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Comment on ``Quantum mechanics of an electron in a homogeneous magnetic field and a singula

Comment on “Quantum mechanics of an electron in a homogeneous magnetic ?eld and a singular magnetic ?ux tube”
R. M. Cavalcanti?
Instituto de F?sica, Universidade de S?o Paulo, Cx. Postal 66318, 05315-970, S?o Paulo, SP, ? a a Brazil (February 1, 2008)

arXiv:quant-ph/0003148v1 30 Mar 2000

Abstract

Recently Thienel [Ann. Phys. (N.Y.) 280 (2000), 140] investigated the Pauli equation for an electron moving in a plane under the in?uence of a perpendicular magnetic ?eld which is the sum of a uniform ?eld and a singular ?ux tube. Here we criticise his claim that one cannot properly solve this equation by treating the singular ?ux tube as the limiting case of a ?ux tube of ?nite size.

Typeset using REVTEX

? E-mail:rmoritz@fma.if.usp.br

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The Pauli Hamiltonian for an electron (of mass M, charge ?|e| and g-factor 2) moving in the (x, y)-plane under the in?uence of a magnetic ?eld pointing in the z-direction is given by 1 H= 2M |e| p+ A c
2

+

|e|? h Bz Sz . Mc

(1)

Thienel [1] has recently investigated the eigenvalue problem for this Hamiltonian in the case of a magnetic ?eld which is the sum of a uniform ?eld and a singular ?ux tube, Bz (r) = B + αΦδ 2 (r) (B > 0; Φ ≡ 2π? c/|e|) . h (2)

He claims that standard approaches to this problem fail. The purpose of this Comment is to show not only that they do work, they are also simpler than his alternative method. As Thienel, we choose the vector potential in the symmetric gauge, A(r) = Br αΦ e? , + 2 2πr

and use magnetic units (where the unit of length is λ = (Φ/πB)1/2 and the unit of energy is hω, with ω = |e|B/Mc the Larmor frequency). Then we can rewrite (1) as ? ? 1 ? r H =? 4r ?r ?r 1 ? 2 4r ? + iα ??
2

?

i 2

? 1 α + iα + r 2 + 1 + δ(r) Sz ?? 4 2r

(3)

(our r corresponds to his r ). Since H, Lz and Sz commute with each other they can be ? diagonalized simultaneously, so we can write the eigenfunctions of H as ΨE,m,σ (r, ?) = ψE,m,σ (r) eim? |σ , with m an integer and Sz |σ = σ |σ , σ = ±1/2. Solving the resulting di?erential equation for ψE,m,σ (r) and demanding that ψE,m,σ (r) → 0 as r → ∞ we ?nally obtain ΨE,m,σ (r, ?) = N r |m+α| e?r
2 /2

U(ξ, |m + α| + 1, r 2 ) eim? |σ ,

(4)

where U(a, b, z) is one of Kummer’s functions [2], N is a normalization constant and ξ≡ 1 (|m + α| + m + α + 1 + 2σ) ? E. 2 (5)

In order to determine the possible values of E we need to know the correct boundary condition at the origin. This problem was examined by Hagen [3] and G?rnicki [4] in the o case of a pure Aharonov-Bohm potential (i.e., with B = 0). By treating the singular ?ux tube as the limiting case of a ?ux tube of ?nite size, they obtained the following result: the eigenfuntions corresponding to the spin component which “sees” a repulsive delta-function potential at the origin (i.e., σ = +1/2 if α > 0, σ = ?1/2 if α < 0) must be regular there. (By contrast, Thienel requires, without justi?cation, that they vanish at the origin. This is the reason why a vacancy line occurs in his (E, m + σ)-plane — see FIG. 1 of Ref. [1] — for integer α = 0, whereas in our solution no such vacancy line occurs.) The supersymmetry of 2

the Pauli Hamiltonian [1,5] determines the eigenfunctions corresponding to the other spin component. A few remarks are in order here: (i) although such a boundary condition was derived for the the Dirac equation, one can easily show that it also holds for the Pauli equation (with g = 2) [5,6]; (ii) the presence of background smooth magnetic ?eld does not alter the boundary condition at the origin, as it does not add any singular term to the Hamiltonian. Let us ?rst consider the case α > 0. Then ΨE,m,1/2 must be regular at the origin, which occurs only if ξ = ?n (n = 0, 1, 2 . . .), for then U(?n, |m + α| + 1, r 2) = (?1)n n! L|m+α| (r 2 ), n (6)

where La (z) is the associated Laguerre polynomial [2]. Combining this with (4) and (5) and n normalizing ΨE,m,1/2 to unity we thus obtain Ψn,m,1/2 (r, ?) = En,m,1/2 Γ(n + 1) 2 r |m+α| e?r /2 L|m+α| (r 2 ) eim? |+ , n π Γ(|m + α| + n + 1) 1 = n + 1 + (|m + α| + m + α) (n = 0, 1, 2 . . . ; m = 0, ±1, ±2 . . .). 2 (7) (8)

For each of these states there is a superpartner with the same energy and opposite spin, obtained by applying the supercharge Q? (Eq. (10) of Ref. [1]) to (7): Ψn,m+1,?1/2 (r, ?) = En,m,1/2 Q? Ψn,m,1/2 (r, ?) = 1 2 Γ(n + 1) π En,m,1/2 Γ(|m + α| + n + 1) m+α ? 2 + + r r |m+α| e?r /2 L|m+α| (r 2 ) ei(m+1)? |? . n ?r r (9)
?1/2

× ?
?1/2

The factor En,m,1/2 ensures proper normalization:
?1 Ψn,m+1,?1/2 |Ψn,m+1,?1/2 = En,m,1/2 Ψn,m,1/2 | QQ? |Ψn,m,1/2 ?1 = En,m,1/2 Ψn,m,1/2 | (H ? Q? Q) |Ψn,m,1/2 = 1.

The eigenstates with zero energy are anihilated by both supercharges, Q |E = 0 = Q? |E = 0 = 0. They are given by ΨE=0,m,?1/2 (r, ?) = r ?(m+α) e?r
2 /2

eim? |?

π Γ(?m ? α + 1)

;

(10)

square integrability requires m + α < 1. If α < 0 it is ΨE,m,?1/2 which must be regular at r = 0. Thus 3

Ψn,m,?1/2 (r, ?) = En,m,?1/2 = n +

Γ(n + 1) 2 r |m+α| e?r /2 L|m+α| (r 2 ) eim? |? , n π Γ(|m + α| + n + 1) (n = 0, 1, 2 . . . ; m = 0, ±1, ±2 . . .).

(11) (12)

1 (|m + α| + m + α) 2

The zero modes are already included among these states (n = 0, m + α ≤ 0). The spin-up states are obtained by applying the supercharge Q (Eq. (9) of Ref. [1]) to the spin-down ?1/2 states with nonzero energy (the factor En,m,?1/2 ensures proper normalization): Ψn,m?1,1/2 (r, ?) = En,m,?1/2 Q Ψn,m,?1/2 (r, ?) = 1 2 × Γ(n + 1) π En,m,?1/2 Γ(|m + α| + n + 1) m+α ? 2 + + r r |m+α| e?r /2 L|m+α| (r 2 ) ei(m?1)? |+ . n ?r r (13)
?1/2

The same results can be obtained by treating the singular ?ux tube as the limiting case of a ?ux tube of ?nite size in the presence of a background homogeneous magnetic ?eld. The calculations, however, are more complicated and not illuminating. Here I shall only point out to the origin of Thienel’s wrong conclusion that such an approach fails. First of all, we note that the correct form of Ψ outside the ?ux tube is given by (4). Then, by demanding continuity of ?Ψ/?r at the border of the tube, one is led to what is essentially 2 his Eq. (17) multiplied by R|m+α|?1 e?R /2 U(ξ, |m + α| + 1, R2 ). As a function of ξ, U has an in?nite number of zeros, which were completely overlooked by Thienel. [One can show, in particular, that the zeros ξn of U satisfy limR→0 ξn = ?n (n = 0, 1, 2 . . .). This follows from the asymptotic behavior of U for small R [2], U(ξ, |m + α| + 1, R2 ) ? ?
R→0

Γ(|m + α|) ?2|m+α| R Γ(ξ)

(valid for ξ = ?n and m + α = 0), the fact that Γ(?n + ?) and Γ(?n ? ?) have opposite signs for n = 0, 1, 2 . . . and 0 < ? < 1, and the continuity of U as a function of ξ.]
ACKNOWLEDGMENTS

I thank Adilson Jos? da Silva and Marcelo Gomes for a critical reading of this paper. e This work was supported by FAPESP.

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REFERENCES
[1] H. P. Thienel, Ann. Phys. (N.Y.) 280 (2000), 140. [2] M. Abramowitz and I. A. Stegun (Eds.), “Handbook of Mathematical Functions,” Chap. 13, Dover, New York, 1965. [3] C. R. Hagen, Phys. Rev. Lett. 64 (1990), 503. [4] P. G?rnicki, Ann. Phys. (N.Y.) 202 (1990), 271. o [5] Y. Aharonov and A. Casher, Phys. Rev. A 19 (1979), 2461. [6] R. Jackiw, in “M. A. B. B?g Memorial Volume” (A. Ali and P. Hoodbhoy, Eds.), World e Scienti?c, Singapore, 1991.

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