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5 December 1994

Physics Letters A 195 (1994) 221-226

PHYSICS LETTERS A

Spectral degeneracy of a Dirac electron in a uniform magnetic field

Chang Jae Lee Department of Chemistry, Sunmoon University, Asan 337-840, South Korea

Received 27 June 1994; accepted for publication 23 September 1994 Communicated by V.M. Agranovich

Abstract

In this paper we completely account for the spectrum of a relativistic electron in a uniform magnetic field using the conven- tional normal ordering, and demonstrate the equal "legitimacy" of the two prescriptions of normal ordering and supersymmetrization.

1. Introduction

The calculat ion o f the spec t rum o f a Dirac electron in a uni form magnet ic field is an old problem [ 1 ]. As shown in Fig. la , apar t f rom the cont inuous degeneracy due to free motion, the energy levels for the (posi t ive energy) electron exhibit two-fold degeneracy except for the nondegenerate ground state. In the nonrelat ivis t ic l imit this corresponds to the Landau levels [ 2 ]. The Landau levels are also doubly degenerate ( in addi t ion to being cont inuously degenerate as ment ioned above) except for the ground state, which is a singlet. The spectral

U,,_ U~ Un_ U~ Vn. V,._

w

E=0

Vn, Vn_

(~ (b)

Fig. I. (a) Schematic energy levels of a relativistic electron in a uniform magnetic field. If continuous degeneracy is not considered, both the positive and the negative energy states are doubly degenerate except the nondegenerate ones that are nearest to the zero of energy. (b) After interpretin_4 the negative energy electron as the positive energy positron both the electron and the positron have exactly the same energy levels. There is four-fold degeneracy except for the ground states that are doubly degenerate.

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222 CJ. Lee/Physics Letters A 195 (1994) 221-226

degeneracy of both the relativistic and the nonrelativistic electron is characteristic of a supersymmetric quantum mechanical system [ 3 ], and this naturally has led researchers to apply supersymmetry quantum mechanics (SSQM) to account for the spectral degeneracy of the electron [4-7]. However, relativity introduces energy levels that are equal in magnitude but opposite in sign to the positive energy. Interpreting as usual the negative energy electron as the positive energy positron, one has the energy levels shown in Fig. lb. The ground state is doubly degenerate and the rest of the energy levels are four-fold degenerate. Refs. [ 5 ] and [ 7 ] explain the two- fold degeneracy of the positron as well as the electron states as a manifestation of supersymmctry. In the previous work [8 ], the author elucidated the symmetries underlying the degeneracy between the electron and the positron states in addition to the degeneracy of the electron (also the positron) states employing SSQM.

In the above SSQM arguments the Dirac Harniltonian minus the rest energy is identified as the supersym- mettle Hamiltonian, so that the ground state energy is zero. Note that the supersymmetry arguments hinge upon the specific choice of the zero of energy. In the ordinary theory of the Dirac electron, where the rest energy is not removed from the Hamiltonian, the supersymmetry arguments are not tenable. Then what symmetry (other than supersymmetry) arguments can be made to account for the aforementioned spectral degeneracy in the spectrum of the Dirac electron? The aim of this paper is to address this problem by elucidating all symmetries of the Dirae Hamiltonian. In Section 2 we give a brief discussion of the Dirac equation for an electron in a uniform magnetic field. The complete set of symmetries that gives rise to the spectral degeneracy of the relativ- istic electron is identified in Section 3. Section 4 concludes with a summary of the main results of this paper along with some discussions.

2. The Dirae electron in a uniform magnetic field

The Dirae equation for an electron with minimal electromagnetic coupling is ( h = c = 1 )

(iTUDu- m)(o= 0, (2.1)

where Duff Ou+iqA u with q= - ]el. Especially, if the electron is placed in a uniform magnetic field applied along the z-axis, B= (0, 0, B), A u is chosen asAU= (0, 0, Bx, 0). We seek the solution for a stationary energy in the form ~o(x, t) = e x p ( - i E t ) exp i(kyy+k~z)~t(x) = e x p ( - i E t ) 7t(x). Then the Dirac equation reduces to

~"~D ~( X ) = ( Otx Px + l elBotyx+ otyky + otzkz +tim) ~( x ) = E ~ ( x ) , (2.2)

where

0o) ,410 o) are Dirac matrices with Pauli matrices tr. We will assume ky = kz = 0 for simplicity - this removes the free motion of the electron - and then we are left with the one-dimensional equation

(OtxPx + l el Botyx+flrn)~t(x) = E ¥ ( x ) . (2.4)

Eq. (2.4) has both positive and negative energy solutions. Let us denote them as u (x) and v(x), respectively,

YgDU,(X)=E,u,(x), ~ v , ( X ) = - E , v , ( x ) . (2.5)

The energy levels are given by

E,=~/m2+21e lBn (n=0, 1, 2, ...). (2.6)

The energy levels are doubly degenerate (remember that we have removed the continuous degeneracy in ky). The exception is the ground state energy that is nondegenerate. The energy levels are shown in Fig. 1 a.

C.J. Lee /Physics Letters A I95 (1994) 221-226 223

The presence of degeneracy implies the existence of underlying symmetry. For the system under consideration neither parity nor spin is conserved. However, the combined operation of parity transformation fi and spin projection & is conserved: [ tin, &Zz] = 0. Consequently, the states may be classified according to the eigenvalues of both the Hamiltonian and the symmetry operator

(2.7)

In the matrix notation the Dirac Hamiltonian may be written as

-i]e]Bx+p, 0 0

-m

,u 0 0 -b

(2.8)

In the above, p=rn/idm and we used iIeIBx+p,=i,/&@b+, -ilelBx+p,= -idmb. The eigen- functions are

(2.9)

where (xl n) is an eigenfunction of a harmonic oscillator with frequency o= I el B/m and the coefficients are CY)= [ (E,+m)/2E,] li2. The eigenfunctions in Eq. (2.9) are identical to those given in Ref. [ 71, but note that there the spin projection labels are incorrect. It is easy to check that

83u,+ = +%d, cs3v,* = fV,k . (2.10)

Thus, the label s= + 1 signifies the eigenvalue of S3 and the degenerate pairs of states are {u,,+, u,_} and {v,+, v,_}. The nondegenerate ground states are &,_ =‘(O, IO), 0, 0) and vo+ =‘(O, 0, 0, IO) ), where t denotes transpose.

Let us now interpret, as usual, the electron with the quantum number p as the positron with the quantum number p, where the collective quantum number p= (E,, s) and p= ( -E,, -s). The new energy levels are depicted in Fig. lb. We see that the ground state, which is equal to the rest energy m, is doubly degenerate and all other states are four-fold degenerate. To see this more formally consider the quantized Dirac field

@(x, t)= c [C,P#, t) +c#$44 t) I, 9+(x, t) = c [4&(x, 1) +cfe;k t) 1 9 (2.11) P P

224 C.J. Lee / Physics Letters A 195 (1994) 221-226

where it is to be understood that q~p(x, t )=exp( - iE . t )u ,~ (x ) , ~Oa(X , t )=exp(+iE. t )v , ._s(x) with Ep=E,, Ea= - E , , and c~ (cu) are creation (annihilation) operators for the electronic state. They are quantized accord- ing to Fermi-Dirac statistics: {Cp, c~}= J~, {Cp, cq} = {c~, C*q } = 0. Introduce dp= c*~ and d~-= c a. Then the Dirac Hamiltonian becomes in the second-quantized form

HD= J d3x :(/*(x, t)o.~Dfft(x, t): = ~ EpCtpCp+ ~ Epdtpdp. (2.12) P P

In the above : : denotes normal ordering, by which the infinite zero-point energy is removed. The first and the second terms are energies for the electron and the positron, respectively, and are equal in both magnitude and sign. Since there are already two electron (positron) states corresponding to s = + 1 for the energy E., the spec- trum is four-fold degenerate. The degenerate states are {u.+, u._, v.+, v._}. The exception is the ground state with doubly degenerate states { Uo_, Vo+ }.

3. Symmetries of the Dirac Hamiltonian

We have already given a symmetry operator, Ja, of the Dirac Hamiltonian in the previous section. In this section we elucidate other symmetry operators. First we note that, even though spin is not conserved, helicity is [ 9] [~D, 27" (p+ l elA) ] = 0, where 27= (g o is the spin operator. For the one-dimensional problem considered here the helicity operator reduces to A = 27x Px + Sr [ e I Bx, which in the 4 × 4 matrix notation is

0 - i l e l B x + p x 0 0 ) i lelBx+px 0 0 0

A = | O 0 0 - i l e l B x + p x \ o 0 i lelBx+px 0

~ [ / 0 - I n - 1 ) ( n l 0 =ix-, I n ) ( n - l l 0 0

0 0 0 0 0 I n ) ( n - l l

o) 0

- I n - 1 ) ( n l ' 0

(3.1)

where we used the relation following Eq. (2.8) along with b* = Z, I n) ( n - 1 I, b = Y, I n - 1 ) ( n I. Note further that the reinterpreted Dirac Hamiltonian in Eq. (2.1 2 ), which in the matrix notation is [ 7 ]

I n - l ) ( n - 1 1 0 0 0 )

0 In ) (n l 0 0 (3.2) H v = ~ 0 0 I n - l > < n - l [ 0 '

0 0 0 In><nl

also commutes with the helicity operator. It is easy to check that the states transform under A as (for n > 0)

A A . _-:--Tr'~-'~_ u,,+ = + un~ , _ - (3.3) ix/2lelBn - i ~ v"+ = +v,~ .

The helicity operator annihilates the ground states, so it is an exact symmetry of the system. Therefore, the spectral behavior of either the electron or the positron can be attributed to the symmetry of the Dirac Hamilto- nian (both ~D and Hv) under the transformation by the helicity operator A.

What about the degeneracy between the electron and the positron states? To answer this question note that this time the degeneracy and hence the underlying symmetry stems from interpreting the electron with quantum

C.J. Lee / Physics Letters A 195 (1994) 221-226 225

number/i as the positron with quantum number p as is done in Section 2. Thus the relevant Hamiltonian is HD, not the original Hamiltonian YfD. We found that the combined operation of charge conjugation and time reversal

y2~j,= _ i (~ O 1 ) = -p2 (3.4)

commutes with HD, but not with YfD. Actually, P2 anticommutes with Yfv and with the symmetry operator J3 introduced in the previous section. Consequently, the symmetry generator P2 gives rise to the degeneracy be- tween the electron and the positron states with opposite s values. It couples the states as

P2 \vn+ / \ - un~: /

For the ground state we have p2uo_ = -Vo+, and hence the ground state degeneracy may be regarded as being due to the spontaneous breaking of the symmetry P2.

Finally, we have to account for the degeneracy between the electron and positron states with the same s values. This can be done with the product of operators Ap2, since both commute with Hr . Explicitly, the states transform under Ap2 as

ix/21elBnkv~+_l \ u,,+_- /

and Ap2uo_ =Ap2vo+ = 0. Thus, Ap2 is also an exact symmetry. Are there other symmetry operators? One can verify that among the remaining Dirac matrices only

or3 = ( 0 ~) commutes with HD. However, a3 = iJ3P2, so the effects of or3 are not independent of and easily de- duced from the other symmetry generators already discussed. The results are

a 3 ( u ~ + - ~ = i ( - v , , ~ . (3.7) \v~+ I \ u,,~: I

Also, the transformation of the states under Aot3 is

Aa3 ( u , , + ~ = + i ( v,,+ ~ ( n > 0 ) (3.8) i~/2le[Bn" " - / - k-u, ,+

and Aot3uo_ =Aot3Vo+ = O. Therefore, the degeneracy of the system can be completely explained with either the set {A, P2, Ap2} or the set {A, a3, Act3}. Fig. 2 summarizes how the states transform under the former set of symmetry generators.

t .~ t~_ Vr,, V._

U,_ Vo.

Fig. 2. Degenerate eigenstates of the Dirac I-/amiltonian and symmetry transformations (denoted as arrows) that couple the states. Only the set {A, P2, Ap2} is used to illustrate the connection. See text for the explicit form of the states and the symmetry operators.

226 C.J. Lee/Physics Letters A 195 (1994) 221-226

4. Condusions and discussions

We have shown in this paper that the discrete spectral degeneracy of a Dirac electron in a uniform magnetic field can be completely accounted for in terms of three symmetry operators {A, P2, Ap2} or {A, t~a, Aot3}. In particular, we were able to explain not only the degeneracy for the electron (or positron) states, but also the degeneracy between the electron and the positron states. The degeneracy in the spectrum arises from the mini- mal substitution or, equivalently, setting the gyromagnetic ratio g = 2. In reality, the anomalous magnetic mo- ment of the electron destroys the degeneracy. So the symmetries given are good for the "pure" Dirac electron and approximate for a real one.

The viewpoint that the degeneracy of the spectrum of the Dirac electron is a manifestation of supersyrnmetry hinges upon the choice of the zero of energy. In general supersymmetric theories the spectrum is semipositive definite, so if supersymmetry is unbroken the ground state energy is exactly zero. Hence, the relevant Hamilto- nian is not HD but HD-- m. This in turn means that not only the divergent zero-point energy is removed but also the zero of energy is chosen to be the ubiquitous rest energy. In nonrelativistic theories the rest energy is taken off when calculating energies. We note that such a choice of the zero of energy is behind the supersymmetry theories [ 4-7 ] to explain the two-fold degeneracy of the Landau levels, which is the nonrelativistic counterpart of Eq. (2.6). On the other hand, it was shown in this paper that the spectrum of the relativistic electron can also be completely explained in the more conventional context of normal ordering, which also removes the infinite zero-point energy, resulting in the finite ground state energy m. Consequently, it seems that the normal ordering prescription is as "legitimate" [ 10 ] as the supersymmetry consideration along with the aforementioned choice of the zero of energy.

Acknowledgement

This work was supported by KOSEF under Grant No. 921-0300-005-2.

References

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[ 10 ] L.E. Gendenshtein and I.V. Krive, Usp. Fiz. Nauk 146 (1985) 553 [ Soy. Phys. Usp. 28 ( 1985 ) 645].