- Email: [email protected]
Violation of the Schiff theorem for unstable atomic states
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
+
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
= + ~d• =
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> (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,
~_
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~
(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
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
+
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
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>
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,
~_
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~
(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