The Heisenberg versus the Schrödinger picture in quantum field theory. Dan Solomon Rauland-Borg Corporation 3450 W. Oakton Skokie, IL USA

Transcription

1 1 The Heiseberg versus the chrödiger picture i quatum field theory by Da olomo Raulad-Borg Corporatio 345 W. Oakto kokie, IL 677 UA Phoe: da.solomo@raulad.com PAC 11.1-z March 15, 24

2 2 Abstract The Heiseberg picture ad chrödiger picture are supposed to be equivalet represetatios of quatum mechaics. However this idea has bee challeged by P.A.M. Dirac [4]. Also, it has bee recetly show by A. J. Faria et al [3] that this is ot ecessarily the case. I this article a simple problem will be worked out i quatum field theory i which the Heiseberg picture ad chrödiger picture give differet results.

3 3 I. Itroductio The Heiseberg picture ad chrödiger picture are supposed to be equivalet represetatios of quatum theory [1][2]. However A.J. Faria et al[3] have recetly preseted a example i o-relativistic quatum theory where they claim that the two pictures yield differet results. Previously P.A.M. Dirac [4] has suggested that the two pictures are ot equivalet. I order to further ivestigate this problem we will examie quatum field theory i both the Heiseberg ad chrödiger pictures ad will show that, eve though they are formally equivalet, they yield differet results whe a actual problem is worked out. I quatum field theory a quatum system, at a give poit i time, is specified by the state vector Ω ad field operator ˆ x ψ. We will write this as the pair Ω, ψ. Let the state vector Ω ad the field operator ψˆ x be defied at some iitial poit i time, say t=. This may be take as the iitial coditios of the quatum system. Now there are two ways to hadle the time evolutio of the system. I the chrödiger picture it is assumed that field operator ψˆ x is costat i time ad the time depedece of the system goes with the state vector Ω t. I the Heiseberg picture the time depedece is assiged to the field operator ψˆ x,t ad the state vector Ω remais costat i time. Both pictures are supposed to give equivalet results i that the expectatio values of operators are the same. Note that at the iitial time, t=, both pictures are idetical. Therefore the time idepedet chrödiger field operator ψˆ x is equal to ψ ˆ x,, which is the time depedet Heiseberg field operator at t=.

4 4 imilarly, the time idepedet Heiseberg state vector Ω equals Ω, which is the time depedet chrödiger state vector Ω t at t=. For example, let the iitial state of the system, at t=, be represeted by the pair, x, picture this iitial state evolves ito, x,t ψ. state evolves ito Ω t, x, Ω ψ. I the Heiseberg Ω ψ. I the chrödiger picture the From Chapter 9 of Greier et al [5] the Dirac Hamiltoia i the presece of a classical electromagetic field is give by, where, 1 H ˆ ψ ˆ = ˆ,Hˆ d 2 ψ ψ x ξ H = H qα A+ qa r 1 2 ad H = α +β i m ad ξ is a reormalizatio costat. I the above expressio A,A r 3 is the classical electric potetial ad is take to be a uquatized, real valued quatity. Also q ad m are the charge ad mass of the electro, respectively, ad the 4x4 matrices α ad β are defied i [5]. Note that i the above equatios we use brackets i 1 is defied by the followig expressio, = c= 1. The term i the T T ψˆ ψ ˆ = ψˆ ψˆ ψˆ ψ ˆ = ψˆ ψˆ ψˆ ψˆ,h H H α Hαβ β Hαβ β α 4 where the summatio is over the spi idices. Equatio 1 ca also be rewritte as,

5 5 ˆ ˆ H H J A x,t dx A x,t dx ψ ˆ = ˆ ψˆ ψˆ + ρˆ ψˆ 5 where H ˆ ψ ˆ = 1 ˆ,Hˆ 2 ψ ψ dx ξ r 6 ad ˆ q q J ψ ˆ = ψˆ, αψ ˆ ad ρψ 2 ˆ ˆ = ˆ, ˆ 2 ψ ψ Hˆ ψ ˆ is the free field Hamiltoia operator, J ψˆ ˆ 7 is the curret operator, ad ˆ ˆ ρψ is the charge operator. I the Heiseberg picture the evolutio of the field operator is give by, ψˆ x,t t i ˆ = H ˆ x,t, ˆ ψ ψ x,t 8 The field operator obeys the equal time ati-commutator relatioship, 3 ψˆ x,t ψ ˆ x,t +ψˆ x,t ψ ˆ x,t =δ δ x x α β β α αβ 9 with all other equal time ati-commutators beig equal to zero. It is show by Greier see Chapt. 9 of [5] or ectio 8 of [1] that whe these are used i 8 we obtai, ψˆ x,t i = Hˆ ψ t x,t 1 To covert to the chrödiger picture defie the operator Û t which satisfies the differetial equatio, Û t i Hˆ = ψ x, U t ˆ ˆ t ; t Û ˆ ˆ i = U t H ψˆ x, 11 t

6 6 where Û t is subject to the iitial coditio Û = 1. From the above expressios we have that ˆ ˆ U t U t t =. From this, ad the iitial coditio o Û t we obtai Uˆ t Û t =1 which yields, = ˆ Uˆ t U -1 t 12 The time depedet chrödiger state vector is defied by, ˆ Ω t = U t Ω ad t Uˆ Use this ad 11 to show that Ω t satisfies, t Ω = Ω t 13 Ω ˆ i = H ψˆ x, Ω t 14 t II. The Vacuum state I this sectio we will develop some of the tools of quatum field theory i the chrödiger picture. I particular we will reach some coclusios cocerig the expectatio value of the operator Ĥ i the chrödiger picture. Defie φ x as beig basis states solutios of the followig equatio, φ =λ φ H x E where the eergy eigevalues λ x 15 E ad ca be expressed by for a positive eergy state E =+ p + m, λ = 1 for a egative eergy state 16 ad where p is the mometum of the state. The φ x ca be expressed by,

7 7 ip x ue x φ = 17 where u is a costat 4-spior. The φ x form a complete orthoormal basis i Hilbert space ad satisfy φ x φ m x dx =δ m 18 Defie the idex so that it is positive for positive eergy states ad egative for egative eergy states. The time idepedet chrödiger field operator ca, the, be defied as follows, ψ ˆ x = bˆ φ x + dˆ φ x ; ˆ x ˆ b x dˆ ψ = φ + φ x 19 > > where the ˆb ˆb are the destructiocreatio operators for a electro associated with the state φ x ad the ˆd ˆd are the destructiocreatio operators for a positro associated with the state φ { d,d ˆ ˆ j k} =δj k ; ˆ ˆ { j k} The vacuum state is defied by, x. They satisfy the aticommutator relatioships, b,b =δ jk ; all other ati-commutators are zero 2 dˆ = bˆ = ad j j dˆ = bˆ = 21 j j Usig the above results the free field Hamiltoia operator becomes, 1 ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ Hˆ ψ = λ E bb bb +λ E dd dd ξ 22 r 2 > Use 2 i the above to yield, ˆ ˆ ˆ ˆ ˆ Hˆ ψ = E b b + E d d E ξ vac r > 23

8 8 where we have used E =, λ = 1, ad λ = 1 for > ad, E E vac = E > Next defie ξ = E to obtai, r vac ψ ˆ = ˆ ˆ + Hˆ E bb dd ˆ ˆ > The vacuum state is a eigestate of the free field Hamiltoia Hˆ ψ ˆ with eigevalue ε =, i.e., H ˆ ψ ˆ = 26 New eigestates, ca be produced by operatig o with various combiatios of the electro ad positro creatio operators ˆb ad ˆd, respectively. ice electros ad positros are particles with positive eergy the ew states are positive eergy states with respect to the vacuum. The total set of eigestates which icludes the vacuum state form a orthoormal basis that satisfies the followig relatioships, ad ˆ Hˆ ψ =ε where ε >ε = for 27 m =δ m 28 Ay arbitrary state Ω ca be expaded i terms of these basis states, Ω = c 29 where c are the expasio coefficiets. Use the above to obtai,

9 9 2 ˆ c Ω Hˆ ψ Ω = ε 3 ice the ε are all o-egative we obtai the relatioship, ˆ Ω Hˆ ψ Ω for all Ω 31 III. Comparig Expectatio values I this sectio we will compare the expectatio of the operator Hˆ ψˆ i the Heiseberg ad chrödiger pictures. For the quatum system Ω, ˆ ψ the expectatio ˆ Hˆ ˆ value of H is give by t=, is, ˆ x Ω ψ Ω. Now cosider a system whose iitial state, at Ω ψ ad let the electric potetial at this time be zero. Now apply a electric potetial for some period of time ad the remove it at time A,A = for t < A,A for t t ; 1 ; t1 > A,A = for t > t 1 so that, 32 We wat to determie the quatity Hˆ ˆ Ω ψ Ω at some fial time tf > t 1. I the Heiseberg picture the state evolves ito Ω, ψˆ x,t f where x,t ˆψ satisfies Eq. 1 with the iitial coditio ψ ˆ x, = ψˆ x. I the chrödiger picture the state evolves ito Ω t f, ψˆ x where Ω t satisfies Eq. 14. Therefore, i the Heiseberg picture, at the fial time, the expectatio value of Ĥ ˆ t f ψ is give by, H t = Ω H ψ x,t Ω ˆ ˆ,eH f f 33 ad i the chrödiger picture the expectatio value is give by, ˆ ˆ H t = Ω t H ψ Ω t 34,e f f f If the two pictures are equivalet the we should have,

10 1,eH = t H t H,e for all t 35 This relatioship is obviously true at the iitial time t =. I the Appedix a formal proof is preseted which shows that this relatioship is true for all time. However, whe the above problem is actually worked out it is show that the above relatioship does ot ecessarily hold at the time t f. Now i geeral it is ot possible to fid a solutio to Eq. 1 for most o-zero electrical potetials. However we will cosider a electric potetial for which a exact solutio ca be calculated. Durig the time iterval t t1 the electric potetial is o-zero. Let it be give by, χ A,A =, χ ; t t t 1 where χ x,t is a arbitrary real valued fuctio that satisfies the followig iitial coditio at t=, 36 χ x,t χ x, = ; = 37 t t= Refiig to Eqs. 1, 32, ad 36 we obtai, ψˆ x,t i t χ = H + qα χ+ q ψˆ x, t for t t t 1 38 ad, ψˆ x,t i = Hψ ˆ x,t for t > t 1 t 39 ice the time derivative is to the first order the boudary coditio at t = t 1 is, ψ ˆ +δ = ψ x,t x,t ˆ 1 δ 1 δ 4

11 11 The solutio to 38 is, iqχ x,t iht ψ ˆ x,t = e e ψˆ x for t t1 41 where we have used the iitial coditio ψ ˆ x, =ψˆ x as well as 37. The solutio to 39 is, ih tf t1 ψ ˆ x,tf = e ψ ˆ x,t 1 for tf > t1 42 Usig the boudary coditios 4 we obtai, ih tf t1 iqχ x,t1 ih1 t ψ ˆ x,tf = e e e ψˆ x 43 Rewrite 43 as, ih tf t1 iqχ x,t ˆ 1 ψ x,t = e e ψ ˆ x,t f 1 44 where, ψ ˆ x,t = e ih t ψˆ x Note that ψ ˆ x,t is the field operator that ψˆ would evolve ito if the electric x 45 potetial was zero. Use these results i 6 to yield, ˆ 1 iq x,t1 iq x,t1 H ˆ x,tf + χ χ ψ = ψˆ x,t1 e,he ψ x,t1 dx 2 ξ Use the followig result, iq iq H e χ ˆ x,t e χ ψ = qα χ+ H ψˆ x,t r i 46 to obtai, ˆ 1 H ψ ˆ x,tf = ˆ x,t 1, q x,t1 H ˆ x,t1 dx 2 ψ α χ + ψ ξ r 48 Use 7 i the above to obtai,

12 12 ˆ ˆ ˆ H ψ ˆ x,tf = H ψˆ x,t1 J ψˆ x,t1 χ x,t1 dx 49 Next sadwich the above expressio betwee Ω ad obtai, H t H x,t = Ω ˆ ψˆ Ω ˆ Ω ψˆ 1 Ω χ 1,eH f 1 Defie the curret expectatio value by The quatity ˆ,e Ω ad refer to 33 to J x,t x,t dx 5 ˆ J x,t Ω J ψ x,t Ω J,e x,t 51 is the curret that would exist at time t if the electric potetial was zero durig the iterval from to t. Use this defiitio i 5 ad assume reasoable boudary coditios ad itegrate by parts to obtai, H t = Ω H ψ x,t Ω + χ x,t J x,t dx ˆ ˆ,eH f 1 1,e 1 52 I the above expressio examie the quatities o the right of the equals sig. Note that Ω H ψ x,t Ω ˆ ˆ 1 ad J x,t,e 1 are idepedet of χ x,t 1. This meas that χ x,t 1 ca be varied i a arbitrary maer without affectig either of these quatities. Therefore, if J, e x,t 1 is ozero, it is always possible to specify a χ x,t 1 χ such that the quatity o the left, x,t = f J x,t 1,e 1 H, eh, is a egative umber. For example, let where f is a real umber. Use this i 52 to obtai, H t = Ω H ψ x,t Ω f J x,t dx ˆ ˆ 2,eH f 1,e 1 53

13 13 Ω H ψ x,t Ω This quatity will be egative for large eough f sice ˆ ˆ J x,t,e 1 1 do ot vary as f is icreased. This result depeds o J model of the real world the there must exist quatum states where J x,t,e 1 x,t,e 1 beig o-zero. How do we kow that this is the case? If quatum mechaics is a accurate is ozero because there are may examples i the real world where this is the case. For example, J x,t,e 1 ad will be o-zero for a localized electro wave packet travelig with a o-zero velocity. o that by properly preparig the iitial state Ω we ca always esure that J x,t,e 1 will be o-zero. Now refer back to equatio 35. From 53 we see that the left had side ca be egative. However it ca be see by examiig 31 ad 34 that H,e t which meas that the right side of 35 must be o-egative. Therefore the relatioship give by 35 does ot hold i this case ad the Heiseberg picture ad chrödiger picture gives differet results for the expectatio value i questio. Thus there appears to be a mathematical icosistecy i quatum field theory. O oe had it is possible to show, formally, that the Heiseberg picture ad chrödiger picture are equivalet. O the other had whe a actual problem is worked out we see that they give differet results. The reaso for this is that i the chrödiger picture the Ω ψ Ω quatity Hˆ ˆ x must always be o-egative. It is ot possible to fid a Ω ψ Ω state vector Ω where Hˆ ˆ x is egative. I order to elimiate this icosistecy it is ecessary to redefie the vacuum state so that eigestates exist

14 14 where the ε <. This will allow for the existece of states Ω where ˆ x Ω Hˆ ψ Ω is egative. The way this ca be achieved is examied i [6] ad [7]. V. ummary ad Coclusio It is geerally assumed that the Heiseberg ad chrödiger pictures are equivalet ad that the expectatio values of operators should be the same i both pictures. However it has bee show that they do ot produce equivalet results whe a actual problem is worked out. Here the iitial quatum state, ˆ x Ω ψ evolves forward i time uder the actio of the electric potetial give by 32 ad 36. I the Heiseberg picture this evolves ito the state, ˆ x,t Ω ψ. It was show that it is possible for the expectatio value of Ĥ of this quatum state to be egative. If the same problem is examied i the chrödiger picture the iitial state evolves ito t, ˆ x Ω ψ. As has bee show see Eq. 31 the expectatio value of for this state must be o-egative. Therefore the Heiseberg ad chrödiger pictures do ot Ĥ yield that same result whe the expectatio value of Ĥ is calculated. This result is cosistet with previous research [3] ad cofirms the commets by Dirac that the Heiseberg picture ad chrödiger picture are ot equivalet [4]. Appedix We will formally show that,,eh = t H t H,e 54 tart with 34 ad covert to the Heiseberg picture by usig 13 ad 12 to obtai, ˆ 1 ˆ ˆ ˆ H t = Ω U t H ψ U t Ω 55,e

15 15 Next use = -1 Uˆ t Uˆ t 1 ad refer to 6 ad use the fact that Û t commutes with H to obtai, ˆ -1 ˆ ˆ 1 ˆ -1 ˆ ˆ -1 U t H U t U t U t,h U t Uˆ t dx ξ r ψ ˆ = ψˆ ψˆ 2 ˆ -1 = H Uˆ t ψˆ ˆ U t 56 Next prove that, ˆ 1 ψ ˆ x,t = U t ψˆ Uˆ t 57 To show this take the time derivative of ˆ 1 U t ψˆ Uˆ t to yield, ˆ 1 U t ˆ -1 U ˆ t Uˆ t Hˆ ˆ ˆ ˆ U t = i t ˆ 1 U t ψˆ Hˆ ψˆ Uˆ t ψ ψ ψ 58 Use 57 ad the fact that ˆ ˆ -1 U t U t = 1 to obtai, 1 U t ψˆ U t ψ -1-1 ψ 1-1 ˆ ˆ ˆ ˆ Hˆ Uˆ t ˆ Uˆ t Uˆ t ˆ Uˆ t = i t Uˆ t Uˆ t Hˆ Uˆ t Uˆ ψ ψ t -1-1 where we have used the relatioship U ˆ t H ˆ ˆ U ˆ t H ˆ U ˆ t ˆ U ˆ t ψ = ψ. 59 Compare this to 8 to show that ˆ 1 U t ψ ˆ Uˆ t ad ˆ x,t ψ obey the same differetial equatio. Also, due to the fact that Û = 1 they are equal at the iitial time t=. Therefore equatio 57 is true. Now use 57 i 56 ad 55 to obtai Ω ˆ ˆ H t = Ω H ψ x,t,e. Compare this to 33 to show that 54 is true.

16 16 Refereces 1. W. Pauli. Pauli Lectures o Physics Vol 6 elected Topics i Field Quatizatio, MIT Press, Cambridge, Massachusetts, A. Messiah, Quatum Mechaics Vol 1. North Hollad Publishig Compay, Amsterdam, A.J. Faria, H.M. Faca, C. P. Malta, R. C. pochiado, Physics Letters A, P. A. M. Dirac. Physical Review, Vol B684 B W. Greier, B. Muller, ad J. Rafelski. Quatum electrodyamics of strog fields. priger-velag, Berli, D. olomo. Ca. J. Phys., 81, 1165, 23. see also quat-ph/ D. olomo, Ca. J. Phys. 76, see also quat-ph/99521.