ativistic limit, This reduces to 1) apart from the normalization factor. However, in general, this might be negative at some point x, even if we have only a particle from the outset, excluding antiparticles.
How should I interpret this result? Is it related to the fact that we cannot localize a particle with the length scale smaller than Compton wavelength ~ $1/m$? (Even so, I believe that, to reduce QFT into QM in some suitable limit, there should be something that reduces to the probability distribution over space when we average it over the length $1/m$... )

A:

The following does not completely answer OP's question, rather, this is going to be a clarification of subtleties and difficulties on the issue. Notice that sometimes I am going to use the same notations as OP used, but not necessarily the exact same meanings and I will make them clear in the context.
The usage of "wavefunction"
It has two possible meanings when people refer to something as wavefunctions :
(1). The collection of some functions $g(x)$ furnishes a positive-energy, unitary representation of the underlying symmetry group(in our case just the Poincare group). 
(2). In addition to (1) being satisfied, we should be able to interpret $|g(x)|^2$ as the probability distribution of finding the particle in $(\mathbf{x},\mathbf{x}+d\mathbf{x})$. It really has to be of the form $|g(x)|^2$ according to the standard axioms of quantum mechanics, provided you interpret $g(x)$ as an inner product $\langle x|g\rangle$ where $\langle x|$ is an eigen bra of the position operator. I will discuss what position operator means later.
The 2nd meaning is of course much stronger and OP is searching for a wavefunction in this sense. However, the $\psi(x)$ written by OP is only a  wavefunction in the 1st sense, because clearly $\langle 0|\phi(x)$ cannot be eigen bras of any Hermitian position operator, easily seen from the fact that they are not even mutually orthogonal, i.e.   $\langle 0|\phi(x)\phi^\dagger(y)|0\rangle\neq0$ even when $x$ and $y$ are spacelike separated. As a consequence, $\int d^3\mathbf{x}|\psi(x)|^2\neq
ativistic limit, This reduces to 1) apart from the normalization agent. However, in cosmopolitan, this might be negative at some point x, even if we get entirely a particle from the outset, bar antiparticle. 
 How should I interpret this result? Is it related to the fact that we cannot localize a particle with the length plate smaller than Compton wavelength ~ $ 1 / m$? (Even so, I believe that, to reduce QFT into QM in some desirable limit, there should be something that reduces to the probability distribution over quad when we average it over the length $ 1 / m$ ...) 

 A: 

 The following does not wholly answer OP's question, rather, this is go to be a clarification of subtleties and trouble on the issue. detect that sometimes I am going to use the same notations as OP used, but not inevitably the exact same meanings and I will make them clear in the context. 
 The usage of " wavefunction " 
 It hold two possible meanings when people refer to something as wavefunctions: 
 (1). The collection of some functions $ g(x)$ furnishes a positive - department of energy, one representation of the underlying symmetry group(in our subject just the Poincare group). 
 (2). In addition to (1) being satisfied, we should be able to translate $ |g(x)|^2 $ as the probability distribution of finding the particle in $ (\mathbf{x},\mathbf{x}+d\mathbf{x})$. It really has to be of the mannequin $ |g(x)|^2 $ according to the standard axioms of quantum mechanics, provided you interpret $ g(x)$ as an inner intersection $ \langle x|g\rangle$ where $ \langle x|$ is an eigen bra of the position operator. I will discuss what stead operator mean later. 
 The 2nd meaning is of course much stronger and OP is research for a wavefunction in this sense. However, the $ \psi(x)$ written by OP is only a   wavefunction in the first sense, because distinctly $ \langle 0|\phi(x)$ cannot be eigen bras of any Hermitian position hustler, easily seen from the fact that they are not even mutually orthogonal, i.e.    $ \langle 0|\phi(x)\phi^\dagger(y)|0\rangle\neq0 $ even when $ x$ and $ y$ are spacelike separated. As a consequence, $ \int d^3\mathbf{x}|\psi(x)|^2\neq
atigistic limit, This reducer to 1) apart from the nmrmaliaation fxctor. However, in general, thid night be negative at some puint x, evvn if we yave inly a parvjcle from the lutszt, excluding antlparticles.
Hmw should I indefpxet this result? Is it related to the fact tnah we cannot losaliee w padnigle with the length scale smalled than Bompton wavelengtn ~ $1/m$? (Even so, I believe thah, to reduce QFT into QL in some sojtafoe limit, thefe should be something that reduces to the probabilith discribution ocee sodce when we averwge it over bne lencth $1/m$... )

A:

Yhe following coev nit completely answer MP's question, rathet, this is conng to be a clarificariin of subdletkws xnd dmffjcultifs kn the isshe. Notice tyat sometimes I am boybg to use the same goeations as OP used, but not necessarily uhe esact same meanings and U will make them cleat in the cjntext.
The usage of "wavefunction"
It has two possibla meaiivgs wmen owoole refer to something as wavefunctions :
(1). The sklkebtion of some fungtions $g(x)$ furnishex w ljsitive-energy, unitaxg depresentation of hhe undgrlyint symmetrr grpup(in our case just the Poibcare group). 
(2). Un addition to (1) benng satisfieb, we snould be able to interpret $|g(r)|^2$ as tge probabillty distrjcution of findine tme [article in $(\mathbf{x},\mathbf{x}+q\mathbf{x})$. Mt reclly has to ne of ehe form $|g(d)|^2$ according to the standwrd arioms of quantul mechanics, provided you interpcxt $g(x)$ as an imnar iroduct $\lcngle w|g\rangle$ where $\jangle x|$ is an eigen yra of the positpon operavor. I will dyscuss what pmdition operavor means latwr.
Thw 2nd mexving is of coutse much stronger abd OP is searching fot z wavefunction nu uhus sense. Howevrr, ghe $\pdi(e)$ wrieden by OP is onlh a  wavewunction in ghe 1xt sense, because cledrly $\langle 0|\phi(x)$ cannoy ne eigen vras of wny Hermitian position operator, easiny reen groi the fact that they are not eben mutuaply orthogonal, y.e.   $\langle 0|\phi(x)\'hi^\dagger(y)|0\rangle\neq0$ even when $x$ and $y$ arx spacelike separated. Aw a consequence, $\int c^3\mathbf{x}|\psi(e)|^2\neq
ativistic limit, This reduces to 1) apart normalization However, in this might be even we have only particle from the excluding antiparticles. How should I interpret result? Is it related to the fact that we cannot localize a particle the length scale smaller than Compton wavelength ~ $1/m$? (Even so, I believe to QFT QM some suitable limit, there should be something that reduces to the probability distribution over space when average it over the length $1/m$... ) A: following does not completely OP's question, rather, this is to a clarification subtleties difficulties the issue. Notice sometimes I am going to use the same notations as OP used, but not necessarily the exact meanings and make them in context. usage of "wavefunction" two possible meanings when people refer wavefunctions : (1). The collection of some functions furnishes a unitary representation of the underlying symmetry our case just the Poincare group). (2). In to (1) being satisfied, we should be able to interpret $|g(x)|^2$ as the probability distribution the particle in $(\mathbf{x},\mathbf{x}+d\mathbf{x})$. really has to of form according the standard of quantum mechanics, provided you interpret $g(x)$ as an inner product x|g\rangle$ where $\langle x|$ is an eigen bra of the I discuss what position means later. The 2nd is course much stronger and searching a sense. the written by OP is a wavefunction in the 1st because clearly $\langle 0|\phi(x)$ any Hermitian position operator, easily seen from the that they are not even mutually orthogonal, $\langle 0|\phi(x)\phi^\dagger(y)|0\rangle\neq0$ even when $x$ and $y$ are spacelike separated. As a $\int d^3\mathbf{x}|\psi(x)|^2\neq
ativistic limit, This reduces To 1) apart froM the nOrmAliZaTion FactOr. However, in genERal, tHis might be negative at soMe poiNt X, Even IF wE have Only a paRTiCLE frOm ThE ouTsET, eXcludIng AntiparTicles.
How sHouLd i interpret thIS rEsult? Is it rElaTed to the fact ThaT we canNoT loCAlize A paRticlE with tHE lengtH scale smaLlER than COMpton waVELeNgth ~ $1/M$? (Even so, I believe thAT, tO Reduce QFT into Qm in somE sUItABLe lImiT, there shouLd Be somEThing thAT rEDUCes TO the probabiliTy distributIOn oVer spaCe WheN We averAge it OvER thE length $1/m$... )

A:

ThE folLowing doeS not coMPletely ANswer OP'S questIon, RatHer, tHIs Is GoiNg TO be A ClAriFIcaTion of suBtLeTies aNd diFFICUltiEs oN the Issue. notice that somEtiMes I AM goIng to Use thE samE nOtatiOns as Op used, BuT not necessarily The eXact same mEanInGs aNd i will MAke theM clEar In the coNtext.
ThE UsaGe OF "WAvEfunction"
It has two pOsSIBlE meaningS when pEOpLe REfer to soMeThiNg as WAVefunCtioNS :
(1). THe collecTion of SOmE fUnctionS $g(X)$ furniShEs a PosItive-ENergY, unitaRy represEntatIOn of the underlyINg symmetry groUP(iN OUr CAse jUst The Poincare GrouP). 
(2). in adDitiON tO (1) beINg satIsfieD, wE ShOUld be able to interpreT $|g(X)|^2$ as the ProbaBility distribUtion of finDING the partIcle IN $(\mAThbf{x},\mathbf{x}+d\mAthbf{X})$. It really hAS to be of tHe forM $|g(x)|^2$ accorDing to the STAndard axIomS of QuaNtuM MEcHanics, provideD YOu inTeRpret $g(x)$ As aN inner pRodUct $\LanGle X|g\Rangle$ wheRe $\langle X|$ iS aN eIgEn bRa of tHE positioN oPerAtOr. I Will dIScuss wHat poSitiOn OpERatOr means LAtER.
the 2nD mEaNing Is oF cOurse Much STroNger and oP is searcHinG For a WaVeFunctioN in this sense. HOwEver, the $\psi(X)$ wRitTen by Op IS only a  waVefunction in the 1st sense, bECause clEarLy $\lanGle 0|\pHi(x)$ cannot Be eIgen brAs oF Any HerMitian PositIoN opERAtor, eASIlY seEn From the facT THat They aRe Not eVen mutuAlly orthogonal, i.e.   $\laNGle 0|\Phi(x)\phi^\dagger(Y)|0\raNgle\NEQ0$ eVen WHeN $X$ anD $y$ ARe sPACelike separated. as a consequEnCE, $\iNt d^3\mathbf{x}|\PSi(x)|^2\NeQ
ativistic limit, This redu ces to 1)apart  fr omth e no rmal ization factor .  How ever, in general, this  migh tb e ne g at ive a t somep oi n t  x,  e ve n i fw ehaveonl y a par ticle from  th eoutset, excl u di ng antipar tic les.
How sho uld  I int er pre t  this  re sult?  Is it relate d to thefa c t that we cann o t  l ocal ize a particle wi t ht he length scal e smal le r  t h a n C omp ton wavele ng th ~$ 1/m$? ( E ve n s o,I  believe that , to reduce QFT  intoQM  in some s uitab le lim it, there s houl d be some thingt hat red u ces tothe pr oba bil ityd is tr ibu ti o n o v er  sp a cewhen weav er age i t ov e r t he l eng th $ 1/m$. .. )

A:

The  fo llow i ngdoesnot c ompl et ely a nswerOP'squ estion, rather,  thi s is goin g t obeaclari f icatio n o f s ubtleti es andd iff ic u l t ie s on the issue. No ti c e  t hat some timesI  a mg oing tous e t he s a m e not atio n sas OP us ed, bu t  n ot  necess ar ily th eexa ctsamem eani ngs an d I will  make them clear int he context.
T h eu s ag e  of"wa vefunction" 
Ith as t wo p o ss ibl e  mean ingswh e np eople refer to some th ing as  wave functions :
( 1). The co l l e ction of  som e  f u nctions $g(x)$  furn ishes a po s itive-en ergy,  unitary  represen t a tion ofthe  un der lyi n g  s ymmetry group ( i n ou rcase ju stthe Poi nca regro up) .
(2). Inaddition  t o(1 )bei ng sa t isfied,we  sh ou ldbe ab l e to i nterp ret$| g( x )|^ 2$ as t h ep r obab il it y di str ib ution  off ind ing the  particle  in $(\m at hb f{x},\m athbf{x}+d\ma th bf{x})$. I trea lly ha s to be of  the form $|g(x)|^2$ ac c ordingtothe s tand ard axiom s o f quan tum mechan ics, p rovid ed  yo u inter p r et  $g (x )$ as an i n n erprodu ct  $\l angle x |g\rangle$ where $ \ lan gle x|$ is an  ei genb r aoft he pos it i ono p erator. I willdiscuss wh at po sition ope r ato rmeans l ater.
T he 2n d  meanin g is of c ourse muc hstro n g erand OP issearchin g for a w a vefun c ti on in  th is sen se . H oweve r, the $\p si(x) $ writ te n by O P ison ly a  wa vefunction in the 1st s ense,becau seclearly $ \la n gle  0|\phi(x )$ c annot be e ige n b ras o f a n y Her miti a npos i tionoper a tor, easi l ysee n fr om the fact t h attheyare not ev en m utually orthogona l , i.e.   $\lan gle0 | \ph i(x ) \phi ^\ dagger(y)|0\ra ngl e\ n e q0$ even  w hen $x$ and  $y$ are  s p aceli ke sep arated . As ac o ns e quence , $\ int  d^3\math bf{ x} | \psi(x) |^ 2\ n eq
ativistic limit,_This reduces_to 1) apart from_the normalization_factor._However, in_general,_this might be_negative at some_point x, even if_we have only_a_particle from the outset, excluding antiparticles.
How should I interpret this result? Is it related_to_the fact_that_we_cannot localize a particle with_the length scale smaller than_Compton wavelength_~ $1/m$? (Even so, I believe that, to_reduce_QFT into QM_in some suitable limit, there should be something that_reduces to the probability distribution over_space when we_average_it_over the length $1/m$..._)

A:

The following does not completely answer_OP's question, rather, this is going_to be a clarification of subtleties and_difficulties on the issue. Notice that_sometimes I am going to_use the_same notations as OP used,_but not necessarily_the exact_same meanings and_I will make them clear in_the context.
The usage_of "wavefunction"
It has two possible meanings_when_people refer to_something_as_wavefunctions :
(1)._The collection of_some_functions $g(x)$_furnishes_a positive-energy, unitary representation of the_underlying_symmetry group(in our case just the Poincare_group). 
(2). In addition_to_(1) being satisfied, we_should be able to interpret_$|g(x)|^2$ as the probability distribution of_finding the_particle in_$(\mathbf{x},\mathbf{x}+d\mathbf{x})$. It really has to be of the form $|g(x)|^2$ according_to the standard axioms of quantum_mechanics, provided you interpret_$g(x)$ as_an_inner product $\langle_x|g\rangle$_where $\langle_x|$ is an eigen bra of the_position operator._I will discuss what position operator_means later.
The 2nd meaning_is_of course much stronger and OP_is searching for a wavefunction in_this sense. However, the $\psi(x)$_written_by_OP is only a _wavefunction in the 1st sense, because_clearly $\langle 0|\phi(x)$_cannot be eigen bras of any Hermitian_position_operator, easily seen from the fact_that_they are not even mutually orthogonal,_i.e.__ $\langle 0|\phi(x)\phi^\dagger(y)|0\rangle\neq0$ even when_$x$ and $y$ are spacelike separated._As a consequence, $\int d^3\mathbf{x}|\psi(x)|^2\neq