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Against Supersymmetry

Against Supersymmetry
Against Supersymmetry

a r X i v :h e p -t h /0606029v 1 5 J u n 2006

Against Supersymmetry

Dan Radu Grigore

1

Dept.Theor.Phys.,Inst.Atomic Phys.,Bucharest-M?a gurele,MG 6,Rom?a nia

G¨u nter Scharf

2

Institute of Theor.Phys.,University of Z¨u rich

Winterthurerstr.,190,Z¨u rich CH-8057,Switzerland

Abstract

We consider the massless supersymmetric vector multiplet in a purely quantum framework

and propose a power counting formula.Then we prove that the interaction Lagrangian for a massless supersymmetric non-Abelian gauge theory (SUSY-QCD)is uniquely de-termined by some natural assumptions,as in the case of Yang-Mills models,however we do have anomalies in the second order of perturbation theory.The result can be easily generalized to the case when massive multiplets are present,but one ?nds out that the massive and the massless Bosons must be decoupled,in contradiction with the standard model.Going to the second order of perturbation theory produces an anomaly which cannot be eliminated.We make a thorough analysis of the model working only with the component ?elds.

PACS:11.10.-z,11.30.Pb

1Introduction

A quantum?eld theory should provide two items:the Hilbert space of the physical states and the(perturbative)expression of the scattering matrix.In perturbation theory the Hilbert space is generated from the vacuum by some set of free?elds i.e.it is a Fock space.In theories describing higher spin particles one considers a larger Hilbert space of physical and unphysical degrees of freedom and gives a rule of selection for the physical states;this seems to be the only way of saving unitarity and renormalizability,in the sense of Bogoliubov.In this case one should check that the interaction Lagrangian(i.e.the?rst order of the S-matrix)leaves invariant the physical states.If the preceding picture is available in all detail then one can go very easily to explicit computations of some scattering process.Other constructions of a quantum?eld theory,as those based on functional integration,are incomplete in our opinion if they are not translated in the operatorial language in such a way that the consistency checks can be easily done.

The construction of the QCD Lagrangian in the causal approach goes as follows[7],[16].The Hilbert space of the massless vector?eld vμis enlarged to a bigger Hilbert space H including two ghost?elds u,?u which are Fermi scalars of null mass;in H we can give a Hermitian structure such that we have

v?μ=vμu?=u,?u?=??u.(1.1) Then one introduces the gauge charge Q according to:

Q?=0,Q?=Q,

[Q,vμ]=i?μu,

{Q,u}=0,{Q,?u}=?i?μvμ;(1.2) here?∈H is the vacuum state.Because Q2=0the physical Hilbert space is given by H phys=Ker(Q)/Im(Q).

The gauge charge is compatible with the following causal(anti)commutation relation: [vμ(x),vν(y)]=i gμνD0(x?y){u(x),?u?(y)}=?i D0(x?y)(1.3) and the other causal(anti)commutators are null;here D m(x?y)is Pauli-Jordan causal dis-tribution of mass m≥0.In fact,the?rst relation together with the de?nition of the gauge charge,determines uniquely the second relation as it follows from the Jacobi identity [vμ(x),{?u(y),Q}]+{?u(y),[Q,vμ(x)]}={Q,[vμ(x),?u(y)]}=0.(1.4) We can then assume that all the?elds vμ,u,?u have the canonical dimension equal to1so the gauge charge raises the canonical dimension by1.It is usefull to convince the reader that the gauge structure above gives the right physical Hilbert space.We do this for the one-particle Hilbert space.The generic form of a stateΨ∈H(1)?H from the one-particle Hilbert subspace is

with test functions fμ,g1,g2verifying the wave equation equation.We impose the condition Ψ∈Ker(Q)??QΨ=0;we obtain?μfμ=0g2=0i.e.the generic element Ψ∈H(1)∩Ker(Q)is

Ψ= fμ(x)vμ(x)+ g(x)u(x) ?(1.6) with g arbitrary and fμconstrained by the transversality condition?μfμ=0;so the elements of H(1)∩Ker(Q)are in one-one correspondence with couples of test functions(fμ,g)with the transversality condition on the?rst entry.Now,a generic elementΨ′∈H(1)∩Im(Q)has the form

Ψ′=QΦ= ? ?μf′μ(x)u(x)+ ?μg′(x)vμ(x) ?(1.7) so ifΨ∈H(1)∩Ker(Q)is indexed by(fμ,g)thenΨ+Ψ′is indexed by(fμ+?μg′,g??μf′μ). If we take f′μconveniently we can make g=0.We introduce the equivalence relation f(1)μ~f(2)μ??f(1)μ?f(2)μ=?μg′and it follows that the equivalence classes from Ker(Q)/Im(Q)are indexed by equivalence classes of wave functions[fμ];we have obtained the usual one-particle Hilbert space for the photon.The preceding argument can be generalized to multi-particle Hilbert space[6];the idea comes from Hodge theory and amounts in?nding an homotopy operator?Q such that the spectum of the“Laplace”operator{Q,?Q}can be easily determined.

By de?nition quantum chromodynamics assumes that we have N copies vμj,u j,?u j j= 1,...,N verifying the preceding algebra for any j=1,...,N.

The interaction Lagrangian t(x)is some Wick polynomial acting in the total Hilbert space H and verifying the conditions:(a)canonical dimensionω(t)=4;(b)null ghost number gh(t)=0 (where by de?nition we have gh(vμj)=0,gh(u j)=1gh(?u j)=?1and the ghost number is supposed to be additive);(c)Lorentz covariant;(d)gauge invariance in the sense:

[Q,t(x)]=i?μtμ(x)(1.8) for some Wick polynomials tμof canonical dimensionω(tμ)=4and ghost number gh(tμ)=1.

The gauge invariance condition guarantees that,after spatial integration the interaction Lagrangian t(x)factorizes to the physical Hilbert space Ker(Q)/Im(Q)in the adiabatic limit, i.e.after integration over x;the condition(1.8)is equivalent to the usual condition of(free) current conservation.Expressions of the type

d Q b+?μtμ(1.9) with

ω(b)=ω(tμ)=3gh(b)=?1gh(tμ)=0(1.10) are called trivial Lagrangians because they induce a null interaction after space integration

(i.e.the adiabatic limit)on the physical Hilbert space.One can prove that the condition

(1.8)restricts drastically the possible form of t i.e.every such expression is,up to a trivial Lagrangian,equivalent to

where the(real)constants f jkl must be completely antisymmetric.In the the rest of the paper we will skip the Wick ordering notations.Going to the second order of the perturbation theory produces the Jacobi identity.So we see that,starting from some very natural assumptions,we obtain in an unique way the whole structure of Yang-Mills models.We expect the same thing to happen for more complex models as supersymmetric theories.

If we want to generalize to the supersymmetric case we must include all the?elds vμ,u,?u in some supersymmetric multiplets.By de?nition[9]a supersymmetric multiplet is a set of Bose and Fermi?elds b j,f A together with the supercharge operators Q a such that the commutator (resp.the anticommutator)of a Bose(resp.Fermi)?eld with the supercharges is a linear combination of Fermi(resp.Bose)?elds;the coe?cients of these linear combinations are partial derivative operators.We must also suppose that the supercharges are part of an extension of the Poincar′e algebra called the supersymmetric algebra;essentially we have(for N=1 supersymmetry):

Q a?=0,ˉQˉa?=0ˉQˉa=(Q a)?(1.12)

Pμ=0(1.13)

{Q a,Q b}=0,{Q a,ˉQˉb}?2σμ

aˉb

and

Q a U A=A a b Q b.(1.14)

[Q a,Pμ]=0,U?1

A

Here U A is a unitary representation of the Poincar′e group and Pμare the in?nitesimal generators of the space-time translations.There are not many ways to do this.We will show that for the vμwe must use the vector multiplet and for the ghost?elds we must use chiral multiplets.Then we must impose that t is also supersymmetric invariant.A natural de?nition is:

[Q a,t]=d Q s a+?μtμaω(s a)=7/2gh(s a)=?1ω(tμa)=7/2gh(tμa)=0;(1.15) this means that after space integration(i.e.the adiabatic limit)we obtain on the physical Hilbert space an expression commuting with the supercharges.

In the supersymmetric framework one usually makes a supplementary requirement,namely that the basic supersymmetric multiplets should be organized in super?elds[1],[8],[17],[18]i.e.?elds dependent on space-time variables and some auxiliary Grassmann parametersθa,ˉθˉa.It is showed in[9]that there is a canonical map w→sw≡W mapping a ordinary Wick monomial w(x)into its supersymmetric extension

W(x,θ,ˉθ)≡exp iθa Q a?iˉθˉaˉQˉa ;(1.16) in particular this map associates to every?eld of the model a super?eld.Moreover,one postu-lates that the interaction Lagrangian t should be of the form

t(x)≡ dθ2dˉθ2T(x,θ,ˉθ)(1.17)

One can hope to have an uniqueness result for the coupling if one?nds out a supersymmetric generalization of(1.8).A natural candidate would be the relation:

[Q,T(x,θ,ˉθ)]=D T(x,θ,ˉθ)?H.c.=D T+ˉDˉT(1.18) where

D a≡?

?ˉθˉa

+iσμ

bˉa

θb?μ.(1.19)

We have

(D a T)?=±ˉDˉa T?,

{D a,D b}=0,{ˉDˉa,ˉDˉb}=0,{D a,ˉDˉb}=?2iσμ

aˉb

?μ(1.20) where in the?rst formula the sign+(?)corresponds to a super-Bose(-Fermi)?eld.The last relations is used to eliminate space-time divergences?μTμ(x,θ,ˉθ)in the right-hand side of the relation(1.18).It is clear that(1.18)implies(1.8).

The most elementary and general way of analyzing supersymmetries is to work in compo-nents and to see later if the solution can be expressed in terms of super?elds.We will prove that in the massless case there is an unique solution for SUSY-QCD if we consider a weaker form of(1.15)i.e.we require that this relation is true only on physical states:

<Ψ1,([Q a,t]?d Q s a??μtμa)Ψ2>=0(1.21) whereΨ1,Ψ2∈Ker(Q)modulo Im(Q).We remark that(1.15)implies(1.21)but not the other way round.

In the next Section we give the structure of the multiplets of the model and we present the gauge structure.In Section3we determine the most general form of the interaction Lagrangian compatible with gauge invariance and prove that we have supersymmetric invariance also. The expression for the ghost coupling seems to be new in the literature.The details of the computation are given in the Appendix.

We also investigate in what sense one can rephrase the result using super?elds.An im-mediate consequence of the analysis in terms of component?elds is that one cannot impose (1.18).However we can establish a contact with traditional literature based on the so-called Wess-Zumino gauge.

In Section4we extend the result to the massive case hoping to obtain the minimal su-persymmetric extension of the standard model.We obtain a curious obstruction,namely the sector of massive gauge?elds and the sector of massless gauge?elds must decouple;this does not agree with the standard model.

Unfortunately,if we proceed to the second order of perturbation theory,we obtain a su-persymmetric contribution to the anomaly which cannot be eliminated by rede?nitions of the chronological products.

In Section5we do the same analysis for the new vector multiplet[9]working in components

2The Quantum Super?elds of the Model

2.1The Vector Multiplet

The vector multiplet is the collection of?elds C,φ,vμ,d,χa,λa where C,is real scalar,φis a complex scalar,vμis a real vector andχa,λa are spinor?elds.We suppose that all these?elds are of mass m≥0.We can group them in the super?eld

V=C+θχ+ˉθˉχ+θ2φ+ˉθ2φ?+(θσμˉθ)vμ+θ2ˉθˉλ+ˉθ2θλ+θ2ˉθ2d.(2.1) It is convenient to de?ne the new?eld:

λ′a≡λa+i

4

C(2.2)

and then the action of the supercharges is given by

i[Q a,C]=χa

{Q a,χb}=2i?abφ

{Q a,ˉχˉb}=?iσμ

aˉb

(vμ+i?μC)

[Q a,φ]=0

i[Q a,φ?]=λ′a?iσμ

aˉb

?μˉχˉb

i[Q a,vμ]=σμ

aˉb ˉλ′ˉb?i?μχ

a

{Q a,λ′b}=2i?ab d′?2iσμρ

ab

?μvρ

{Q a,ˉλ′ˉ

b

}=0

[Q a,d′]=?

1

4

c1+c2 D m(x?y)

[φ(x),d(y)]=

m2

16c1D m(x?y)

2

{χa(x),χb(y)}=2(c4?ic3)?ab D m(x?y), {χa(x),ˉχˉb(y)}=c1σμ

aˉb

?μD m(x?y) {λa(x),λb(y)}=?

m2

4c1σμ

aˉb

?μD m(x?y)

{χa(x),λb(y)}=?2i c2?ab D m(x?y),

χa(x),ˉλˉb(y) =?i(c4+ic3)σμaˉb?μD m(x?y)(2.4)

and the rest of the(anti)commutators are null;here c j j=1,...,4are some real coe?cients. However,if we want that the commutation relations for the vector?eld remain unchanged(1.3) then we must require c1=0,c2=?1

2

D m(x?y)

[d′(x),d′(y)]=?

im2

2D m(x?y)

[vμ(x),vν(y)]=i gμνD m(x?y)

{χa(x),λ′b(y)}=i?ab D0(x?y),

λ′a(x),ˉλ′ˉb(y) =σμaˉb?μD0(x?y)(2.5) and the rest of the(anti)commutators are null.More compactly

[V(X),V(Y)]=?

1

2ω(λ′)=

3

2.2The Ghost Chiral Multiplets

We require that the ghost?elds are also members of some multiplets of chiral type.We admit that these ghost multiplets are also massless.The generic forms of a chiral ghost and anti-ghost super?elds are

U(x,θ,ˉθ)=a(x)+2iˉθˉζ(x)+i(θσμˉθ)?μa(x)+ˉθ2g(x)+ˉθ2θσμ?μˉζ(x)(2.10) and respectively

?U(x,θ,ˉθ)=?a(x)?2iˉθˉ?ζ(x)+i(θσμˉθ)?

μ

?a(x)+ˉθ2?g(x)?ˉθ2θσμ?μˉ?ζ(x)(2.11) where a,g,?a,?g are Fermi scalar?elds andζa,?ζa are Bose spinor?elds.Let us remind that we choose them such that[10]

(ζa)?=ˉζˉa(?ζa)?=?ˉ?ζˉa(2.12) -see(1.1);the chirality condition means

D a U=0D a?U=0.(2.13)

The action of the supercharges on these?elds is

{Q a,a}=0{Q a,a?}=2ζa

{Q a,g}=?2iσμ

aˉb

?μˉζˉb{Q a,g?}=0

[Q a,ζb]=?ab g?i[Q a,ˉζˉb]=σμ

aˉb

?μa(2.14) and respectively:

{Q a,?a}=0{Q a,?a?}=2?ζa

{Q a,?g}=2iσμ

aˉb ?μˉ?ζ

ˉb

{Q a,?g?}=0

[Q a,?ζb]=?ab?g?i[Q a,ˉ?ζˉb]=?σμ

aˉb

?μ?a.(2.15) It is convenient to work with the Hermitian(resp.anti-Hermitian)?elds

u≡a+a?v≡?i(a?a?)

?u≡?a??a??v≡?i(?a+?a?)(2.16) such that we have

u?=u v?=v?u?=??u?v?=??v.(2.17) Then we have the following action of the supercharges:

{Q a,u}=2ζa{Q a,v}=2iζa

{Q a,g}=?2iσμ

aˉb

?μˉζˉb{Q a,g?}=0

and respectively:

{Q a,?u}=?2?ζa{Q a,?v}=?2i?ζa

{Q a,?g}=2iσμ

aˉb ?μˉ?ζ

ˉb

{Q a,?g?}=0

[Q a,?ζb]=?ab?g?i[Q a,ˉ?ζˉb]=?

1

2ω(?u)=1ω(?v)=1ω(?g)=2ω(?ζ)=

3

?The gauge charge commutes with the action of the Poincar′e group;in this way the Poincar′e group induces an action on the physical space H phys≡Ker(Q)/Im(Q).

?The gauge charge anticommutates with the supercharges:

{Q,Q a}=0;(2.24)

in this way the supersymmetric algebra induces an action on the physical space H phys.

?The gauge charges squares to zero as in the Yang-Mills case

Q2=0.(2.25) If one makes the most general ansatz for the action of Q compatible with the preceding conditions one?nds out an important result:making some convenient rescaling of the?elds, the gauge charge is uniquely determineded preceding assumptions.In the massless case we have d′=d and Q is uniquely determined by:

Q?=0Q?=Q(2.26)

and

[Q,C]=i v[Q,vμ]=i?μu[Q,φ]=?g?[Q,φ?]=g[Q,d]=0

{Q,χ}=2iζ{Q,ˉχ}=?2iˉζ{Q,λ′}=0 {Q,u}=0{Q,v}=0{Q,g}=0{Q,g?}=0[Q,ζa]=0[Q,ˉζˉa]=0

{Q,?u}=?i?μvμ{Q,?v}=2i d{Q,?g}=0{Q,?g?}=0

[Q,?ζa]=?1

2

σμ

bˉa

?μλ′b.(2.27)

One can express everything in terms of super?elds also[8]:

[Q,V]=U?U?{Q,U}=0

{Q,?U}=?

1

where b=C,φ,φ?,vμ,d′,χ,ˉχ,λ′,ˉλ′and f=?u,?g,?g?,?ζ,ˉ?ζ;if we take into account the particular choice we have made for the causal(anti)commutation relations of the vector multiplet one ?nds out[10]that we have

{a(x),?a?(y)}=i

4

σμ

aˉb

?μD0(x?y)(2.30)

and all other causal(anti)commutators are null.So,like for ordinary Yang-Mills models,the causal(anti)commutators of the ghost?elds are determined by the corresponding relations of the vector multiplet?elds.

Finally we mention that we can impose that the vector?eld is part of the so-called rotor multiplet[10].The?elds of this are(d,vμ,λa)and the supercharges de?ned through

[Q a,d]=σμ

aˉb

?μˉλˉb.

{Q a,λb}=?i(?ab d+σμν

ab

Fμν)

{Q a,ˉλˉb}=0

i[Q a,vμ]=σμ

aˉb

ˉλˉb.(2.31) where Fμν≡?μvν??νvμ.

However in this case the general form of the gauge charge

[Q,vμ]=i?μu[Q,d]=0{Q,λ′a}=ασμ

aˉb

?μˉζˉb(2.32) is not compatible with the relation{Q,Q a}=0.Also if we consider that the ghost multiplets are Wess-Zumino we obtain a contradiction.

3Supersymmetric QCD

3.1Supersymmetric QCD in Terms of Component Fields

It is better to illustrate the method we use to?nd the most general gauge invariant Lagrangian on the simplest case,namely ordinary QCD,i.e.we will brie?y show how to obtain the expres-sion(1.11)as the unique possibility.So,we consider a Wick polynomial t which is tri-linear in the?elds vμj,u j,?u j has canonical dimension4and null ghost number,is Lorentz covariant and gauge invariant in the sense(1.8).First we list all possible monomials compatible with all these requirements;they are:

f(1)=f(1)

jkl vμj vνk?μv lμf(2)=f(2)

jkl

vμj v kμ?νvνl(3.1)

and

g(1)=g(1)

jkl vμj u k?μ?u l g(2)=g(2)

jkl

?μvμj u k?u l g(3)=g(3)

jkl

vμj?μu k?u l.(3.2)

We now list the possible trivial Lagrangians.They are total divergences of null ghost number

t(1)μ=t(1)

jkl vνj v kνv lμt(1)

jkl

=t(1)

kjl

and the co-boundary terms of ghost number?1:

b(1)=b(1)

jkl vμj v kμ?u l b(1)

jkl

=b(1)

kjl

b(2)=b(2)

jkl u j?u k?u l b(2)

jkl

=?b(2)

jlk

.(3.4)

Now we proceed as follows:using?μt(1)μit is possible to make

f(1) jkl =?f(1)

lkj

;(3.5)

using d Q b(1)we can make

f(2)

jkl

=0;(3.6) using?μt(2)μit is possible to take

g(3)

jkl

=0;(3.7)?nally,using d Q b(2)we can make

g(2) jkl =g(2)

kjl

.(3.8)

So we are left only with three terms.If we compute d Q t and use the known identity:?2f j=0,j=1,2,3=?(?μf1)(?μf2)f3=

1

the?rst equation,together with(3.5)amounts to the total antisymmetry of the expression

f jkl≡f(1)

jkl .The second equation from the preceding system gives then g(2)

kjl

=0.As a result we

obtain the(unique)solution:

t(1)=f(1)

jkl

(vμj vνk?νv lμ?vμj u k?μ?u l);(3.16) it can be easily be proved,using the formula(3.9)that d Q t(1)is indeed a total divergence.

The supersymmetric case goes on the same lines only the computational di?culties grow exponentially because now we have many more terms of the type f,g,tμ,b.

We would expect to obtain the expression(3.16)together with terms with supersymmetric partners.The details are given in the Appendix.It is remarkable that we obtain only two possible solutions namely the usual Yang-Mills solution for QCD t(1)given above by(3.16) and

t(2)=f′jkl[(λ′jσμˉλ′k)vμ

l

+2i(λ′j?ζk)u l+2i(ˉλ′kˉ?ζj)u l].(3.17) We impose now supersymmetric invariance condition(1.15)on

t≡t(1)+t(2)(3.18) and we hope to obtain a non-trivial solution.If we are successful we must also go the the second order of perturbation theory.First we consider only the terms bilinear in vμand linear inλ′from the commutator[Q a,t]and impose that condition that they are a sum of a co-boundary

and a total divergence.It is not very hard to prove that we obtain the restriction f(2)

jkl =?i f(1)

jkl

i.e.the interaction Lagrangian should be:

t=f jkl[vμj vνk?νv lμ?i(λ′jσμˉλ′k)vμ

l

?vμj u k?μ?u l+2(λ′j?ζk)u l+2(ˉλ′kˉ?ζj)u l](3.19)

where we have simpli?ed the notation:f jkl≡f(1)

jkl .We note that this interaction Lagrangian is

R-invariant.The?rst two terms are standard in the literature-see for instance[14],[15].The next contribution is the standard ghost coupling from the Yang-Mills theory[6],[16].The last two contributions to the ghost coupling in seems to be absent from this analysis.So,gauge invariance in the?rst order is not true for the expression from[15].

We can prove after some computation that the preceding expression veri?es the following equation:

[Q a,t]=d Q s a+?μtμa?2f jkl(vμj u k?μ?ζla+2iσμν

ab

v jν?μ?ζb k u l)(3.20) where

s a≡f jkl[?2σμν

ab v jμv kν?ζb l?iχja vμ

k

?μ?u l?2i(λ′j?ζk)ζla?2i(ˉλ′kˉ?ζj)ζla

+i?v jσμ

aˉb

ˉλ′ˉb

k

v lμ?2?v j?ζka u l?2λ′jφk u l](3.21)

and

tμa≡f jkl ?μχja u k?u l?χja?μu k?u l+χja vνk?μv lν?i

So,its seems that the last two terms from(3.20)-which cannot be rewritten as d Q s a+ total divergence-are apparently spoiling the supersymmetric invariance condition(1.15).

However,as we have said in the Introduction,there is a natural way to save supersymmetric invariance namely to impose instead of(1.15)the weaker form(1.21)

<Ψ1,([Q a,t]?d Q s a??μtμa)Ψ2>=0

withΨ1,Ψ2∈Ker(Q)modulo Im(Q).We?rst takeΨto be generated from the vacuum by the physical?elds vμj,λ′j,d′j;if we consider as before,the terms bilinear in vμand linear inλ′from the commutator[Q a,t]we obtain the interaction Lagrangian should be given by(3.19). It follows that we also have(3.20);however,if we apply this relation on a physical statesΨj we obtain that(1.21)is true;indeed the extra terms give zero in this average.If we substitute in(1.21)Ψj→Ψj+QΦj the relation stays true(one has to use the anticommutativity of Q with the supercharges.)So we have obtained an unique solution for supersymmetric QCD as in the Yang-Mills case.

3.2Supersymmetric QCD in Terms of Super?elds

From the preceding Subsection we conclude that we cannot express the interaction Lagrangian in terms of super?elds such that(1.17)and(1.18)are true;indeed if this would be true then we would have(1.15)with s a=0.Even the possibility of obtaining the expression(3.19)in the form(1.17)is very unlikely.Nevertheless let us start from the super?eld expression of the interaction suggested by the classical analysis[10].

Arguments from classical?eld theory suggests that the interaction should be an expression of the type

t classical=t(1)

classical +t(2)

classical

(3.23)

where

t(1) classical ≡?

i

2

(χjσμ?μˉχk)d l?

1

2

(χj?μλ′k)vνl?

1

the third line is suggested in the literature e.g.[13]see formula (C.1c)and appears in (3.19)also.The expression of the ghost coupling is much more complicated:

t (2)classical =f jkl

?v μj u k ?μ?u l +v μj ?μv k ?v l ?2d j u k ?v l ?2φj g k ?v l ?2φ?j g ?k ?v

l +2i φj u k ?g l +2i φ?j u ?k ?g l +2i C j g ?

k ?g l +2i C j g k ?g ?l ?i (χj σμˉζk )?μ?u l +i (ζj σμˉχk )?μ?u l +2i (λ′j ζk )?v l +2i (ˉλ′j ˉζk )?v l ?(?μχj σμˉζk )?v l +(χj σμ?μˉζk )?v l +(?μζj σμˉχk )?v l ?(ζj σμ?μˉχk )?v l

+2(λ′j ?ζ

k

)u l

+2(ˉλ′j ˉ?ζl

)u l

+i (?μχj

σμˉ?ζl

)u k

?i (χj

σμ?μˉ?ζl

)u k

+i (?μ?ζl σμˉχj )u k ?i (?ζl σμ?μˉχj )u k ?(χj σμˉ?ζl )?μv k +(?ζl σμˉχj )?μv k ?4i φ?j

(ζk ?ζ

l )+4i φj (ˉζk ˉ?ζl

)?2C j (?μζk σμˉ?ζl

)+2C j (ζk σμ?μˉ?ζl

)?2C j (?μ?ζl σμˉζk )+2C j (?ζl σμ?μˉζk )?2i v μj (ζk σμˉ?ζl )?2i v μj

(?ζl σμˉζk )+2(χj ζk

)?g l

+2(ˉχj ˉζk

)?g ?l

?2(χj ?ζl

)g k

+2(ˉχj ˉ?ζ

l

)g ?k

+total divergence (3.26)

Now one can eliminate trivial terms following the procedure given in the table from the

Appendix.As a result the total expression is of the following form:

t classical =t +d Q b +?μt μ+?t

(3.27)

where t is given by (3.19)and

?t

≡f jkl i

2

(χj σν?νˉχk )?μv μl +φ?j (χk σμ?μˉλ′l )?φj (?μλ′l σμˉχ

k )+2i φj u k ?g l +2i φ?j u k ?g

?l +2i C j g ?

k ?g l +2i C j g k ?g ?l +i (?μχj σμˉ?ζl )u k ?i (χj σμ?μˉ?ζl

)u k +i (?μ?ζl σμˉχk )u k ?i (?ζl σμ?μˉχk )u k +(χj σμ?μˉ?ζl

)v k ?(?μ?ζl σμˉχk )v k +2(χj ζk )?g l +2(ˉχj ˉζk )?g ?l

(3.28)

is not trivial.So we do not have an exact match between classical and quantum analysis.

However,there is a way of eliminating the last term used in the literature,namely to compute t classical in the so-called Wess-Zumino gauge [20].This means that the super?eld V can be written as

V =A +A ?+V ′(3.29)where:V ′=(θσμˉθ)v μ+θ2ˉθˉλ′+ˉθ2θλ′+θ2ˉθ2d ′(3.30)

and

A =

1

2

(θσμˉθ

)?μC ?i

The conclusion is that our expression for the interaction Lagrangian t can be obtained from (3.23)-(3.26)if we substitute V j→V′j.

Gauge invariance and supersymmetric invariance are not very conveniently expressed in terms of V′j so one must work with the interaction Lagrangian(3.23)in every order of the perturbation theory and make at the end V j→V′j.However,going to the second order of per-turbation theory might be more di?cult in this super?eld formalism than working in component ?elds.

3.3Second Order of Perturbation Theory

One can go to the second order of perturbation theory very easily in components and compute the anomaly as in[7];we have in(1.8)

tμ=f jkl 12u j u k?μ?u l?i(λ′jσμˉλ′k)u l .(3.33) From here we obtain

d Q[t(x),t(y)]=i ?

?yμ

[t(x),tμ(y)](3.34)

The anomalies are obtained in the process of causal splitting of this identity;we obtain in

general

d Q T(t(x),t(y))=i ?

?yμ

T(t(x),tμ(y))+A(x,y)(3.35)

where T(a(x),b(y))is the chronological product associated to the Wick monomials a and b and A(x,y)is the anomaly which spoils the gauge invariance in the second order.One?nds out that

A(x,y)=A YM(x,y)+A susy(x,y)(3.36) where the?rst term already appears in the pure Yang-Mills case and can be eliminated(as a co-boundary plus a total divergence)if and only if one imposes Jacobi identity on the constants f jkl.The second term from the anomaly is of purely supersymmetric nature:

A susy(x,y)=a susy(x)δ(x?y)(3.37)

with

a susy=f jkl f mnl ?i

space of the massive vector?eld vμis enlarged to a bigger Hilbert space H including three ghost ?elds u,?u,h.The?rst two ones are Fermi scalars and the last is a Bose real scalar?eld;all the ghost?elds are supposed to have the same mass m>0as the vector?eld.In H we can give a Hermitian structure such that we have

v?μ=vμu?=u,?u?=??u h?=h(4.1) and we also convene that gh(h)=0.Then one introduces the gauge charge Q according to:

Q?=0,Q?=Q,

[Q,vμ]=i?μu,[Q,h]=imu

{Q,u}=0,{Q,?u}=?i(?μvμ+mh)(4.2) and,because Q2=0,the physical Hilbert space is given,as in the massless case,by H phys= Ker(Q)/Im(Q).

The gauge charge is compatible with the following causal(anti)commutation relation:

[vμ(x),vν(y)]=i gμνD m(x?y){u(x),?u?(y)}=?i D m(x?y)[h(x),h(y)]=?i D m(x?y)

(4.3) and the other causal(anti)commutators are null.We see that the canonical dimension of the scalar ghost?eld must beω(h)=1.It is an easy exercise to determine the physical space in the one-particle sector:the equivalence classes are indexed by wave functions of the form

Ψ0= fμ(x)vμ(x)?μfμ=0.(4.4)

The general argument can be found in[7].

To go to the supersymmetric case we must include the?eld h into a supersymmetric mul-tiplet.Again,the natural candidate is a chiral?eld we take:

B(x,θ,ˉθ)=b(x)+2ˉθˉψ(x)+i(θσμˉθ)?μb(x)+ˉθ2f(x)

m2

?iˉθ2θσμ?μˉψ(x)+

θ2ˉθ2u(x)and m2

4

These relations are compatible with the following canonical dimensions:

ω(b)=1ω(ψ)=3/2ω(f)=2.(4.8)

It is convenient to work with the self-adjoint bosonic ghost?elds:

h≡b+b?h′≡?i(b?b?).(4.9)

Next we must?nd the general form of the gauge charge.We proceed as in Subsection1.2 but we do not assume that the(anti)commutator with the gauge charge raises the canonical dimension by an unit;we?nd out that in the massive case Q is uniquely determined by:

Q?=0Q?=Q(4.10)

and

[Q,C]=i v[Q,vμ]=i?μu[Q,φ]=?g?[Q,φ?]=g[Q,d′]=0

{Q,χ}=2iζ{Q,ˉχ}=?2iˉζ{Q,λ′}=0 {Q,u}=0{Q,v}=0{Q,g}=0{Q,g?}=0[Q,ζa]=0[Q,ˉζˉa]=0

{Q,?u}=?i(?μvμ+mh)

{Q,?v}=i(2d′+mh′+m2C)

{Q,?g}=?m2φ??imf{Q,?g?}=?m2φ+imf?

[Q,?ζa]=?1

2

m2χa[Q,ˉ?ζˉa]=?

1

2

m2ˉχˉa

[Q,h]=im u[Q,h′]=im v[Q,f]=?img{Q,ψa}=mζa{Q,ˉψˉa}=mˉζˉa.(4.11) One can express everything in terms of super?elds also[10]:

[Q,V]=U?U?{Q,U}=0

{Q,?U}=?

1

2D m(x?y)[ζa(x),ˉ?ζˉb(y)]=?

1

and all other causal(anti)commutators are null.So,like for ordinary Yang-Mills models,the causal(anti)commutators of the ghost?elds are determined by the corresponding relations of the vector multiplet?elds.So we see that the causal(anti)commutators are uniquely?xed by some natural requirements.We also can prove that it is not possible to use rotor multiplet instead of the full vector multiplet and it is not possible to use Wess-Zumino multiplets instead of the ghost multiplets.

First,we eliminate terms independent on the?elds b,f,ψa,of the type d Q b+?μtμexactly as in the massless case.Now comes an important observation:suppose that we have a Wick monomial t B which has at least one factor b,f,ψa and it is of canonical dimensionω(t B)≤4; if we consider the expression d Q t B then from(4.11)it follows thatω(d Q t B)≤4.

This means that the new(b,f,ψ-dependent terms)from the interaction Lagrangian cannot produce terms of canonical dimension5when commuted with the gauge charge.So,?rst we can proceed with the elimination of trivial Lagrangians of canonical dimension4like in the Appendix.This in turn means that the general solution of the gauge invariance problem must be a sum of two expressions of the type t(1)and t(2)from the preceding Section-see(3.16) and(3.17)-to which one must add b-dependent terms such that gauge invariance is restored. These new terms are easy to obtain.For the Yang-Mills coupling t they are well-known[7]

t(1)

massive

≡f jkl(v jμv kν?μvνl?vμj u k?μ?u l)

+f′jkl(h j?μh k vμ

l ?m k h j v kμvμ

l

?m k h j?u k u l)+f′′jkl h j h k h l(4.14)

and for the second coupling the generic expression is

t(2)massive≡?i f jkl[(λ′jσμˉλ′k)vμ

l

+2i(λ′j?ζk)u l+2i(ˉλ′kˉ?ζj)u l]

+p(1)

jkl (λ′jψk)h l+p(2)

jkl

(ˉλ′jˉψk)h l+p(3)

jkl

(λ′jχk)h l+p(4)

jkl

(ˉλ′jˉχk)h l.(4.15)

The gauge invariance condition gives well-known constraints on the constants f′,f′′and:

p(3) jkl =

i

2

p(2)

jkl

m k

i p(1)

jkl m l=2f jkl m k i p(2)

jkl

m l=?2f jkl m k.(4.16)

We see from the second set of relations that we must have

f jkl=0for m k=0m l=0(4.17) which implies that the massive and massless gauge?elds must decouple;this does not agree with the standard model.

Regarding supersymmetric invariance,this means that the argument which prevents the equation(1.15)to be true remains valid but(1.21)should stay true.

We note that the anomaly(3.38)will remain in this case also.

Remark The solution found in[12]has in fact terms of canonical dimension5:indeed

5The New Vector Multiplet

The new vector multiplet was introduced in[9];it is the second possibility of a multiplet which contains a vector?eld.First we give the de?nition of a Wess-Zumino multiplet:such a multiplet contains a complex scalar?eldφand a spin1/2Majorana?eld f a of the same mass m.The supercharges are de?ned in this case by:

[Q a,φ]=0,[ˉQˉa,φ?]=0

i[Q a,φ?]=2f a,i[ˉQˉa,φ]=2ˉfˉa

{Q a,f b}=?i m?abφ,{ˉQˉa,ˉfˉb}=i m?ˉaˉbφ?

{Q a,ˉfˉb}=σμ

aˉb ?μφ,{ˉQˉa,f b}=σμ

bˉa

?μφ?.(5.1)

The?rst vanishing commutators are also called(anti)chirality condition.The causal (anti)commutators are:

φ(x),φ(y)? =?2i D m(x?y),

{f a(x),f b(y)}=i?ab m D m(x?y),

f a(x),ˉfˉb(y) =σμaˉb?μD m(x?y)(5.2)

and the other(anti)commutators are zero.

To construct the new vector multiplet one and adds a vector index i.e.makes the substitu-tionsφ→vμf a→ψμa;here vμis a complex vector?eld.In the massless case the action of the supercharges is:

[Q a,vμ]=0,[ˉQˉa,v?μ]=0

i[Q a,v?μ]=2ψμa,i[ˉQˉa,vμ]=2ˉψμˉa

{Q a,ψμb}=0,{ˉQˉa,ˉψμˉb}=0,

{Q a,ˉψμˉb}=σν

aˉb

?νvμ,{ˉQˉa,ψμb}=σνbˉa?νv?μ.(5.3) The gauge structure is done by considering that the ghost and the anti-ghost multiplets are also Wess-Zumino(of null mass)i.e.we can take in the analysis from Subsection2.2 g=0?g=0.

The gauge charge is de?ned by

Q?=0Q?=Q(5.4) and

[Q,vμ]=i?μu,[Q,ψμa]=?μζa,

{Q,a}=0,[Q,ζ]=0

{Q,?a}=?i?μvμ,[Q,?ζ]=??μψμa(5.5) and the rest of the(anti)commutators are zero.It is more convenient to work with two real

struggle的用法和短语例句

struggle的用法和短语例句 【篇一】struggle的用法 struggle的用法1:struggle的基本意思是“奋斗”,即为实现某一目的而尽力做某事。常指遇到有力的反抗而在逆境中拼搏或努力从 束缚中解脱出来。有时含有“挣扎”的意味。 struggle的用法2:struggle也可表示“(与某人)争斗,搏斗,打斗”,还可表示“艰难地(朝某方向)行进”。 struggle的用法3:struggle是不及物动词,与介词against连用,表示“同与之对立或对抗的人或物实行斗争”; 与介词for连用,表示“为…而斗争”。 struggle的用法4:struggle可接动词不定式作目的状语。 struggle的用法5:struggle的基本意思是“斗争”,表示抽象的行为,用作不可数名词; 表示具体的“打斗,搏斗,战斗,斗争”时,可用作可数名词。 struggle的用法6:struggle作“努力,奋斗”解时,一般用单数形式。 【篇二】struggle的常用短语 struggle against (v.+prep.) struggle along1 (v.+adv.) struggle along2 (v.+prep.) struggle for (v.+prep.) struggle in (v.+prep.) struggle on (v.+adv.)

struggle out (v.+adv.) struggle with (v.+prep.) 【篇三】struggle的用法例句 1. It's a constant struggle to try to keep them up to par. 要让他们达标,需要持续努力。 2. Curiously, the struggle to survive has greatly improved her health. 奇怪的是,她拼命求生的抗争使得她的健康状况大有好转。 3. He grandly declared that "international politics is a struggle for power". 他一本正经地宣称“国际政治是一场权力之争”。 4. This age-old struggle for control had led to untold bloody wars. 这场由来已久的对控制权的争夺已经引发了无数流血的战争。 5. It is only a hobby, not a life or death struggle. 这仅仅一项爱好,不需要拼死拼活地努力。 6. There is a ceaseless struggle from noon to night. 从中午到夜晚,争斗没有停歇。 7. He is currently locked in a power struggle with his Prime Minister. 他当前陷入了一场同的权力之争当中。 8. He praised her role in the struggle against apartheid.

品牌文化与企业文化的异同

品牌文化与企业文化的 异同 集团标准化工作小组 [Q8QX9QT-X8QQB8Q8-NQ8QJ8-M8QMN]

品牌文化与企业文化的异同管理的概念可以说是层出不穷,是最近业界谈的比较多的一个概念。和都有“文化”,但是不是说他们两者就是相同的这个问题值得我们深入地探讨。 文化是一个社会和群体形成的共同的信念、和行为方式,具有三个要素:精神、载体和群体。世界有三大文化圈,文化圈、基督教文化圈和伊斯兰教文化圈。这三大文化圈的历史都很悠久,人口都在10亿以上,影响非常深远。基督教文化强调“赎罪”,要拼命工作才能减轻罪过,所以为什么西方(美国、德国、以色列)这些国家的法律很规范、职业道德水平高、比较敬业,跟传统文化是分不开的。伊斯兰的意思是“和平、顺从”,只有按照真主的意愿去生活和工作,才能升入天国,所以这些国家的人都非常的虔诚,宗教色彩浓厚,这也导致了这些国家大都在经济上比较落后,思想上较为保守。文化圈包括中国、日本、韩国、东南亚国家,其精神内涵是道德、和谐、和睦、亲情,这些深刻地影响着我们每一个中国人。 另外,文化都需要载体,比如春节、端午节等很多节日都是亲情和家族文化的代表,中国人的婚丧嫁娶、衣食住行等很多风俗、仪式也都体现着文化的精神内涵,另外,很多故事、典故、寓言和英雄也都从不同层面反映和传承着文化,岳飞代表精忠报国,关羽代表忠义。文化的第三个要素是群体,没有群体也自然无法形成文化,而且,这个群体可以按民族(汉、回、满等)形成独特的民族文化,还可以按地区(南方、北方、西

北)形成地域文化,这些都是中华文化下面的亚文化。这些精神、风俗、仪式和群体结合在一起,就构成了从深层到表层的中华文化。 与都脱离不了文化,他们的形式和内容与文化都密切相关,细想一下,的塑造是不是也分成三个层次,核心理念(精神)、制度与行为(载体)、文化群体(不同职能部门)是不是也包括品牌精神、品牌传播(载体)、目标(群体)三个方面 笔者认为,与都是文化的一种表现形式,都是一种亚文化现象。我们可以从相同和不同两个方面来进行深入阐释。 首先,与有相通的地方。 一个企业的文化,是这个企业的、信念和行为方式的体现,打个形象的比喻,如果我们把企业当作一个人,当你第一次见到这个人,那么他的衣着打扮会给你第一印象,这就是公司的VI(视觉识别),包括公司的建筑、办公环境、办公器具、LOGO等表面的直观的有形实体;通过他的言行举止你又能了解到他的做事风格,这是的具体表现,但是究竟是什么决定了这个人的言行举止这就取决于他内心深处的和信念了,同样的,对于企业来说,是文化决定了这个企业的制度和行为,这个文化的核心,就是我们常说的和企业核心。那么什么是我们说日久见人心,你跟一个人经过长久交往,你发现了他可能是一个诚实、活泼、开朗的人,你对他的个性有一个认知,你觉得喜欢跟他交往,也许会成为知己。如果对于企业来说,这个企业给的心理感受和心理认同,就是或者叫品牌内涵,他是联系心理需求与企业的平台,是的最高阶段,目的是使在消费公司的产品和服务时,能够产生一种心理和情感上的归属感,并形成。比如,当我们提到,

品牌文化概述

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bring的详细用法

1. Bring 带来;拿来;领来?Did you bring an umb rella? 你带伞了吗? bring sb/sth to sb/sth ?Is it OK if I bring some friends to the party? 我带几个朋友来参加聚会行吗? bring sb/sth with you ?For some reason, Jesse had brought a tape recorder with him. 不知为什么,杰西带了一台盒式录音机来。 bring sb sth ?Can you bring me another beer? 你再给我拿一杯啤酒好吗? bring sb/sth to sb/sth ?He expects me to bring everything to him. 他指望我什么东西都给他带去。 2.引起〔某种情况或情感〕,造成,导致?efforts to bring peace to the region 给这个地区带来和平的种种努力?The strikes are expected to bring chaos. 预计罢工会造成混乱。 使处于某种状况?It was the war that first brought him to power (= make him have power over a country ) . 是战争首次将他推上权力宝座。 ?Bring the sauce to the boil (= heat it until it boils ) . 把调味汁煮沸。 bring sth to an end/a close/a halt/a conclusion 使某事结束?The trial was swiftly brought to an end. 庭审匆匆就结束了。 3.使朝某个方向移动bring sth up/down/round etc ?Bring your arm up slowly until it’s level with your shoulder. 慢慢举起手臂到齐肩的高度。?The storm brought the old oak tree crashing down. 暴风雨把这棵老橡树刮倒了。 4.促使某人去…what brings you here? (=used to ask why someone is in a particular place)什么风把你给吹来了? 5.带来〔可供人们使用、拥有或享受的东西〕;使得到;创造 ?The expansion of state education brought new and wider opportunities for working class children. 公立教育的普及为工人阶层出身的儿童带来了更多新的机会。 bring sth to sb/sth?The government is launching a new initiative to bring jobs to deprived areas. 政府正在出台一个新的方案,为贫困地区创造就业机会。 bring sb sth?It’s a good sign –let’s hope it will bring us some luck. 这是一个好征兆——但愿会给我们带来好运。 6.〔某段时间〕带来;使发生 ?Who knows what the future will bring? 谁知道未来会发生什么?

品牌文化vs企业文化(1)

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另外,文化都需要载体,比如春节、端午节等很多节日都是儒家亲情和家族文化的代表,中国人的婚丧嫁娶、衣食住行等很多风俗、仪式也都体现着文化的精神内涵,另外,很多故事、典故、寓言和英雄也都从不同层面反映和传承着文化,岳飞代表精忠报国,关羽代表忠义。文化的第三个要素是群体,没有群体也自然无法形成文化,而且,这个群体可以按民族(汉、回、满等)形成独特的民族文化,还可以按地区(南方、北方、西北)形成地域文化,这些都是中华文化下面的亚文化。这些精神、风俗、仪式和群体结合在一起,就构成了从深层到表层的中华文化。 企业文化与品牌文化都脱离不了文化,他们的形式和内容与文化都密切相关,细想一下,企业文化的塑造是不是也分成三个层次,核心理念(精神)、制度与行为(载体)、文化群体(不同职能部门)?品牌文化是不是也包括品牌精神、品牌传播(载体)、目标消费者(群体)三个方面? 笔者认为,品牌文化与企业文化都是文化的一种表现形式,都是一种亚文化现象。我们可以从相同和不同两个方面来进行深入阐释。 首先,企业文化与品牌文化有相通的地方 一个企业的文化,是这个企业的价值观、信念和行为方式的体现,打个形象的比喻,如果我们把企业当作一个人,当你第一次见到这个人,那么他的衣着打扮会给你第一印象,这就是公司的VI(视觉识

2020年考研英语词汇:against和but的含义与用法

2020年考研英语词汇:against和but的含义与用法 在考研英语考卷中,against的意项比较丰富,与其他词语的搭配比较灵活。 1. 表对比 One is the weakness of the euro, which has lost 15% against the Chinese yuan, 19% against the Japanese yen and 9% against the dollar since the start of 2020. 一方面,2020年伊始欧元持续走低,兑人民币、日元、美元分别 贬值15%、19%和9%。 2. 表示与某个方向或过程相反 Try not to row against the current. 尽量不要逆水行舟。 3. 表示轻轻地接触或是依靠 She leans herself against the jade door. 她斜靠在翡翠门上。 4. 表示猛烈的碰撞 During the storm, waves dash against the boat. 暴风雨中, 破浪猛烈地拍打着小船。 5. 表示反抗或者抗争 fight against AIDS 6. 表示为……而作准备 He is preparing money against the school. 他一直在咱学费。 7. 表示兑换、提取

Mike will draw a check against his bank balance. 迈克准备开支票支取自己的结余。 接下来,小编再带领大家探讨一下but的特殊用法,but在考研英语中是一个很重要的单词,在它的附近往往设置有重要的题设。准确理解but的含义,对于在考研英语中获取高分相当重要。 1. 用作副词,表示"仅仅,只有" The takeover is but a step towards the long-forecast consolidation of a crowded industry. 视频行业混乱无序,本次收购不过是向并购整合的行业趋势靠近了一步而已。 I don't like it. Still, I can but try. 我不喜欢这事儿。但是,也不妨试试。 2. 用作连词,在非正式用语中表示"比起来",类似于"than" We had no sooner arrived but we turned around and left. 我们离开后不久就到达了目的地。 We had no sooner arrived than we turned around and left. 我们离开后不久就到达了目的地。 3. 用作介词,表示"除了" This problem is anything but difficult. 这个问题一点儿也不难。 Who but you can show me the love? 除了你,谁还能爱我? 4. 作名词,表示"借口",常常用复数形式 No buts! We must win. 没有借口。我们必须成功。

品牌精神文化的内涵与特征

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advise用法

advise的常用句式1. advise doing sth 建议做某事。如:He advised leaving early. 他建议早点动身。I advise waiting till proper time. 我建议等到适当时机。注:不能直接后跟不定式作宾语。所以不能说:* He advised to leave early. 2. advise sb to do sth 建议某人做某事。如:He advised me to buy a computer. 他建议我买台电脑。The doctor advised me to take a complete rest. 医生建议我完全休息。注:以上意思也可用后接从句的形式表达(注意从句用should+v. 这样的虚拟语气)。如上面第一句也可说成:He advised that I (should) buy a computer. 3. advise sb against doing sth 劝告某人不要做某事。如:He advised her against going out at night. 他劝她晚上不要出去。Her father advised her against marrying the worker. 她父亲劝告她不要嫁给这个工人。她父亲劝告她不要嫁给这个工人。注:该句型有时可与advise sb not to do sth 替换。如上面第一句也可说成:He advised her not to go out at night. 4. advise sb of sth 通知或告知某人有某情况。如: Please advise us of the arrival of the goods. 货物到达时请通知我们 advise 是个常用动词,意为“劝告”,“忠告”;“建议”。由于它的用法比较复杂,有些同学在使用时经常出错。为了帮助大家正确理解和使用advise ,现将它的几种常见用法归纳如下。 1. advise + 名词 / 代词例如: Li Ming advised a swim after school. 李明建议放学以后去游泳。 She will advise you about the right thing to do. 她会帮你出主意该怎么办。 What would you advise ? 你有什么建议? 2. advise + sb. + 不定式短语 在这个句式中,不定式短语作 advise 的宾语补足语。例如: He often advised people to use their brains . 他常常劝人多用脑子。 His wife advised him to give up smoking , but he wouldn't. 他妻子劝他把烟戒了,但他不肯。 We strongly advised him not to do such a thing . 我们竭力劝他不要这样做。 3. advise + sb. + 特殊疑问词 + 不定式短语 在这个句式中,“特殊疑问词 + 不定式短语”作 advise 的宾语补足语。例如:

汽车公司品牌文化的内涵及其意义

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against的用法

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误:He against the tree. 这个计划你是赞成还是反对呢? 正:Are you for or against the plan? 误:Do you support or against the plan? 3. 正因为against是介词,所以后接动词时要用动名词。如:我反对在警察到达之前采取任何行动。 正:I’m against doing anything till the police arrive. 误:I’m against to do anything till the police arrive.

不可小看的against用法总结

在考研英语的文章中,会出现一些至关重要的小词,比如but作为连词并转折,作为 介词表“除了......”,for作为介词是“对于.......”“为了......”,而作为连词表原因; 因此积累并掌握这些重要的小词在文章中的用法,对于解题是非常关键的。 由于against这个词会频繁出现在考研的文章中,而其跟在不同的动词后含义大相径庭,那么下面老师就against这个小词的含义用法做一个详细总结。 (一)表示“不利,不顺”,常见搭配如下: turn against (对...不利), run against (对...不利); (二)表示“防备、预防”,常见搭配如下: prepare against(防备),provide against(准备),guard against(防备),warn against(谨防),lay up against(贮存以备),take precaution against fire(采取防火措施),guard against error(谨防错误); (三)表示“诉讼、控告”,常见搭配如下: charge against(控告),inform against(告发) (四)表示“反抗、对抗”,常见搭配如下: rise against(起而反抗),stand against(反抗),protest against(抗议),set against(对抗),rebel against(反叛、不服从); (五)作“抵抗、阻止、斗争”解,常见搭配如下: fight against(与...斗争),stand up against(抵抗),hold out against(斗争),defend against(抵抗),strive against(与...斗争),come against(攻击),protect one against(防御),strike against(攻击),march against(攻击),contend against(抵御),pit against(与...竞争),run against(向...撞击)。

品牌力要依托于品牌的文化内涵。

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常用介词及副词的搭配用法归纳

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常用英语介词用法小结

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