Gauge invariance and the vacuum state. Dan Solomon Rauland-Borg Corporation 3450 W. Oakton Skokie, IL Please send all correspondence to:

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1 Gauge invaiance and he vacuum sae 1 Gauge invaiance and he vacuum sae by Dan Solomon Rauland-Bog Copoaion 345 W. Oakon Skokie, IL 676 Please send all coespondence o: Dan Solomon 164 Bummel Evanson, IL 622 Phone: ex dsolomon@nohshoe.ne May 4, 1999

2 Gauge invaiance and he vacuum sae 2 Absac Quanum field heoy is assumed o be gauge invaian. I is shown ha fo a Diac field he assumpion of gauge invaiance impacs on he way he vacuum sae is defined. I is shown ha he convenional definiion of he vacuum sae mus be modified o ake ino accoun he equiemens of gauge invaiance.

3 Gauge invaiance and he vacuum sae 3 I. Inoducion Quanum field heoy is assumed o be gauge invaian (see Schwinge [1]). I will be shown in his aicle ha in Diac field heoy he equiemen of gauge invaiance pus some equiemens on he way he vacuum sae veco is defined. I will be shown ha he convenional definiion of he vacuum sae mus be modified o ake ino accoun he equiemens of gauge invaiance. Thoughou his discussion naual unis ae used so ha c = h = 1. In he Schodinge epesenaion of Diac field heoy he ime evoluion of he sae veco Ω Ω i and is dual Ωaf is given by af Ω Haf Ωaf = i = Ωaf Haf $ ; $ (1) whee H $ af is he Hamilonian opeao and is given by $ $ $ H = Ho + J x A x, dx ρ $ x Ao x, dx (2) (See, fo example, Rayski[2] o secions 17 and 19 of Pauli[3].) In he above expession d A o x,, A x, i is he elecic poenial. In his aicle he elecic poenials ae $ assumed o be unquanied, eal valued funcions. J x and ρ $ axf ae he cuen and chage opeaos, especively and $ H o is he fee field Hamilonian, which is he Hamilonian opeao when he ineacions ae uned off (i.e. he elecic poenial is eo). Thoughou his discussion i is assumed ha Ω Ωaf Ωaf = 1. is nomalied, i.e.,

4 Gauge invaiance and he vacuum sae 4 The fee field enegy is defined by ξ o af = Ωaf H$ o Ωaf (3) Now given he quanum sae Ω a b g, defined a = a, he applicaion of Eq. (1) will evolve he quanum sae fowad in ime o poduce he quanum sae Ω b = b. In geneal Ωbbg geneal, ξobag ξo b b g b ag no hee is a lowe bound o ξ o b g a Ω and, if he elecic poenial is no eo, hen, in. The quesion ha will be addessed in his aicle is whehe o. I will be shown ha if quanum heoy is gauge. O o invaian hen he answe o his quesion is no. Thee is no lowe bound o ξ o pu he poblem moe pecisely, i will be shown ha, given he quanum sae Ωb a g, defined a = a, i is possible o find an elecic poenial da o x,, A x, i ha, when used in Eq (1), evolves he quanum sae ino Ω b hen ξ o b a g by an abiaily lage amoun. II. Gauge invaiance b g a = b such ha ξ o b b g is less The elecomagneic field is given in ems of he elecic poenial accoding o F I A E = + Ao ; B = A (4) HG KJ A change in he gauge is a change in he elecic poenial ha poduces no change in he elecomagneic field. Such a change is given by χ A A = A χ; Ao Ao = Ao + (5)

5 Gauge invaiance and he vacuum sae 5 whee χax, f is an abiay eal valued funcion. Now, in geneal, when a change in he gauge is inoduced ino Eq. (1) his will poduce a change in Ω. Howeve, an impoan equiemen of a physical heoy is ha i be gauge invaian which means ha he change in he gauge poduces no change in he physical obsevables. These include he cuen and chage expecaion values which ae defined by $ J x = Ω J x Ω (6) e, and b g b g b g b g ρe x, = Ω ρ$ x Ω (7) The equiemen fo gauge invaiance can be expessed as follows. Le hee exis wo sae vecos Ω 1 af and Ω 2 af which obey Eq. (1). The elecic poenial ha acs on Ω 1 e is A o 2 is Aa1 o f e a2f ax fj x,, A, ansfomaion, ha is a1f ax fj x,, A, and he elecic poenial ha acs on Ω 2 af. Assume he elecic poenials ae elaed by a gauge F HG e a Ao 2f j a A o 1f 2 χ x, 1 x,, A x, = x, +, A ax, f χ ax, f (8) Assume ha a = a we have ha Ω1bag = Ω2bag (9) I KJ and χ x, a χbx, ag b g = and = (1)

6 Gauge invaiance and he vacuum sae 6 Ohe hen his consain χax, f is an abiay eal valued funcion. We have defined wo saes, which ae equal a ime = a bu evolve fowad in ime unde he acion of elecic poenials which ae elaed by a gauge ansfomaion. Then, even hough in geneal Ω1 Ω2 fo > a, gauge invaiance equies ha he physically obsevable quaniies fo each quanum sae be equal, ha is, and 2 1 J x, = J x, fo a (11) ρ e e x, = ρ x, fo a (12) 2 1 e e a whee Je f ax, f and ρa e fax, f ae he cuen and chage expecaion value, especively, fo he quanum sae Ω af whee =1,2. Nex we will deive an expession fo he ime deivaive of ξ o af which will be used lae in he discussion. Use Eq. (2) in (3) o obain ξ RST F H o o = Ω H$ $ J x A x, dx + ρ$ x A x, dx Use (6) and (7) in he above o obain ξ I K UVW af Ω H$ af Ω { d J ax, f Aax, fdx ρ ax, fa ax, fdxi} + = o e e o Ω (13) (14) Take he deivaive of he above expession wih espec o ime o obain dξoaf d J e x ρ e x Haf R Aax f = S F dx + Aoax fdx d d T HG Ω $,, Ω,, F A x A o x Jeax f, dx ρeax f a, f I, +, dx HG KJ IU KJ V W (15)

7 Gauge invaiance and he vacuum sae 7 Now he ime deivaive of Ω d d Ω H$ Ω Ω H$ $H af Ω is given by F = H G I K J + + F H G H $ Ω Ω Ω Ω H$ Ω I KJ (16) Use Eqs (1) and (2) in he above o yield d d R S F T HG H A x A o x Ω H$ $ $,, Ω = Ω Ω = Ω J x dx + ρ$ x dx Ω (17) IU KJ V W Use (6) and (7) in he above o yield d d A x A o x Ω H$,, af Ω Jeax, f dx eax, f = + ρ dx (18) Use his in Eq. (15) o obain dξ d (19) J x, ρ x, = A x, dx A o e e III Negaive fee field enegy o x, dx In his secion i will be shown ha if quanum heoy is gauge invaian hen hee is no lowe bound o he fee field enegy. e A o 1 Assume he exisence of a quanum sae Ω 1 a1f ax fj x,, A, e A o 1 a1f, acing on Ω 1 af is eo, i.e., ax fj x,, A,. The elecic poenial, = (2) Assume ha a ime = b he chage expecaion value fo Ω 1 af saisfies he condiion ρa 1 e fax, f = b (21)

8 Gauge invaiance and he vacuum sae 8 Now le us ceae anohe quanum sae Ω 2 af which is deived fom Ω 1 af in he following manne. A = a, whee a < b, we have ha Ω2bag = Ω1bag (22) This may be consideed an iniial condiion on Ω 2 af. Then Ω 2 af evolves fowad in ime accoding Eq. (1) in he pesence of he elecic poenial e A o 2 a2f ax fj x,, A, F HG ax, f d ax fi I KJ χ =, χ, (23) whee χax, f is a eal valued funcion subjec o he following condiion χ x, a χbx, ag b g = and = (24) Ohe hen hese consains χax, f is abiay. Now how do we know ha a quanum sae can be found whee Eq. (21) is ue? If ou heoy is a coec model of he eal wold, ha is eal elecons, hen hee mus be quanum saes whee Eq. (21) holds because hee ae numeous examples in he eal wold whee he ime deivaive of he chage densiy is no eo. Given he above, we will pove he following. If quanum heoy is gauge invaian hen i is possible o find a χ x, such ha he fee field enegy ξa2 o faf of he quanum sae Ω 2 af a ime = b is a negaive numbe wih an abiaily lage magniude. Fis, noe ha, fom Eq. (22) we have ha ξ b g b g 2 1 o a = ξo a (25)

9 Gauge invaiance and he vacuum sae 9 Also, efeing o Eq. (19), if he elecic poenial is eo, hen ξ o is consan in ime. Theefoe fo he quanum sae Ω 1 af, ξ b g b g 1 1 o b = ξo a Also, noe ha A o 1 e a1f ax fj x,, A, e and A o 2 a2f ax fj x,, A, (26) ae elaed by a gauge ansfomaion (compae Eqs. (2) and (23)). Theefoe, if quanum heoy is gauge invaian, hen J x, = J x, ; ρ x, = ρ x, (27) e e e e Nex use Eq. (23) in (19) o obain fo he fee field enegy of Ω 2 dξ d J ax f ρ x χ x χax f, a, f a, f =, dx dx o e e Assume easonable bounday condiions a x and inegae by pas o obain d d J x ρ x χ x χax f =,,,, dx dx ξ o e e This can be wien as d d (28) (29) J x ρ x ρ x χax f dx χax f,, a, f,, dx + χax, fdx ξ o e e e = Reaange ems o obain dξo d F H (3) 2 ρ e x 2 ρ e x χax f J e ax f = G + J,,,, dx χax, fdx (31) 2 2 Use his esul in he expession I K

10 Gauge invaiance and he vacuum sae 1 ξ b 2 dξo ξ b g b g = + d (32) d 2 2 o b o a a o obain ξ 2 ρ e x b b g ξb g b, gχ x b dx b g =, b 2 ρ e x 2 d χax f F + G + J e ax f I,,, J dx 2 2 o b o a a H K (33) whee we have used he fac ha χbx, a g= (see Eq. (24)). Use Eqs. (25), (26), and (27) in (33) o obain ξ 1 ρ e x b b g ξb g b, gχ x b dx b g =, b 1 L x d χax f a, f +, dx 2 1 o b o b a (34) whee L a1f 1 ρ e x 1 ax f a, f J, + e ax, f b g b g In Eq. (34) he quaniies ξo b, ρe x, b and L x, ae independen of χ. Noe ha La1fa x, f = (36) is he coninuiy equaion. Since chage consevaion is an esablished expeimenal fac we could a his poin use Eq. (36) o simplify (34). Howeve, he poof is no dependen on he uh of he coninuiy equaion, only on gauge invaiance. Theefoe we shall conside wo possibiliies. (35)

11 Gauge invaiance and he vacuum sae 11 In one case assume ha Eq. (36) is ue o moe specifically ha L a1 fax, f = (37) Use his in Eq. (34) o obain ξ 1 ρ e x b b g = ξb g b, gχ x b dx b g, (38) 2 1 o b o b Now i is possible o find a χ x, such ha ξa2 f o b b g is a negaive numbe wih an abiaily lage magniude. Fo example le ρa 1f e x, b χbx, bg b g = f (39) whee f is a consan. Ohe hen his χax, f is abiay (excep ha i mus also saisfy he iniial condiions in Eq. (24)). Use Eq. (39) in (38) o obain ξ 1 ρ e b ξ x b g b g b, g = f G 2 1 o b o b F H I K J 2 dx (4) The quaniy unde he inegal sign is always posiive. Theefoe, given Eq. (21), he inegal is geae han eo so as f hen ξ o a2 f b b g. Nex conside he possibiliy ha L a1 fax, f (41) Then we can se χ x, 1 L x, = f fo a < < fo o a b b (42)

12 Gauge invaiance and he vacuum sae 12 Use his in Eq. (34) o yield ξ b 1 ξ L x b g b g a, f = f d S 2 1 o b o b a R T U V W 2 dx Once again he inegal is always posiive so ha as f hen ξ o sae Ω 1 Ω 1 IV Discussion a2 f b b g. Le us eview he esuls of he pevious secion. We sa ou wih a quanum such ha he condiion given by Eq. (21) is ue a some poin in ime = b. saisfies Eq. (1) whee he elecic poenial is eo. Then a some ime = a < b 2bag = Ω1bag. Nex, use Eq. (1) o evolve Ω 2 af define he quanum sae Ω fowad in ime o b in he pesence of an elecic poenial ha is elaed o he elecic poenial of Ω 1 by a gauge ansfomaion. These leads o Eq. (34) fo he fee field enegy of Ω 2 af 1 a = b. In his expession he quaniies J e and ρa1 e f ae independen 2 of χ. Theefoe χ may be defined as discussed above o make ξ o b g less han b g by an abiaily lage amoun. Theefoe, hee is in pinciple, no lowe bound 1 ξ o b o he fee field enegy fo Diac field heoy. Now his may seem like a supising esul because he geneally held belief is ha he vacuum sae is he minimum value of he fee field enegy (see Chap. 9 of Geine e al[4] o Chap. 1 of Pauli[3]). Wha has been shown hee is ha hee mus exis quanum saes whose fee field enegy is less han ha of he vacuum sae. In he nex secion we will discuss he popeies of he vacuum sae as hey ae nomally defined in quanum b

13 Gauge invaiance and he vacuum sae 13 field heoy and hen, in he following secion, i will be shown how he vacuum sae can be edefined o be consisen wih hese esuls. V. The vacuum sae Define he field opeao in he fom of he following expansion, ψ$ $ φ ; ψ$ axf = a x x a$ φ x n n = n n n n (43) whee he $a n ($a n ) ae he desucion(ceaion) opeaos fo a paicle in he saeφ n x The φ n x. They saisfy he anicommuao elaion m n n m mn a$ a$ + a$ a$ = δ ; all ohe anicommuaos = (44) ae eigenfuncions of he fee field single paicle Diac equaion wih enegy eigenfuncion λ n E n. Tha is, λ E φ x = H φ x (45) n n n o n whee Ho = i α + β m (46) and whee E n = + p n + m = + 1 fo a posiive enegy sae 2 2, λ n (47) 1 fo a negaive enegy sae whee p n is he momenum of he sae n. Soluions of (45) ae of he fom = ip x φ n x u n e n whee u n is a consan 4-spino. The φ n x space and saisfy (48) fom a complee ohonomal basis in Hilbe

14 Gauge invaiance and he vacuum sae 14 and 3 φ n x φ n y = δ abδ x y n e j b g aaf = φ x φ x dx δ abf n m mn whee a and b ae spino indices (see page 17 of Heile [5]). (49) (5) Following Geine [4] define he sae veco,bae which is he sae veco ha is empy of all paicles, i.e., a$ n, bae = fo all n (51) Fo he index n we will define he following n< efes o negaive enegy saes. (52) n> efes o posiive enegy saes. The vacuum sae veco is defined as he sae veco in which all negaive enegy saes ae occupied by a single paicle. Theefoe = $ a, bae n< n (53) whee, as defined above, n< means ha he poduc is aken ove all negaive enegy saes. Fom his expession, and Eqs. (44) and (51), can hen be defined by a$ = fo n > ; a$ = fo n < n n (54) (Noe, i is possible, fo n<, o eplace he elecon desucion and ceaion opeaos, a$ and a$, wih he posion ceaion and desucion opeaos, b $ and b $, n n especively. Howeve, in his aicle, i will be convenien o sick wih he pesen noaion). n n

15 Gauge invaiance and he vacuum sae 15 Nex define he opeaos $ H o, $ J, and $ρ in ems of he field opeaos. Thee ae o a numbe of ways of peceding. Fo example $ρ can be expessed as q $ ψ x, ψ$ x whee q is he elecic chage. All hese expessions ae basically q: $ ψ x ψ$ x : equivalen and simply epesen diffeen ways of subacing ou he infinie vacuum chage. In his discussion i will be convenien o use ρ$ = qψ$ ψ$ ρ R $ $ J = qψ αψ$ J R H$ $ = ψ H ψ$ dx ξ o o R Whee he enomaliaion consan ρr, JR, and ξr (55) (56) (57) ae used o make he vacuum expecaion values of he above opeaos equal o eo. Any abiay sae Ω can be expanded in ems of a se of ohonomal basis saes ϕ n which ae ceaed by he acion of he opeaos $a n and a n on (see Chap. 3 of Iykson and Zube [6]). Theefoe Ω can be wien as Ω = c ϕ (58) whee he c ae expansion coefficiens and whee ϕm ϕn = δ nm (59) The basis saes ϕ n ae eigensaes of $ H o,i.e., $H o ϕ n = ε n ϕ n (6)

16 Gauge invaiance and he vacuum sae 16 The eigenvalues ε n ae all posiive (o eo fo he vacuum sae). This is because he acion of he opeaos $a n and a n on is o eihe ceae a peviously unoccupied posiive enegy sae o o desoy an occupied negaive enegy sae, especively. If n= efes o he vacuum sae we have ha ε n > ε = fo n ( 61) o Using Eqs. (6) and (59) in (3) we have ha he fee field enegy fo a nomalied quanum sae Ω is ξ b g= 2 o Ω cm εm m ( 62) Use Eq. ( 61) in he above o yield ξob Ω g> ξob g= if Ω ( 63) This equaion saes ha he fee field enegy of he vacuum sae is lowe han ha of any ohe sae. This is because, as jus discussed, any abiay sae Ω can be expessed as a combinaion of he sae and he ohe basis saes ϕ m which ae fomed fom he acion of he opeaos $a n o a n acing on. These ohe basis saes have fee field enegy geae han ha of. Theefoe he fee field enegy of Ω is always geae hen ha of (unless Ω = ). VI. Redefining he Vacuum sae In Secion III i was shown ha gauge invaiance equies ha quanum saes mus exis whose fee field enegy is less han ha of he vacuum sae. Howeve when is defined as in he pevious secion i is seen ha has he lowes enegy fee field of

17 Gauge invaiance and he vacuum sae 17 any sae. Hence we have an inconsisency. Diac quanum field heoy canno be boh gauge invaian and have as he vacuum sae. In he following i will be shown how his inconsisency can be eliminaed by a modes modificaion o he definiion of he vacuum sae. The immediae poblem wih is ha he basis saes ha ae deived fom ae all posiive enegy saes. As we have seen his means he fee field enegy of any abiay sae mus be posiive. Theefoe a minimum equiemen fo a consisen heoy is ha basis saes mus exis wih less enegy hen he vacuum sae. To see how his can be done, sa by defining he sae veco, E w as follows. The op of he negaive enegy band has an enegy of m. Define, E w as he mulipaicle quanum sae in which each single paicle sae in he band of negaive enegy saes fom enegy m o (m+ E w ) is occupied and all ohe single paicle saes ae unoccupied. Le he noaion n band mean ha n is a single paicle quanum sae wih enegy in he ange m o (m+ E w ). Then, E w can be defined by, Ew a$ n, bae (64) n band Le n<band efe o he single paicle quanum sae wih enegy less han (m+ E w ). Recall, also, ha n> efes o posiive enegy saes. Theefoe, E w can be defined by a$, E = fo n > n n w a$, E = fo n band w a$, E = fo n < band n w (65)

18 Gauge invaiance and he vacuum sae 18, E w is an eigenveco of he opeao H $ o. The eigenvalue can be se equal o eo by pope selecion of he consan ξ R in Eqs.(57) heefoe we wie, H$, E = (66) o w We edefine he vacuum sae as follows. Le he vacuum be he sae, E w which is he sae, E w in he limi ha E w. In calculaions involving, E w, E w is assumed o be finie wih he limi E w aken a he end of he calculaion. I has been shown in Secion III ha quanum saes exis whose fee field enegy is less han ha of he vacuum sae. This is no possible if is he vacuum sae. Bu using, E w as he vacuum sae allows his because hen basis saes will exis ha have less enegy han he vacuum sae. Le and n<band. The opeao pai n m a$ a$ acing on, E w will desoy a paicle wih he quanum numbe m wihin he band and poduce a paicle wih quanum numbe n undeneah he band. This will poduce a quanum sae wih enegy less hen, E w by he amoun E n E m (noe ha E n >E m because n is a single paicle quanum sae wih less enegy (moe negaive) han m ). An abiay sae can, hen, be expanded in ems of basis saes whose fee field enegy can be geae han o less han ha of he vacuum. Theefoe he fee field enegy of an abiay sae can be less han he fee field enegy of he edefined vacuum sae, E w. Theefoe he quanum sae, E w is consisen wih he equiemens of gauge invaiance.

19 Gauge invaiance and he vacuum sae 19 One poblem ha mus be addessed is does he sae, E w give sabiliy o posiive enegy paicles? An impoan ole ha he vacuum sae plays is o peven he scaeing of posiive enegy paicles ino unoccupied negaive enegy saes by a peubing poenial. Does, E w allow posiive enegy paicles o scae ino unoccupied saes ha exis undeneah he boom edge of he negaive enegy band? A song agumen ha, E w povides he sabiliy needed is o noe ha in peubaion heoy ems of he following fom will appea: φ n ( x, )V( x, ) Φ( x, ) dxd whee Φ( x, ) is some posiive enegy wave funcion, V( x, ) poenial, and φ n ( x, ) band fo which p n. Now φ is some peubing (67) is he unoccupied negaive enegy sae undeneah he negaive λ n n ip x x u e n (, ) = e i ne n Since p n and E n he wave funcion φ n ( x, ) (68) oscillaes a a ae appoaching infiniy in boh space and ime. If we assume ha he oscillaoy behavio of he posiive enegy wave funcion Φ( x, ) and he peubing poenial V( x, ) inegand is dominaed by he apid oscillaion of φ n ( x, ) is finie hen he and he inegal will appoach eo as p n. Theefoe he ansiion pobabiliy fom he posiive enegy wave funcion ino saes undeneah he negaive enegy band will become abiaily small as E w. Anohe feaue of, E w is ha i is consisen wih posion heoy. In his egad i is idenical o he sae in ha a hole will look like a posion. Theefoe

20 Gauge invaiance and he vacuum sae 2, E w mees all equiemen fo a vacuum sae. I sabilies posiive enegy paicles agains ansiions o negaive enegy saes, is consisen wih posion heoy, and is consisen wih he equiemens of gauge invaiance. VII. The vacuum sae and gauge invaiance In his secion i will be shown by diec calculaion ha using he sae as he vacuum sae will desoy he gauge invaiance of he heoy. Refeing o Eq. (1) we have ha e j Ω + = Ω ih$ 2 Ω + O (69) e j Ω + = Ω + i Ω H$ 2 + O (7) Use his in Eq. (6) o obain = + L 2 N + M QP e j $ Je x, + = Ω + J x Ω + $ Ω J x Ω i Ω H$, J $ x O Ω O (71) Le he elecic poenial be given by x, da o ax, f, Aax, fi F χ =, d ax, fi I χ (72) Use his and Eq. (2) in (71) o obain ea + f = e + L o N M HG O QP LF I O NHG 2 KJ Q e j J x, J x, i Ω H$, J $ x Ω $ y, + i M χ Ω J y χ y, dy + ρ$ y dy, J $ x P Ω + O Reaange ems o obain KJ (73)

21 Gauge invaiance and he vacuum sae 21 O QP R + L N M O + L QP N M O T U QP W a e + f = e + L o N M J x, J x, i Ω H$, J $ x Ω $ y i J y, J $ x y, $ y, J $ χ, S Ω Ω χ Ω ρ x Ω Vdy+ Oe 2 j (74) If quanum heoy is gauge invaian hen Je ax, + f mus no be dependen on χ because he elecic poenial of Eq. (72) is a gauge ansfomaion fom he case whee he elecic poenial is eo. Theefoe, since χ is abiay, he quaniy L$ y, J $ ρ x O Ω Ω NM QP mus equal eo fo any Ωaf. Check he case fo which Ωaf =. Define I x, y $ y, J $ x $ $ $ L N ayfjaxf Jaxf$ M O = ay QP f a, Iax, yf mus equal eo. Use (43) in (55) and (56) o obain ρ$ ay f = q a$ a$ φ y φ y n m n m ρr ρ ρ ρ (75) I x yf is called he Schwinge em. If quanum field heoy is gauge invaian hen he Schwinge em (76) n, m and $ J x = q a$ a$ φ x αφ x J (77) n m n m n, m R Fom he definiion of he vacuum sae (Eq. (55) and (56)) and he anicommuao elaionships (Eq. (44)) we have ha

22 Gauge invaiance and he vacuum sae 22 n m n m n> m< $ J x = q a$ a$ φ x αφ x J (78) whee J 1 is given by J = J q φ x αφ x 1 R n n n< 1 (79) Fom Eq. (48) i is seen ha J 1 is consan. Similaly ρ$ y = q a$ a$ φ y φ y m n m n ρ1 (8) n> m< whee ρ 1 is a consan given by ρ = ρ q φ y φ y 1 R n n n< (81) Use Eqs. (78) and (8) in (75) o obain F I G e je jj H m< K 2 I x, y = q φ m y φ n y φ n x αφ m x h. c. G n J > whee (h.c.) means o ake he hemiian conjugae of he peceding em. Now if Iax, yf = hen x Iax, y f =. Fom he above equaion we obain F I b g G b g b g d b g b gij b g > H m < K 2 x I x, y = q φ m y φ n y φ n x αφ m x G J h. c. n (82) (83) To evaluae his expession use he following, F H b g b gi K = e b gj b g+ b g b g (84) φ αφ α φ φ φ α φ n x m x n x m x n x m x

23 Gauge invaiance and he vacuum sae 23 Nex, fom he definiion of H o (Eq. (46)), we obain d nb g mb gi { c o nb gh mb g nb g o mb g} (85) φ x αφ x = i H φ x φ x φ x H φ x Now use Eq. (45) in he above o obain e n mj b n n m mg n m φ x αφ x = i λ E λ E φ x φ x (86) Use his in Eq. (83) o obain F I G b g J H m< K x I x, y = iq m n n n m m n m G 2 φ y φ y λ E λ E φ x φ x J h. c. n> Use Eq. (47) o obain F I G b g J H m< K x I x, y = iq n + m m n n m G 2 E E φ y φ y φ x φ x J h. c. n> Evaluae his a y = x o obain F I G b g J H m< K x I x, y = iq + G 2 En Em φ m x φ n x φ n x φ m x J h. c. y= x n> This yields F I G b g J > H m < K 2 2 = 2i q ben + Emgφ maxfφn axf 2 2 x I x, y = iq En + Em φ m x φ n x G J h. c. y= x n n> m < (87) (88) (89) (9)

24 Gauge invaiance and he vacuum sae 24 Each em in he sum is posiive heefoe he above expession is no equal o eo. Theefoe Iax, yf so ha he heoy is no gauge invaian when he vacuum sae is. Now le us wok he same poblem using he quanum sae, E w. Define L N M O w QP I x, y, E ρ $ y, J $ x, E (91) w Fom Eqs. (56) and (65) we obain $ J x, E = qφ x αφ x a$ a$, E w n m n m n> + qφ x αφ x a$ a$, E J, E n< band n m n m w w 1 w (92) whee J 1 is a consan given by Similaly J1 = JR qφ m x αφ m x (93), $ E ρ y =, E qa$ a$ φ y φ y w w m n m n n> +, E qa$ a$ φ y φ y, E ρ w m n m n n< band w 1 (94) whee ρ 1 is a consan given by Theefoe ρ = ρ q φ φ (95) 1 R m m

25 Gauge invaiance and he vacuum sae 25 R S U V k p T n> n< band W I x, y = q 2 φ + m y φ n y φ n x αφ m x φ m y φ n y φ n x αφ m x h. c. (96) Nex define Theefoe (97) F φ y φ y φ x αφ x 2 m n n m n band F φ y φ y φ x αφ x 2 (98) n m m n n band One can inechange he dummy indices n and m in Eq. (97) o obain F = F F F = (99) Theefoe he quaniy q F F = obain 2 e 2 2j can be added o he igh side of Eq. (96) o R S U V kh.c. p T n> n< band W I x y = q y y x x F y y, φ m φ n φ n αφ m 2 φ m φ n φ n x αφ m x Use he expession fo F 2 fom Eq. (97) in he above expession o obain (1)

26 Gauge invaiance and he vacuum sae 26 a f I x, y = q R S T 2 n> + φ mayfφ nayfφ naxfαφ maxf + φ mayfφn ayfφn axfαφ maxf φ y φ y φ x αφ x m n n m n< band n band U V W k h.c. p (11) Nex use = + + all n n> n band n< band (12) in he above expession o yield R S U V k p T all n W 2 I x, y = q φ m y n y n x m x φ φ αφ h.c. (13) In he above expession do he summaion ove he index n fis and use Eq. (49) o yield R U S m m V T W k p 2 3 I x, y = q φ y αφ x δ x y Now use he elaionship whee f y is an abiay funcion o obain R U 2 3 I ax, yf = S q φ maxfαφ maxfδ ax yfv T W k p h.c. (14) f y δ 3 x y = f x (15) Use he fac ha h.c. (16) eφ m x αφ m x j = φ maxfαφ maxf (17)

27 Gauge invaiance and he vacuum sae 27 o obain (18) I x, y = Theefoe he heoy is gauge invaian if, E w is used as he vacuum sae. Now wha is he diffeence beween and, E w ha accouns fo he fac ha Iax, yf bu ha I ax, yf =. The key diffeence can be seen by compaing Eq. (82) and (96). In he expession fo Iax, yf (Eq. (82)) he only ansiions ae fom occupied negaive enegy saes o unoccupied posiive enegy saes. Fo I ax, yf (Eq. (96)) hee is an analogous conibuion due o ansiions fom occupied saes wihin he negaive enegy band o unoccupied posiive enegy saes. Howeve, hee is also an addiional em due o ansiions fom saes wihin he band o unoccupied negaive enegy saes undeneah he band. I is he conibuion made by his addiional em ha. makes I x, y = VIII. Gauge Invaiance and he vacuum cuen. The fac ha hee ae poblems wih gauge invaiance and quanum field heoy has a long hisoy. A diec calculaion, using peubaion heoy, of he vacuum cuen, induced by an exenally applied elecic poenial does no poduce a gauge invaian esul. This has been shown by many auhos (see Pauli and Villas[7], secion 22 of Heile[5], chap. 14 of Geine[4], and Sakuai[8]). The calculaion in Secion VII showing ha Iax, yf is consisen wih hese pevious esuls. This has always been ecognied as a poblem because quanum mechanics mus be gauge invaian. The main appoach o dealing wih his poblem seems o be o

28 Gauge invaiance and he vacuum sae 28 assume hee is somehing devian abou he mahemaics involved in his calculaion. The non-gauge invaian ems ae hen isolaed and emoved fom he calculaion. This may involve some fom of egulaiaion (Ref [7]) whee ohe funcions ae inoduced ha happen o have he coec behavio so ha he non-gauge invaian ems ae cancelled. Howeve, as has been poined ou by Pauli and Villas [7], hee is no physical explanaion fo inoducing hese funcions. They ae mahemaical devices used o foce he desied (gauge invaian) esul. In his secion i will be shown ha fo he vacuum cuen o be gauge invaian he Schwinge em Iax, yf mus vanish. And, as was shown in he las secion, his em does no vanish if is used as he vacuum sae. Fom Eq. 8.3 of Pauli [3] he fis ode change in he vacuum cuen due o an exenal peubing elecic poenial is given by L F H I O NM KQP $ Jax, f and ρ$ ax, f ae he cuen and chage 1 $ $ Jvac x, = i M J x,, dy d J y, A y, + ρ $ y, Ao y, P (19) In he above expession he opeaos opeaos, especively, in he ineacion epesenaion. They ae elaed o he Schodinge opeaos by b g b g b g b g $ ih$ $ $ $ $ o iho J x, e J x e $ iho x, e $ iho = and ρ = ρ x e (11) Accoding o Eq of Pauli [3] he above ineacion opeaos saisfy ρ $ x, $ = Jax, f (111)

29 Gauge invaiance and he vacuum sae 29 The change in he vacuum cuen δ Ja1f vacax, f due o a gauge ansfomaion is obained by using Eq. (5) in (19) o yield L F IO NM HG KJ QP 1 $ $ χ y, δ J vac x, = i M J x,, dy d J y, χ y, + ρ$ y, P (112) If quanum field heoy is gauge invaian hen a gauge ansfomaion of he elecic poenial should poduce no change in any obsevable quaniy. Theefoe δ Ja1f vacax, f should be eo. To veify his we will solve he above equaion as follows. y y d y y y d y = χ, ρ $, ρ$, ρ$, χ, χ, (113) Assume ha χ y, = a. Use his and Eq. (111) in he above expession o obain χ y, d y y y d y J y = + ρ$, $,, ρ χ χ,, (114) Subsiue his ino Eq. (112) o obain L NM vac M e 1 $ $ δ J x, = i J x,, dy d J y, χ y, + χ y, J y, Reaange ems o obain L NM $ + i J x,, ρ$ y, χ y, dy O j QP vac M e 1 $ δ $ J x, = i J x,, d dy J y, χ y, $ + i J x,, ρ$ y, χ y, dy O j QP (115) (116)

30 Gauge invaiance and he vacuum sae 3 Assume easonable bounday condiions a y so ha F I dy J $ H y, χ y, K = (117) Use his o obain vac 1 $ δ J x, = i J x,, ρ$ y, χ y, dy (118) $ = i J x,,$ ρ y, χ y, dy Use Eq. (11) and he fac ha H $ o = in he above o obain (119) vac 1 $ δ J x, = i J x,$ ρ y χ y, dy = i I x, y χ y, dy Theefoe fo δ Ja1f vacax, f o be eo, fo abiay χ y,, he Schwinge em Iax, yf Iax, yf is mus be eo. When is eplaced by, E w in Eq. (19) hen eplaced by I ax, yf in Eq. (119). Since I ax, yf has been shown o be eo hen he vacuum cuen will be gauge invaian if, E w is used as he vacuum sae. IX. Conclusion In his aicle we have shown ha he sae veco is no consisen wih he equiemens fo gauge invaiance. One consequence of using fo he vacuum sae is ha he Schwinge em is no eo. This em mus vanish fo he heoy o be gauge invaian. The esul is ha non-gauge invaian ems appea in he calculaion of he vacuum cuen. This has led pevious eseaches o speculae ha he eason fo hese ems was due o some difficul o devian mahemaical behavio in he calculaion of he vacuum cuen. Hee we have shown ha he eason fo hese non-gauge invaian ems

31 Gauge invaiance and he vacuum sae 31 is no devian mahemaical behavio bu simply ha, as was shown in Secion III, gauge invaiance equies he possibiliy ha quanum saes can exis whose fee field enegy is less han ha of he vacuum sae. Based on his fac one would expec ha a calculaion of he vacuum cuen using as he vacuum sae would, indeed, poduce a non-gauge invaian esul. The non-gauge invaian ems ae no a esul of devian o difficul mahemaics bu ae he coec esul of a saigh fowad mahemaical calculaion. On he ohe hand, as has been shown, he sae veco, E w is consisen wih he equiemens of gauge invaiance. I is simila o in ha is sabilies posiive enegy paicles agains ansiions ino negaive enegy saes and i is consisen wih posion heoy. Howeve, i allows fo he exisence of negaive enegy saes by including negaive enegy basis saes which is necessay fo a gauge invaian heoy. Refeences 1. J. Schwinge, Physical Review, Vol 81, p.664 (1951). 2. J. Rayski, Aca Physica Polonica, Vol. IX, p.129 (1948). 3. W. Pauli, Pauli Lecues on Physics. Vol 6 Seleced Topics in Field Quaniaion, MIT Pess, Cambidge, Mass. (1973). 4. Geine, W. & Mulle, B. & Rafelski, J. Quanum Elecodynamics of Song Fields, Spinge-Velag, Belin (1985). 5. W. Heile, The Quanum Theoy of Radiaion, Dove Publicaions, Inc., New Yok (1954).

32 Gauge invaiance and he vacuum sae Iykson, C. & Zube, J., Quanum Field Theoy, McGaw-Hill, Inc., New Yok (198). 7. W. Pauli and F. Villas, Reviews of Moden Physics, Vol. 21, p. 434 (1949). 8. J. J. Sakuai, Advanced Quanum Mechanics, Addison-Wesley Publishing Co., Redwood, Calif (1967).