Why do non-gauge inariant terms aear in the acuum olarization tensor? Author: Dan Solomon. Email: dsolom@uic.edu Abstract. It is well known that quantum field theory at the formal leel is gauge inariant. Howeer a calculation of the acuum olarization tensor will include non-gauge inariant terms. These terms must be remoed from the calculation in order to get a hysically correct result. One common way to do this today is the technique of dimensional regularization. It has recently been noted [] that at one time a suersymmetric-like solution to the roblem was exlored that is, the right combination of fields would cause the offending terms to cancel out. I will examine some of this early work and ose the question why do the non-gauge inariant terms aear in the first lace? I will show that this is due to an imroer mathematical ste in the formulation of the erturbatie exansion. I will then show that when this ste is corrected the result is gauge inariant. Howeer a suersymmetric-like solution is still required to cancel out a diergent term.. Introduction. Suersymmetry, as currently conceied, can trace its origins back to the 97s []. Howeer a recent aer by C. Jarlskog [] reeals that a suersymmetric-like solution to a roblem related to the calculation of the acuum olarization tensor was suggested as far back as the late 9 s by arious researchers (Refer to [] for references). The roblem in question was the fact that calculations of the acuum olarization tensor roduced a term that was both diergent and non-gauge inariant. This term was obiously non-hysical and had to be remoed from the solutions in order to obtain a hysically correct theory. It was shown that this term would cancel for the right combination of fields with the correct masses. One thing that is disturbing about this result is that fact that these calculations gie non-gauge inariant results. This is desite the fact that quantum field theory is suosed to be gauge inariant. I will reiew some of this work and show that the non-gauge inariant term is a result of an imroer mathematical ste and that when this is corrected all terms are gauge inariant. Howeer there will still be a diergent term which can be eliminated by the correct combination of fields so that a suersymmetric solution is still required. In the following discussion c. Also I will use a somewhat old fashioned notational conention of Ref. [] where ab ab ab ab ab and x ix where x t. This conention is used because it is consistent with most of the references cited herein. Also in this conention the osition of the indices does not matter. That is a is the same as a. For consistency I hae made all indices as subscrits.
. The roblem of acuum olarization. The electromagnetic field tensor is Fu A A where A is the electric otential. A change in the gauge is a change in electric otential which does not roduce a change in where x gien by A A is an arbitrary function. F u. Such a change is In order for quantum field theory to be gauge inariant a change in the gauge cannot roduce a change in any hysical obserable such as the current and charge exectation alues. For examle the first order change in the acuum current, due to an alied electromagnetic field, can be shown to be, j,ac x x x A xd x (.) where is the acuum olarization tensor. We can write this equation in momentum sace as,,ac (.) j A In this case the gauge transformation takes the form, A A i (.) Using the aboe the change in the acuum current due to the gauge transformation is, j i (.) g, ac For QFT to be gauge inariant j must be zero. Therefore the acuum olarization tensor must satisfy the relationshi, g, ac (.5) Desite this is has been well established that when the acuum olarization tensor is calculated nongauge inariant terms aear in the result so that (.5) does not hold. (See discussion in D. Solomon []). For examle a calculation of the acuum olarization tensor by W. Heitler (see of [5]) yields, The term G NG (.6) G e u z m z m z z z m z m dz (.7) m NG e dz z m (.8) G G is gauge inariant because. Howeer the term NG NG inariant because is not gauge. In order to obtain a hysically alid result it is necessary to correct equation (.6) by remoing the non-gauge inariant term NG from the result. (Note the term does not aear in [5]. I hae added this because Heitler works in Gaussian units and the work here will be in natural units).
Another examle is gien in section 6. of Nishijima []. Here it is shown that the acuum olarization includes a non-gauge inariant term which must be remoed to obtain the correct gauge inariant result. According to Nishijima (ages 8- of []) the term that is discarded is NG A B where A is a quadratically diergent constant and B is a finite constant. Sakurai (Section -7 of Ref [6]) argues that based on requirements of orentz coariance the acuum olarization tensor should be of the form, D (.9) Here D is a constant which would corresond to a hoton mass which means that D must be zero. Sakura then shows that D is gien by, D ie d m (.) m i Sakurai writes that It is not difficult to conince oneself that almost any honest calculation gies D. Another examle is from Section. of Greiner et al [7] who obtain the following exression for the acuum olarization tensor, u u g g s where and s inariant unless s is zero. In [7] it is shown that s (.) are gien in [7]. It is eident that the aboe exression is not gauge. Therefore, in order to obtain the hysically correct gauge inariant result this term must be droed. This is just a small samle of a ast literature on the subject. Virtually all deriations of the acuum current show the existence of a non-gauge inariant term. Of course in order to get a result that ultimately agrees with exeriment these non-gauge inariant terms must be remoed. Different authors use different aroaches. As mentioned aboe an early aroach used a suersymmetric-like solution. Jost and Rayski [8] showed that the non-gauge inariant terms from Dirac and scalar fields tend to cancel out. The combination of fields that will gie exact cancelation is, where M i is the mass of the scalar field and N n i i i i (.) N n; M m m i is the mass of the Dirac field. Sakata and Umezawa [9] extend this formula to include ector fields. This aroach was eentually abandoned (and forgotten) because, at the time, it was thought that there was no reason to beliee that such a relationshi existed. The history of this aroach is discussed in more deth by Jarlsog []. Another ossibility is to simly dro the offending term from the final result. Heitler[5], Nishijima[] and Greiner [7] take this aroach. The other aroach is to deelo mathematical methods to remoe these terms. One such technique is dimensional regularization and is widely used today. An examle of this is gien by []. The obious question to ask is the following - Why do non-gauge inariant terms aear in a theory that is suosed to be gauge inariant? Why don t we obtain a result that is manifestly gauge
inariant? By manifestly gauge inariant I mean that the gauge inariance of some deried quantity such as the acuum olarization tensor is obious. Why is an additional ste to needed to remoe the non-gauge inariant term? In the rest of this aer I will examine this roblem and try to demonstrate what must be done at the fundamental leel to roduce a theory that is manifestly gauge inariant. My hilosohical oint of iew is as follows A scientific theory must be a correct model of the real world in that it must agree with exeriment, but it also must be mathematically consistent. Quantum theory claims to be gauge inariant. Howeer this roerty of gauge inariance is not born out when the acuum olarization tensor is calculated. To make the theory gauge inariant an extra ste is required to remoe the offending terms. This extra ste may be a regularization method such as dimensional regularization or, as is often the case, the offending non-gauge inariant terms are simly remoed by hand, that is, just droed from the result. The result is that the theory, in the end, does agree with exeriment. Howeer my concern is that the theory is not mathematically consistent. A theory that claims to be gauge inariant should roduce manifestly gauge inariant results and not roduce non-gauge inariant terms that hae to be remoed at the end of the calculation by an additional ste. I will start by examining a relatiely straightforward calculation of the acuum olarization tensor by W. Pauli []. Here an equation for the first order change in the acuum current of a Dirac field due to an external alied electromagnetic otential is obtained. Pauli shows that this equation is not gauge inariant. I will look in detail at Pauli s calculation and how show how to modify the field oerator in order to achiee a gauge inariant result and then deelo an exression for the acuum current, j, for the Dirac field which is manifestly gauge inariant. I will then calculate the acuum D current in momentum sace for the simle case where the four momentum,, of the alied electric otential satisfies and m where m is the mass of the Dirac field. The result is that j consists of a sum of three terms. The first is a finite term that is consistent with the standard result. D The second is a logarithmically diergent term that can be identified as a charge renormalization term. This is also consistent with the standard result. The third term is a highly diergent term which is obiously non-hysical. Een though it is non-hysical it is mathematical correct. That is, it follows from the mathematical calculation. It is shown that this term can be eliminated through suersymmetry. If we assume that there are two charged scalar fields with identical mass to the Dirac field then it is shown that this diergent term cancels out when the contribution to the acuum current by the scalar fields is included.. Vacuum Polarization from Pauli s lectures. In this section we will examine a calculation of acuum olarization as discussed by W. Pauli []. In Section 8 of [] the acuum olarization tensor for a Dirac field is calculated in the resence of an external field. The calculation is done in the Heisenberg icture so that the time deendence is associated with the field oerator and the state ector is time indeendent. Referring to Section 8 of [] the Dirac equation in the resence of a classical electromagnetic otential is,
where m ˆ x ie A x x x ˆ ˆ x is the field oerator. The free field solution is the solution where A x Section of Ref [] as, where the aˆr x e a q u q e a q u q V q r r iqxet r iqxet r ˆ ˆ ˆ r r (.). It is gien in (.) q obey the usual anticommutation relationshis. The current oerator is gien by, ˆ ie j ˆ ˆ x x, x, (.) x. Howeer, if A x x In general we cannot sole (.) for an arbitrary A then a solution is ossible. In this case the solution to the Dirac equation is, ˆ x ie x e ˆ x ; ˆ x iex e ˆ x (.) If A x x the electromagnetic field tensor A x x is simly a gauge transformation from zero electromagnetic field. When (.) is used in (.) we obtain, ˆ j ˆ ˆ x ie x, x,. Note that x does not aear in this exression. Therefore a change in the gauge did not roduce a change in the current oerator which is a requirement of a gauge inariant theory. So far it has been shown that Dirac field theory is gauge inariant. So why is there a roblem with gauge inariance in erturbation theory? To start an examination of erturbation theory first consider an exansion of the exonential in (.) and to obtain, ˆ ˆ F so that ˆ ˆ (.5) x ie x x ; x x ie x Use this in (.) and retain the lowest order term to obtain, ˆ ie j ˆ ˆ x x ie x, ie x x, (.6) ie ˆ ˆ x, x, Oe This result shows that the first order exansion of the theory is gauge inariant. This is, of course, what we would exect from a gauge inariant theory. Howeer the situation changes when we consider the first order exansion for arbitrary A x. In this case an exact solution is normally not ossible and we hae to use the erturbation theory. From Section 8 of [] the exansion of the field oerator is, where ˆ x is gien by, ˆ x ˆ x e ˆ x (.7) 5
m e ˆ x ie A x ˆ x (.8) x When (.7) is substituted into (.) we obtain that the first order change in the current oerator is, ˆ ie j ˆ ˆ ˆ ˆ x x, x x, x, (.9) The first order change in the acuum current is then, ˆ j x j x (.) D where is the acuum state and the subscrit D on j x D indicates that this is the acuum current for a Dirac field. From Pauli [] the solution to (.8) is, ret e ˆ x ie ˆ S x x A x x d x (.) ˆ ˆ ad e x ie x A xs x xd x (.) where gien by, ret ad S x and S x are the retarded and adanced Greens function, resectiely and are ad, ret S x t S x S x t S x (.) sint S x mx, x, t ex ik x d k, k m x (.) When (.) and (.) are used in (.9) and (.) the first order change in the acuum current is, This can be shown to be, ret e S x x S x x jd x A x Tr d x ad (.5) S x x S x x ˆ S x x x, x ˆ cost ;, ex (.6) S x x m x x x t ik x d k x (.7) Pauli [] shows that (.5) can be written as, jd x e Ku x x Ku x xa xd x (.8) K x x can also be exressed as, u u K x m (.9) u t Ku x x t t Ku x x; t (.) t 6
K x x x x x x x m x x (.) x t x (.) The quantities,, and satisfy the following relationshis, m m ; m x ; (.) Note that x x x x x x x x Substitute A x x where x. into (.8) and integrate by arts to obtain, jd x e x Ku x x Ku x xd x (.) In order for the theory to be gauge inariant this equation must equal zero. For this to haen u u we must hae K x K x. It can be shown that K x inariance requires that u u. Therefore gauge K x. According to Pauli this is not the case. Pauli obtains, K x x x x Pauli states that this exression is not zero because of the highly diergent nature of (.5) x. Therefore the first order exansion of the acuum current is not gauge inariant. Notice that we hae erformed two different calculations of the acuum current due to the resence of the otential A x x. In the first calculation (Eqs. (.) through (.6)) it was shown that the current oerator is not affected by this otential which is consistent with a gauge inariant theory. In the second calculation (Eqs. (.7) through (.5)) it was shown that the acuum current is changed by this otential (because K x ) which is not consistent with a gauge u inariant theory. So what went wrong? Why do two methods of calculation roduce different results? I consider this an indication that the theory is either mathematically inconsistent in some way or an error was erformed in Pauli s calculations leading u to the second result. The oint of iew will be exlored further in the next section.. Exressing the acuum current in manifestly gauge inariant form. In order to better understand this roblem I will attemt to write the exression for the acuum current j x D in a form that is manifestly gauge inariant. This will be the case if the electric otential terms A always aear in the form F A A. To roceed define the quantities, Therefore, ; (.) I x K x x A x d x I x K x x A x d x D j x e I x I x (.) 7
First examine x I. Use (.) in the (.) to obtain, x x x x x x x x x x x x m x x x x I x A xd x (.) Through reeated use of integration by arts and Eq. (.) it is shown in Aendix that this becomes, x x x x F x (.) m x x x x A x I x x x x x F x d x Next consider I x. I x can be deried from I x, as gien in Eq. (.), if x x relaced by x x. Therefore I x is equialent to (.) with x x x x. Use this fact and m x x to obtain, is relaced by (.5) I x x x x x F x x x x x F x d x From the aboe we see that x I is manifestly gauge inariant because the deendence on the electric otential is contained in terms of the form F. Howeer if we examine (.) it seen that this is not the case for term is, I x If we set A x x. It is eident that there is a term that aears to be non-gauge inariant. This (.6) I, NG x m x x x x A x d x in the aboe and use we obtain, (.7) x is an arbitrary function this I, NG x m x x x x x d x This quantity must be zero if the theory is gauge inariant. Since means that the term m x x x x must be zero. Use m x x x x to obtain, m x x x x x x x x Comaring this exression to (.5) we see hae run into the same roblem that Pauli encountered when trying to determine if that theory is gauge inariant. At this oint I will focus the discussion on whether or not (.8) is zero and if it is not how to modify the theory so that it is. Too start this analysis I will do a detailed examination of, B x m x x (.8) (.9) 8
which must be zero for the theory to be gauge inariant. Use t t t (.) to obtain, t m x, t t x, t t To ealuate this refer to (.) and use, to obtain, Next ealuate i t t B x where,,, along with (.) and t x t sint sint t t cost t t (.) (.) m x, t t ex ik xd k (.) i. Use (.7) along with tcost t 6 iki Bi x t ex ik xd k ex ik xd k in (.9). So the question is does this equal zero? It will simlify the discussion at this oint to change from + dimensions to + dimensions where z will be the sace dimension. In + dimensions this becomes, (.) ik z ik ikz Bi z, t;, t e dk e dk Note that I hae ut in limits of integration, and, where and. This is necessary in order to ealuate the integrals. Take the Fourier transform of the aboe to obtain, Note that, (.) iz ik Bi, t;, Bi z, t;, e dz t dk k k dk Use this in (.5) to obtain, (.5) k k dk k (.6) Bi, t;, t k dk (.7) Gauge inariance requires that B z, t;, equals zero which means that B, t;,. et s see if this is true. Ealuate, ;, i i ik B t for the case and assume to obtain, ik ik Bi t t dk k t dk, ;, (.8) To ealuate this use, i 9
Therefore, m (.9) ik i i i dk k m m i Bi, t;, t In order for the theory to be gauge inariant we must hae B, t;,, therefore this result seems to show that the theory is not gauge inariant. Howeer there is a way to correct this roblem. Go back to (.7) and let integration and. In this case k. In this case (.7) becomes, ik B t t dk i for all k within the limits of (.), ;, (.) i Therefore if the limits of integration are correctly secified we can obtain a gauge inariant result. The next ste is to figure out how to ut these correct limits of integration into the beginning of the calculation so that they will automatically aear at the end. First, since we can relace with in (.). Next we can also relace limit of integration with roided we add k to the integrand. The result is, where the deendence ik B z t t dk k dk ikz ikz, ; e e (.) i B i on is exlicitly shown. but is a retained in the calculation as a finite, but arbitrary large ositie number. Recall that this result alies to the + dimensional sace. In + dimensions we hae, ikx iki ik x Bi x; t e d k e k d k At this oint it and been shown that we achiee a gauge inariant result if we can add the cutoff into the exression for x. We define x factor k If we relace where does x by x (.) by, cost, ex x t ik x k d k (.) in Eqs. (.9) and (.) then the result will be gauge inariant. Now come from? It comes from Eq. (.7) which is a mathematical ealuation of which in turn is defined in Eq. (.6). We can, then, achiee a gauge inariant theory by modifying the definition of the field oerator. Define the modified field oerator by, S x
ˆ ˆ ˆ V q r r iqxet r iqxet ; r x q e a r q u q e ar q u q (.5) where. When this is done then The result of these changes is that that non-gauge inariant term, In this case (.) becomes, S x becomes, S x x ; m x x x I x instead of x x,ng (.6) will aear in Eq. (.). The result of this is, gien by (.6), will be zero. x x x x F x I, ac x; d x (.7) x x x x F x 5. Discussion and Further analysis. It was shown in the last section there is an exact solution to the Dirac equation where A x x and that the current oerator is inariant with resect to this solution. This is consistent with a gauge inariant theory. Howeer when erturbation theory is used it is found that if A x x will be a non-gauge inariant term. This term can be eliminated if the quantity x the quantity x. This, in turn led to a redefining of the field oerator from ˆ x there is redefined as to ˆ x ;. The roblem with this analysis, in my oinion, is that we had to get to the end of a long detailed calculation to realize that we had made a mistake that required correction. I will argue that is should hae been ossible to hae detected this roblem earlier on in the calculation. I will show that the mistake occurs in the way the first order term e ˆ x is defined in Eq. (.) which I reroduce below for conenience, e ˆ x ie S x x ˆ A x x d x (.) ret We hae already shown that if A x x then ˆ x ie x e ˆ x to obtain ˆ x ie x ˆ x. Therefore if A x x we should obtain e ˆ x ie x ˆ x (.) and integrate by arts to obtain,. Exand the exonential is substituted into (.). et s see if this is true. Substitute A x x ret into e ˆ x ie x ˆ S x x x d x (5.) ret Use S x x t ts x x and t t t t t ˆ to obtain, t t ˆ S x x x (5.) t t S x x ˆ x e x ie x d x
It can be shown that S x x ˆ x. Also, for conenience, set t to obtain, e ˆ x, ie x, S x x, ˆ x, d x (5.) Using (.) it can be shown that, S x, ik x dk e (5.) The field oerator ˆ x was gien in Eq. (.) as an infinite sum. Here we shall rewrite it as an integral to obtain, Use all this in (5.) to obtain to obtain, ˆ r r iqx, ˆ r x d qe a q u q (5.5) d k d q eˆ x, ie d xx, e e a q u q r ik xx iqx ˆ r r (5.6) For the moment dro the requirement that x, Next erform the satial integration to obtain, is a real alued function and set x, ix. d q d k eˆ x, ie a u q, r e q k As before we will simly this exression by conerting from three to one sace dimension and add limits of integration. For this roblem we obtain, ik x ˆ qr, r (5.7) dq dk eˆ z,;, ie a u q, r e q k r ikz ˆqr, (5.8) where the limits,. Use the following relationshi to obtain, Use this in (5.8) to obtain, dk e ikz q k q e iq z (5.9) dq eˆ z,;, ie a r u q, r q e Ealuate this for and assume to obtain, iq z ˆqr, (5.) dq i q z iz dq iqz e ˆ z,;, aˆ,, ˆ q ru q r e ie e aq, ru q, r e r r For this one sace dimensional examle, z, (5.) iz e and dq e ˆ ie a u q r e iqz z,; ˆqr,, r (5.) e
Recall that for a gauge inariant theory we want e ˆ z,;, to equal ie z, ˆ z,;. If we examine (5.) and (5.) we see that this is not the case. The roblem is due to the lower limit of integration in (5.) which is instead of. The solution to this roblem is to secify that. When this is done then q in the integrand of (5.) can be relaced by. Therefore (5.) becomes, dq e ˆ e ie a u q r e When this result is comared with (5.) we obtain, iz iqz z,;, ˆqr,, r (5.) ˆ e z,;, ie z, z,; ˆ (5.) Therefore we see in order to correctly do the erturbatie exansion the limits of integration must be correctly chosen. In our one dimensional examle this is achieed by making sure that. In + dimensions this is ensured by modifying the field oerator by adding the factor q definition of the state ector so that (5.5) becomes, ˆ iqx dq,, ˆqr, into the x a u q r q e (5.5) r where. In other words for the exansion in erturbation theory to gie the correct result the field oerator must be defined using a cutoff function. 6. Polarization tensor in momentum sace. In this section we will examine the acuum olarization tensor in momentum sace. We could use the results obtained aboe howeer it will be more informatie to use the exression for the acuum olarization tensor gien by G. Kallen []. According to Kallen the acuum olarization tensor is, k k Tr i k m i k m e d k d k 6 (6.) k m f k k m f k f k P i k k m (6.) k m Kallen shows that this exression is not gauge inariant. Howeer based on the reious discussion we know that in order to achiee a gauge inariant theory we hae to modify the field oerator by including a cutoff er Eq. (.5). When this is done (6.) becomes, k k Tr i k m i k m e ; d k d k 6 k m f k k k m f k k (6.)
As will be shown this will result in an exression that is manifestly gauge inariant. This, of course, differs from Kallen who showed that was not gauge inariant (See discussion in section of []). Ealuate the trace as follows, Tr i k m i k m kk k k k k m Use this in (6.) and relabel dummy indices to obtain, k k kk e kk m k k f k, d k d k 6 k m k k k f k (6.5) Next use f k f k to obtain e k k kk, d k d k k k k m f k k (6.6) kk m Integrate with resect to k to obtain, k k m k k k k e, d k k m f k k (6.7) Rearrange terms and relabel the dummy indices to obtain, e kk k k, d k k m f k k (6.8) k k m The current in momentum sace is gien by, j ;, D A To show that this is gauge inariant we will ut, A inariant. To do this I want A to always aear in the form F i A A shown in Aendix that, (6.) (6.9) in a form that is manifestly gauge. It is j if d k k k m f k k (6.) e kk D ; which is manifestly gauge inariant. Use (6.) in the aboe to obtain, ; ; ; j j j (6.) D DAu DBu e kk jdau ; if d k k k m P k k m (6.) e k k i k jdbu ; if d k k k m k (6.) k m
7. Ealuation of the acuum current. If we can always do a orentz boost to a reference frame where. In this section we will ealuate the acuum current for this case. Therefore the electric otential has no satial deendence and is only deendent on time. We will ealuate (6.) assume the following: A, A t A x, t,,,, and m. This last assumtion will mean the denominator in the integrand will neer be zero so that we can dro the P (indicating rincile art ). With this assumtions the only non-zero terms for F is F i A and F i A where i. For this roblem it is shown in Aendix that, where to obtain, k z m m z e A DA ; 6 z m z m j dz k (7.). Next use the following in the aboe, z z z z z z z ; ; ; DA DAA DAB (7.) j j j (7.) z m z m m z z e A jdaa ; dz k 6 (7.) z m z m j dz k (7.5) e A DAB ; 6 z m In terms of the acuum olarization tensor we use jdaa ; DAA ; A jdab ; DAB ; A ; ; ; DAA DAB and, and so that the total olarization tensor for this examle is e z m z m DAA ; 6 m z z dz k (7.6) DAB 6 z m e z m z m ; dz k (7.7) Comare this with Heitler s results which are gien in Eqs. (.7) and (.8). It can be readily shown that if we dro the term k G from the integrand that ; where G DAA is 5
obtained by setting in (.7) along with we ignore the term k. Now comare to NG ; DAB the quantity in the integrands is the same. The only difference in the non-gauge inariant factor that aeared in the term for NG. This term is what made Heitler s result non-gauge inariant. Howeer in the aroach taken here we hae a fully gauge inariant theory. The result is that this non-gauge inariant term is now fully gauge inariant with by. Thus the theory is mathematically consistent. Howeer from the oint of iew of hysics this creates a roblem because the term Next use (7.) in (7.) to obtain, is highly diergent. ; DAB ; ; ; j j j (7.8) DAA DAAA DAAB m e A z m z m DAAA ; 6 z z j dz (7.9) e A z m z m DAAB ; 6 z m j dz k (7.) Note that the cuttof term k conerges fast enough that k j DAAB ; was droed from Eq. (7.9). This is because the integrand has no effect as. It is readily aarent that is logarithmically diergent. Howeer it can be readily shown that this is just a charge renormalization term because, from Maxwell s equations, current. Finally j DAAA ; shown Aendix that, A is roortional to the alied is finite and is the standard result for the acuum current. Also it is i e F m m jdb m ; Howeer this is zero for the range m so we will not discuss it further.. If (7.) So at this oint the total acuum current for this examle in the range m is gien by, where jdaaa ; is the finite standard result, jdaab ; renormalization term, and j j ; j ; j ; j ; (7.) D DAAA DAAB DAB DAB ; is the logarithmically diergent charge is a highly diergent term which must be eliminated. It will be shown that this term will be cancelled out if there are two scalar fields with the same mass as the Dirac field. 6
8. Polarization tensor for the scalar field. In this section I will examine gauge inariance in scalar field theory. The acuum olarization current for a charged scalar field is gien in Section 9 of [] as, js x e u x x u x x A x d x (8.) x x x x x x x x x x x (8.) and, x x x x x x x x x (8.) Note the in the aboe exressions I hae relaced with x shown in Aendix that scalar acuum current can be exressed as,. Using integration by arts it is K x x xx js x e A x x x F x d x K x x (8.) x x Next, refer to (.8) to obtain, js x jd x e x x x x x x F x d x (8.5) From Maxwell s equations F x j x where j x A A is the alied current. Also define, S x; x x x x ja x d x (8.6) and, S x; x x x x ja x d x (8.7) Use all this in (8.5) to obtain, js x jd x e Sx; S x; (8.8) In momentum sace this becomes, js jd e S ; S ; (8.9) This will be ealuated for the case, m. From Aendix 5, ; ; ; S f j S f f j (8.) A a b A 7
Write, z m fa dz 8 m z z i m f m (8.) 6 z ; m fb dz k 8 z (8.) m j j j j j (8.) S Sa Sb Sc D j Sa e fa ja ; jsb e fb ja ; jsc e f ja For the range m jsc will be zero, jsb is logarithmically diergent but is roortional to the alied current so is simly a charge renormalization term, and j (8.) The diergent terms are all included in the j Sa is finite. term. Therefore as in the case of the Dirac field D we hae highly diergent terms which must be remoed if the theory is to be consistent with the real world. To do this we can inoke a Suersymmetric solution. We assume that there are two charged scalar fields for eery Dirac field and both fields hae the same mass. In this case the total acuum current is, j j j j j j (8.5) T D S Sa Sb Sc Therefore we are left with a logarithmically diergent charge renormalization terms and two finite terms, one of which is zero for the examle m. 9. Summary and Conclusion. In the beginning of this aer I noted that calculations of the acuum current in quantum field theory roduce non-gauge inariant results. In an attemt to understand why this roblem exists I examined in detail a calculation of the acuum current by Pauli. It was shown that this roblem could be corrected by including a cutoff function in the definition of the field oerator (see Eq. (.5)). When this is done the acuum current of the Dirac field will be gauge inariant. Howeer there is a otential roblem in that a diergent term which is nonhysical is included in the result. The diergent can be eliminated if the acuum current due to a scalar field is included in the result. It was shown that if two scalar fields exist with the same mass as the Dirac field then this diergent term is eliminated. Aendix. In the following we will rearrange terms in attemt to obtain an exression that is exlicitly gauge inariant in order to derie Eq. (.). To simlify notation we will dro the exlicit deendence on the coordinates and write and for x x and x x, resectiely. Also we will be reeatedly using integration by arts. In all cases the limits at infinity are assume to go to zero. In order to simlify 8
notation instead of writing ealuate the following quantity. u w d x u wd x I will write m A x A x u w u w. First (9.) where I hae used (.). Integrate by arts to obtain, A x A x (9.) Next integrate by arts the term A x to obtain, A x A x A x Use this in (.) to obtain, Define, so that, (9.) I x A x A x A x d x ; u F x A x A x S x x (9.) I x S A x F x d x (9.5) Perform the following sequence of mathematic stes, S x x A x S x x A x S x x A x (9.6) S x x A x S x x A x S x x A x (9.7) A x A x S x x A x A x (9.8) S x x A x Rearrange terms and relabel dummy ariables to obtain, S x x A x A x F x (9.) Next erform the following integrations by arts, (9.9) A x A x A x (9.) x x A A x A x A Use (.) in the aboe to obtain the following sequence of integration by arts, (9.) A x m A x A x (9.) A x A x A x (9.) A x A x A x m A x (9.5) 9
A x m A x A x (9.6) S x x A x m A x F x F F x F x S x x A x m A x F x (9.7) (9.8) (9.9) The last term in this last exression is zero due to the fact that F is antisymmetric. Use this in (9.5) to obtain, I x m A x F x F x d x (9.) Integrate by arts to obtain, I x m A x F x F x d x (9.) which is equialent to Eq. (.). Aendix. In this aendix we will derie Eq. (6.). Define,, S k k k k k k k m (.) u We want to examine S A. Use basic algebra to obtain the following two exressions, k k A u kk if kk A A kk A k k A (.) Use these results along with (.) to obtain, k k A k A A k if (.) kk if k k A Su k, A k A k if k m A (.) Use the following result, k k k m k A k A k A k A k m in (.) to obtain, k k m k if k A Su k, A k if k m A k m Rearrange terms to yield, (.5) (.6)
u, S k A Next use A A if Therefore, kk if k A k m k if k A A A k m in the aboe to obtain,,,, u, GI, NGI (.7) S k A C k C k (.8) k k C k k if k m,, GI A if C, NGI k, k m k k m k (.9) (.) e C, GI k, jd ;, A d k k m f k k (.) C, NGI k, x x to obtain, Use the relationshi A C, NGI k, k m f k k k m (.) Therefore the otentially non-gauge inariant art is, e A d k k k m k (.) Because the integrand is odd in k this term equals zero. Using the aboe results (along with x x ) we obtain Eq. (6.). Aendix. Ealuate (6.) for the simle case where the simle case where the electrical otential has no satial A and Ax, t A t,,. In this case the only non-zero term for F is F ia where i k F i k F k F deendence. It is only deendent on time and sace F i A k m k m k m k. In momentum and. Also and ikf and. Use all this in (6.) to obtain, e F k jda ; d k k k m P k (.) k Next use,
k k E k E k k (.) m E k where k to obtain, E k m ef dkek k jda ; P k (.) Ek Ek In the aboe using symmetry arguments we can relace k with k k k k k k. In addition use d k k d k k E k de E E m de, k E m and Ek k Ek m along with Next define E k k F i A A in the aboe to yield, ea E k m Ek m jda ; de k P k (.) m Ek z to obtain, ea DA ; 6 m z z m z m j dz k (.5) In the aboe k z m. Next calculate j the first aragrah of this section. We obtain, DB ; (Eq. (6.) for the assumtions gien in e F k i k j x d k k k m k Next use (.) in the aboe to obtain, Next substitute k to obtain, DB ; (.6) k k i Ek E k ef dk Ek jdb ; k (.7) Ek k i Ek Ek Ek k along with k E m and d k E E m de in the aboe k k k k
E k m i Ek Ek Ek E i Ek k m Ek Ek E m de (.8) e F k k DB ; m j Next use E E and obtain, k k E E to k k m m i e F m de k jdb ; (.9) m i m Recall the definition of (Eq. (.)) and rearrange terms to obtain, Aendix. i e F m m jdb m In this aendix we derie Eq. (8.). Define, Therefore obtain, ; ; ; ; N x x x A x d x N x x x A x d x u u (.) (.). First I will examine Nu x x. Use (.) to j x e N x x N x x ˆS u u x x x x A x m x x x x A x In the following we will write integrate by arts to obtain, and similarly for x x and x x for x x. Next m A x m A x A x A x Using integration by arts we hae, Use this in (8.) to obtain, A x A x A x A x A x A x (.) (.) (.) (.5) A x m A x x x A x A x A x A x Rearrange terms to obtain, (.6)
A x m A x A x F x x x A x (.7) Next use, to obtain, m A x A x (.8) A x x x A x m A x F x (.9) Refer to Eq. (.) x x A x K x x A x x xx x F x (.) N x; K x xa x x x x xf xd x (.) Also it can be shown that, N x; K x xa x x x x xf xd x (.) Therefore, K x x ˆ xx js x e A x x x F x d x K x x (.) x x Aendix 5. We want to ealuate S x and S x; ; momentum sace we hae, which are gien by Eqs. (8.6) and (8.7), resectiely. In ; ; S ; f J S f J (.) f k kd k; f k k d k; and where (see Ref. []), (.) i k P ; k k k m ; k k m k m. Also, in momentum sace, k k k Sole for the case where m and. Use the aboe relationshis to obtain, (.)
Next use (.) to obtain, k m k ; (.) f d k k d k d k f k k Ek E Ek E k Ek k (.5) Use d k E E m to obtain, where This becomes, k k k E m. Use z E k to obtain, k so that Ek m f de k k (.6) E m k z m f dz k 8 z (.7) m ; a b f f f (.8) a b S f f J z ; m fb dz k z m fa dz k 8 m z z Next calculate f where I hae used (.) to obtain, Use d k E E m. Use (.) to obtain, 8 i z (.9) f d k k k k m k 8 (.) k m k m k k m and set. Use m Ek Ek i d k f k 8 (.) Ek Ek Ek k k and k k Ek Ek to obtain, This yields, E E and E E i Ek m f dek k 8 m Ek Ek (.) k k 5
Recall that S ; f J. i m f m (.) 6 References. [] J. H. Schwarz, String Theory Origins of Suersymmetry, Nucl. Phys. Proc. Sul.,, 5 () [arxi:he-th/78]. [] C. Jarlskog. Suersymetry Early Roots That Didn t Grow. Adances in High Energy Physics. Article ID 76875. May 5. Also - arxi:5.769. [] K. Nishijima. Fields and Paricles: Field theory and Disersion Relations. 969 (W.A. Benjamin, New York). [] D. Solomon. A new look at the roblem of gauge inariance in quantum field theory. Phys. Scr. 76 6 (7). [5] W. Heitler. The quantum theory of radiation. 95 (Doer Publications, New York). [6] J.J Sakurai. Adanced Quantum Mechanics. 967. (Addison-Wesley Publishing Co. Redwood City, California). [7]. W. Greiner, B. Muller, and T. Rajelski; Quantum Electrodynamics of Strong Fields. 985. (Sringer, Berlin). [8] R. Jost and J. Rayski, Remarks on the roblem of the acuum olarization and the hoton selfenergy. Hel. Phys. Acta (99), 57. [9] S. Sakata and H. Umezawa. On the Alicability of the Method of the Mixed Fields in the Theory of Elementary Particles. Prog. Theor. Phys. Vol. 5, 95. [] M. E. Peskin and D.V. Schroeder. An introduction to quantum field theory. 995 (Addison-Wesley, Reading MA) [] W. Pauli. Pauli ectures on Physic: Volume 6. Selected Toics in Field Quantization. (976) MIT Press, Cambridge, Massachusetts and ondon, England). [] G. Kallen. Quantum Electrodynamics. Sringer-Verlag, Berlin, 97. 6