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Physics LettersA 158 (1991) 295—299
North-Holland
PHYSICS LETTERS A
Violation of the Schiff theorem for unstable atomic states
E.V. Stambuichik and O.P. Sushkov
Institute of NuclearPhysics, 630090 Novosibirsk, USSR
Received 23 May 1991; revised manuscript received 28 June 1991; accepted for publication 3 July 1991
Communicated by V.M. Agranovich
We discuss the screening of the external static electric field on the nucleus of the neutral atom. It is shown that for the excited
atomicstates the screening is not complete.
Due to the well-known Schifftheorem [1] for the
neutral atom the external static homogeneous electric field is exactly screened on the nucleus by the
polarization of the electronic shells. The theorem is
valid for relativistic electrons [2,3]. Radiation corrections do not violate the theorem [21. One can
easily understand this theorem: the homogeneous
electric field does not accelerate the neutral atom.
Therefore the field acting on the nucleus is equal to
zero.
The physical arguments as well as formal proof of
the theorem are valid only for an atom in a stationary state. For the excited states which decay due to
photon emission the situation is not obvious. This
problem is connected with the radiation correction
to First
the energy
levels.
of all let
us demonstrate simple physical arguments in favor of the Schiff theorem violation for
unstable states
In the present work we will consider the hydrogen atom with an infinitely heavy nucleus to avoid the recoil. Let us consider the 2s
1,2
state which decays via Ml -transition to the 1 s1,2 state
(in the present work we are interested in one-quanturn transitions only). In the
external
electricdecays
field
2P1/2
state which
there
a mixing(fig.
with1).
theWe neglect the mixing with
via
Elistransition
the 2P3/2 state. The decay amplitude corresponding
~‘.
~
~
p
a
b
Fig. 1. Amplitude ofthe 2s
112-state decay in the external electric
field. The cross corresponds to the states mixing in the field.
c
z.
/ i
JI
—
\
S112
y
i
S112
+ <lsi,21h512p1,2
2Pi,~I—e~rI2s
><
1,2>
E2~112—E2~112+ ~ir~
1
Here .~‘is the external
2P1/2 electric
state, field, f’~is the radiation width
of
the
r—
h~=V2it/wec~ee”
(2)
is the operator of the radiation of the photon with
momentum k and polarization e, ~ is the Dirac matrix. The use of relativistic notation is technically
convenient. A simple linear calculation in the ! approximation gives the angular distribution of the
photons
averaged
overthe polarizations ofthe atomic
states
(see
ref. [41),
dW(k)_F
2s(1+2k)T~,
(3)
to fig. 1 equals
where
~‘
We are grateful to V.V. Flambaum in discussions with whom
these arguments were formulated.
Elsevier Science Publishers B.V.
DT’P
,~.
—
w Ml (E~ E~
) 2 + 1f2
—
295
Volume 158, number 6,7
PHYSICS LETTERS A
Here El and Ml are the amplitudes of the y-transitions 2p1,2—+lsl,2 and 2s1,2—*ls112. D= <2P1/2,
ez 2s
1,2,3 I~>
Mlis12the
is the
amplitude
one-photon
of thewidth
2s—2pofmixthe
ing,
2s , F2~=~w
2-state. Let us stress that correlation of flight direction with the electric field k•~in (3) is T-odd.
Just therefore )~is proportional to T’~.With the angular distribution (3) the photon takes away the average momentum directed along the electric field. The
recoil force is equivalent to the unscreened electric
field at the nucleus,
9 September 1991
We have used the representation of Hd in the cornmutator form (6).
—
(5)Now
mean
we the
can shift
ask the
of the
question:
2s energy
Doeslevel
the which
correction
can
be observed in the resonant scattering of light on the
ls state of hydrogen? In the leading order in h~the
amplitude of the resonant scattering is shown in fig.
2, and due to eq. (8) it equals
f
(1 +ik1 d/e)(l —ik2 ~d/e)
><
(4)
3e
If the nucleus has an electric dipole moment d then
one may suppose that a correction to the energy level
should arise,
~
,
(9)
w+E15—E2~+iO
where k1 and k2 are the momenta of the initial and
final photons. The amplitude (9) depends on d, but
this dependence is not
with
any
shift of
2 isconnected
independent
ofd.
However,
energy.
Moreover
If]
it is obvious beforehand that in the leading order in
SE
2, = ~‘N
(5)
Now we would like to understand what the energy
shift (5) means and how it can be observed. Let us
emphasize once more that SE~-~
i.e. SE arises
just due to the instability of the levels. At infinite
mass ofthe nucleus the only probe ofan electric field
at the origin can be the electric dipole moment of the
nucleus. The interaction ofthis dipole moment with
the external field and with the electron is equal to
—
~.
r’pj’2~,
3—d.f= !d.[p,HJ,
h~the shift SE (5) cannot arise, since SE-.J’2,F2~.
One should consider at least the Lamb shift (fig. 3a),
and even the second order in the Lamb shift (figs.
3b—3d). Nevertheless one can easily verify that in
any order in the radiation correction there is no dependence on d in the scattering amplitude except the
trivial one (9). Indeed, let us consider for example
the amplitude in fig. 3a. The insertion is the self-energy operator
(6)
Hd_-retFr/r
where
H=rt~p+flm—e2/r—&f•r
(7)
is the Dirac electron Hamiltonian. The external electric field is included into the Hamiltonian (7). To
emphasize this point we will denote its eigenstates by
a bar: HI ti> =E~I fl>. The eigenstates of the total
Hamiltonian H+Hd will be denoted by the tilde:
Fig. 2. Amplitude of the photon resonant scattering in leading
order in H,.
~21çJ1
(H+Hd)Ifl> =E~Iñ>.Due to the Schiff theorem
=
Treating Hd as a perturbation one can easily
calculate the matrix element of the radiation operator (2) between the tilde states,
I ~2~d:il_l:~: 3. ~:zJI:I: ’
b
~.
~c
<thIh~In>= <rnlh,Ifl> + ~d•<rnI[h1,p] In>
=
296
(1
—
ik~d/e)<,n I h~I ,~>
.
(8)
~
d
Fig. 3. Amplitude of the photon resonant scattering with the
Lamb-shift insertion.
Volume 158, number 6,7
f
£ E
~
=
J
~
PHYSICS LETFERS A
d 3q <2~I h7 ( q) In> <n I h.~(q) I 2~>
(2x )
E—
+iO
‘
(10)
s’,,
~
but due to eq. (8), the dependence on d in the matrix element of h7 exactly compensates that in the
matrix element h~.In the same way we can prove
the independence of d of the insertions in the diagrams presented at figs. 3b—3d. Thus, there is no energy shift proportional to d which can be observed
in the resonant scattering of light in the atomic
ground state. Then the question arises: what does
formula
(5) mean?
To answer
this question let us first of all answer a
more simple one: What is the usual pressure oflight?
Thus without any external static electric field the laser
shines on an atom in resonance with the transition
1s1 2 2s1 ‘2• The interaction of an electron with the
classical electromagnetic wave is ofthe form (cf. with
eq. (2))
c= v
Vt~
+
-‘-
—
I
v(
eA.~ei*.~_~~f
V~—~
( V~~ ÷
—‘
/
(11 ~
/
A is the wave vector potential. The rescattenng is
isotropic and therefore light pressure anses. This is
equivalent to a static electric field acting on the nucleus. Due to balance of momentum at a small saturation parameter <2p I VI ls> /f’~,~ 1,
I
I
2
(w—w0) 2 +~f’~
2P1/2, al ~
1s
<
Just in such an experiment the recoil electric field
(4), (12) can be measured and exactly in this sense
the Schiff theorem is violated for the unstable quantum states.
The shift of the NMR frequency due to the nucleus electric dipole moment is equal to
SE= Tr(Hdp). Here p is the density matrix of an
atom in the laser field (11) and Hd is defined by eq.
(6). We will solve the equation for the density matrix by iterations in the perturbation P (see e.g. ref.
[5]),
.
9/ôt—Wjk)pIk+1f’jk(pIk—plk(e)
(1
t
p~ is the equilibrium density
)=
[V, P11k,
(13)
matrix. In our case
~ (e) corresponds to the equal population ofthe states
I 1 Si,
2, ±
~>. In first approximation the positive and
negative frequency components of p arise:
[V
=
(+)
—
,
P
(e)
(14)
Ilk
±(O—O)1k+iI’~k
The effect we are interested in arises in the second
approximation. The time-independent components
)
—
e
9 September 1991
1,2,fl>
2,
(12)
(2)
ofp
are equal to
(2)
Pk
=
~ry7(+)
—
V
,
P
(1—)i
J 1k
rrz(—)
LV
,
P
(1+)i
uk
(15)
The further calculation is straightforward:
SE=Tr(H2))=!d~
e 1k ~p,H]1~p~)
t~)]p”~+[p, V~~1p”~)
=
Tr( [p, V
—
w
0 =E2~1,2—E15112, a, fl= ±~ are the projections of
the angular momentum. Similar to (5), an energy
shift proportional to d must arise. However, we argue above (eq. (10)) that there are no corrections
to
the photon
amplitude
proportional
to
d. Thus
we canscattering
conclude that
the photon
which prepares the unstable quantum state cannot measure by
itself the recoil electric field (4), (12) on the flucleus. However, a different experiment is possible.
Let the laser field (11) prepare the unstable quanturn state and the other field probe the atom. Say,
using the microwave field one can search for the dependence of the nuclear-magnetic-resonance (NMR)
frequency on the nucleus electric dipole moment d.
=—
.
~
(16)
After the substitution
I) from
(14) we really
8’N with ofp
8’N from
eq. eq.
(12).
getWe
SE=
d
now return to the Schiff theorem (formulae
(4) and (5)). Here the Situation ~5 very similar to
the consideration of the light pressure. However, in
this case the indices i, k in the density matrix P1k flUmerate not only the states of an atom, but the states
ofa photon as well. This is a rather unusual situation
and therefore we emphasize it once more. Usually in
the density-matrix description one averages over all
photon states and keeps explicit the electron degrees
—
297
Volume 158, number 6,7
PHYSICS LETTERS A
of freedom only. To spot the Schiff-theorem violation (eqs. (4) and (5)) we should keep explicit the
states of an atom with one photon and average over
the states with rnore than one photon.
Let the laser (11) be tuned to the transition ls
1,2
—*2s
1,2.
It
produces
some
population
ofthe
2s112
level
which corresponds to a stationary density matrix p(O).
Saythat at saturationthepopulationsofailfourstates
Its
1,2, ±~>, I 2s1,2, ±
~> are equal. We will solve
eq.
(13) the
starting
from p withFirst
all let usstatic
takeelecinto
account
interaction
the of
external
tric field U= ef~r which mixes 2s
2Pi/2
1,2 and
levels,
~
—
(17)
9 September 1991
SE= Tr(H~p(3))
=
—
=
—
x
!Tr[d.q(H.~+)p(2_)
_pt2~~H~~)J
e
3q
i
d 3 dq
e J (2x)
f
(hkI[h+~P”)]Ik
~
[h~P~1kih,~)
(22)
1k
W O)ik + iO
W Wk1 + iO
Wk1 the
One
can
verify
that
due
to
the
relation
real parts of the Green functions in (22) are cancelled out, and only ô-functions survive:
0:k=
—
~
(w—wo—iOY’--’ixd(w—wo). Usingeq. (18) SEcan
be transformed to the form
SE= ~ReJ
WIk—if’,k
—
(
2)3dqd(wwo)
e
More explicitly,
<sI UIp>
~_
Psp
—
(0)
~
=
(~4~))+.
(18)
wsp_lirpPss
the angular momentum. Comparing with eqs. (1)—
Here s, p denotes 2s1,2 and 2p1,2, and p,~l)still is a
matrix in the projections of angular momenta.
The interaction with the photon with momentum
q and polarization a due to eq. (2) is of the form
H1(q, a) = H$ ~ + H~
—),
~
=h a
1
7
q,e~
~
+h~a~.
(19)
I
Here a + and a are the creation and annihilation operators of the photon. In the second approximation
(2±) —
P-k
—
[H~
±Cl)
),
p” ~]Ik
— W,k
(20)
+ Iflk
Similarly to eq. (15) the time independent part of
p (3) is equal to
2~]lk+~
=
—
0~1k
([H~,p~
.
(21)
Analogously to eq. (16) the correction to the energy
is of the form
298
xTr(p~<pIhI1s><lsIh~Is)).
(23)
The trace in this formula is over the projections of
(5) we see that expression (23) identically coincides with the energy shift (5) which is derived from
the balance of momenta.
In conclusion we formulate the results of the present work. The Schiff theorem (screening of an external static homogeneous electric field on the nucleus of a neutral atom) is violated for the excited
(unstable) atomic states. As a matter of principle this
violation cannot be observed in the scattering of a
photon on the ground state ofan atom. In other words
there is no effect if one uses as a probe the photon
which itself prepares the unstable quantum state. The
violation takes place if the photons (laser field) are
used to prepare the unstable quantum state and the
other field probes the atom. Say using the microwave
field one can observe the dependence of nuclear
magnetic response frequency on the nucleus electric
dipole
d. Just
in this sense
the Schiff
theorem
is moment
violated for
the unstable
quantum
states.
The main subject of the present paper is to point
out the Schiff theorem violation for unstable atomic
states. Besides that we would like to draw the attention of the reader to an interesting principal possibility. We mean the use of the effective electric field
[12] due to the light pressure for the experimental
search for the electric dipole moment of the nucleus
Volume 158, number 6,7
PHYSICS LETFERS A
in nuclear magnetic resonance experiments. This
method is more suitable for light atoms.
We are grateful to V.F. Dmitriev, M.G. Kozlov, J.
Sucher, V.B. Telitsin, V.F. Yezhov, and especially to
V.V. Flambaum and I.B. Khriplovich for the numerous helpful discussions and stimulating questions.
9 September 1991
References
[1] LI. Schiff, Phys. Rev. 132 (1963) 2194.
[2] I.B. Khriplovich, in: Proc. LNP Winter school in Nuclear
physics and elementary particles, Leningrad, 1982, p. 137.
[3] V.A. Dzuba et al., Phys. Lett. A 118 (1986)177.
[4] Ya.I. Azimov et al., Zh. Eksp. Teor. Fiz. 67 (1974)17.
[5] R.H. Pantell and H.E. Puthoff~Fundamentals of quantum
electronics (Wiley, New York, 1969).
299