington and used in the discussion ofγγ→ππ[41].
Fig.2shows a sketch of the absorptive part of the isospin zero,S-wave amplitude forππscattering as calculated from our data set.From this we can clearly see that the f0(400?1200)and the f0(1370)appear as peaks,while the f0(980)and the f0(1500) show up as sharp dips.Thus these latter two states,and in particular the f0(980),will largely decouple from the spectral density and if saturation of the sum rules is achieved with either of our two Coupling Schemes,it will not be due in any great part to the f0(980)or f0(1500).By varying both M2and s0it may be possible to determine which of the f0(400?1200)and the f0(1370)plays the dominant role in saturation,but for the evaluation of the Wilson coe?cients in the OPE to be trustworthy,we would expect s0 to be too high to exclude the f0(1370).Nevertheless,as the decaying exponential in the weight functions of our sum rules suppresses the integrand for s>M2,it is still possible to determine which resonance is most important for saturation.
12
5.2Normalisation
A value for the normalisation factor f,can be determined by making a comparison with the spectral density obtained using an Omn`e s representation[46]of the form factor.This representation is given by
d(s)=d(0)exp s s′(s′?s) ,(28) whereφ(s)is the phase of I=J=0ππscattering.The form factor at zero momentum transfer,which normalises the Omn`e s function,can be determined from the Feynman-Hellman Theorem[47]and is given by
?m2π
d(0)=m q
qq =?m2πf2π[48],and thus we see that
d(0)=m2π.(30) Going to next-to-leading order inχPT would introduce a correction to Eq.(30)of the order of1%(see for example[49]),which is safe to neglect.
To evaluate the integral in Eq.(28)exactly requires knowledge of the scattering phase in the high energy region where it is unknown.However,the integrand is dominated by the region around s.So if we restrict our comparison to the low energy region then the behaviour of the phase above some point s1,where s1?s,will not greatly a?ect the value of the integral.We take s1to correspond to the last data point and approximate the phase above this point as a constant,φ1.Thus,the spectral density,as calculated from the Omn`e s representation,is written
ImΠ(s)
β(s) s1π? s14m2πφ(s′)d s′
32π2
01
2
3
4
5
6
7
I m
Π
/
π
(10-
6
G e V
4)Energy (GeV)
Figure 3:The spectral density below 400MeV as calculated from our Coupling Scheme (points)and the single channel Omn`e s representation (solid line).
6Results
In any sum rule calculation,we hope to ?nd regions of the parameter space where the theoretical and phenomenological sides are equal.We measure this saturation of the sum rules by introducing the double ratios
D k (s 0,M 2)=[T
k +1/T k ]had
Parameter
0.334
qq From GMOR relation[48]
a s G2
m s?(195MeV)4
159MeV
m20Sum rule Estimate[50]
V vs
ρc
νTo be determined
Table1:Standard values of all parameters used to evaluate the sum rules.
6.1The Instanton Parameters
The?rst task is to determineν2,the scale of the quark mass in the instanton contribution. To evaluate Eq.(20),we introduce a new parameter,λ,which is equal to the ratio of m2q(ν2)to m2q(1GeV2),i.e.,
λ= m q(ν2)β(a′)d a′ ,(33) where the second equality follows from Eq.(18).In Fig.4we show the stability curves obtained for various values of this parameter using Coupling Scheme II and s0=3.7GeV2. We have found that the saturation of the sum rules is practically independent of s0in the region3GeV2 s0 4GeV2and so no signi?cance should be attached to this choice of s0.However,it was found that saturation could not be achieved with the standard value ofρc=(600MeV)?1,and in Fig.4we pre-empt the results of Figs.5and6by using the non-standard valueρc=(480MeV)?1.
It is immediately clear that the sum rules are far from satis?ed whenλ=0,i.e.when the instanton contribution is removed.However,the sum rules are well saturated for λ≈1.2and we have found that the best value isλ=1.25,which we will use from now on.This value ofλcorresponds to a scaleν=0.96GeV.
As stated earlier saturation can only be achieved by use of a non standard value of ρc.In Figs.5and6we show the e?ect that varying this parameter has on the saturation curves.We see that good saturation is achieved for1/ρc=470?490MeV,with the‘best’value beingρc=(480MeV)?1≈0.42fm.This value is slightly larger(25%)than the estimate[32]based on the size of the gluon condensate which has become the standard value employed in the Instanton Liquid Model.However,it is entirely consistent with current lattice determinations which tend to?nd sizes in the range0.32–0.43fm and, unless stated otherwise,we will use this‘best’value ofρc from now on.
15
D
M2 (GeV2)
Figure4:The saturation curves obtained for various values ofλ,withρc=
(480MeV)?1and all other parameters as listed in Table1.
6.2Saturation
As shown in Fig.6,once the instanton parameters have been determined the quality of the saturation of our sum rules is very good.This means that the f0(980),being largely decoupled from our Coupling Schemes,does not play an important role in this saturation. On the other hand,the saturation curves are seen to be?at from quite low values of M2. With our weight function suppressing the portion of the integrand s>M2relative to the low energy region,the shape of the curves implies that saturation is achieved long before the f0(1370)contributes signi?cantly to the phenomenological integral.This suggests that it is the f0(400?1200)that plays the most important role in saturating our sum rules.
As the values of the condensates have not been?xed de?nitively,we have performed a number of tests to determine how sensitive the saturation of our sum rules is to variations of these input values.To test the dependence on the size of the gluon condensate,we write
a s G2 =c a s G2 0(34) where a s G2 0is the standard value listed in Table1.In Fig.7we show the saturation curves for various values of this factor,c.We see that the saturation is not strongly dependent on the size of the gluon condensate,but the best results are obtained in the
16
D M 2 (GeV 2)Figure 5:The saturation curves obtained for various values of ρc ,with λ=1.25and
all other parameters as listed in Table 1.D M 2 (GeV 2)Figure 6:As Fig.5but for ?ner steps in ρc .
17
D
M2 (GeV2)
Figure7:The saturation curves,D0,for various values of the gluon condensate.
D
M2 (GeV2)
Figure8:The saturation curves,D0,for various values of the vacuum saturation violation parameter,V vs.
18
range 0≤c ≤2.In Fig.8we see how deviations from vacuum saturation a?ect our results.We ?nd reasonable agreement for 0≤V vs ≤2,suggesting that vacuum saturation may be violated by 100%with little e?ect.For the remaining OPE parameters,varying their values from 0to 200%was found to have no noticeable e?ect.However,the best results are obtained with the strong coupling taken to be close to the central value quoted in Table 1.
6.3The Average Light Quark Mass
Another way to measure whether the sum rules are saturated is by making a prediction for a physical quantity.As M 2is an unphysical parameter,for ‘sensible’values of s 0this prediction should be e?ectively independent of M 2over a wide range.From Eq.(18),we
see that the theoretical side of our sum rules contains an overall factor of m 2q (1GeV 2),
we can thus make an estimate for the average light quark mass from
m q (1GeV 2)= [T k ]calc .(35)
If the graph of this quantity against M 2is found to be ?at then we can say that the sum
4.84.9
5
5.1
5.25.3
5.4
5.5
00.51 1.52 2.53 3.54
m q (M e V )M 2 (GeV 2)Figure 9:The quark mass curves for our two Coupling Schemes.The dashed line shows Coupling Scheme I and the solid line corresponds to Coupling Scheme II.
19
rules are saturated.Although the actual value of the quark mass extracted will depend on the constant f ,its M 2dependence will not,thus as a test of saturation Eq.(35)is independent of any uncertainties in the normalisation of the hadronic side.Once again we will use only the lowest sum rule,i.e.T 0,to estimate the quark mass.
4.84.9
5
5.1
5.25.3
5.4
5.5
00.51 1.52 2.53 3.54m q (M e V )M 2 (GeV 2)ρc = (470 MeV)-1ρc = (480 MeV)-1ρc = (490 MeV)-1(a)Average Instanton Size
4.84.95
5.15.25.35.45.500.51 1.52 2.53 3.54
m q (M e V )M 2 (GeV 2) c = 0c = 1c = 2(b)The Gluon Condensate 4.84.9
5
5.1
5.25.3
5.4
5.5
00.51 1.52 2.53 3.54
m q (M e V )M 2 (GeV 2)V vs = 0V vs = 1V vs = 2(c)Vacuum Saturation Violation Parameter
Figure 10:The e?ect of various parameters on the quark-mass curves.
In Fig.9we show the curve obtained from Eq.(35)using the instanton parameters determined in Sect.6.1for our two Coupling Schemes.We notice immediately that the curve is extremely ?at,in particular for Coupling Scheme II.In both cases the variation is less than 1%in the range 0.6GeV 2≤M 2
20
The journey my daughter Cathy has had with her swimming is as long as it is beautiful. Cathy suffered some terrible 16 in her early childhood. After years of regular treatment, she17became healthy. Two years ago, while Cathy was watching the Olympics, a dream came into her sweet little head— to be a swimmer. Last summer, she wanted to18our local swim team. She practiced hard and finally19it. The team practice,20 ，was a rough start. She coughed and choked and could hardly21 her first few weeks. Hearing her coughing bitterly one night, I decided to22her from it all. But Cathy woke me up early next morning, wearing her swimsuit23to go! I told her she shouldn’swimt after a whole night’coughing,s but she refused to24 and insisted she go . From that day on, Cathy kept swimming and didn ’ t 25 a single practice. She had a 26intention within herself to be the best she could be. My ten-year-old was growing and changing right before my eyes, into this27 human being with a passion and a mission. There were moments of28of course: often she would be the last swimmer in the race. It was difficult for Cathy to accept that she wasn29---ever. But that’didnta’ t stop her from trying. Then came the final awards ceremony at the end of the year. Cathy didn’ t expect any award but was stil 30 her friends and praise their accomplishments. As the ceremony was nearing the end, I suddenly heard the head coach31 ,“ The highest honor goes to Cathy!” Looking around, he continued,“ Cathy has inspired32 us and enthusiasm.33skills and talents bring great success, the most valuable asset(财富 )one can hold is the heart.” It was the greatest34of my da ughter’ s life. With all she hadbeen35in her ten years, this was the hour of true triumph( 成功 ). 16.A. failure B. pressure C. loss D. illness 17.A. usually B. finally C. firstly D. frequently 18.A. improve B. train C. join D. contact 19.A. increased B. found C. created D. made 20.A. however B. therefore C. otherwise D. instead 21.A. use B. survive C. save D. waste 22.A. pull B. tell C. hide D. fire 23.A. afraid B. nervous C. ready D. free 24.A. take off B. set off C. give up D. show up 25.A. attend B. miss C. ban D. Start 26.A. rich B. weak C. firm D. kind 27.A. trusted B.determined C.experienced D. embarrassed 28.A. frustration B. delight C. excitement D. surprise 29.A. beginner B.learner C. partner D. winner 30.A. cheer on B. compete with C. respond to D. run after 31.A. admitting B.explaining C.announcing D. whispering 32.A. humor B. will C. honesty D. wisdom 33.A. Although B. Since C. Once D. Because 34.A. discovery B. choice C. influence D. moment 35.A. through B. under C. across D. around My fiance (未婚夫) and I were excited about shopping for our first home. But our funds were 16 , and none of the houses in our price range seemed satisfactory. One agent 17 a house in particular. Although her description sounded wonderful, the price was18our range, so we declined. But she kept urging us to have a look 19. We finally did and it was20at first sight. It was Our Home, small and charming, overlooking a quiet lake. Walking through the rooms and talking with the owners, a nice elderly couple, we felt the warmth and21of the
关于毕业演讲稿 尊敬的各位领导、各位老师、各位同学： 大家好！我是06级2班的争气的败家子，非常荣幸代表我们班48名毕业生发言。四年过去了，学校的学习和生活为我们奠定了坚实的基础，明天我们就要离开曾经憧憬向往的大学生涯，走向我们的最终归宿——社会。服务社会才是我们的最终目标，我们会投身在社会的大课堂中不断进步，在社会的大舞台上大展鸿图。再此，我代表我们班的全体毕业生，感谢母校四年来对我们的培养和教育，感谢各位领导和老师对我们的关爱和教诲，感谢家人对我们的付出和鼓励，感谢身边朋友带给我们的快乐和帮助。 毕业，就像一个大大的句号。从此，我们告别了一段纯真的青春，一段年少轻狂的岁月，一个充满幻想的时代……毕业前的这些日子，时间过的好像流沙，想挽留，一伸手，有限的时光却在指间悄然溜走，毕业答辩，散伙席筵，举手话别，各奔东西……一切似乎都预想的到，一切又走的太过无奈。 还记得入学第一天我们的自我介绍么？ 还记得为次日的比赛挑灯做准备么？ 还记得我们一起逛街，一起喝酒，一起聊天，一起唱歌么？ 自习室、野游、考试、获奖……一幕幕的场景就像一张
张绚烂的剪贴画，串连成一部即将谢幕的电影，播放着我们的快乐和忧伤，记录着我们的青春和过往，也见证着我们的情深义重。从大一开始第一次上讲台的激动，第一次加入社团的好奇，第一次考试的紧张……到此时在为工作各种选择里彷徨，每一个人都忙忙碌碌，一切仿佛一首没写完的诗，匆匆开始就要匆匆告别。这些岁月里，大学是我们的资本，也是我们的慰藉。 班级聚餐的时候，所有的同学都在那里举杯，为过去的日子和情感，为将来的分别和感伤。昔日笑声不断的整个宿舍楼就这样在几天之内变回空楼，变成一个无限伤感的符号。想起四年以前，我们拎着简单的行李来到这里，而明天，我们重新拎起新的行李，将要开始下一站的生活。 再见了，我的宿舍，再见了，我的兄弟，再见了，我的青春，再见，我的大学。 毕业，又像一个长长的省略号。青春散场，我们等待下一场开幕。等待我们在前面的旅途里，迎着阳光，勇敢地飞向心里的梦想；等待我们在前面的故事里，就着星光，回忆这生命中最美好的四年，盛开过的花……道一声离别，送一声祝福，无论再过多少年，无论我们走到哪里，我们也不会忘记，曾经孕育过我们的这一片深情的土地。 大学时光只是人生路途中的一个小小的驿站，毕业并不代表结束，而是欢呼开始，不是庆祝完成，而是宣布进步。
完形填空高考真题解析 一、高中英语完形填空 1．阅读下面短文，掌握其大意，然后从各题所给的四个选项A、B、C、D四个选项中，选 出最佳选项。 I started volunteering at a soup kitchen several years ago. The original reason I was going was to 1 community service hours for school. My plan was to 2 go there a few times and get my service hours, but it taught me a lot. The typical volunteer there served 3 to people. Basically, I was 4 serving bread and juice to whoever wanted it, which was a simple task. Some of the people were homeless, and some of them were 5 families. All of them were people in need of a hot meal and a place to 6 for an hour or several minutes. 7 some of them looked like they weren't behaving well, we always took care of them. The first time I went there was right before Christmas. For the people coming to the soup kitchen, it was not exactly a 8 time. It made me think about my happy Christmas and made me feel how 9 I was. Unlike them, I have a home and I don't 10 cold or hunger. At that point, I decided that I 11 wanted to go back there. I couldn't offer them much, but I could always offer my time and 12 . The experience also gives me a feeling of 13 . Whenever I go there, people are 14 that I showed up again. They know my name and they know that I am more than happy to 15 them. It truly feels good to know that you can 16 someone's day. I've realized that the feeling of doing good for people can be a better 17 than any amount of money. You can't buy that feeling. I have never 18 a single second of my volunteering. It 19 me that dozens of cities have made it illegal to set up a soup kitchen. But I will continue my volunteer work and find more ways to show my 20 to people in need. 1. A. reduce B. avoid C. complete D. cancel 2. A. yet B. just C. even D. still 3. A. food B. work C. time D. money 4. A. tired of B. worried about C. responsible for D. free from 5. A. busy B. serious C. experienced D. struggling 6. A. hide B. rest C. live D. study 7. A. Although B. If C. Because D. Until 8. A. available B. strange C. pleasant D. painful 9. A. wise B. honest C. curious D. fortunate 10. A. turn down B. suffer from C. pass down D. learn from 11. A. definitely B. gradually C. equally D. hardly 12. A. reason B. effort C. chance D. patience 13. A. stability B. guilt C. loss D. appreciation 14. A. grateful B. confident C. proud D. shocked 15. A. change B. leave C. forget D. help
英语完形填空用法详解 一、高中英语完形填空 1．阅读下面的短文，从短文后各题所给的A、B、C和D四个选项中，选出可以填入空白 处的最佳选项。 Owura Kwadwo Hottish teaches computers in the school he works in. I think it is a 1 school except for the fact that the school didn't have 2 . Owura became famous after he posted photos of him on the Internet. In the picture, people could see he was teaching his students by 3 an entire computer on the blackboard. The photos showed the 4 level of education for children in Ghana. People were 5 that Owura made sure each button (按钮)was drawn correctly. Owura wanted the students to 6 what life with a computer could be like someday. He would come to school half an hour ahead of the 7 every day. He drew the computer on the blackboard, but at the end of his class, it was 8 off to start the next class, so he had to 9 it the next day! Owura's efforts were 10 when Microsoft (微软公司)took 11 of his act. They first took him to an international educators' meeting in Singapore. He made a 12 about his teaching methods at the meeting. He 13 a standing ovation (致敬)after the speech. 14 , Owura got the thing he always wanted for his students—some companies 15 computers to the school. Not a single child in the 16 had seen a real computer in their lives. Thanks to their teacher's 17 , the world took notice and responded with 18 to them. "Your work has really made us feel 19 about the world. At Microsoft, we believe that educators are heroes. They 20 influence the lifelong skills of their students." said Anthony Salcito, Vice president at Microsoft. 1. A. final B. dusty C. normal D. personal 2. A. computers B. playgrounds C. classrooms D. managers 3. A. operating B. drawing C. describing D. repairing 4. A. clear B. high C. ancient D. poor 5. A. worried B. disappointed C. afraid D. amazed 6. A. start B. imagine C. rebuild D. harm 7. A. line B. culture C. schedule D. judge 8. A. shown B. called C. cut D. rubbed 9. A. improved B. repeated C. ruined D. calculated 10. A. rewarded B. selected C. forgotten D. affected 11. A. care B. notice C. place D. charge 12. A. plan B. medal C. decision D. speech 13. A. received B. replaced C. decreased D. contained 14. A. Suddenly B. Hopelessly C. Importantly D. Strangely 15. A. gave B. sold C. lent D. applied
有关六年级毕业演讲稿集锦5篇 尊敬的学校领导、老师，亲爱的同学们： 大家好！ 我是六年级82班的杨茜。今天，我代表我们全班43位同学，向在场的各位恩师，致以最崇高的敬意！ 六年前，是您，敬爱的老师，将我们领进知识的大门，不仅让我们领略了知识的无穷魅力，更教会了我们做人的道理。我们就是您亲手栽种的桃李。在您谆谆的教导下，我们一天比一天懂事，一天比一天成熟，一天比一天勇敢！ 六载春秋，两千多个日日夜夜，恩师如山啊！您阳光雨露般的教育，给予了我们生命的色彩，让我们收获了生命的精彩！这些，我们都将一辈子铭记于心！请允许我们道一声：“老师，您辛苦了！” “六年磨一剑”再过两周，我们就要迎来人生中的第一次大项的考试——小学毕业考试。这次考试既是献给母校最好的礼物，又是自己学习道路上的小结，更是一张成为合格初中生的名片。我们深知，它的重要性不亚于中考，甚至是高考。 为了能给母校献上一份最美的答卷，为了给自己的人生打下坚实的基础，我们要发扬百米赛跑冲刺的精神，努力奋斗，潜心复习，力争考出六年来最优异的成绩！ 如何在短时间内有针对性地进行复习，提高成绩呢？我们准备从下面几点入手： 一、全面研读教材 我们的考试内容都来自教材，从教材的第一页到最后一页，每个部分都可能考到。我们只有充分准备，在考试时才能游刃有余。 二、善于总结 就是在仔细看完一篇教材的前提下，我们一边看书，一边作总结性的笔记，把教材中的要点列出来，从而让厚书变薄，并理解其精华所在。 三、对症下猛药 在老师、同学的指导下，我们要找出自己在学习上的薄弱环节，集中精力，重点攻破。这样就能以较小的投入获得较大的考试收益，在考试中力于不败之地。
【英语】完形填空高考真题解析 一、高中英语完形填空 1．阅读下面短文，从短文后各题所给四个选项（A、B、C和D）中，选出可以填入空白处 的最佳选项。 Have you ever seen a miracle happen? Two winters ago, I did. It was 1 that day. The road was covered with snow. I asked my mom if we could go to our neighbor's hill and take my dog, Buddy. She said, "Sure. 2 don't go any farther than that." After a while, we walked over to another hill by a pond (池塘) not far away. Then suddenly I saw Buddy walk out onto the ice. I was 3 when he walked to the middle of the pond. The 4 was thin there. We 5 Buddy, but he didn't listen. Then 6 the ice broke, and my dog 7 into the water. I asked my friend if I should go in and get him out. She said, "NO!" 8 , I ran to my house to tell my mom. It was 9 for me to run on the snowy ground. I was tired, but I had to 10 Buddy. When my mom heard what 11 , she said, "Get in the car." When we got there, she told me to run home 12 and get her phone. But back at the pond, my mom 13 to get Buddy out of the water and fell in. 14 , my friend saw it happen. She ran to a neighbor's house for help. Soon they got my mom out of the water, but Buddy was still 15 for his life. The fire department (部门) put a ladder (梯子) out onto the ice to get him, but it didn't 16 . I was back again by then, and someone said I should 17 inside the neighbor's house. It didn't look 18 for Buddy. I went inside and 19 and hoped for a miracle. Then, my uncle walked into the house with Buddy in his arms! I was so 20 , I hugged Buddy and thanked the firefighters (消防员). 1. A. dry B. rainy C. warm D. cold 2. A. Unless B. Or C. So D. But 3. A. dissatisfied B. moved C. frightened D. excited 4. A. light B. air C. voice D. ice 5. A. looked for B. laughed at C. shouted at D. woke up 6. A. suddenly B. slowly C. naturally D. possibly 7. A. jumped B. fell C. ran D. walked 8. A. Still B. Besides C. However D. Instead 9. A. difficult B. fun C. wrong D. important 10. A. find B. watch C. help D. train 11. A. followed B. happened C. came D. appeared 12. A. again B. once C. too D. together 13. A. stopped B. agreed C. tried D. decided 14. A. Luckily B. Firstly C. Strangely D. Sadly 15. A. swimming B. dreaming C. hunting D. fishing
高中英语完形填空解题技巧大全 开篇练习 My son Joey was born with club feet. The doctors said that with treatment he would be able to walk， but would never run very well. The first three years of his life was 1 in hospital. By the time he was eight，you wouldn‘t know he has a problem when you saw him 2 . Children in our neighborhood always ran around 3 their play， and Joey would jump and ran and play，4 . We never told him that he probably wouldn‘t be 5 to run like the other children. So he didn’t know. In 6 grade he decided to join the school running team. Every day he trained. He ran more than any of the others， 7 only the top seven runners would be chosen to run for the 8 . We didn‘t tell him he probably would never make the team， so he didn’t know. He ran four to five mile every day - even when he had a fever. I was 9 ，so I went to 10 him after school. I found him running 11 . I asked him how he felt. “Okay，” he said. He has two more miles to go. Yet he looked straight ahead and kept 12 . Two weeks later， the names of the team 13 were caked. Joey was number six on the list. Joey had 14 the team. He was in seventh grade - the other six team members were all eighth graders. We never told him he couldn‘t do it … so he didn’t know. He just 15 it. 1. A. spent B. taken C. cost D. paid 2. A. talk B. sit C. study D. walk 3. A. after B. before C. during D. till 4. A. either B. too C. though D. yet 5. A. able B. sorry C. glad D. afraid 6. A. sixth B. seventh C. eighth D. ninth 7. A. so B. if C. then D. because 8. A. neighborhood B. familyC. school D. grade 9. A. excited B. tiredC. pleased D. worried 10. A. think about B. hear fromC. agree with D. look for 11. A. alone B. away C. almost D. already
小学生毕业演讲稿合集5篇 演讲稿具有逻辑严密，态度明确，观点鲜明的特点。在快速变化和不断变革的新时代，演讲稿对我们的作用越来越大，那么问题来了，到底应如何写一份恰当的演讲稿呢？下面是收集整理的小学生毕业演讲稿5篇，希望对大家有所帮助。 尊敬的老师，亲爱的同学： 大家好! 我是六(4)班xx同学，很高兴能在开学典礼上发言。 高效课堂，既然是高效，那么高效课堂之效高在哪里?我认为高效课堂是面对所有学生的课堂，是老师讲得好、学生学得好、学生有热情、学生满意。第一、要使一节课成为一节高效课堂,我觉得首先课前的准备是最重要的。“台上一分钟，台下十年功”，要实现课堂高效，必须下足课前准备功夫。第二、优化时间安排，努力创建学习型课堂。时间就是效率。抓紧时间，用好时间才能保证课堂的高效率。 高效课堂就是指高效率，高效益，高效果的课堂。就是要改变一块黑板，一张嘴，一支粉笔，一讲到底。 (1)三高：高效率、高效益、高效果 (2)三动：身动、心动、神动 (3)三量：思维量、信息量、训练量 (4)三特点：立体式、快节奏、大容量 (5)三学：肯学、想学、会学 (6)减负：轻负担、高质量;低耗时、高效益 通过一年的学习，我们收获到的有哪些呢? 1、学会评价别人; 2、学会认识自己的不足; 3、学会与别人合作。然后取长补短，相得益彰。打造出一个更优秀的自己。我认为，高效课堂就是让课堂成为学生的讲台，让教室成为学生的天下。 以上是我的开学典礼发言，谢谢大家! 尊敬的各位家长、老师们，亲爱的同学们：你们好！今天，我们欢聚在这美丽的校园举行20xx届小学毕业典礼。向为了同学们健康成长而奉献心血与智慧的老师们和家长们致以崇高的敬意！
（一）2012年全国卷一 Body language is the quiet, secret and most powerful language of all! It speaks 1 than words. According to specialists, our bodies send out more 2 than we realize. In fact, non-verbal communication(非言语交际) takes up about 50% of what we really 3 , And body language is particularly 4 when we attempt to communicate across cultures（文化）.Indeed, what is called body language is so 5 a part of us that it's actually often unnoticed. And misunderstandings occur as a result of it . 6 , different societies treat the 7 between people differently. Northern Europeans usually do not like having 8 contact(接触)even with friends, and certainly not with 9 . People from Latin American countries 10 , touch each other quite a lot. Therefore, it's possible that in 11 . it may look like a Latino is 12 a Norwegian all over the room. The Latino, trying to express friendship, will keep moving 13 . The Norwegian, very probably seeing this as pushiness, will keep 14 - which the Latino will in return regard as 15 . Clearly, a great deal is going on when people 16 .And only a part of it is in the words themselves. And when parties are from 17 cultures, there's a strong possibility of 18 . But whatever the situation, the best 19 is to obey the Golden Rule: treat others as you would like to be 20 . 1. A. straighterB. louder C. harder D. further 2. A. sounds B. invitations C. feelings D. messages 3. A. hopeB. receive C. discover D. mean 4. A. immediate B. misleading C. important D. difficult 5. A. well B. far C. much D. long 6. A. For example B. Thus C. However D. In short 7. A. trade B. distance C. connections D. greetings 8. A. eye B. verbal C. bodily D. telephone 9. A. strangersB. relatives C. neighboursD. enemies 10. A. in other words B. on the other hand C. in a similar way D. by all means 11. A. trouble B. conversation C. silence D. experiment 12. A. disturbing B. helping C. guiding D. following 13. A. closer B. faster C. in D. away 14. A. stepping forward B. going on C. backing away D. coming out 15. A. weakness B. carelessness C. friendliness D. coldness 16. A. talk B. travel C. laugh D. think 17. A. different B. European C. Latino D. rich 18. A. curiosity B. excitement C. misunderstanding D. nervousness 19. A. chance B. time C. result D. advice 20. A. noticed B. treated C. respected D. pleased (二)2012年全国卷二 Around twenty years ago I was living in York. 1 I had a lot of experience and a Master’s degree, I could not find 2 work. I was 3 a school bus to make ends meet and 4 with a friend of mine, for I had lost my flat. I had 5 five interviews (面试) with a company and one day between bus runs they called
与毕业有关的经典演讲稿 与毕业有关的经典演讲稿1 尊敬的各位领导、各位老师、各位同学： 大家好!我是06级2班的争气的败家子，非常荣幸代表我们班48名毕业生发言。四年过去了，学校的学习和生活为我们奠定了坚实的基础，明天我们就要离开曾 经憧憬向往的大学生涯，走向我们的最终归宿——社会。服务社会才是我们的最 终目标，我们会投身在社会的大课堂中不断进步，在社会的大舞台上大展鸿图。 再此，我代表我们班的全体毕业生，感谢母校四年来对我们的培养和教育，感谢 各位领导和老师对我们的关爱和教诲，感谢家人对我们的付出和鼓励，感谢身边 朋友带给我们的快乐和帮助。 毕业，就像一个大大的句号。从此，我们告别了一段纯真的青春，一段年少轻狂 的岁月，一个充满幻想的时代……毕业前的这些日子，时间过的好像流沙，想挽留，一伸手，有限的时光却在指间悄然溜走，毕业答辩，散伙席筵，举手话别， 各奔东西……一切似乎都预想的到，一切又走的太过无奈。 还记得入学第一天我们的自我介绍么? 还记得为次日的比赛挑灯做准备么? 还记得我们一起逛街，一起喝酒，一起聊天，一起唱歌么? 自习室、野游、考试、获奖……一幕幕的场景就像一张张绚烂的剪贴画，串连成 一部即将谢幕的电影，播放着我们的快乐和忧伤，记录着我们的青春和过往，也 见证着我们的情深义重。从大一开始第一次上讲台的激动，第一次加入社团的好奇，第一次考试的紧张……到此时在为工作各种选择里彷徨，每一个人都忙忙碌碌，一切仿佛一首没写完的诗，匆匆开始就要匆匆告别。这些岁月里，大学是我 们的资本，也是我们的慰藉。 班级聚餐的时候，所有的同学都在那里举杯，为过去的日子和情感，为将来的分 别和感伤。昔日笑声不断的整个宿舍楼就这样在几天之内变回空楼，变成一个无 限伤感的符号。想起四年以前，我们拎着简单的行李来到这里，而明天，我们重 新拎起新的行李，将要开始下一站的生活。 再见了，我的宿舍，再见了，我的兄弟，再见了，我的青春，再见，我的大学。
一、计划实施的目标：1.提高自己在英语、数学（物理）的学习能力； 2.加强运动，提高身体素质； 3.学会做简单的家常菜。 二、具体措施： 1.组织好学习小组，以三个人一组为好，进行合作性学习。（学习水平相当的同学组成小组比较合适） 2.按教材知识前后的顺序，整理上半学期的英语、数学（物理）练习卷和考试卷。（或者可以根据老师的推荐去买一两本合适提高学科能力的辅导材料） 3.把练习卷上做正确的题目进行整理，确认自己已经掌握了哪些知识，具备了哪些运用能力，树立自己对本学科的信心。 4.把练习卷上做错的题目进行整理，打开教科书，逐题进行分析，找到错误的关键之处，进行认真的订正后，再到教材上找到相关类型的题目，进行练习、强化。 5.无论在订正试卷习题，还是在做辅导材料时，遇到困难，三人小组一起分析，或者到图书馆去找参考书，可以去找老师，也可以进行挑战性竞赛，看谁先解决问题。 6.把错题订正完成以后，就把错题订正的全过程背出来，进行接替方法的积累。 7.到办学质量比较好的社会业余学校去读英语口语训练班，提高自己的英语会话能力，和英语学习素质。 8.对于新初三的学生，可以在暑期中参加一些竞赛辅导班，为新学年的各类竞赛打好基础。（也可以超前读一些新初三的课本，提前了解初三的学习内容）9.每周打两次球，游三次泳，增加运动，提高体能。 10.每天晚上，听音乐，上网浏览，看书读报，和同学聊天等，做自己有兴趣的事。 11.每周跟着父母学做两次家常菜，如炒茄子、蒸鱼之类，再做一些力所能及的家务。 三、时间安排： 1.一星期学习五天，上午小时，下午 2.5小时。按一小时一节课安排好课表。 2.每天下午3点以后是运动或做家务的时间。 3.晚上两个小时是发展兴趣的时间。 4.双休日去看一些有益的展览会，或参加一些有益的社会活动。 四、督促自律： 1.设计好每天的生活学习评价表，对自己每天的生活、学习作好评价。 2.合作学习小组的同学互相督促，执行暑期生活、学习计划有始有终。 1、重视学习习惯的培养 培养良好的学习习惯，不仅对初三学生必要，实际上对所有的初中学生都有着重要的意义。学习的计划性、目的性一定要强，不要搞突击战；听课要讲究效率，在课堂上无论会与否，都要紧跟着老师的节奏，与老师多交流，尤其是有不懂的地方要及时问。 老师要帮助学生做好成绩分析，这对提高每个同学的成绩是非常有效的，初三的同学不妨学习一下，通过反复的成绩分析，对自己的学习做到心中有数。在做成绩分析时一定要细致，哪里知识掌握不牢固，哪一部分考试经常丢分，然后适当练习弥补不足，并且给自己定下目标，避免今后再次丢分。 2、跟着老师走