Date post: | 28-Dec-2015 |
Category: |
Documents |
Upload: | julianna-butler |
View: | 214 times |
Download: | 0 times |
Vacuum energy in quantum field theory status, problems and recent
advances
M. Bordag
(Leipzig University)
a aP
s ddgggggg ggga ba f ra s
• some historical remarks• what can be calculated and what cannot• new representation of Casimir force
some historical remarks
1911 Planck introduced his 2nd quantization hypothesisas a consequence the energy of the quantum
oscillator became a instead of zero at T=0Planck called it remaining energy; afterwards it
was renamed simply zero-point energy
1912-1928 consequences of zero-point energy were found by:Nernst and Lindemann (1911) on specific heat
dataLindemann and Aston (1919) on no isotope
separation of Neon Hartree and collaborators (1928) on X-ray
diffractionpatterns on rock salt
hº
what we should look on:in terms of quantum mechanics,
En = ¹h!µn +
12
¶
the zero-point energy is
E0 =¹h2!
it is the same for all energy levels, hence it can be measuredonly through its dependence on external parameterslike the mass of isotopes
zero-point energy must be associated to any oscillatory de-gree of freedom (but not, for example to the rigid rotator),hence the formula reads in fact
E0 =¹h2
X
J! J
T he concept of zero-point energy was not generally ac-cepted:Pauli in 1933 critized the general validity of the concept ofzero point energy. He argued, whence attributed to each de-gree of freedom, it must be large and because of its gravita-tional ¯ eld 'the radius of the world would not extend beyondthe moon'.
Statusof zero-point energye®ectsby1943wasdiscussed in areviewpaper byK. Clusius, especially hediscussed theknownexperimental con rmations
(for more details see Rechenberg's contribution in the QFEXT98proceedings)
Thework by Casimir
'preliminary': in 1948, Casimir and Polder published theirpaper on The in°uence of retardation on the London-vander Waals forcesfor the interaction of two neutral atoms they found
E = ¡23
4¼
¹h®1®2
R 7
for oneatom in front of a conducting plane even simpler,
E = ¡3
4¼
¹h®
R 4
Casimir considered2metallicplatesand thezero-point energyof theelectromagnetic ¯eld con ned in between
E 0 =¹hc
2
Zd2kjj(2¼)2
1X
n=0
s
k21 + k22 +µ¼n
L
¶ 2
hemanaged toget ridof thedivergencesandobtaineda¯niteanswer for the force
F = ¡@
@LE 0 = ¡
¼3
480
¹hc
L 4
(scalar case, forceper unit area)
alternativeapproach using Lifshitz formula
E (a) =¹h
4¼2
Z 1
0k? dk?
Z 1
0d»
nln£1 ¡ r 2TM (i»; k? )e
¡ 2qa¤
+ ln£1 ¡ r 2T E (i»; k? )e
¡ 2qa¤o
re°ection coe±cients
rTM (i»; k? ) ="(i»)q(i»; k? ) ¡ k(i»; k? )
"(i»)q(i»; k? ) + k(i»; k? );
rT E (i»; k? ) =q(i»; k? ) ¡ k(i»; k? )
q(i»; k? ) + k(i»; k? );
with notations
q2 = q2(i»; k? ) = k2? +»2
c2; k2 = k2(i»; k? ) = k2? + "(i»)
»2
c2:
sometimes considered as theonly approach,however onemay consider the vacuumenergy of a quantum¯eld ' (x) in thebackground of someclassical ¯eld Á(x),
E = E class+ E 0;
with theclassical energy
E class =1
2
Zdr Á(r )
h¡ r 2+ M 2+ ¸Á2(r )
iÁ(r )
and thevacuumenergy
E 0(s) =¹ 2s
2
X
J
! 1¡ 2sJ
for example
S = ¡1
2
Zd4x Á(x)
³¤ + M 2+ ¸Á2(x)
´Á(x)
¡1
2
Zd4x ' (x)
³¤ + m2+ ~Á2(x)
´' (x):
Statuswhat can be calculated depends on the handling of theultraviolet divergencesin general, weneed to introducea regularization,e.g. zetafunctional regularization, s > 3
2, s ! 0
E vac(s) =¹ 2s
2
X
n
! 1¡ 2sn
or frequency damping function, ±> 0, ±! 0
E vac(±) =1
2
X
n
! n e¡ ±! n
and express thedivergent part of thevacuumenergy in termsof theheat-kernel coe±cientsin zetafunctional regularization
E divvac(s) = ¡
a232¼2
µ1
s+ ln ¹ 2
¶
or with frequency damping function
E divvac(±) =
3a02¼2
1
±4+
a1=24¼3=2
1
±3+
a18¼2
1
±2+
a216¼2
ln ±
example: hkks for theconducting sphere
a0 =4¼
3R 3; a1
2=½¡ 1
1
¾2¼3=2R 2; a1 = ¨
½2
14
¾2¼
3R ;
a32=½23
7
¾¼3=2
6; a2 = ¨
½1
7
¾16¼
315R:
In general, one needs 5 structure in the classical energy toaccomodate thecounterterms
E class = pV + ¾S + h1R + h2+ h31
Rthis is not always possible, for exampledielectric ball: insidepermittivity ² 6= 1, outside ² = 1here we have the known matching conditions across thesurface: ²E n; E t; B n; 1
¹ B t
resulting in a hkk
a2 = ¡2656¼
5005R
(c1 ¡ c2)3
c22+ O
h(c1 ¡ c2)
4i
similar problem in bagmodel
now consider the limitof makingthebackground singulular
~Á(x)2 = V (r ) ! ~±(r ¡ R )
what happens to theheat kernel coe±cients?
a1 =Rdx V (r ) ! a1 = 4¼®R
a3=2 = ¼3=2®2
a2 =Rdx V (r )2 6! a2 = 2¼
3®3
R
theclassical background becomes singular, too
in someliterature, theinsertionof aboundary intoaquantum¯eld was considered a 'unnatural act'
There is an example for handling such situation(M.B., N. Khusnutdinov, PRD 77 (2008) 085026)
classical background: spherical plasma shell, radiusRthis is the hydrodynamic model used by Barton to describethe¼-electrons of graphenequantum¯eld: elm°uctuationsinteraction viamatching conditions ( - plasma frequency)
limr ! R+0
f l;m(kr ) ¡ limr ! R ¡ 0
f l ;m(kr ) = 0;
limr ! R+0
(r f l;m(k; r ))0¡ lim
r ! R ¡ 0(r f l ;m(k; r ))
0 = R f l ;m(kR );
limr ! R+0
(rgl;m(k; r ))0¡ lim
r ! R ¡ 0(rgl ;m(k; r ))
0 = 0;
limr ! R+0
gl ;m(kr ) ¡ limr ! R ¡ 0
gl ;m(kr ) = ¡
k2R(Rgl ;m(k; R ))
0
theseareequivalent of a±resp. ±0 function on thesurface
Weallow for radial vibrations (breathingmode) of theplasmashell. In C60 thesearedetermined by theelastic forces actingbetween thecarbon atoms.
wedescribethesevibrationsphenomenologicallybyaHamiltonfunction
H class =p2
2m+m
2! 2b (R ¡ R 0)
2+ E rest
with a momentum p = m _R . Herem is the mass of theshell, ! b is the frequency of the breathing mode, R 0 is theradius at rest and E rest is the energy which is required tobring the pieces of the shell apart, i.e., it is some kind ofionization energy.
thecompleteenergy is
E tot = E class(R ) + E vac(R )
and we can perform the renormalization by rede ning theparameters entering (! b, R 0, E rest) provided all heat kernelcoe±cientshavethe 'right' structure, i.e., dependenceon theradius
indeed, they dok l = 0; 1; : : : l = 1; 2; : : :0 0 01=2 0 01 ¡ 4¼ R 2 ¡ 4¼ R 2
3=2 ¼3=2 2R 2 ¼3=2 2R 2
2 ¡ 23¼
3R 2 ¡ 23¼
3R 2+ 4¼
k l = 0; 1; : : : l = 1; 2; : : :0 0 01=2 8¼3=2R 2 8¼3=2R 2
1 ¡ 4¼3 R
2 ¡ 4¼3 R
2
3=2 143¼
3=2 ¡ 103¼
3=2
2 ¡ 8¼ + 2¼15
3R 2 ¡ 4¼ + 2¼15
3R 2
with thesecoe±cients therenormalization canbecarried out,especially contributions growingwith can be removed by a¯nite renormalizationand a ¯nitevacuumenergy can becalculated: E ren
vac =E( R )R
it has for ! 1 theexpected limit E(1 ) = 0:0461766
New representation for the Casimir force acting betweenseparated bodies
In general, this force, F = ¡ (@=@L )E 0, is always ¯nitesince it is ameasurablequantity
in representations like E 0 = ¹h2
P
J! J we need to calculate
the eigenvalues and to subtract out several asymptoticcontributions,after that a numerical approach is quite hopeless. In fact,there is no successful attempt in the literature to calculatethe Casimir force for a complicated geometry numerically inthis way.
A new approach was found only quite recently. As shown byBulgac et al. (2006) and Emig et al. (2006), it is possible torewrite a representation of the vacuumenergy in terms of afunctional determinant whichdoesnot contain anyultravioletdivergences and which is ¯nite in all intermediatesteps.
before that ¯rst weneed aspeci¯c representation of the propagator with boundaryconditions(M.B., D.Robaschik, E.Wieczorek, 1985)
functional integral representation of generating functional
Z [ ] =Z
D 'Y
x2S
±(' (x)) eiS[' ]:
Y
x2S
±(' (x)) = CZ
Db eiRS d¹ (´) b(´)' (u(´));
rewrite theexponentialZ
S
d¹ (´) b(´)' (u(´)) =Z
S
d¹ (´)Z
d4x b(´)H (´; x)' (x)
with a newly de ned kernel H (´; x) = ±4(x ¡ u(´))Gaussian path integral, diagonalization
Z [0] = C (det K )¡ 1=2³det ~K
´ ¡ 1=2
and thevacuumenergy (distancedependent part) is
E 0 =1
2¼
Z 1
0d»Tr ln ~K »:
~K (´; ´0) =Z
d4xZ
d4x0H (´; x)G(x; x0)H (´0; x0)
= G(u(´); u(´0)):
is a new kernel for theb-¯eldIt is the projection of the free space propagator onto theboundary surfaceS
nowweassumethesurfaceS toconsist of twononintersectingparts, SA and SB , with
S = SA [ SB ; SA \ SB = 0:
thekernel takes a block structure
~K (b)» =
Ã~K »;AA (´ A ; ´
0A ) ~K »;AB (´ A ; ´
0B )
~K »;B A (´ B ; ´0A ) ~K »;B B (´ B ; ´
0B )
!
thenext essential step is a factorization of theparts resultingfromthe individual surfaces
~K (b)» =
Ã~K »;AA 0
0 1
! Ã1 0
0 ~K »;B B
!
£
Ã1 ~K ¡ 1
»;AA~K »;AB
~K ¡ 1»;B B
~K »;B A 1
!
using
det
Ã1 B
C 1
!
= det(1 ¡ B C ) = det(1 ¡ CB ):
and dropping thedistance independent parts wecometo
E =1
2¼
Z 1
0d»Tr ln
³1 ¡ ~K ¡ 1
»;AA~K »;AB ~K ¡ 1
»;B B~K »;B A
´:
which is thedesired representation
Note: all integrations and summations entering do converge
Example: Application to cylindrical geometry
RA RBx
A
B0b b
y
calculateall quantities in thecorresponding basis
jl >=ei l 'p2¼
;
³~K ° ;AB
´
l l0= (¡ 1) l
0I l(°R A )K l¡ l0(° L )I l0(°R B ):
and ³~K ° ;AA
´
l l0= ±l l0I l(°R A )K l(°R A ):
¯nally onecomes to theenergy in the form
E =1
4¼
Z 1
0d° ° Tr ln (1 ¡ M ° )
with thematrix elements
M ° ;ll0=1
K l(°R )K l+l0(° L ) I l0(°R )
(cylinder in front of a plane)
Limiting cases1. large separationsDirichlet boundary conditions
E D =1
4¼L 2
1
ln RL
+ Oµln¡ 2
R
L
¶:
Neumann boundary conditions
E N = ¡5R 2
2¼L 4+ O
ÃR 4
L 4lnR
L
!
:
2. small separationsproximity forceapproximation and correction beyond(M.B. 2006)
E D = ¡¼3
1920L 2
rR
2L
"
1+7
36
L
R+ O
ÃL 2
R 2
! #
E N = ¡¼3
1920L 2
rR
2L
"
1+µ7
36¡
40
3¼2
¶L
R+ O
ÃL 2
R 2
! #
similar results were obtaind also for a sphere in front of aplane(M.B., V. Nikolaev 2008)
E sphereDirichlet = ¡
¼3
1440
R
L 2
½1+
1
3² + O
³²2´¾
E sphereNeumann = ¡
¼3
1440
R
L 2
½1+
µ1
3¡10
¼2
¶² + O
³²2´¾
(² = a=R )
Conclusions
² Basic problems areclari¯ed
² Someproblemswith renormalization still persist
² Computational tools for the calculation of Casimir forcesareavailable