1
The stability of the QED vacuum in the temporal gauge
by
Dan Solomon
Rauland-Borg Corporation 3450 W. Oakton Skokie, IL USA
Email: [email protected]
August 7, 2004
PACS: 11.10.-z, 03.70.+k, 03.65.-w 2
Abstract
The stability of the vacuum for QED in the temporal gauge will be examined. It is generally assumed that the vacuum state is the quantum state with the lowest energy.
However, it will be shown that this is not the case for a system consisting of a fermion field coupled to a quantized electromagnetic field in the temporal gauge. It will be shown that for this situation there exist quantum states with less energy than the vacuum state. 3
I. Introduction.
In this article the stability of the QED vacuum in the temporal gauge will be
examined. In the temporal gauge the gauge condition is given by the relationship A00 =
[1,2,3,4] where A0 is the scalar component of the electric potential. The advantage of
the temporal gauge is due to the simplicity of the commutation relationship between the
electromagnetic field quantities which are given below. In the coulomb gauge, for
instance, these are more complicated. Due to this fact the temporal gauge is particularly
useful in the treatments of QED which use the functional Schrödinger equation [3,4].
We will examine a field consisting of fermions coupled to a quantized
electromagnetic field. It is generally assumed that that there is a lower bound to the
energy of quantum states and that the quantum state with the lowest energy is the vacuum
state. We will show that that this is not the case for QED in the temporal gauge and that
there must exist quantum states with less energy than the vacuum state. Throughout this
discussion we will use =c=1. Also vectors are indicated by bold text.
We will work in the Schrödinger picture. In this case the field operators are time
independent and the time dependence of the quantum system is reflected in the state
vector Ω()t which evolves in time according to the Schrödinger equation,
∂Ω()t i=ΩHˆ ()t (1.1) ∂t where the Hamiltonian Hˆ is [3],
ˆˆ ˆ ˆ ˆ HH=+0,D H0,M −∫ Jx()⋅A(x)dx (1.2)
The quantities in the above expression are defined by, 4
ˆ 1 † H0,D =ψ∫ ˆˆ()xx,H0,Dψ()dx; Hi0,D = −⋅α ∇+βm (1.3) 2
ˆˆ1 22ˆ ˆ ˆ Hd0,M =∫ EB+x; Bx( ) =∇×Ax( ) (1.4) 2 ()
q Jxˆ ()=ψˆ† ()x,αψˆ()x (1.5) 2
In the above expressions m is the fermion mass, α and β are the usual 4x4 matrices, q is
ˆ ˆ the electric charge, H0,D is the Dirac Hamiltonian, H0,M is the Hamiltonian for the electromagnetic field, and Jxˆ () is the current operator. The Schrödinger picture time independent fermion field operators are ψˆ (x) and ψˆ † (x) . The field operators for the electromagnetic field are Axˆ () and Exˆ ( ). The electromagnetic field operators are real so that Axˆ† ()= Aˆ(x) and Exˆˆ† ()= E(x).
The field operators obey the following relationships [2,3],
A,ˆ ijx Eˆ y =−iδ δ3x −y ; A,ˆˆijxyA = Eˆˆix,Ejy= 0 (1.6) () () ij ( ) () () () () and
ˆˆ† ˆˆ†† ˆˆ {ψψab()x , (y)} =δabδ()x −y ; {ψa()xy,ψ=b ()} {ψa()xy,ψb()} =0 (1.7) where “a” and “b” are spinor indices. In addition, all commutators between the electromagnetic field operators and fermion field operators are zero, i.e.,
Axˆˆ,,ψˆˆy =Eˆˆx ψy =Ax,ψ=ˆ†y Ex,ψˆ†y=0 (1.8) () () () () () () () ()
Define,
Gˆ ()xE=∇⋅ˆ ()x−ρˆ (x) (1.9) 5 where the current operator ρˆ (x) is defined by,
q ρ=ˆˆ()xψ† ()x,ψˆ()x (1.10) 2
The state vector Ω()t must satisfy the gauss’s law constraint [3],
Gˆ ()x Ω(t)=0 (1.11)
ˆ ˆ It can be shown [1] that H,G ()x = 0 therefore if (1.11) is valid at some initial time it will be valid for all time.
The energy of a normalized state Ω is given by,
E()Ω=ΩHˆ Ω (1.12) where the normalization condition is Ω Ω=1. The commonly held assumption is that
there exists a state Ωvac , called the vacuum state, which is a state of minimum energy so that all other states have an energy that is greater than the vacuum state, i.e.,
E ()Ω−E ()Ωvac >0 for all Ω≠Ωvac (1.13)
We will show that the above statement cannot be true if there exists a normalized state
Ω1 which satisfies Gauss’s law and for which the divergence of the current expectation value is non-zero, that is,
∇⋅ Ω11Jx()Ω ≠0 in some region of space. (1.14)
Before proceeding we must ask the question “how do we know that a state Ω1 can be found where the above condition holds?”. The answer is that if quantum mechanics is a correct model of the real world then there must exist many states where the above 6 condition holds because in the real world there are many examples where the divergence of the current is non-zero over some region of space.
Now given some initial state new states can be defined by acting with operators on the initial state [6]. With this in mind define the state,
−iCˆ Ω=2e Ω1 (1.15) where the operator Cˆ is defined by,
Cˆ =⋅∫ Exˆ ()∇χ()xdx (1.16) and where χ(x) is an arbitrary real valued function. Note that dual state is,
+iCˆ† +iCˆ Ω=21Ωe=Ω1e (1.17) where we have used Cˆ† = Cˆ since Exˆ ( ) and χ(x) are both real. From this we have that
ΩΩ22=Ω1Ω1=1 where we use the relationship,
ˆˆ ee+−iC iC =1 (1.18)
Now is Ω2 a valid state, i.e., does it satisfy (1.11)? Based on the commutator relationships (1.6) and (1.8) we see that the operator Cˆ commutes with both Exˆ () and
ˆ−iCˆ ˆ ρˆ (x). Therefore Ge()xΩ=21G()xΩ=0 so that Ω2 satisfies (1.11) since Ω1
ˆ has been assumed to satisfy G0()x Ω1 = .
Next we want to evaluate the energy of the state Ω2 . To do this use (1.2) and
(1.12) to obtain,
ˆˆ ˆˆ EH()Ω=22Ω 0,DΩ2+Ω2H0,MΩ2−Ω2∫ Jx( )⋅A(x)dxΩ2 (1.19) 7
ˆ ˆ Consider first the term Ω20H ,DΩ2. To evaluate this use the fact that Ex(), and thereby Cˆ , commutes with the fermion field operators ψˆ (x) and ψˆ † (x) . Use this fact along with (1.18) to obtain,
ˆˆ ΩΩ20H,D2=Ω1H0,DΩ1 (1.20)
ˆ Next consider the term Ω20H ,MΩ2. From (1.6) we obtain,
ˆˆ Ax(),C =−i∇χ(x) (1.21)
Use this result to obtain,
ˆ ˆˆˆ Β()xA,C =∇×()x,C=−i∇×∇χ()x=0 (1.22)
ˆ ˆ Therefore C commutes with H0,M so that,
ˆˆ ΩΩ20H,M2=Ω1H0,MΩ1 (1.23)
Now for last term in (1.19) use the fact that Cˆ commutes with Jxˆ ( ) to obtain,
ˆˆˆ+−iCˆˆˆiC Ω⋅22∫∫Jx()A(x)dexΩ=Ω1J()x⋅( A()xe)dxΩ1 (1.24)
To evaluate the above expression further use the Baker-Campell-Hausdorff relationships
[7] which states that,
ˆˆ 1 eO+−OO11ˆˆe =+OOˆ,Oˆ+Oˆ,Oˆ,Oˆ+… (1.25) 22122 112
ˆ ˆ where O1 and O2 are operators. Use this relationship along with (1.6) and (1.21) to obtain,
ˆˆ ee+−iCAxˆˆ() iC =−Ax() ∇χ(x) (1.26)
Use this result in (1.24) to obtain, 8
ˆˆˆˆ Ω⋅22∫∫Jx()A(x)ddxΩ=Ω1J(x)⋅(A()x−∇χ()x) xΩ1 (1.27)
Use the above results in (1.19) to yield,
ˆˆ ˆˆ ˆ EH()Ω=21Ω()0,D+H0,M−∫∫Jx()⋅A()xdx Ω1+Ω1J()x Ω1⋅∇χ()xdx
(1.28)
Next use (1.2) and (1.12) in the above and integrate the last term by parts, assuming reasonable boundary conditions, to obtain,
ˆ EE()Ω=21()Ω−∫ χ()xJ∇⋅Ω1(x) Ω1dx (1.29)
Next subtract the energy of the vacuum state, E( Ωvac ) , from both sides to obtain,
ˆ EE()Ω−2 ()Ωvac =()E()Ω1 −E( Ωvac ) −∫ χ()xJ∇⋅Ω1 ()xΩ1 dx (1.30)
ˆ Now in the above expression the quantities (EE( Ω−1) ( Ωvac)) and ΩΩ11Jx() are independent of χ(x). Recall that we have picked the quantum state Ω1 so that
ˆ ∇⋅ Ω1Jx()Ω1 is nonzero. Based on this we can always find a χ(x) so that
(EE()Ω−2(Ωvac)) is a negative number. For example, let
ˆ χ=()xλ∇⋅Ω1J()xΩ1 where λ is a constant. Then (1.30) becomes,
ˆ 2 EE()Ω−2 ()Ωvac =()E()Ω1 −E()Ωvac −λ∫ (∇⋅Ω1 Jx()Ω1 ) dx (1.31)
ˆ Now, since ∇⋅ Ω1Jx()Ω1 is nonzero, the integral must be positive so that as λ→∞ the quantity (EE()Ω−2v()Ωac)→−∞. Therefore the energy of the state Ω2 is less than that of the vacuum state Ωvac by an arbitrarily large amount. Therefore there is no lower bound to the energy of a QED quantum state in the temporal gauge. 9
II. Interaction with Classical fields
In the previous section we have shown that if there exists a state Ω1 that satisfies (1.14) then there exists a state Ω2 whose energy is less that that of Ω1 by an arbitrarily large amount. This suggests the possibility that it would be possible to extract an arbitrarily large amount of energy from a quantum state through the interaction with an external field. It will be shown in this section that this is theoretically possible.
In the absence of external interactions the energy of a quantum state remains constant. In order to change the energy we must allow the field operators to interact with external sources or fields. This is done by adding an interaction term to the Hamiltonian.
Let this term be,
ˆˆ ˆ H,int =−∫Sx()t⋅Ax()dx−∫J(x)⋅R(x,t)dx (2.1)
In the above expression Sx(,t) is a classical field that interacts with the quantized electromagnetic field and Rx(,t) is a separate classical field that interacts with the fermion current operator. It should not be assumed that the classical fields Sx(,t) and
Rx(,t) correspond to physical fields that actually exist. For the purposes of this discussion these fields are fictitious. They have been introduced for the purposes of perturbing the Hamiltonian in order to change the energy of some initial state. It will be shown that for properly applied fields Sx( ,t) and Rx( ,t) an arbitrarily large amount of energy can be extracted from some initial state. Therefore even though these fields do not correspond to actual physical objects we believe that the following results are mathematically interesting. The reason we pick these fields is because for particular 10 values of the interaction we obtain an exact solution to the Schrödinger equation. This is demonstrated in the following discussion.
When the interaction is included the Schrödinger equation becomes,
∂Ω()t i=Hˆ Ω()t (2.2) ∂t T where,
ˆˆˆ ˆ ˆ ˆ HHTi=+Hnt=−H∫Sx(),t⋅Ax( )dx−∫J(x)⋅R(x,t)dx (2.3)
Now we will solve (2.2) for the following interaction,
Rx(),t =<0for t t1; Rx(),t = −∇g(t) χ(x) for t12≤≤t t ; Rx( ,t) =>0for t t2 and,
Sx(),t =<0for t t1 ; Sx(), t = g(t)∇χ(x) for t12≤ t ≤ t ; Sx( ,t) =>0for t t2 where the double dots represent the second derivative with respect to time. In addition to the above gt() satisfies the following relationship at time t,2
gt ()2 = 0 and gt()2 =−1 (2.4)
According to the above expressions the interaction is turned on at time t1 and turned off at time t> t2 . During this time energy is exchanged between the quantized fermion- electromagnetic field and the classical fields Sx( ,t) and Rx( ,t) . At some initial time ti< t1 the state vector is given by Ω(ti ) . We are interested in determining the state vector Ω(tf ) at some final time tf> t2. Based on the above remarks the state vector
Ω()t satisfies, 11
∂Ω()t iH=Ωˆ ()t for t ∂Ω()t Hgˆ −∇ ()t∫ χ(xA) ⋅ˆ (x)dx it=Ω() for t≤t≤t (2.6) ∂t ˆ 12 +⋅gt()∫ Jx( )∇χ(x)dx ∂Ω()t iH=Ωˆ ()t for t>t (2.7) ∂t 2 Since these equations are first order differential equations the boundary conditions at t1 and t2 are, Ω+()tt11ε=Ω−(ε) and Ω(t2+ε) = Ω(t2−ε) (2.8) ε→0 ε→0 The solution to (2.5) is, ˆ −−iH(t ti ) Ω=()te Ω()ti for t It is shown in Appendix A that the solution to (2.6) is, ˆ ˆˆ ig()t C ig ()t D iw()t −−iH(t t1) Ω=()te e ee Ω()t12 for t≥t≥t1 (2.10) where the operator Dˆ is defined by, Dˆ =⋅∫ Axˆ ()∇χ()xdx (2.11) and t 2 gt ()′ 22 wt()=∇∫∫χdx+ g()t′ g()t′∫∇χdxdt′ (2.12) 2 t1 The solution to (2.7) is, ˆ −−iH(tf2t ) Ω=()tefΩ()t2 where tf>t2 (2.13) Use the boundary conditions (2.8) in the above to obtain, 12 ˆˆˆ ˆ −−iH(tf2t )ig()t2C ig ()t2D iw()t2−iH(t2−ti) Ω=()tefiee ee Ω(t) (2.14) Use (2.4) in the above to obtain, ˆ ˆ −−iH(tf2t ) −iC iw(t2) Ω=()tefee Ω0(t2) (2.15) where Ω02(t ) is defined by, ˆ −−iH(t2it ) Ω=02()te Ω(ti) (2.16) Ω02(t ) is the state vector that the initial state Ω(ti ) would evolve into, by the time t2 , in the absence of the interactions. Use (2.14) in (1.12) to show that the energy of the state Ω(tf ) is, iCˆˆˆ −iC Et()Ω=()f0Ω()t2eHeΩ0(t2) (2.17) From the discussion leading up to equation (1.29) we obtain, ˆ Et()Ω=()f0E()Ω()t2−∫ χ()xJ∇⋅Ω0(t2) ()xΩ0(t2)dx (2.18) Now, as before, assume that we select an initial state Ω(ti ) so that ˆ ∇⋅ Ω02()tJx()Ω02(t) is non-zero. Recall that Ω02(t ) is the state that Ω(ti ) evolves into in the absence of interactions. Therefore EtE(Ω=02( )) (Ω(ti)) and Ω02(t ) is independent of χ(x). The function χ(x) can take on any value without ˆ ˆ affecting ∇⋅ Ω02()ttJx()Ω02( ). Let χ=(x) λ∇⋅Ω02(tt) J(x)(Ω02) so that (2.18) becomes, ˆ 2 Et()Ω=()fiE()Ω()t−λ∫ (∇⋅Ω0()t2Jx()Ω0(t2)) dx (2.19) 13 Define ∆Eext as the amount of energy extracted from the quantum state due to its interaction with the classical fields. From the above equation, ˆ 2 ∆=Etext λ∫ (∇⋅Ω0 ()2 Jx()Ω0 (t2 ))dx (2.20) Obviously as λ→∞ then ∆→Eext ∞. In conclusion, it has been shown that there exist quantum states with less energy than the vacuum state for QED in the temporal gauge. In fact there is no lower bound to the energy of quantum states. If an initial state interacts with properly applied classical fields then it is possible to extract an arbitrarily large amount of energy from the initial state. The classical fields that were applied in this article are mathematical objects that are not assumed to correspond to real physical objects. A possible next step in this research would be to determine if these same results can be obtained through the interaction of real existing fields. Appendix A It will be shown that (2.10) is the solution to (2.6). Take the time derivative of (2.6) and multiply by “i” to obtain, ∂ itΩ=() −g Cˆ +w Ω()t−Ω+Ω (A.1) ∂t () ab where, ˆ ˆ ˆ igC ˆ igD iw(t) −−iH(t t1) Ω=aeg Dee e Ω(t1) (A.2) and ˆ ˆ ˆ igC igD iw(t) ˆ −−iH(t t1) Ω=b eee He Ω(t1) (A.3) To evaluate (A.2) we will use the following relationships. 14 ˆˆˆˆ eDigC ˆˆ= (eDigC e−igC )eigC (A.4) Use (1.6) and (1.25) to obtain, igCˆˆˆˆ−igC ˆ ˆˆ 2 eDe =+DigC,D=−Dg∫ ∇χdx (A.5) Use these results in (A.2) to yield, ˆ 2 Ω=a gD( −g∫ ∇χdx) Ω()t (A.6) Next evaluate (A.3). Use (1.25) and the commutation relationships to obtain, 2 ˆˆ g eHigD ˆˆe−igD =+Hig Dˆ,Hˆ− Dˆ,Dˆ,Hˆ (A.7) 2 where, ˆˆ ˆˆˆ ˆ ˆ D,H =⋅∫∫Ax,H ∇χd =A,H0,M ⋅∇χdx=−iC (A.8) and, D,ˆˆD,Hˆ =−i D,ˆCˆˆ=−i Ax⋅∇χd , Eˆ⋅∇χdx=− ∇χ2 dx (A.9) ∫∫∫ Therefore, 2 ˆˆ g 2 eHigD ˆˆe−igD =+Hg Cˆ + ∫ ∇χdx (A.10) 2 Use this in (A.3) to obtain, 2 ˆ g ˆ ˆ Ω=eHigC ˆ +g Cˆ + ∇χ2 dx eigD eiw()t e−−iH(t t1) Ωt (A.11) b ∫ (1 ) 2 To evaluate this further use, igCˆˆˆˆ−igC ˆ ˆˆˆ eHe =+HigC,H=+Hg(∫ J⋅∇χdx) (A.12) where we have used, 15 ˆ ˆˆ C, H =−i∫ J⋅∇χdx (A.13) and the fact that C,ˆˆC, Hˆ = 0 . Therefore, g 2 Ω = Hgˆˆ+ Jx⋅∇χd + g Cˆ + ∇χ 2 dxΩ t (A.14) b ()∫∫() 2 Use this along with (A.11) and (A.6) in (A.1) to obtain, 2 g 2 Hgˆˆ+ ∫∫Jx⋅∇χd + g Cˆˆ+ ∇χ dx− gC+ w ∂ ()2 () itΩ=() Ω()t (A.15) ∂t 2 −− gDˆ g ∇χdx ( ∫ ) Rearrange terms and do some simple algebra to obtain, ()Hgˆˆ+⋅∫∫Jx∇χd− gAˆ ⋅∇χdx ∂ itΩ=() g 2 Ω()t (A.16) ∂t −−wd ∇χ2xx− gg∇χ2d ∫∫ 2 Now let 2 g 22 wd =∇∫χx+ gg∫∇χdx (A.17) 2 to obtain ∂ itΩ()= Hˆˆ+ g∫∫Jx⋅∇χd− gAˆ ⋅∇χdxΩ()t (A.18) ∂t () which is (2.6) in the text. This completes the proof. 16 References 1. G. Leibbrandt, Rev of Modern Phys, 59, 1067 (1987). 2. M. Creutz, Ann. Phys., 117, 471 (1979) 3. B. Hatfield. “Quantum Field Theory of Point Particles and Strings”, Addison-Wesley, Reading, Massachusetts (1992). 4. C. Kiefer and A. Wiepf , Annals Phys., 236, 241 (1994). 5. W. Greiner, B. Muller, and T. Rafelski, “Quantum Electrodynamics of Strong Fields”, Springer-Verlag, Berlin (1985). 6. S. Weinberg. The Quantum Theory of Fields. Vol. I. Cambridge University Press, Cambridge, 1995. 7. W. Greiner and J. Reinhardt. “Field Quantization”, Springer-Verlag, Berlin (1996).