On vacuum-vacuum amplitude and Bogoliubov coefficients

20 Pages·0.19 MB·English

by  A. I. Nikishov

hep-th/0207085
FIAN/TD/06/12
June 2002
On vacuum-vacuum amplitude and Bogoliubov
coefficients
3
0
0
2
A.I.Nikishov †
n
a
J I.E.Tamm Department of Theoretical Physics, Lebedev Physical Institute,
†
9 Leninsky Prospect 53, 119991, Moscow, Russia

e-mail: [email protected]
2
v
5
Abstract
8
0
We consider theproblemof fixingthephasesof Bogoliubov coefficients in Quantum
7
0 Electrodynamics so that the vacuum -vacuum amplitude can be expressed through
2
them. For the constant electric field and particles with spin 0 and 1/2 this is done
0
/ starting from the definition of these coefficients. Using the symmetry between electric
h
and magnetic fields the result is extended to the constant electromagnetic field. It
t
-
turns out that for the case of a constant magnetic field it is necessary to distinguish
p
e the in- and out-states although they differ only by a phase factor. For spin-1 particle
h
with gyromagnetic ratio g = 2 this approach fails and we reconsider the problem by
:
v
the proper-time method.
i
X
r
a 1 Introduction
Even if the electromagnetic field does not create pairs, virtual pairs lead to the appearance
of a phase in vacuum-vacuum amplitude. This makes it necessary to distinguish the in-
and out-solutions even when it is commonly assumed that there is only one complete set of
solutions as, for example, in the case of a constant magnetic field. Then in- andout-solutions
differ only by a phase factor which is in essence the Bogoliubov coefficient. The propagator
in terms of in- and out-states takes the same form as the one for pair creating fields.
We use the solutions with conserved quantum numbers and do not consider radiation
processes. Then the events in a cell with quantum numbers n are independent of events in
cellswithdifferent quantumnumbers. Inotherwordsweworkinthediagonalrepresentation.
The knowledge of Bogoliubov coefficients is sufficient for obtaining the probability of any
process in an external field (disregarding the radiation processes) [1-3]. But the real part of
action integral W, defining the vacuum-vacuum amplitude
< 0 0 >= eiW, W = d4x , (1)
out in
| L
Z
1
is not directly expressed through Bogoliubov coefficients. At the same time some effects
connected with ReW are observable. So the Lagrange function of a slowly varying field
L
defines the dielectric permittivity and magnetic permeability of the field [4-5].
The Lagrange function of the constant electromagnetic field in one loop approximation
was obtained in [6-8] and in two loop approximation in [9]. Studying a model of particle
production, B.DeWitt noted that ReW can be expressed through Bogoliubov coefficients
with the natural choice of their phases [10]. Our purpose is to choose these phases in such a
way that ReW can be expressed through them. We show that for the constant electric field
and particles with spins 0 and 1/2 the natural choice would be sufficient if it were not for
the necessity to make renormalizations. For vector boson with gyromagnetic ratio g = 2 the
situation is more complicated even for the case of a constant electric field.
We note here that transition amplitude for an electron to go from an in-state to out-state
turns out to be unity. To show this we write down the Bogoliubov transformations and the
relation between < 0 and < 0 [2] (n is the set of quantum numbers)
nout nin
| |
a = c a c∗ b+ ,
nout 1n nin − 2n nin
b+ = c a +c∗ b+ ,
nout 2n nin 1n nin
< 0 =< 0 (c∗ c a b ), c 2 + c 2 = 1. (1a)
nout| nin| 1n − 2n nin nin | 1n| | 2n|
Here a (b+ ) is the particle (antiparticle) annihilation (creation) operator, a 0 >=
nin nin nin| nin
0 and similarly for out-states. 0 > is the vacuum state in the cell with quantum number
nin
|
n. c , c are Bogoliubov coefficients and star means complex conjugation.
1n 2n
From the third relation in (1a) we have eq. (28) below and from the first one
a+ = c∗−1[a+ +c b ].
nin 1n nout 2n nin
Using this relation and anticommutator {an′out,a+nout} = δn′,n, we find [2]
< 0 a a+ 0 >= c∗−1 < 0 0 >= 1. (1b)
nout| nout nin| nin 1n nout| nin
The Pauli principle prohibits virtual pair creation in the state occupied by the electron. So
even the phase of scattering amplitude remains unchanged. In particular, for the constant
magnetic field c = 0 but we cannot assume c = 1 without violating eq. (1b) and eqs.
2n 1n
(28-29) below because W = 0 [4-5]. In other words, even when c = 0 the in- and out-
2n
6
vacuum are different. (This is in contrast with the remark after eq. (15) in [10].) So the
Bogoliubov coefficient c have to be coordinated with vacuum-vacuum amplitude. For the
1n
constant electromagnetic field we represent the action integral W = d4x (x) = W as
L n n
a sum over the set of quantum numbers n. Then W define (in general complex) phase of
n
R P
Bogoliubov coefficient.
In Sec. 2 and 3, starting from the definition of Bogoliubov coefficients, we consider the
phase fixing for particles with spin 0 and 1/2 correspondingly. In Sec. 4-6 we reconsider the
problem by a more general proper-time method for spins 0, 1/2 and 1.
2
2 Scalar particle in the constant electromagnetic field
For the case of set of wave functions with conserved quantum numbers n the Bogoliubov
transformation have the form
ψ = c +ψ +c −ψ ,
+ n 1n n 2n n
ψ = c∗ +ψ +c∗ −ψ ;
− n 2n n 1n n
c 2 c 2 = 1. (2)
1n 2n
| | −| |
Here ψ (+ψ ) is the positive-frequency in- (out-) solution and similarly for the negative-
+ n n
frequency states.
We are free to choose the phase of c by redefining ψ . Indeed, if we substitute
1n n
ψ = e±if ψnew, ±ψ = e∓if±ψnew, c = ei2fcnew,
± n ± n n n 1n 1n
then eq. (2) and the propagator [2, 11]
+ψ (x) ψ∗(x′), t > t′,
G (x,x′) = i c∗−1 n + n (2a)
0 1n ψ (x)−ψ∗(x′), t < t′
n (cid:26) − n n
X
will have the same form in terms of redefined quantities.
For definiteness we assume that the particle charge is e′ = e,e = e . For a constant
− | |
electric field we have [2] (n = (p ,p ,p ), A = δ Et)
1 2 3 µ µ3
−
√2π πκ π π m2 +p2 +p2
c = exp( +i ), c = exp( πκ i ), κ = 1 2. (3)
1n Γ(1 iκ) − 2 4 2n − − 2 2eE
2 −
We now note that in a weak electric field, c is exponentially small and may be neglected.
2n
| |
Then in- and out-states differ only by a phase factor. The same should be true for the
magnetic field where c = 0 exactly and lnc∗ should be determined.
2n 1n
The probability amplitude that the vacuum in the state n remains vacuum is [2]
< 0 0 >= c∗ −1. (4)
nout| nin 1n
The total vacuum-vacuum amplitude is
< 0 0 >= c∗ −1 = eiW0, W = W , W = ilnc∗ (5)
out| in 1n 0 0n 0n 1n
n n
Y X
Only ReW contain divergences and as we shall see below, c∗ in (4) and (5) should be
0 1n
replaced by C∗ren. This is the renormalization of c∗ . From (3) we have
1n 1n
1 πκ iπ 1
lnc∗ = ln2π lnΓ( +iκ). (6)
1n 2 − 2 − 4 − 2
As shown in [2], the vacuum-vacuum probability < 0 0 > 2 obtained from (5) and (3)
out in
| | |
agrees with Schwinger result [8]. This means that ImW is given correctly by (5) and (3).
0
To find ReW , we first consider the asymptotic representation, see eq. (1.3.12) in [12]
0
1 1 B (1)
lnΓ( +iκ) = iκ[ln(iκ) 1]+ ln2π + 2k 2 (iκ)1−2k. (7)
2 − 2 2k(2k 1)
k=1 −
X
3
(Letting k run to , we may say that the r.h.s. of (7) in a certain sense exactly represent
∞
the l.h.s.; the information encoded in the r.h.s. can be decoded [13].) From (6) and (7) it
follows
π ( 1)kB (1)
lnc∗ = i κ(lnκ 1)+ + − 2k 2 . (8)
1n − { − 4 2k(2k 1)κ2k−1}
k=1 −
X
This asymptotic expansion contains only the imaginary part of lnc∗ or only real part of
1n
W . It is seen from (8) that as a first step we have to go from lnc∗ to
0n 1n
π
lnC∗ = lnc∗ +i κ(lnκ 1)+ (9)
1n 1n { − 4}
in order to have lnC∗ 0 for κ (i.e. for E 0). Due to the necessity of charge
1n → → ∞ →
renormalization we have to make the second step and introduce
π 1
lnC∗ren = lnc∗ +i κ(lnκ 1)+ + . (10)
1n 1n { − 4 24κ}
In other words we include in lnC∗ren also the term with k = 1 in (8). Then we have the
1n
following asymptotic representation
( 1)kB (1)
lnC∗ren = i − 2k 2 . (11)
1n − 2k(2k 1)κ2k−1
k=2 −
X
Summing (11) over n according to the rule
d3pL3
, dp eET, (12)
→ (2π)3 3 →
k Z Z
X
and making renormalization [8], we obtain the correct asymptotic representation for Re
0
L
1 (eE)2 ( 1)kB (1) m2
Re = E2 + − 2k 2 , κ = . (13)
L0 2 16π2 k(k 1)(2k 1)κ2k−2 0 2eE
k=2 − − 0
X
To simplify formulas andminimize confusion with T ineq.(50) we oftenputL = T = 1inthe
expressions like (12). Besides, in the following we drop the Maxwell part of the Lagrangian
(1E2 in this case).
2
Now we show that the expression (9) can be brought to the formsuggested by the proper-
time formalism:
1
lnC∗ ln√2π+η(lnη 1) lnΓ( +η) = F(η),
1n ≡ − − 2 −
∞
1 dθ 1 1
F(η) = e−2ηθ[ ]. η = iκ. (14)
2 θ sinhθ − θ
Z
0
Differentiating (14) with respect to η and using eq. (2.4.225) in [16], we see that the results
on the left- and the right-hand sides coincide. Besides, both sides have the same asymptotic
behavior for η . So we have
→ ∞
1 ∞ ds sinhθ
lnC∗ = e−is(m2+p2⊥)[1 ],θ = eEs,p = p2 +p2. (15)
1n −2 ssinhθ − θ ⊥ 1 2
Z0
4
Next, we note that the term i in (10) can be written as
24κ
i 1 ∞
= dθe−i2κθ. (16)
24κ −12
Z0
Hence
1 ∞ ds 1 θ
lnC∗ren = e−is(m2+p2⊥)R(θ), R(θ) = 1 sinhθ . (17)
1n −2 ssinhθ − θ − 6
Z0 (cid:18) (cid:19)
Here R(θ) is a ”regularizer.” It is independent of quantum numbers n and is the same as in
the proper-time representation of the Lagrange function [8].
Now we consider the case when there is a constant magnetic field collinear with the
constant electric field. Then
1 ∞ ds
lnC∗ren(E,H) = e−is(m2+eH(2l+1))R(θ,τ),τ = eHs,l = 0,1, (18)
1n −2 ssinhθ ···
Z0
and we assume that R(θ,τ) may be obtained by the same reasoning as in [8] (or simply
taken from [8])
1 1H2 E2
R(θ,τ) = 1 sinhθsinτ + − , τ = eHs, θ = eEs. (19)
− θτ 6 EH
(cid:18) (cid:19)
Integrating over p , we get (see (12) with T = 1)
3
1 ∞ ds
dp lnC∗ren(E,H) = eE e−is(m2+eH(2l+1))R(θ,τ). (20)
3 1n −2 ssinhθ
Z Z0
In this expression we can turn off the electric field
1 ∞ ds
dp lnC∗ren(E = 0,H) = e−is(m2+eH(2l+1))R(0,τ),
3 1n −2 s2
Z Z0
1 τ
R(0,τ) = 1 sinτ + . (21)
− τ 6
(cid:18) (cid:19)
To remove the integration over p , we write the factor s−2 as s−3/2s−1/2 and note that 1/√s
3
should be due the integration over p :
3
1 eiπ/4 ∞
= dp e−isp23. (22)
3
√s √π
Z0
So
eiπ/4 ∞ ds
lnC∗ren(E = 0,H) = e−is(m2+eH(2l+1)+p23)R(0,τ). (23)
1n −2√π s3/2
Z0
(Substituting s it we see that expression (23) is purely imaginary.) From here, or from
→ −
(21) we obtain in agreement with [8, 9]
∞
dp dp
i lnC∗ren(E = 0,H) = i 2 3 lnC∗ren(E = 0,H) = =
1n 2π 2π 1n L0
n Z Z l=0
X X
5
eH ∞ ds
e−ism2R(0,τ), (L = T = 1) (24)
−16π2 s2sinτ
Z0
Relation (39) was used here and the sum over l was performed with the help of formula
∞
1
e−iseH(2l+1) = . (25)
2isin(eHs)
l=0
X
3 Electron in the constant electromagnetic field
The Bogoliubov transformation has the form
ψ = c +ψ +c −ψ ,
+ n 1n n 2n n
ψ = c∗ +ψ +c∗ −ψ ;
− n − 2n n 1n n
c 2 + c 2 = 1. (26)
1n 2n
| | | |
For the constant electric field we have
c∗ = i 2π e−π2κ , c = e−πκ, n = (p ,p ,p ,r). (27)
1n − κ Γ(iκ) 2n 1 2 3
r
These Bogoliubov coefficients are independent of the spin state index r = 1,2.
As in the scalar case, we start with [2]
< 0 0 >= c∗ , (28)
nout| nin 1n
and
< 0 0 >= c∗ = eiW1/2, W = W , W = ilnc∗ (29)
out| in 1n 1/2 1/2;n 1/2;n − 1n
n n
Y X
From (27) we have
iπ 1 2π πκ
lnc∗ = + ln lnΓ(iκ). (30)
1n − 2 2 κ − 2 −
The asymptotic expansion for Γ(iκ) is, see eq.(8.344) in [14] or eq. (6.1.40.) in [15]
1 1 B
lnΓ(iκ) = (iκ )ln(iκ) iκ + ln2π +i ( 1)k 2k (κ)1−2k. (31)
− 2 − 2 − 2k(2k 1)
k=1 −
X
From (30) and (31) we obtain
π B
lnC∗ lnc∗ +i(κlnκ κ + ) = i ( 1)k 2k (κ)1−2k, (32)
1n ≡ 1n − 4 − − 2k(2k 1)
k=1 −
X
i B
lnC∗ren lnC∗ = i ( 1)k 2k (κ)1−2k. (33)
1n ≡ 1n − 12κ − − 2k(2k 1)
k=2 −
X
As in the scalar case we find
1 ∞ dx 1
lnC∗ = e−i2κx(cothx ), (34)
1n −2 x − x
Z0
6
1 ∞ dx 1 x
lnC∗ren = e−i2κxcothx[1 tanhx( + )], (35)
1n −2 x − x 3
Z0
Eq. (2.4.22.6) in [16] was used to verify (34), cf. the text before eq.(15).
The generalization of (35)for the presence of a constant magnetic field is straightforward.
We rewrite it in the form (x = θ = eEs)
1 ∞ dθ
lnC∗ren(E,H) = e−is(m2+eH2l)cothθR(θ,τ),
1n −2 θ
Z0
n = (p ,p ,p ,r); l = l ,l +1, , l = 0 for r = 1, l = 1 for r = 2. (36)
1 2 3 min min min min
···
R(θ,τ) may be taken from the Lagrange function [8, 9] (τ = eHs)
1 E2 H2
R(θ,τ) = 1 tanτ tanhθ( + − ). (37)
− θτ 3EH
Integrating over p with the help the second equation in (12) we find
3
dp eE ∞ dθ
3 lnC∗ren = e−is(m2+eH2l)cothθR(θ,τ). (38)
2π 1n −4π θ
Z Z0
The subsequent integration over p is performed according to formula similar to (12) [2]
2
dp = eHL. (39)
2
Z
To sum up over r and l in (36), we use the formula obtainable from (25)
2 ∞
e−is2eHl = icot(eHs). (40)
−
Xr=1 Xlmin
So in agreement with the Lagrange function for the constant electromagnetic field [8, 9] we
have
e2EH ∞ dθ
lnC∗ren = i e−ism2cothθcotτR(θ,τ), (L = T = 1). (41)
1n 8π2 θ
n Z0
X
Now returning to (38), we can switch off the electric field
dp 1 ∞ ds
3 lnC∗ren = e−is(m2+eH2l)R(0,τ),
2π 1n −4π s2
Z Z0
1 τ
R(0,τ) = 1 tanτ( ); (42)
− τ − 3
l are given in (36). Using (22) we obtain as in the scalar case
eiπ/4 ∞ ds
lnC∗ren(E = 0,H) = e−is(m2+p23+eH2l)R(0,τ),
1n −2√π s3/2
Z0
n = (p ,p ,p ,r); l = 0,1,2, for r = 1; l = 1,2, for r = 2. (43)
1 2 3
··· ···
Inthe subsequent Sections we give the heuristic derivationof lnC∗ren not resorting toc∗ ,
1n 1n
but using the proper-time method The main problem here is due to the necessity to make
renormalizations. We know how to renormalize as a whole, but we have to renormalize
L
a contribution to it from a particular state n. To do this we assume as before that the
regularizer does not depend on n.
7
4 Scalar particle
We take the vector-potential of a constant electromagnetic field in the form
A = δ Hx δ Et, (44)
µ µ2 1 µ3
−
but start with the particle in a constant magnetic field, E = 0 in (44). The propagator with
coinciding x and x′ has the form (see for example [11])
eH ∞ ∞ dp ∞ dp ∞ dp0
2 3
G (x,x E = 0,H) = i
0
| π 2π 2π 2π ×
r l=0 Z−∞ Z−∞ Z−∞
X
∞ D2(ζ) p
ds l exp is[m2 +eH(2l+1)+p2 p2] , ζ = √2eH(x + 2 ). (45)
l! {− 3 − 0 } 1 eH
Z0
According to (1) we have to integrate and hence G (x,x) over d4x. The integration over
0 0
L
x is done with the help of formula
1
∞ ∞
π 1/2
dζD2(ζ) = √2πl!, or dx D2(ζ) = l!. (46)
l 1 l eH
−Z∞ −Z∞ (cid:16) (cid:17)
Integrating over p0 and x , we obtain
1
∞ exp[i3π] ∞ dp ∞ dp ∞ ∞ ds
dx G (x,x) = 4 2 3 e−is[m2+eH(2l+1)+p23]. (47)
1 0
2√π 2π 2π √s
Z−∞ Z−∞ Z−∞ l=0 Z0
X
As noted in [3] (see eq. (2.12) there), it follows from Schwinger results [8] that for scalar
particle (boson)
∂W ∞
i b = d4xG (x,x), or W = i dm˜2 d4xG (x,x m˜2). (48)
− ∂m2 b b − b |
Z Zm2 Z
This means that can be obtained from (47) by inserting 1/s in the integrand. Inserting
0
L −
also the regularizer from (21), we get
exp[iπ] ∞ dp ∞ dp ∞ ∞ ds
iW (E = 0,H) = i (E = 0,H) = 4 2 3
0 L0 2√π 2π 2π s3/2×
Z−∞ Z−∞ l=0 Z0
X
∞ dp ∞ dp ∞
e−is[m2+eH(2l+1)+p23]R(0,τ) = 2 3 lnC∗ren. (L = T = 1) (49)
− 2π 2π 1n
Z−∞ Z−∞ l=0
X
Now for the constant electromagnetic field, described by the vector-potential (44), we
substitute in (2a) the expressions for the wave functions (see [2] with the modifications for
e′ = e = e ) and use relation (93) in [11] (or relation similar to (96) below). Than we
− −| |
find
∞ ∞ ∞
ei3π/4 dp dp ∞ eH 1/2 D2 dθ
G (x,x E,H) = 2 3 l √2
0
| 2√πeE 2π 2π π l! √sinh2θ×
Z Z l=0 (cid:18) (cid:19) Z
−∞ −∞ X 0
8
T2 p
exp[ i2κθ i ], θ = eEs, T = √2eE(t 3 ). (50)
− − 2cothθ − eE
Integrating over x (see (46)) and t, we get
1
∞ ∞
dx dtG (x,x E,H) =
1 0
|
Z Z
−∞ −∞
∞
i ∞ dp ∞ dp ∞ ds
2 3 exp[ is(m2 +eH(2l+1))]. (51)
2 2π 2π sinh(eEs) −
Z−∞ Z−∞ l=0 Z
X 0
Passing from G (x,x) to is realized by inserting the factor ( 1/s) in the integrand in
0 0
L −
(51). Inserting also the regularizer R(τ,θ), see eq. (19), we obtain
∞ ∞
∞
dp dp
W (E,H) = i 2 3 lnC∗ren
0 2π 2π 1n
Z Z l=0
−∞ −∞ X
∞ ∞ ∞
∞
i dp dp ds
= 2 3 exp is[m2 +eH(2l+1)] R(τ,θ). (52)
−2 2π 2π ssinhθ {− }
Z Z l=0 Z
−∞ −∞ X 0
5 Spinor particle
First we consider the electron in the constant magnetic field, E = 0 in (44). The squared
Dirac equation can be brought to the form
d2 ζ2 p2 p2 1 σ 0
+ 0 − 3 Σ Z = 0, Σ = 3 . (53)
{dζ2 − 4 2eH − 2 3} 3 0 σ
3
(cid:18) (cid:19)
Here ζ is the same as in (45). We see that Z can be written as follows
Z = diag(f ,f ,f ,f )ei(p2x2+p3x3−p0t) (54)
1 2 1 2
and f , f have to satisfy the equation
1 2
d2 ζ2 p2 p2 1
+ 0 − 3 f = 0. (55)
{dζ2 − 4 2eH ∓ 2} 1,2
We choose f = D (ζ), f = D (ζ) in order that p2 = 2eHl in both cases. The solutions
1 l−1 2 l ⊥
of the Dirac equation are obtained as the columns of the matrix [2]
(m iΠˆ)Z, Πˆ = γ Π , Π = i∂ +eA . (56)
µ µ µ µ µ
− −
Using the γ matrices in the standard representation [4] we have
−
m+Π0 0 Π Π +iΠ
3 1 2
− −
0 m+Π0 Π iΠ Π
(m iΠˆ) =  − 1 − 2 3 . (57)
− Π Π iΠ m Π0 0
3 1 2
− −
 Π +iΠ Π 0 m Π0 
1 2 3
 − − 
 
9
In terms of ζ we get
d ζ d ζ
Π +iΠ = i√2eH , Π iΠ = i√2eH + . (58)
1 2 1 2
− dζ − 2 − − dζ 2
(cid:18) (cid:19) (cid:18) (cid:19)
Using also the relations
d ζ d ζ
+ D (ζ) = lD (ζ), D (ζ) = D (ζ), (59)
l l−1 l l+1
dζ 2 dζ − 2 −
(cid:18) (cid:19) (cid:18) (cid:19)
we find (the exponential factor in (54) is dropped here for brevity)
ˆ
(m iΠ)Z =
−
(m+p0)D (ζ) 0 p D (ζ) il√2eHD (ζ)
l−1 3 l−1 l−1
−
0 (m+p0)D (ζ) i√2eHD (ζ) p D (ζ)
 l − l 3 l . (60)
p D (ζ) il√2eHD (ζ) (m p0)D (ζ) 0
3 l−1 l−1 l−1
− −
 i√2eHD (ζ) p D (ζ) 0 (m p0)D (ζ) 
 l − 3 l − l 
 
Choosing the second and the first columns as ψ and ψ (subscripts 1 and 2 indicate spin
1 2
states) and normalizing them, we get
0
(m+p0)D (ζ) eH 1/4 1
ψ = N  l eiq·x, N = ,
+ 1 n il√2eHD (ζ) n π 2p0(p0 +m)l!
l−1 s
− (cid:18) (cid:19)
 p D (ζ) 
 − 3 l 
 
p0 = m2 +2eHl+p2, q x = p x +p x p0t,
3 · 2 2 3 3 −
q p
n = (p ,p ,l,r), ζ = √2eH(x + 2 ), (61)
2 3 1
eH
(m+p0)D (ζ)
l−1
0
ψ = N √l eiq·x, l = 0,1,2, (62)
+ 2 n p3Dl−1(ζ) ···
 i√2eHD (ζ) 
 l−1 
 
As seen from(62) in this state l begins actually fromunity. The negative-frequency solutions
ψ are obtained from (61-62) by substitution q q. We note here that eqs. (61-62) differ
− n
→ −
from eq. (10.5.9) in [4] because there the authors assumed the charge of a spinor particle to
be positive.
Having obtained the wave functions, we are going to find the contribution to from
1/2
L
each state ψ . For the field which does not create pairs, the propagator has the usual form
n
ψ (x) ψ¯ (x′), t > t′,
G (x,x′) = i + n + n ψ¯ = ψ∗β. (63)
1/2 ψ (x) ψ¯ (x′), t < t, n n
− n − n
n (cid:26) −
X
In the standard representation
I 0 1 0
β = , I = . (64)
0 I 0 1
(cid:18) − (cid:19) (cid:18) (cid:19)
10