Electron scattering by a solenoid

0.3 MB·

by  I. V. Tyutin

Preprint of P.N. Lebedev Physical
Institute,No. 27,1974,unpublished
Electron Scattering by a Solenoid
I.V. Tyutin1
Abstract
The quantum-mechanical problems of electron scattering by an infinitely thin solenoid and by
8 a half of an infinitely thin solenoid are examined from the viewpoint of constructing a self-adjoint
0 Hamiltonian. It is demonstrated that in both problems there exist unique self-adjoint operators
0 with a “non-singular” domain, that, due to physical reasons, are identified with the corresponding
2 Hamiltonians. In thecase of quantized values of magnetic flow along thesolenoid, the electron does

n not experience any scattering by the string. It is shown that the scattering amplitude and wave
a function of an electron in the problem of scattering by an infinitely long solenoid of radius a in the
J limit a→0turnintothecorrespondingexpressionsfortheproblem ofan infinitelythinsolenoid. In
7 particular, at a quantized valueof magnetic flow along the solenoid, scattering disappears at a→0.
2

] 1 Introduction
h
p
- In this article, we examine two quantum-mechanical problems: the scattering of an electron by the field
nt of an infinite, infinitely thin solenoid, A(c)(~r)= µε3kiri, ~r =(x,y,z), ρ2 =x2+y2, and the scattering
k − ρ2
a of an electron by the field of a semi-infinite, infinitely thin solenoid, A(D)(~r) = µε3kiri . Our interest
u k − r(r z)
q in these problems is due to the following reasons. −
[ As is well-known, Dirac [1] proposed a Lagrangian theory of electrodynamics with two charges: the

electricandmagneticones,subjecttoachargequantizationcondition;see[1]. InDirac’stheory,however,
2
v the electromagnetic potentials contain singularities on certain lines (Dirac’s strings) coming out of the
7 magnetic charges (in principle, Dirac’s strings may also come out of the electric charges, as well as out
6 of the chargesof both types). The motion of these strings is not determined by the Lagrangianand may
1
be defined arbitrarily. Dirac presented some arguments in favour of the fact that physical results must
2
actually be independent of the form and specific motion of strings, since the electromagnetic potentials
.
1
fortwodifferentpositionsofastringarerelatedbyagradienttransformation(anywhereoutsidestrings).
0
Thisargumentisnotveryconvincing,because,infact,twopotentialsfortwodifferentpositionsofstrings
8
cannot be related by a gradient transformation in the entire space (for more details, see [1]).
0
: Thesimplestproblemconcerningtheinteractionofanelectronwithamagneticchargeistheproblem
v
of the scattering of a non-relativistic electron by the static field of an infinitely heavy magnetic charge.
i
X In addition, the electromagnetic potential (suggested by Dirac) of a magnetic charge is identical with
r the electromagneticpotential ofa semi-infinite, infinitely thin solenoid. According to Dirac’s hypothesis,
a
physicalresults(inparticular,thecross-sectionofelectronscattering)mustnotdependonthedirectionof
a string (the aboveexpressionfor A(D) correspondsto the rectilineal formof a string; of course,physical
k
resultsmustnotdependontheformofastring,either). Suchaproblemhasbeenexaminedbynumerous
authors[1]. Finally,intheworkbyZwanziger[2]itwasfoundthatthecross-sectionofelectronscattering
by the potential of a magnetic charge is, in fact, independent of the direction of a string.
If the electron “does not feel” the magnetic field along the string, then it must not experience any
scattering by the potential of an infinitely thin and infinitely long solenoid (in case the magnetic flow
along the string takes quantized values, which is in agreement with the quantization condition for the
electricandmagneticcharges). Thisproblemhasbeenexamined(inadifferentconnection)byAharonov
and Bohm [3], who have found that in the case of quantized values of the magnetic charge of a solenoid
(c)
(this corresponds to µ = 0, 1, 2,... in the above expression for A ), the electron, indeed, does not
± ± k
experience any scattering by the string.
1E-mail: [email protected]
1
If,however,onelooksatthestructureofsolutionsofthisproblem,itturnsoutthatthewavefunction
of an electron has the form ψ = eiµϕei~k~r, where ~k is the momentum of a moving electron, and ϕ is the
angle between the projections of the vectors ~k and ~r on the plane xy. This form of a wave function
suggests the idea that the potential A(c) is a pure gradient: A(c) ∼ µ∂ ϕ. If, indeed, one formally (i.e.,
k k k
withouttakingintoaccountthefactthatϕisadiscontinuousfunction)calculatesthegradientofϕ,then
(c)
one obtains the equality A = µ∂ ϕ. It is clear, nonetheless, that such an equality cannot take place,
k k
since the rot of its r.h.s. is equal to zero, while at the same time rotA(c) is the magnetic field (with
k
a finite flow µ) along the string. Then, there arises the question as to the correctness of the solution
presented by Aharonov and Bohm for this problem.
In the case of electron scattering by a semi-infinite, infinitely thin solenoid, Zwanziger [2] has found
that the scattering amplitude has the form f ~k,~k = eiΩf ~k,~k (~k and~k are the respective initial
′ 0 ′ ′
and final momenta of an electron; f0 is the sc(cid:16)atteri(cid:17)ng amplitu(cid:16)de in(cid:17)case the string is directed along the
momentum of a moving electron) and depends, as a consequence, merely on the angle between~k and~k ;
′
Ω is a certain function (see Section 4). As will be shown in Section 4, the wave function ψ =eiΩψ has
0
the same form, where ψ is the wave function in case the string is directed along ~k. Once again, there
0
arises the suspicionthat sucha form of solutionis due to the fact that the difference of vector potentials
A(D) A(D,k) (A(D,k)(~r)= µ εiljklrj isthepotentialofasemi-infinite,infinitelythinsolenoid,withthe
i − i i − r(kr ~k~r)
−
stringbeingdirectedalong~k)isapuregradient. IfoneformallycomputesthegradientofΩ(i.e.,without
taking into account the fact that Ω is a discontinuous function), then one obtains A(D) A(D,k) = ∂ Ω.
i − i i
However, this equality cannot take place, either, because the rot of the r.h.s. is equal to zero, while at
the sametime the rotofthe l.h.s. is the flowofmagnetic field(withfinite flows)alongthe strings. Thus,
in this case there also arises the question as to the correctness of the solution.
ThequestionofcorrectnessofthesolutionsfoundbyAharonov–BohmandZwanzigerfortheproblems
in question arises also in connection with another reason. The point is that the Hamiltonians of these
problems are extremely singular and thus cannot be defined immediately in the class of functions for
which they would be self-adjoint operators (in particular, the natural domain of a Hamiltonian does
not include all differentiable functions). In the natural domain, the Hamiltonians prove to be merely
symmetric, with non-vanishing deficiency indices. In addition, it is well-known [4, 5] that the problem
of constructing a self-adjoint Hamiltonian from a given symmetric operator (a self-adjoint extension of
a symmetric operator) admits more than one solution. Let us remind that the self-adjoint character of
an operator is necessary for the corresponding operator of evolution to be unitary and uniquely defined
in the entire Hilbert space. Different self-adjoint extensions lead to different solutions of the scattering
problem, corresponding to the same Schr¨odinger equation.
InSections2and4,inthecasesofinfiniteandsemi-infinitesolenoidsrespectively,witharbitraryvalues
ofµ,itisshownthatinbothproblemsthereexistuniqueself-adjointextensionsoftheHamiltonian,such
that they obey the usual physical condition: the Hilbert-space functions for which the Hamiltonian is
defined must not be singular (by the way, it is natural to call this condition the “principle of minimal
singularity”). Giventhis, it is shownthat for the integervalues of the parameterµ any scatteringby the
string is absent and the solution of the scattering problem is identical with the corresponding solutions
of Aharonov–Bohm and Zwanziger. Therefore, in both problems it is rigorously proved that in the case
of a quantized magnetic flow along the strings the electron does not experience scattering, in agreement
withDirac’s hypothesis,andthat, inconsequence,a semi-infinite, infinitely thin solenoidcanactuallybe
considered as a realization of a fixed monopole (magnetic charge). In the same sections, it is shown how
to solve the (seeming) difficulty that has been described earlier in this Introduction, i.e., the one related
with the gauge transformations of potentials and with the form of solutions.
In Section 3, it is shown that the result of Aharonov–Bohm can be deduced with the help of a
“physical”regularizationofthe potential,whichconsistsin usinga solenoidwitha finite radiusa, rather
than an infinitely thin solenoid, and then making a go to zero.
In Appendix A, in connection with the article [7], it is shown that the problem of scattering by
the potential of an infinitely thin solenoid cannot be solved as a perturbation theory in ∆µ, where
µ=[µ]+∆µ, and [µ] is the integer part of µ.
2
2 Electron scattering by the field of an infinitely thin and in-
finitely long solenoid
The electromagnetic potential A(c) of an infinitely thin and infinitely long solenoid has the form
k
ε r
A(c)(~r)= µ 3ki i . (1)
k − ρ2
It is easy to see that outside the z axis the magnetic field H~ = rotA~(c) is equal to zero everywhere.
Nevertheless, the flow A~(c)d~l through any surface with the boundary being any contour (even an in-
finitely small one) around the z axis does not vanish and is equal to µ.
H
The Schr¨odinger equation for the scattering of an electron by this same potential has the form
i∂ ψ(~r,t)=Hψ(~r,t) , (2)
t
where
2
(c)
H = i∂ +A (~r) , (3)
− k k
(cid:16) (cid:17)
and the mass of an electron is set equal to 1/2.
Eq. (2) has the following solution:
ψ(~r,t)=exp ik2t+ik z ψ(ρ~,t) , (4)
− z z
(cid:8) (cid:9)
where k is the projection of the momentum of the electron on the zaxis, ρ~ = (x,y), and ψ(ρ~,t) obeys
z
the equation
i∂ ψ(ρ~,t)=Hψ(ρ~,t) , (5)
t
2µi µ2
H = ∆+ (y∂ x∂ )+
− ρe2 x− y ρ2
1 1
=e ∂2 ∂ + ∂2 2iµ∂ +µ2 . (6)
− ρ − ρ ρ ρ2 − ϕ− ϕ
(cid:0) (cid:1)
Let us now construct a self-adjoint Hamiltonian corresponding to the differential expression (6).
ItisclearthatH cannotbedeterminedimmediatelyforanydifferentiable(anddecreasingsufficiently
fast as ρ ) functions that belong to L (L is the Hilbert space of square-integrablefunctions ψ(ρ~))
2 2
because o→f t∞he termes µ2 and 1 (y∂ x∂ ).
ρ2 ρ2 x− y
Let us define an operator H as a differential operator (6) on the domain D , where D is the set of
0 0 0
twice-differentiable functions with a compact support that turn to zero at ρ < a for a certain a (a may
be different for different functions). D is dense in L ; (6) is obviously defined for any ψ D , and for
0 2 0
∈
any ψ ,ψ D there holds the equality
1 2 0
⊂
dρ~ψ (H ψ )= dρ~(H ψ )ψ . (7)
1∗ 0 2 0 1∗ 2
Z Z
Therefore, H is a symmetric operator [4, 5]. Besides, H is a positive operator.
0 0
In order to constructself-adjoint extensions of the operatorH , one shouldobtain the eigenfunctions
0
U( ) of the adjoint operator H with the eigenvalues i. In addition, H is defined as the differential
± 0∗ ± 0∗
expression(6); however,the domainis no longerrestrictedby the boundary valuesatρ=0 oratρ= .
∞
This implies that the function U(+) (corresponding to the eigenvalue +i) must obey the equation
HU(+) =iU(+) (8)
on the entire plane, except, perhaps, the poinet ρ=0. Let us searchfor the eigenfunctions U in the form
U(+)(ρ~)=eimϕU(+)(ρ~) . (9)
m
3
m
Note that the function eimϕ = x+iy is infinitely differentiable everywhere except the origin.
ρ
Therefore,theactionoftheoperatorH(cid:16)ona(cid:17)functionoftheform(9)(whichoughttobeknowneverywhere
except the origin) does not require any redefinition. The function U(+) satisfies the equation
m
e
1 (m+µ)2
∂2 ∂ + U(+)(ρ~)=iU(+)(ρ) . (10)
− ρ − ρ ρ ρ2 ! m m
Eq. (10)istheBesselequation. Since(10)musttakeplaceatρ=0,itdoesnotrequireitsredefinition
6
neitherinCartesiannorincylindriccoordinates. Asolutionof(10)beingsquare-integrableat isgiven
∞
by
Um(+)(ρ)=Nm−1Hm(1+)µ eiπ4ρ , (11)
where Nm is a normalization factor (in case the function is(cid:0)squa(cid:1)re-integrable). Function (11) is square-
integrable atzero (with respect to the measuredxdy ∼ρdρ) only if m+µ <1. Therefore,the operator
| |
H hastwoeigenfunctionsthatbelongtoL withtheeigenvalueiatanon-integerµandoneeigenfunction
0∗ 2
[...] an integer µ. H has the same number of eigenfunctions with the eigenvalue i. This follows from
0∗ −
the solution of an equation similar to (10),
Um(−)(ρ)=Nm−1Hm(2+)µ e−iπ4ρ , (12)
as well as from the fact that H is a positive operator. (cid:0) (cid:1)
0
Therefore, H is a symmetric operator with non-vanishing (and equal) deficiency indices: (2,2) at
0
∆µ=0 and (1,1) at ∆µ=0, where µ= [µ] ∆µ, [µ] being the integer part of µ.
6 − − −
All essentially self-adjoint extensions H of the operator H are described as follows [4, 5]:
θ 0
a) the domain is
D =D +F +Fθ, (13)
θ 0
where
F = A1ei([µ]+1)ϕN[−µ]1+1H1(1)∆µ eiπ4ρ +A2ei[µ]ϕN[−µ]1H∆(1µ) eiπ4ρ , at∆µ6=0 (14)
( Aei[µ]ϕH0(1) eiπ4ρ , a−t∆µ(cid:0)=0,(cid:1) (cid:0) (cid:1)
Fθ = B1ei([µ]+1)ϕN(cid:0)[−µ]1+1H(cid:1)1(2)∆µ e−iπ4ρ +B2ei[µ]ϕN[−µ]1H∆(2µ) e−iπ4ρ , at∆µ6=0 (15)
( Aeiθei[µ]ϕρH0(2) e−iπ4−ρ ,(cid:0)at∆µ=(cid:1) 0, (cid:0) (cid:1)
andBi =Ajθji,Ai,Aarearbitrary(cid:0)compl(cid:1)exnumbers;θij andθ are,respectively,anarbitrary(although
fixed for a given extension) unitary matrix and real number;
b)
H D =H D +iF iFθ. (16)
θ θ 0 0
−
Self-adjoint extensions are given by the closure: H : H = H . In addition, the domain of H is
θ θ θ∗ θ
D = D +F +Fθ, where D is the domain of the operator H = H . The functions ψ D are
θ 0 θ 0 0∗∗ ⊂ 0
non-singular.2
Inordertoselectaself-adjointextension,werequirethatthefunctionsfromthedomainofa“physical”
operator should be nonsingular. This fixes the extension in a unique way:
exp i(1 ∆µ)π 0
θij = − 0− 4 exp i∆µπ , for∆µ6=0, (17)
(cid:12) (cid:8) (cid:9) − 4 (cid:12)
θ = (cid:12)0, for ∆µ=0. (cid:12) (18)
(cid:12) (cid:8) (cid:9) (cid:12)
(cid:12) (cid:12)
We must now obtain the complete system of generalized eigenfunctions for the self-adjoint operator
(which shall be denoted as ), corresponding to (17) and (18). Since the matrix θ is diagonal, the
ij
H m
subspaces of functions of the form x+iy f(ρ) reduce , and we must find generalizedeigenfunctions
ρ H
of a “nonsingular” extension of Be(cid:16)ssel’s(cid:17)differential expression on a semiaxis. This problem has been
solved in [6], where all the inversion formulas related to Bessel’s differential expression of an arbitrary
2Thefunctionsψ⊂D0 becomecontinuous aftertheircorrectiononthezero-measuresetψ(0)=0;see[5].
4
index ν 0 have been described. In the case under consideration, the generalized eigenfunctions are
≥
given by
eimϕJ √λρ , m=0, 1, 2,..., (19)
m+µ
| | ± ±
(cid:16) (cid:17)
where λ is an arbitrary real number, and the operator proves to be positive. We note that in case
H
∆µ=0 the operator also proves to be the only positive extension of the operator H .
0
H
Thus,ithasbeenshownthatunderthephysicalconditionof“minimalsingularity”,i.e.,thecondition
that the domain of the Hamiltonian should be composed entirely of nonsingular functions, there exists
a unique self-adjoint operator related to the differential expression (3). It is natural to refer to
H
the operator as the Hamiltonian of the problem in question. The complete system of generalized
H
eigenvectors of this operator is given by the set of functions (19).
Letusnowsolvethescatteringproblem,i.e.,constructawavefunctionψ (ρ~)subjecttotheSchr¨odinger
k
equation
ψ (ρ~)=k2ψ (ρ~) , k2 =E k2 (20)
H k k − z
(with E being the total energy of the electron; ~k being the projection of the total momentum of the
electron on the xy plane), as well as to the following asymptotic condition:
ψk(ρ~) ei~kρ~+ f(ϕ) eikρ−iπ4 . (21)
ρ→ √2πkρ
→∞
Condition (21) implies that in case ρ the wave function is a superposition of an incoming plain
→∞
wave and a scattered divergent cylindrical wave; f(ϕ) is the scattering amplitude.
Using the expansion
ei~kρ~ 1 eikρ−iπ4 ∞ eim(ϕ−δ)+e−ikρ+iπ4 ∞ ( )meim(ϕ−δ) , (22)
→ √2πkρ" − #
m= m=
X−∞ X−∞
where δ is the angle between the vector~k and the x axis, we obtain, due to the well-known asymptotic
behavior of Bessel’s functions,3
ψk(ρ~)=e−iµπ2 eim(ϕ−δ)imJm+µ(kρ)+e−iµπ2 eim(ϕ−δ)( i)mJm+µ (kρ) . (23)
− | |
m+µ 0 m+µ<0
X≥ X
The scattering amplitude (ϕ δ =0) equals to
− 6
sin2∆µπ
f(ϕ)=ei([µ]+21)(ϕ−δ)( )[µ] . (24)
− sin2 ϕ−2δ
(cid:16) (cid:17)
The cross-sectionof scattering equals to (ϕ δ =0)
− 6
sin2∆µπ
σ ∼ f(ψ)2 = , (25)
| | sin2 ϕ−δ
2
(cid:16) (cid:17)
which is identical with the result of [3]. Therefore, it follows from (25) that in the case of a quantized
value of flow along an infinite, infinitely thin solenoid (µ being integer) the scattering on the potential is
absent.
Letus makesomemoreremarksconcerningthe caseofanintegerµ. Namely, wesupposeµ=n. The
wave function then admits the representation
ψ =( )ne in(ϕ δ)e i~kρ~. (26)
k − − −
−
3Note that the wave function (23) and the scattering amplitude (24) are single-valued functions of the coordinates,
whereas the wave function of Aharonov and Bohm is not single-valued. This is related to the fact that, instead of the
asymptoticcondition(21),theauthors of[3]useaconditionthatdiffersfrom(21)bythemultipliereiµϕ,whichresultsin
thenon-single-valuedcharacterofthewavefunctionin[3]. Thesquaredmoduleofthescatteringamplitudein[3],however,
isidenticalwiththesquaredmoduleoffunction(24).
5
(26) has the following asymptotics:
1
ψk(ρ~) ( )nδ(ψ δ)eikρ−iπ4 +δ(ϕ δ π)e−ikρ+iπ4
→ √2πρ − − − −
=ei~kρ~+(( )n (cid:2)1)δ(ϕ δ)eikρ−iπ4 . (cid:3) (27)
− − − √2πkρ
(27) shows that scattering, in pure form, is absent only for even µ. For odd µ, however, the electron
“senses” the presence of the string.
Another remark is as follows. The wave function (26), as mentioned in Introduction, contains the
multiplier e inϕ, being a cause for a certain perplexity, which has been discussed in Introduction. A
−
solution of the mentioned paradox is given by the fact that for integer µ the potential A(c) can actually
k
be transformedoutofthe Schr¨odingerequationwiththe helpofagauge transformation, which,however,
is not a gradient transformation. Namely, the potential A(c) can be represented in the form (for integer
k
µ=n)
n n
x+iy x iy
Ak(c) =−ieinϕ∂ke−inϕ ≡−i ρ ∂k −ρ . (28)
(cid:18) (cid:19) (cid:18) (cid:19)
For non-integer µ, such a representation does not take place!
It can be readily verified that (28) holds true in the entire space, except, perhaps, the origin. Rep-
resentation (28) turns out to be sufficient to eliminate the potential from the Schr¨odinger equation with
the help of the gauge transformation
A A iU 1∂ U, ψ Uψ. (29)
k k − k
→ − →
In order that the transformation of A be a gradient one, it is necessary to impose the subsidiary
k
condition of a continuous differentiability of lnU, which is not necessarily fulfilled in the general case.
This is precisely what happens in the problem under consideration. Representation (28) is not reduced
to a pure gradient, since in this case lnU = iϕ is a discontinuous function (for instance, along the line
−
y =0, x>0).
For the sake of completeness, one should investigate the point ρ = 0 more accurately. We will,
however, restrict ourselves to the above qualitative remark.
3 A thin solenoid as a limit of a thick solenoid
Let us replace the potential A(c) in expression (3) by a regularized potential A(R):
k k
A(R)(~r)=A(c)(~r)f (ρ) , (30)
k k a
f (ρ)=1, ρ a; f (0)=0; (31)
a a
≥
besides, f is continuous at the point ρ = a. Note that the flow of magnetic field corresponding to the
a
potential (30) is equal to µ.
Let us select the eigenfunctions of H as solutions being regular at zero, which is usually postulated
on the basis of physical reasons.4 These functions are
U(m)(ρ~)= cmFm,k(ρ)eimϕ, ρ≤a, (32)
k c A J (kρ)+B N (kρ) eimϕ, ρ a,
(cid:26) m m |m+µ| m |m+µ| ≥
The function F (ρ) is a solutio(cid:0)n, regularat zero (normalized as(cid:1)ρm when ρ 0), of the following
m,k | |
→
equation:
1 1
∂2 ∂ + f (ρ)µ2+2mµf (ρ)+m2 F (ρ)=k2F (ρ) . (33)
− ρ − ρ ρ ρ2 a a m,k m,k
(cid:20) (cid:21)
(cid:0) (cid:1)
4From the mathematical viewpoint, one can say that the Hamiltonian, defined as the differential expression (3) with
potential (30) and a domain consisting of the set of all the twice-differentiable functions with a compact support, is
an essentially self-adjoint operator. This operator is also a unique self-adjoint operator with a “nonsingular” domain
correspondingtothedifferentialexpression(3)withpotential (30).
6
Thecoefficientsc arearbitrary,whereasthe coefficientsA andB arechosenfromtheconditionthat
m m m
U(m)and their first derivatives be continuous at the point ρ=a:
k
πa
A = F (a)N (ka) F (a)N (ka) , (34)
m 2 m,k |′m+µ| − m′ ,k |m+µ|
πah i
B = F (a)J (ka) F (a)J (ka) . (35)
m 2 m′ ,k |m+µ| − m,k |′m+µ|
h i
The solution Ψ (ρ~) of the Schr¨odinger equation with asymptotics (21) is given by
k
b
Ψk(ρ~) = ψk(ρ~)−i eim(ϕ−δ)+imπ−i|m+µ|π2 1+mib H(m1)+µ (kρ) , ρ≥a, (36)
m | |
m
X
( )meim(ϕ−δ)−i|m+µ|π2 Fm,k(ρ)
Ψ (ρ~) = − , ρ a. (37)
k
1+ib A ≤
m m
m
X
The corresponding scattering amplitude φ equals to (ϕ δ =0)
k
− 6
b
φ =f(ϕ) 2i m ( )meim(ϕ δ) im+µπ. (38)
k − −| |
− 1+ib −
m
m
X
In(36)and(38), thefunctionsψ (ρ~)andf(ϕ)aregivenbyformulas(23), (24)andpresentsolutions
k
of the problem for an infinitely thin solenoid; the coefficients b are given by
m
B
m
b = . (39)
m
A
m
Using the notation
F (ρ)=ρmF (ρ) , F (0)=1, (40)
m,k | | m,k m,k
one can present b as follows:
m
e e
bm = J|m+µ|(ka) |m+µ|−|m|− aFeFemm′,,kk((aa)) − akJJ||mm++µµ||+(k1a(k)a) . (41)
(|m|+|m+µ|)N|m+µ|(ka) · 1+ 1 aFem′,k(a) akN|m+µ|−1(ka)
|m|+|m+µ|(cid:20) Fem,k(a) − N|m+µ|(ka) (cid:21)
LetustakeintoaccountanestimationfortheBesselfunction[8],whichisimpliedbyitsrepresentation
as a series:
zν
J (z)= (1+θ) , (42)
ν 2νΓ(ν+1)
z 2
θ <exp | | 1, ν =min(ν+1 , ν+2 ,...) . (43)
0
| | (4(1+ν0))− | | | |
In a similar way, one can obtain an estimation for N , ν 0:
ν
≥
2νΓ(ν)
N (z) = (1+θ ) , ν =0, (44)
ν − πzν 1 6
2 z
N (z) = ln (1+θ ) , (45)
0 2
π 2
where θ and θ are bounded (uniformly in m, ν = m+µ, with a fixed µ) functions in any bounded
1 2
| |
region z; θ 0 as z 0, while θ < C z 2 at ν > 1, the constant C being independent of m,
i 1
θ ∼ z 2ln →z at ν =|1|,→θ ∼ z 2ν,|at|0<ν <| |1, θ ∼1/ln z .
1 1 2
| | | | | | | | | | | | | |
From(39), (41), (34), (35), itis easy to see that allb 0as a 0. Suppose, furthermore,thatthe
m
→ →
quantities
aF (a)
m′ ,k
(46)
F (a)
em′ ,k
e
7
areuniformly boundedin m anda atka<ε for a sufficiently smallε. Then (41) and(42)–(45)imply an
estimation5 for b :
m
π(ka)2|m+µ|
b = Q (a) , (47)
m 22m+µ (m + m+µ)Γ(m+µ)Γ(m+µ +1) m,k
| | | | | | | | | |
where the quantities Q (a) are uniformly bounded in m and a for ka<ε.
m,k
With the help of (47), we easily find
φ (ϕ) f(ϕ) 2 b +a2θ(a) , ϕ δ =0, (48)
k m
| − |≤ | | − 6
|mX+µ|≤1
θ(a) being a bounded function as a 0.
Therefore, φ (ϕ¯) f¯(ϕ) as a →0 for all ϕ, ϕ δ =0, at any value of µ.
k
→ → − 6
It remains to prove that the values aF (a)/F (a) are uniformly bounded in m and a at ka<ε.
m′ ,k m,k
This is done in Appendix B.
Let us examine the quantity Ψk ψek . Usingeestimations (44), (45) and (47), we obtain, for any
| − |
bounded region on the plane xy at ρ a:
≥
m+µ m+µ
Ψ ψ ∼ b N (kρ) + ka | | 1 a | |
| k− k| (cid:12) m |m+µ| (cid:12) 2 (m + m+µ)Γ(m+µ +1) ρ
(cid:12)(cid:12)|mX+µ|<1 (cid:12)(cid:12) |mX+µ|≥1(cid:18) (cid:19) | | | | | | (cid:18) (cid:19)
(cid:12) (cid:12)
(cid:12) (cid:12) a
∼ (cid:12) b N (kρ)(cid:12)+kaφ ∼const a∆µ+const a1 ∆µ+const a, (49)
(cid:12) m |m+µ| (cid:12) ρ · · − ·
(cid:12)(cid:12)|mX+µ|<1 (cid:12)(cid:12) (cid:18) (cid:19)
(cid:12) (cid:12)
(cid:12) (cid:12)
where φ a is a(cid:12)bounded (in a bounde(cid:12)d part of the plane, at ρ a) function is uniform in a as a 0.
ρ ≥ →
Let u(cid:16)s fu(cid:17)rther use estimations (44), (45), (47), (53) and obtain for ρ a:
≤
Ψ ψ ∼ J (kρ) +const a∼const a∆µ+const a1 ∆µ+const a, (50)
| k− k| (cid:12) |m+µ| (cid:12) · · · − ·
(cid:12)m+µ (cid:12)
(cid:12)|X| (cid:12)
Ψ (ρ~) ∼ (cid:12)(cid:12) 1 (cid:12)(cid:12) 1 ρ |m| 0, asa 0. (51)
k (cid:12) (cid:12)
| | m + m+µ · N (ka) a → →
Xm | | | | |m+µ| (cid:16) (cid:17)
Estimations (49), (50) and (51) lead to the following result:6
a)for∆µ=0,Ψ (ρ~)convergesasa 0 to the function ψ (ρ~)uniformly inany boundedpartofthe
k k
6 →
plane xy;
b) at ∆µ = 0 this convergence is uniform only if ρ > δ for an arbitrarily small but fixed δ.
This can be observed, in particular, from the fact that Ψ (0) ∼ an and ψ (0) ∼ 1. However, the
k k
quantity Ψ (ρ~)2dxdy ∼ Ψ (ρ~)2ρdρ, being the probability, “uniformly” converges to the probability
k k
| | | |
ψ (ρ~)2dxdy inthe sensethat Ψ 2 ψ 2 dxdy tends to zeroasa 0uniformly withrespectto
| k | ∆ | k| −| k| →
an arbitrary choice of the integRrat(cid:16)ion region ∆ ⊂(cid:17)D, D being an arbitrary bounded region of the plane
xy.
Thus, it has been proved that the probability distribution in the problem of electron scattering by a
solenoid of radius a, with any distribution of magnetic field inside of it, converges uniformly as a 0
→
and fixed flow of magnetic field µ, in any limited region of the plane (in the integral sense that has been
explainedabove)totheprobabilitydistributionintheproblemofelectronscatteringbyaninfinitelythin
solenoid. Thescatteringamplitudecorrespondingtoathicksolenoidconvergesasa 0tothescattering
→
amplitudecorrespondingtoathinsolenoiduniformlyintheregionε<ϕ δ <2π εforanyfixedε>0.
− −
5(47) gives a correct estimation for m+µ6=0. For m+µ=0, we have b[µ] ∼1/lna. It is important, however, that
b[µ]→0asa→0.
6The first term in (49) is of order a∆µ a ∆µ+a1−∆µ a 1−∆µ ∼a∆µ+a1−∆µ at ∆µ 6= 0. However, at ∆µ = 0
“ρ” “ρ”
(µ=−n)thefirsttermin(49)isoforderlnρ/lna. Atafixedρ,ittendstozeroasa→0;however,thesametermis∼1
atρ=a.
8
4 Electron scattering by the field of a semi-infinite thin solenoid
As a matter of fact, we shall examine, in the first place, the case of two semi-infinite infinitely thin
solenoids, with the corresponding potential
ε r 1 1 ε r z
A(S)(~r)= µ 3ki i = µ 3ki i ,µ =2µ. (52)
k − r r z − r+z − 1 r ρ2 1
(cid:18) − (cid:19)
Potential (52) corresponds to the potential of a monopole in Schwinger’s formulation [1].
The Hamiltonian of the problem is given by
1 1 µ2
H = i∂ +A(S) +V (r)= ∂ r2∂ + J2 1 +V (r) , (53)
− k k −r2 r r r2 − r2
(cid:16)1 (cid:17) 1 (cid:0) (cid:1)
J2 = ∂ (sinθ∂ )+ ∂2 2iµ cosθ∂ +µ2 . (54)
sinθ θ θ sin2θ − θ − 1 θ 1
(cid:2) (cid:3)
We have added to the Hamiltonian the potential V (which in the case of purely electromagnetic
interaction between the electron and monopole is to be set equal to zero).
We shall now construct a self-adjoint operator corresponding to the differential expression (54). Let
usdefine the operatorJ2 asthe differentialoperator(54)withthefollowingdomainD : 1)the functions
0 0
ψ(θ,ϕ) fromD mustbe twice-differentiable;2)ψ and∂ ψ mustbe periodic inϕwith the period2π; 3)
0 ϕ
ψ =0 at θ <ε and θ > π ε for a certain ε, where ε may be different for different functions. Property
−
(2) is necessary due to physical reasons, as well as due to the fact that the Hamiltonian be defined on
these functions in Cartesian coordinates in a natural way, since the singularities of the Hamiltonian are
located only on the z axis. Note also that the functions with properties (1) and (2) form the domain of
a self-adjoint operator corresponding to the differential (in ϕ) expression of second order with constant
coefficients, on the segment [0,2π]. The eigenfunctions of this operator are e imϕ, m=0, 1, 2..., ...,
−
± ±
which are usually selected due to physical reasons.
D isdenseinthespaceL offunctionsonaunitsphere,sothatJ2 isasymmetricoperator. Besides,
0 2 0
J2 is a positive operator. Let us find the eigenfunctions U of the operator J2 ∗. Representing the
0 0
function U in the form U(θ,ϕ)=e imϕU (θ), we find that U (θ) satisfies the equation
− m m
(cid:0) (cid:1)
1
sinθ∂ (sinθ∂ )+m2+µ2 2mµ cosθ U (θ)=λU (θ) . (55)
sin2θ − θ θ 1− 1 m m
(cid:2) (cid:3)
Letusrecallthattheeigenfunctionsoftheoperator J2 ∗ arenotsubjecttoanyboundaryconditions
0
[4, 5]. Eq. (55) has an exact solution, and the eigenfunctions U (θ,ϕ) are given by
m,l
(cid:0) (cid:1)
α β
1 t 2 1+t 2 1 t
Um(1,)l = Nm−,1le−imϕ −2 2 F −l,l+α+β+1;1+α; −2 ,
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
α β
1 t 2 1+t 2 1+t
Um(2,)l = Mm−,1le−imϕ −2 2 F −l,l+α+β+1;1+β; 2 , (56)
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
where
α+β α+β
λ = l+ l+ +1 ,
2 2
(cid:18) (cid:19)(cid:18) (cid:19)
α = m+µ , β = m+µ , t=cosθ. (57)
1 1
|− | | |
F (a,b;c;x) is a hypergeometric function [9]; N and M are normalization coefficients (provided that the
corresponding functions are normalizable). An analysis of the hypergeometric function F (a,b;c;x) as
x 1 shows that for a complex λ the functions U(1) L only on condition that m+µ <1, whereas
→ m,l ⊂ 2 | 1|
Um(2,)l ⊂ L2 only on condition that |−m+µ1| < 1. Thus, for every complex-valued λ the operator J02 ∗
has 4 eigenfunctions from L for ∆µ =0 (µ = [µ ] ∆µ ) and 2 such functions for ∆µ =0.
2 1 1 1 1 1
6 − − (cid:0) (cid:1)
All essentially self-adjoint extensions J2 of the operator J2 are described [4, 5] by the following
T 0
domains D :
T
D =D +A U +A T U , (58)
T 0 i i i ij j
9
where U and U are functions of the form (56) for some fixed complex l:
i i
U =U(2) , m = [µ ] , m = [µ ] 1,
1,2 mi,l 1 − 1 2 − 1 −
U =U(1) , m =[µ ] , m =[µ ]+1, ∆µ >0, (59)
3,4 mi,l 3 1 4 1 1
(1) (2)
U =U , U =U , ∆µ =0. (60)
1 [µ1],l 2 −[µ1],l 1
ThefunctionsU areobtainedformthefunctionsU bythechangel l ,T beinganarbitrary(however
i i ∗ ij
→
fixed for a given extension) unitary matrix of dimension 4 4 for ∆µ =0 and 2 2 for ∆µ =0.
1 1
× 6 ×
The requirement of “minimal singularity” for the self-adjoint extension (which we further denote as
J2) of the operator J2, that is, the condition that the domain of J2 should contain only non-singular
0
functions uniquely determines the extension (the matrix T).
We do not consider the case ∆µ =0. Let us only observe that all the functions from the domain of
1
J2 turntozeroalongthez axis(atlea6st,asθ∆µ1 orθ1−∆µ1,whenθ 0,and(π θ)∆µ1 or(π θ)1−∆µ1,
→ − −
when θ π). For any other extension of the operator J2 they either turn to zero or are singular on the
→ 0
z axis. In any case, this means that for ∆µ =0 the electron experiences scattering by the string.
1
6
We shall now examine the case ∆µ (µ = n). The matrix T for a “non-singular” extension is
1 1
−
determined, as has been observed, in a unique way:
n+i
T = δ . (61)
kj kj
n i
−
It is easy to see that the eigenfunctions of the operator J2 from L for this extension are given by
2
(56),wherelisallowedtotakeonlypositiveintegervalues(foranyotherreall functions(61)aresingular
at θ =0 or at θ =π):
l =0,1,2,... (62)
(another possible set l+α+β+1=0, 1, 2,...gives the same set of eigenfunctions and eigenvalues).
− −
Therefore, J2 is a positive operator.
For integer l, the hypergeometric functions
1 t 1+t
F l,l+α+β+1;1+α; − , F l,l+α+β+1;1+β;
− 2 − 2
(cid:18) (cid:19) (cid:18) (cid:19)
are proportional to each other, as well as to the Jacobi polynomials [9] Pα,β(t) (thus, for integer l the
n
solutionsU(1) andU(2) coincide;thesecondindependentsolutioninthiscaseissingularatoneoftheends
m,l m,l
θ =0orθ =2π). Since inthe caseoffixedαandβ the Jacobipolynomialsformacomplete orthonormal
system in the space L of functions on the segment [ 1,1], with a scalar product determined by the
2
−
weight (1 t)−α(1+t)β, we find as a result that the eigenfunctions Um,l of the operator J2,
−
α β
1 t 2 1+x 2 1 t
Um,l =Nm−.1le−imϕ −2 2 F −l,l+α+β+1;1+α; −2 , (63)
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
form a complete system of vectors in the space of functions on a unit sphere.
This means that functions (63) form a complete system of (generalized) eigenvectors of the operator
J2, whose spectrum is discrete and consists of the points
λ =L(L+1) , L µ , (64)
L 1
≥| |
with each value λ being (2L+1)-times degenerate in m ( L m L).
L
Functions(63)coincide(uptoaphasemultiplier)withthe−gen≤eraliz≤edsphericalfunctionsTm,µ1(ϕ,θ,0)
L
(see [10]). The angularformof solution(63) and its relationwith the groupof rotationshas been known
for a long time (see [11, 12] as well as the discussion in [1, 13]), and we will not dwell on this subject.
Let us only note, once again, that, as has been proved, the operator J2, determined by its eigenvectors
and spectrum (63) (64), is the only self-adjoint extension (of the differential expression (54)) with a
“non-singular” domain.
10