quantum mechanis of strong mag field

8/13/2019 Quantum Mechanis of Strong Mag Field

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HIT Journal of Science and Engineering, Volume 3, Issue 1, pp. 5-55

Copyright C 2006 Holon Institute of Technology

Quantum mechanics of electrons

in strong magnetic field

Israel D. Vagner1,2,3,, Vladimir M. Gvozdikov2,4 and Peter Wyder2

1

Research Center for Quantum Communication Engineering,Holon Institute of Technology, 52 Golomb St., Holon 58102, Israel

2Grenoble High Magnetic Fields Laboratory,

Max-Planck-Institute fr Festkrperforschung and CNRS,

25 Avenue des Martyrs, BP166, F-38042, Grenoble, Cedex 9, France3Center for Quantum Device Technology, Department of Physics,

Clarkson University, Potsdam NY, USA4Kharkov National University, Kharkov 61077, Ukraine

Corresponding author: [email protected]

Received 24 February 2006, accepted 9 March 2006

Abstract

A complete description of the quantum mechanics of an electronin magnetic fields is presented. Different gauges are defined and therelations between them are demonstrated.

PACS: 73.20.Fz, 72.15.Rn

1 Quasiclassical quantization

The classical Larmor rotation of a charged particle in a homogeneous exter-nal magnetic field is presented in Fig. 1

Consider first the quantization in the semiclassical approximation [1, 2].The motion of a charge in a magnetic field is periodic in the plane per-pendicular to the field and, hence, can be quantized by using the standardBohr-Sommerfeld quantization condition which after the Peierls substitution

p pec

A

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yields Imv e

cA

dr= (n + )h (1)

where the vector potential A is related to the magnetic field by

B= A (2)

e and c stand for the electron charge and light speed correspondingly. Wecan take= 12 as is the case in the original Bohr-Sommerfeld quantizationrule, but its real value follows from an exact solution of the Schrdingerequation, which will be discussed later.

Figure 1: The classical Larmor rotation of a charged particle in a homoge-neous external magnetic field.

To estimate the integrals in Eq. (1) we have to take into account that inthe plane perpendicular to the magnetic field the classical orbit of a chargedparticle is a circumference of the Larmor radius L along which the particlemoves with the velocity | v |= L, where the Larmor frequency is givenby relation = eBmc . After that the integrals in Eq. (1) can be calculated as

follows: I A dr =

Z [ A] nds=2LB , (3)I

mv dr = 22LeB

c . (4)

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Now from Eqs. (1), (3) and (4), it is easy to see that the Larmor ra-

dius L,n is quantized, i.e. takes a discrete set of values depending on theinteger n:

L,n=

s2~

m

n +

1

2

. (5)

Figure 2: Larmor orbits in XY plane.

These quantized electron orbits are shown in Fig. 2. In the quasiclassical

limit under consideration (i.e., for n 1), the Larmor orbit size dependsboth on the strength of the magnetic field and the quantum number n. Thisdependence is given by the relation

L,nr

n

B . (6)

Formally, small values of the quantum number n are beyond the scopeof the quasiclassical approximation. On the other hand, it is known that inthe case of harmonic oscillator the Bohr-Sommerfeld quantization rule givesan exact formula for the energy spectrum. Since the motion of a particlealong the circumference with the constant velocity is equivalent to harmonicoscillations, we will see below that one can considernas an arbitrary integeror zero. Thus, puttingn= 0(i.e. in the extreme quantum limit) the Larmororbit radius becomes equal to

L,0 =

rc~

eB

r 0

2B. (7)

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Two fundamental quantities appear in the right hand side of this equa-

tion: the magnetic flux quantum 0 = hc/e which depends only on the

world constants and the magnetic length LH=q

c~eB . The flux quantum is

the lowest portion by which the magnetic flux through some current car-rying loop can be changed. This has a far reaching consequences, as wewill see later. But now let us turn to our problem of the energy spectrumcalculation. With this end in view, consider the above quantization in themomentump-space. The classical equation of motion yields

dp

dt =

e

c

B

dr

dt

. (8)

Figure 3: Quantized orbits in KxKy plane.

One can see from this equation that: (i) the electron orbit in the p-spaceis similar to that in the real space (in x y plane, when field is directedalong the zaxis as it is shown in Figs. 2 and 3), (ii) in the p-space theorbit is scaled by the factor eB/c,and (iii) the orbit is rotated by =/2.

Integrating the above equation of motion we obtain

pn = eB

c L,n =

s2m~

n +

1

2

. (9)

We see that quantization of the orbit radius means also quantization ofthe momentum p which, in turn, implies quantization of the kinetic energy

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of the particle. For the quadratic dispersion relation, E= p2/2me, we have

En = p2n2me

= ~

n +

1

2

. (10)

This result indicates that the kinetic energy of a charged particle in thex y plane is quantized in the external magnetic field and the separationbetween the nearest quantum levels is ~. Equation (10) is known in theliterature as the Landau formula which describes equidistant quantum spec-trum of a charged particle in external magnetic field also called Landaulevels.

Thus, the electron motion perpendicular to the magnetic field is quan-

tized, but the motion along the magneticfi

eld is unaff

ected by the magneticfield and remains free. The corresponding kinetic energy spectrum in thez-direction is given by

Ez = p2z2m

. (11)

Putting together Eqs. (10) and (11) we arrive at the dispersion equation

En(pz) = ~

n +

1

2

+

p2z2m

. (12)

This energy spectrum is shown in Fig. 4.

Figure 4: The Landau levels. Energy dispersion.

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To obtain a more concrete idea about the size of the gap between two

adjacent Landau levels, we estimate itthe Landau gap. In a typical metal inthe fieldB of the order 10Tand taking electron massme' 10

27g, one canestimate the gap roughly as ~= ~eB/mec' 1.5 1015erg '15K. In asemiconductor as GaAs, for example, the effective mass of electron may bemuch smaller , saym ' 0.07me. The Landau gap in this case is of the orderof200K in the field of the same strength B = 10T. In the GaAs/AlGaAsinterface electrons are trapped and behave as a two dimensional electron gas.The energy spectrum of the electrons is completely discrete if the externalmagnetic field is applied perpendicular to the interface. This results inunusual magnetotransport phenomena which we will discuss later in moredetail.

As a result of quantization, all electron states are quenched into theLandau levels. Therefore, each Landau level is highly degenerated. Thedegeneracy of the Landau level g(B) can be calculated as the number ofelectronic states between the adjacent Landau levels

g (B) = g2d~ = 2SB

0, (13)

where

g2d=mS

~2 (14)

is the density of states for a two dimensional electron gas with quadratic

dispersion (see Eq. (12)).A schematic illustration of the density of states is shown in Fig. 5. We

can see from the Eq. (13) that the degeneracy of a Landau level is relatedto the number of the magnetic flux quantum 0 piercing the sample of areaSin a magnetic fieldB . Factor2 in theg (B)comes from the spin degree offreedom. The energy dependence of the 2D density of states shown in Fig.5 can be obtained by counting the number of states within a thin ring of thewidth dk and radius |k| in thek-space.

The density of states in k-space is then 2S/42

2 S

42 2kdk= g(E)dE , (15)

where the factor 2 accounts for the spin. For free electrons with the quadraticdispersionE= ~2k2/2mwe obtain the following expression for the densityof states

g(E) =dN

dE =

Sm

~2 . (16)

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Figure 5: The density of states for a two dimensional electron gas withquadratic dispersion.

It is clear from Eq. (16), that the density of states for a free electrongas in two dimensions is energy independent. Because the Fermi energy EFis obtained by filling electron states up from the lowest energy, the EF isrelated to the areal density ns and the density of states as

EF = nsg2d

=~2

m ns. (17)

We see, therefore, that the Landau spectrum for free electrons in the 2Dcase can be obtained on the basis of elementary quasiclassical consideration.In metals and semiconductors the dispersion relation for the quasiparticles,the conducting electrons, usually far from being a quadratic function of thequasimomentum. The trajectory of the conduction electrons in an externalmagnetic field is determined by cross section of the Fermi surface, which as arule is rather complex. Nonetheless, the quasiclassical quantization method

works well in this case too and we consider them later in this paper.

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1.1 The Landau quantization as a flux quantization prob-

lem. The quasiclassical approach. The Lifshitz-Onsagerquantization rule

We shall discuss in this section a relation between the Landau quantizationand the flux quantization. It is known that the energy of a particle movingalong the closed classical orbit becomes quantized in the quantum limit.A free 2D electron with the dispersion relation E = p2/2m in externalmagnetic field moves along the circle of the Larmor radius with the cyclotronfrequency =eB/mc.Since the energy is the integral of motion in this casethe trajectory of the electron in the momentum space is a circumference ofthe radiusP =

2mE too. The quantization of this motion, as was shown

before by many ways, yields the Landau energy spectrum

En = ~ (n + 1/2) . (18)

Let us rewrite this formula in the following fashion:

Sp(E) =2~eB

c (n + 1/2) (19)

where Sp(E) =P2 is the area of a circle of the radius P =

2mE along

which the electron moves in the momentum space. On the other hand,it follows from the classical equation of motion, Eq. (8), that trajectoriesin the coordinate and momentum spaces are of the same form but turned

related to each other by the right angle and scaled by the factor eB/c. Thelatter means that the radii of the circumferences in the coordinate R andmomentumP spaces are related by the condition R= cP/eB. Taking thisinto account we can rewrite the Landau quantization formula in the followingfashion:

SR(E) =2~c

eB (n + 1/2) (20)

where SR(E) = R2(E) is the area inside the circle of the radius R(E) =

cP(E)/eB in the coordinate space. We can calculate then the flux throughthis circle R(E) =SR(E)B and see that this quantity is quantized

R(E) =

0(n + 1/2) (21)

where 0is the flux quantum. Therefore, we see that both in the coordinateand momentum spaces the Landau quantization means the quantization ofthe area inside the closed classical trajectory but with the different steps:SR = 2~c/eB in the coordinate space and SP = 2~eB/c in the mo-mentum space. In the coordinate space the Landau quantization also means

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the flux quantization through the closed loop with the quantum 0. The

above quantization rules can be easily generalized to the case of an arbi-trary electron dispersion which is the usual case in the crystal solids likemetals and semiconductors. The quantization of theSp(E) is known in theliterature as the Lifshitz-Onsager quantization rule.

The Lifshitz-Onzager quantization rule is a direct consequence of thecommutation rules between the momentum components p= (~/i)/x +(e/c)A in the external magnetic field B directed along the Z-axes of theCartesian coordinate system:

[px,py] =e~

cB, [py,pz] = [px,pz] = 0 (22)

(whereA is the vector-potential). These equations mean that the momen-tum px and the coordinate qx= c py/eB satisfy the stand commutation rule[px,qx] = ~/iso that the quasiclassical quantization rule holdsI

pxdqx= 2~(n + ). (23)

This equation is exactly the Lifshitz-Onzager quantization rule in asmuch as the integral

Hpxdpy =Sp(E, pz) equals to the cross-section of the

Fermi surface by the planepz =constant.

S(E, pz) =2~eB

c

(n + ). (24)

We can obtain Eq. (24) within the Feynman scheme since only oneclassical path connects two arbitrary points p0 and p1 at the trajectorywhich is a cross-section of the Fermi surface. The appropriate Feynmanamplitude is given by

exp

i

Scl~

= exp

i

c

eB~

Z p1p0

px(p0y)dp

0y

, (25)

where Scl denotes the classical action at the segment between the points p0andp1.If the amplitude to find an electron in the pointp0 at the trajectoryisC(p0)then an amplitude to arrive at the same point of the Fermi-surfacecross-section after the one complete rotation is C(p0) exp[i(E, pz) +i]where

(E, pz) = c

eB~

I px(p

0y)dp

0y =

c

eB~S(E, pz) (26)

and is some arbitrary constant. Equating these amplitudes and taking = we arrive at the Lifshitz-Onsager quantization rule of Eq. (24). In

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solids the Fermi surface repeats periodically along the crystal symmetry

directions. This means that in the external magnetic field the cross-sectionof the whole energy surface by the plane pz =constant yields a networkof periodic classical orbits. In some organic conductors and conventionalmetals this 2D network consists of closed orbits connected by the magneticbreakdown centers. We will consider a generalization of the Lifshitz-Onsagerquantization rule to this problem later.

2 Gauge invariant formulation

2.1 The gauge invariance in a classical-analogy approach

In this section we discuss the correspondence between the classical descrip-tion of rotation of a charged particle in external magnetic field and the el-ementary quantum mechanical consideration of this motion. We start withthe classical description of the Larmor orbit.

Figure 6: The classical Larmor orbit where20= x20 + y

20 describes the center

of rotation and is the radius vector directed to the rotation point.

In the quantum mechanical approach, the classical dynamic variablesare generalized to be the quantum operators. In the classical picture, theLarmor radius is given by L =v/ . The classical Larmor orbit is shownin Fig. 6, where20 =x

20+y

20 describes the center of rotation and is the

radius vector directed to the rotation point. These vectors are related by

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the equation of motion

[ ( 0)] = v . (27)

Now turn to the quantum mechanical description in which the classicalvariables we defined above becomes operators. Proceeding in that fashionwe introduce first the operator for the center of orbit

20= x20+ y

20 (28)

where coordinates and velocity components are Hermitian operators

x0= xvy

, y0 = y+

vx

. (29)

Analogous, the Larmor radius operator is given by

2L = 1

2

v2x+ v

2y

. (30)

Let us discuss now some necessary for further commutation relations.Using the commutation identity relation

[A2, B] A[A, B] + [A, B]A (31)

and taking into account that the Hamiltonian of the charged particle in theexternal magnetic field is given by

H= 1

2m

p e

cA2

(32)

with the momentum operator p= ~i, we have

v= [H, r] = 1

m

p e

cA

. (33)

The commutation relation between the components of the velocity op-erator is given by

[vi, vk] = ie~

m2 ikB . (34)

We have obtained this result by direct calculations:

[vi, vk] = ie~

m2

Akxi

Aixk

=

ie~

m2 A= ie~

m2ikB .

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It is straightforward to see that the commutation relation for the coor-

dinate operators describing the center of Larmor rotation is given by

[x0,y0] = iL2H . (35)

The quantityLH in Eq. (35), the magnetic length, which is determinedbyL2H= ~c/eB. The commutation relation between the velocity and coor-dinate operator components is given by

[vi,xk] = i~

mik . (36)

The operators

x0 , y0 , 20 , 2L (37)

are integrals of motion because these operators commute with the Hamil-tonian h

H, x0i

=h

H,y0i

=h

H, 20

i=

hH, 2L

i= 0, [0, L] = 0. (38)

The commutation relations of Eq. (38) are analogous to those of the har-monic oscillator problem. We explore this similarity in what follows forfinding the energy spectrum of the electron in a magnetic field. With thispurpose in mind we discuss here the correspondence between the Larmorrotation and harmonic oscillator in a more detail.

It is natural to start with the Hamiltonian for the harmonic oscillatorproblem written in terms of the momentum Pand coordinate Q

H= P2

2m+

kQ2

2 . (39)

In quantum mechanicsPandQbecome operators Pand Qwhich satisfythe commutation relation

[P , Q] = i~. (40)The energy spectrum for this problem is then given by

En= ~

n +12

, (41)

where is the oscillator frequency

=

rk

m. (42)

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By using the commutation relations Eq. (38) and Eq. (36), we can make

the following correspondence between the Larmor rotation parameters andthe operators in the harmonic oscillator problem

vx P ; vy Q. (43)

The parameters of the Larmor rotation and the harmonic oscillator cor-respond as follows

k 22

; ~ |e|~Bm2c

; m 2

2 . (44)

By using this correspondence relations, it is straightforward to see that

we have the energy spectrum En = ~(n + 1/2). Because the Hamiltonianfor rotating particle is given by

H=mv2

2 =

m(v2x+ v2y)

2 , (45)

and taking into account relation L = v/, the spectrum of the Larmorradius operator is turned out to be discrete and determined by the equation

2Ln

= 2

m2En= L

2H(2n + 1) . (46)

Similarly the discrete spectrum for the center of rotation operator maybe obtained by using the definition of the20 and the commutation relationbetween the coordinates of the center of rotation:

(20)k =L2H(2k+ 1) (47)

The discreteness of the coordinates (L)n and(0)n yield the followingpicture for the electron orbits shown in Fig. 7.One can see the manifoldof concentric circles of discrete radius (0)n centered at the origin of thepolar coordinates and with the Cartesian coordinates x0 and y0 which de-fine 0. These coordinates are analogous to the q and p operators in theone-dimensional harmonic oscillator problem. Therefore, the Heisenberg

uncertainty principle can be applied to them to yield

x0y0L2H

2 . (48)

The distribution of centers of the electron orbit corresponding to a given can be represented geometrically as a manifold of circles with the radius

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Figure 7: The manifold of concentric circles of discrete radius (0)ncenteredat the origin of the polar coordinates and with the Cartesian coordinates x0andy0 which define0.

' L2H(2+ 1) as shown in Fig. 7. The quantum mechanical analog of theLarmor orbit is given by the equation

2L= (x xo)2 + (y yo)2 (49)

and the eigenvalues of the operator 2L are determined by the relation

(2L

)n

= (2n + 1)L2H

, (50)

where n= 0, 1, 2,... The geometric meaning of Eqs. (49) and (50) is illus-trated in the Fig.7.

We can introduce now the angular momentum operator related to theLarmor orbital motion as follows

Lz = ~L2H

2

20 2L

. (51)

Its eigenvalueslz are quantized and given by the relation

lz =sgn(e)~(

n) = ~mz. (52)

Summing up the quasiclassical consideration of the Landau problem wemust say that quantized Larmor orbitals are only an approximations to theLandau orbitals which can be obtained only on the basis of the Schrdingerequation. This will be done in the next section. But before doing this weconsider briefly the gauge invariance in the quantum mechanics.

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2.2 The gauge invariance of the Schrdinger equation in an

external magnetic field

The phase of the wave function should not affects the observable quantitiesin quantum mechanics. In particular, the average of the Hamiltonian of aparticle,

(, H) =

Z~2

2m | |2 +U(r) | |2

dr (53)

should be invariant under the substitution ei, where = (r)is an arbitrary phase. The second term in Eq. (53) is invariant, but thefirst one is not because of the contribution of the gradient. To make itinvariant we must introduce some vectorfieldA compensating the gradientterm (i.e. containing) and require that this field does not change themagnetic field B as a result of the gauge transformations. Both conditionsare satisfied under the following substitution:

~2

2m | |2 1

2m | DA |

2,

where DA = (p ec A) and the magnetic field is related to the vector Aby the standard equation B=rotA, so that B does not changes under thegradient transformationA A + with being arbitrary function ofr.It is easy to check now that the quantity

Z 12m

| DA |2 +U(r) | |2

dr

is invariant under the gauge transformation 0 = ei if we also take= ~c in the gradient transformation A0=A + .Another words,

1

2m | DA0

0 |2= 1

2m | DA |

2 .

We see, therefore, that the formal Peierls substitution p p ec A, bywhich a magnetic field is introduced into Hamiltonian dynamics, finds itstheoretical justification only on the basis of the quantum mechanics. Itis a direct consequence of the gauge invariance under the transformation

0 = ei.

3 The Landau problem in the Landau gauge

Contrary to the elementary consideration of the previous sections whichdeals primarily with the external magnetic field, the solutions of the Schrdinger

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equation depends on the choice of the gauge for the vector potential. Two

gauges are most frequently used in the literature: theLandau gauge andthe symmetric gauge. Let us start with the Landau gauge.

The Schrdinger equation for the charged particle in an external mag-netic fieldB reads

HE=EE, (54)

where the Hamiltonian of a particle is

H= 1

2m

p e

cA2

. (55)

In the Landau gauge A = (

By, 0, 0) only the Ax component of the

vector-potential is nonzero, so thath

H, pxi

=h

H, pzi

= 0. The latter means

that the momentum components are the quantum integrals of motion andEin Eq. (54) should be also the eigenfunction of the operators px, and pz,which means that

E(x,y,z) = E(y)exp

i

~(pxx +pzz)

. (56)

Substituting (56) into (54), we have

H(y) E0(y) = E0E0(y) (57)

with

H(y) = ~2

2m

d2

dy2+

m2

2 (y y0)2 , (58)

E0 =E p2z

2m, (59)

where =eB/mc is the cyclotron frequency, and

y0 = cpx/Be (60)

denote the coordinate of the center of the Landau orbit.Equations (57) and (58) shows the equivalency of the Landau problemto the problem of the quantum oscillator.

Let us introduce a dimensionless variable

q=

rm

~ (y y0) = y y0

LH. (61)

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H0(q) = 1, H1(q) = 2q, H2(q) = 4q2 2.

3.1 The density of states

Consider now another important characteristic of the Landau problem - thedensity of states. According to the definition the density of states can becalculated as a sum over the Landau energy spectrum

g(E) = 2

0

Xn

Z Lzdpz

2~ (EEn(pz)) . (66)

The factor 2 0

appears here because of the degeneracy of the Landau

levels on the spin and orbit position. The integration on pz in Eq. (66) istrivial because of the delta-function. Completing it, we have

g(E) =X

n

gn(E). (67)

The quantity gn(E) is the density of states in three dimensions for theLandau level with the quantum number n

gn(E) =V

2m3/2

2~3~

qE ~ n + 12

, (68)

whereVis the volume of the sample. We see that the density of statesg (E)in the Landau problem has a periodic set of the square-root singularities.This type of the singularity is typical for a one-dimensional system. Thus,for fixed value of the quantum number n a motion of electron is effectivelyone-dimensional. We, therefore, may calculate that external magnetic fieldeffectively reduces the dimensionality of the system.

3.2 The momentum representation

It is well known that the unitary transformation does not change the eigen-values of the Hamiltonian. On the other hand, sometimes a proper choice of

the unitary transformation makes a solution of the eigenproblem much moreeasier. Many eigenvalue problems become simply in the momentum repre-sentation since this approach is, in essence, nothing but a Fourier methodknown in conventional mathematical physics. In this connection it is inter-esting to note that in the Landau problem the Hamiltonian given by Eq.(58) is invariant with respect to the momentum representation. Indeed, in

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the momentum representation the momentum operator becomes just a vari-

able,p=p,while the coordinate becomes a differential operatory= i~/p.Thus, making first the coordinate shifty +y0 ywe can rewrite the Hamil-tonian of Eq. (58) in the momentum representation as follows

H= p2

2m m

2~2

2

2

p2. (69)

This Hamiltonian, after the substitution

q= p/

m~, (70)

and introduction of the operators

a= 1

2

q+

d

dq

, a+ =

12

q d

dq

,

a, a+

= 1

takes exactly the form of Eq. (72). Therefore, there is no need to solve theproblem anew since the energy spectrum and the wave functions remain thesame with the only difference that q in n(q) of Eq. (81) is given now byEq. (70).

For further consideration it is useful to define a couple of Hermitianconjugate operators aand a+

a=

1

2 q+ d

dq

, a+

=

1

2 q d

dq

,

a, a+

= 1. (71)

In terms of these operators the Hamiltonian (58) takes a very simplequadratic form

H= ~

a+a +

1

2

. (72)

We begin the analysis of this Hamiltonian from the definition of thelowest energy eigenstate state, also known in the quantum theory as theground state. To find the ground state let us consider the average of theHamiltonian (58)

E, HE =

~

2 (E, E) + ~ (aE, aE) , (73)

which should be real and positive number since H is Hermitian operator.The ground state wave function must minimize Eq. (73) and one can easilyconclude that this holds under the condition

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a0= 0, (74)

which nullifies the second term in the right-hand-side of the Eq. (73) and

thereby minimizes quantity

E, HE . The differential equation (74) has

a trivial solution, which, after normalization by the condition (0, 0) = 1,yields

0(q) = 1/4 exp

q

2

2

. (75)

One can check then by a direct substitution, that the wave function

n= a+n

n!

0 (76)

is the normalized eigenfunction of the Hamiltonian (72)

Hn = Enn (77)

with the eigenvalue

En= ~

n +

1

2

. (78)

The explicit form for the function n(q) directly follows from Eqs. (71)

and (76) which yield

n(q) = 1n!2n1/2

q d

dq

ne

q2

2 . (79)

One can rewrite the wave function of Eq. (79) in a conventional formwith the help of the Hermitian polynomials Hn(q)

Hn(q) = (1)n

eq2 dn

dqneq

2

. (80)

The final result is:

n(q) = Hn(q)

n!2n1/2e

q2

2 . (81)

Let us summing up the results for the Landau problem.

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3.3 The uncertainty principle in the Landau problem

In this section we shall discuss briefly the uncertainty principle for the coor-dinate and momentum in the Landau problem. Consider for simplicity theground state. The wave function of the ground state both in the coordinateand momentum representation is

0(q) = 1/4 exp

q

2

2

(82)

with q= yy0LH in the coordinate representation and q= pLH

~ in the momen-

tum representation. This means that the coordinate uncertainty (the widthof a strap in the y -axes direction where the probability to find a particle

is appreciable) equals approximately to the y ' LH. The correspondinguncertainty in the momentum of a charged particle in the external magneticfield is p ' ~/LH. Thus, the product of these uncertainties equals to

py ' ~. (83)

Having at hand the wave functions we can calculate the above uncer-tainties exactly. To do this we will proceed in such a fashion. Firstly, itfollows directly from the definition of Eq. (76) that the eigenfunctionsnobey a simple recurrent relations:

a

+

n=

n + 1n+1, (84)

an=

nn1. (85)

On the other hand, from Eq. (71) we have a couple of equations whichexpress the coordinate and the momentum operators in terms of the quan-titiesa+ anda:

y=LH

2(a+ + a), (86)

py = ~

iLH

2(a+ a). (87)

Note that Eqs. (86)-(87) are valid both in the coordinate and momentumrepresentations. Using these equations and taking into account the orthog-onality of the basis n, we can calculate the uncertainties in question withthe help of the formal quantum mechanical definitions:

yn=p

(n,y2n) (n,yn)2, (88)

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= exp | 2 |

2

Xn=0

nn!

n, (96)

where exp ||22

is the normalization coefficient.

One can recast the coherent state vector (96) in a more compact form

= ea+a0. (97)

by taking advantage the well-known operator identity

eA+B =eAeBe

1

2[A,B], (98)

which holds because A= a+ and B =a in our case and the commu-tator

hA, B

i=| |2 is the c-number (not operator).

It is easy to check straightforward that the coherent states are nonorthog-onal

(, ) = exp

1

2

| |2 + | |2 +

. (99)

In the case of | | 1they are approximately ortogonal in as muchas the absolute value of the scalar product (99) is small

| (, ) |= exp12 | | 1. (100)

The set of the coherent states is complete. The completeness means thatfollowing identity holds for the wave vector (q):

1

Z d2(q)(q

0) =

q q0

. (101)

This expression immediately comes out from the Eq. (96) and the com-pleteness of the Landau basis functions n(q):

Xn

n(q)n(q0) = q q

0

. (102)All the equations considered so far are valid both in the coordinate and

momentum presentations. In the coordinate presentation the quantityq isgiven by the Eq. (61) and in the momentum presentation by Eq. (70) .

Using an explicit form for the Landau basis functions n(q) ( see Eq.(81)) and substituting them into the Eq. (96), we have

27


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(q) = 1/4 exp

| |2

2 q2

2

Xn=0

2

n Hn(q)

n! . (103)

The sum in Eq. (103) can be easily calculated with the help of thegeneric function relation for the Hermitian polynomials

e2xtt2

=X

n=0

Hn(x)tn

n!. (104)

This equation simply means that the coefficients of the power series ex-pansion with respect to the variable t for the function standing in the left-hand-side are equal to the Hermitian polynomials given by Eq. (80). This

statement can be easily checked by direct calculations. Thus, taking intoaccount the Eq. (104) one may recast the Eq. (103) into the followingGauss-like form

(q) = 1/4 exp

| |

2

2 +

2

2

exp

q2

2!. (105)

This Gauss-like wave function is known to minimize the uncertainty re-lation for the coordinate and momentum (i.e. makes the right hand side inthe equation yp= ~/2exactly equal to the lowest value ~/2).

Such a wave function was first introduced by Schrdinger under the nameof a coherent state. We see that the Schrdinger definition of the coherentstate and that given by the Eq. (93) are identical in essence.

One of the practical advantages of the coherent state is that thematrix elements of the type

, a

+nam

can be calculated very easy inthe basis of the coherent states. For example, it is easy to check that

, a

+nam

=

an, am

= ()2 m

,

. (106)

With the help of this equation we have from Eq. (72)

, H= ~| |2 + 12 . (107)

On the other hand, the average values of the operators x and p in thecoherent state are equal to

(, x) =

2~

m

1/2Re; (,p) = (2m~)

1/2 Im. (108)

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Combining Eq. (107) and (108) we obtain

, H

=

m2

2 (, x)

2 + 1

2m(,p)

2 +~

2 . (109)

This expression, written in terms of averaged xand poperators, is verysimilar to the energy of the classical oscillator

E=m2

2 x2 +

p2

2m

and demonstrates a closeness of the coherent state description to the clas-sical approach. Another manifestation of this, as was note above, is the

fact that minimizes the uncertainty principle for the coordinate and themomentum. To see this, we can directly calculate these uncertainties in thecoherent state, which yields:

x =

q(,x

2) (,x)2 =

~

2m

1/2, (110)

p =

q(,p

2) (,p)2 =

m~

2

1/2. (111)

Multiplying these quantities, we have

xp = ~

2. (112)

Comparing this result with uncertainty relation in the Landau basis givenby Eq. (92) we see that only the ground staten= 0 minimizes the uncer-tainties product for the coordinate and momentum, since the ground state0(q) is exactly Gaussian in shape, i.e. it is the coherent wave functionaccording to the Schrdinger definition.

In as much as the coherent state was presented above as a series ofthe Landau states (see Eq. (96)) it is easy to write down the probabilitydistribution for the Landau quantum numbern in the coherent state:

W(n) =| C(n) |2=e||

2 | |2n

n! . (113)

Taking into account that | |2= n, where n = (, a+a)standsfor the average of the quantity

qp n in the -state, we see that Eq.(113) is nothing but the Poisson probability distribution function

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W(n) = ennn

n! . (114)

5 The symmetric gauge in the Landau problem

Because the external magnetic field imposes an axial symmetry to the Lan-dau problem, it is natural to solve the Schrdinger equation in the cylindricalcoordinates. The vector potential in the symmetric gauge is defined as fol-lows A = 12 [Br]. The symmetric gauge is very popular, for example, inthe theory of interacting many-body systems in a magnetic field. Thus,we shall use the cylindrical coordinates (,,z), in the Schrdinger equation

for a charged particle of the mass me in a magnetic field described by thesymmetric gauge

~2

2me

1

+

2

z2+

1

22

2

i~

2

+

m2

8 2 =E.

(115)The magnetic field here is assumed to be parallel to the z-axis and we

omit for brevity terms relating to the free motion along the field direction.The above Schrdinger equation can also be obtained in the cylindrical co-ordinates if we choose the vector potential as

A= 12

B , A= Az = 0 , (116)

which is just another form of the symmetric gauge.Because the coefficients in the differential Eq. (115) depend only on

the radial coordinate , the momentum pz = ~kz and angular momentumlz = i~/ are the quantum integrals of motion so that the solution canbe factorized to separate the variables

= zR(). (117)

Here z is the plane wave along the z-axis (the eigenfunction of the

pz = ~kz)z =e

ikzz, (118)

and is the eigenfunction of the operator lz

= eim (119)

30


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28/51

finite value of z, (b) should not take neither zero nor negative integer

values, and (c) is an arbitrary value, (d) F(,,z) is polynomial when is a negative integer. In our case the function u() satisfies Eq. (125) sothat its solution is given by

u= F(,,z) (126)

with

|m| + 1m2

+p2z/2me E

~

, (127)

|m| + 1 , (128)

and z = 2

. Combining these results and normalizingR by the conditionR0 R

2d= 1, we obtain the radial wave function in the form

Rn,m() = 1

L|m|+1H m!

s(|m| + n)!

2|m|n! e

2

4L2H|m|F

n, |m| + 1,

2

2L2H

.

(129)The energy spectrum is determined by the condition of the finiteness of

the wave function, which holds if is a nonzero negative integer, say n.This condition defines the energy levels as follows

E= ~n+ |m| + m + 12

+ ~2k2z

2me. (130)

It is useful to introduce a new quantum number

n=n+|m| + m

2 .

Then the energy spectrum acquire the standard form of the Landauspectrum

En(kz) = ~

n +

1

2

+

~2k2z2me

.

Each level has an infinite degeneracy since for fixed integern the orbitalnumberm takes values from

to n. Putting n = 0, and assumingm to

be positive (which means thatn =m) we note that the radial component ofthe wave function can be rewritten as a function of the complex coordinatez= x + iy in the following form

n(z) =

1

L2H2n+1n!

12

( z

LH)n

e

|z|2

4L2H. (131)

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In the quasiclassical limit (i.e., for large n 1), the electron wave func-

tion localized mainly within a ring of the width LHand radiusLH2n. Onecan see this after writing down the radial coordinate probability distributionfunction

|n|2 =C2

LH

2ne

2

2L2H . (132)

Calculating then the expectation values of the radius and its square, withthe help of this function, we have

hi = 2

Z

0 |n|

2 d= LH

2

n + 32

n + 12

, (133)

h2i = 2Z 0

2 |n|2 d= 2L2H(n + 1) . (134)

The normalization constant Cis given by the equation C2 =L2H/2n+1(n+

1). For largen 1, we obtain

hi ' LH

2n , (135)ph2i ' hi LH . (136)

On the other hand, the radial coordinate probability distribution func-tion2 |n|

2 has a narrow peak of the widthLHcentered at20= L2H(2n+

1). To see this we can do following elementary transformations:

2 |n|2 =C

LH

2n+1exp(

2

2L2H) = Cexp

G(

LH)

, (137)

where

G(

LH) =

2

2L2H (2n + 1) ln(

LH).

This function has a minimum at 2L = L2H(2n+ 1). Expanding then

G( LH)in the power series near the 0,we have

2 |n|2

exp[

( L)2

L2

H

].

The above relations mean that in quasiclassical limit a charged particlemoves most probably within the ring strap of the radius LH

2n and the

width LH


30/51

Figure 8: In a quasiclassical limit a charged particle moves most probablywithin the ring strap of the radius L

H

2nand the widthL

H


31/51

field B, i.e. j = 0. Computing the current in the other two directions of

the cylinder coordinate system in the symmetric gauge, we have

j =

e~m

me e

2B

2mec

|nmkz |

2 , (142)

jz = epz

me|nmkz |

2 . (143)

In the following subsections, we discuss another examples where thesymmetric gauge is convenient to apply for solving the Schrdinger equation.

6 Coherent state in the symmetric gauge

In this section we shall generalize the coherent state introduced previouslyin the Landau gauge to the case of the symmetric gauge. To do this itis necessary again to obtain a solution of the Schrdinger equation in thesymmetric gauge, but this time in the operator form which generalizes theapproach of the section 2.5.

The Hamiltonian of a charged particle in the symmetric gauge

A=1

2(By,Bx, 0) (144)

takes the form

Ht = 1

2m

"Px my

2

2+

Py+

mx

2

2#= (145)

= 1

2m

hPx

2 + Py2i

+m2

2

x2 + y2

+ mLz,

where

Lz = xPy Pxy (146)is the z-component of the angular momentum. (We do not consider a free

motion along the fieldB which is trivial, so that Htstands for the transversepart of the Hamiltonian.) Proceeding in the same fashion as in the section2.5, we can introduce a couple of operators

ax = 1

2

qx+

d

dqx

, ay =

12

qy+

d

dqy

(147)

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with qx = x/LH andqy = y/LH.

In terms of these operators the Hamiltonian (145) takes the form

Ht= ~

a+xax+ a+yay+ 1 i

aya

+x a+yax

. (148)

This quadratic form can be diagonalized with the help of the unitarytransformation

a1= 1

2(ax+ iay) , a2=

12

(ax iay) , (149)

which implies that both left and right hand side operators commute accord-ing to the Bose-like relationsh

ai, a+j

i=ij, [ai, aj] =

ha+i , a

+j

i= 0, (i, j = 1, 2, x , y) . (150)

After the diagonalization, the operators Ht and Lz become

Ht = 2~

a+1a1+

1

2

, (151)

Lz = ~

a+1a1 a+2a2

. (152)

These operators are linear combinations of the operator ni = a+i ai.

On the other hand, from the commutation relations [Ht,Lz] = [Ht, ni] =

[Lz, nj] = 0 it follows immediately that the eigenfunctions of the nj areat the same time the eigenfunctions for the operators Ht and Lz with the

corresponding eigenvalues

En1 = 2~

n1+

1

2

, (153)

Lz = ~(n1 n2) . (154)We see therefore, that the wave function of a charged particle n1n2 is

determined by a couple of quantum numbers n1and n2. It may be obtainedfrom the ground state 00 in the same manner as we have done it in the

section 2.5 for the case of the coherent state in the Landau gauge. Theequations for the ground state generalizing the corresponding definition ofthe section 2.5 given by Eq. (74) yield a100 = 0, a200= 0.

In the coordinate representation these equations read

L2H

x+ i

y

+

1

2(x + iy)00(x, y) = 0, (155)

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L2H x i y +12(x iy)00(x, y) = 0. (156)Solving Eqs.(155), (156) and normalizing the solution, we found

00(x, y) =

m

~

1/2exp

1

2L2H

x2 + y2

. (157)

In full analogy with the equation (76) the wave function belonging tothe energy EN= 2~(N+ 1/2) is given by

Nn= a+1

N

N! a+2

n

n!00. (158)

This wave function is degenerated with respect to the quantum numbernwhich is an arbitrary integer determining the Lz = ~(nN). Introducinga complex coordinate = x+iy one can rewrite Eq. (158) in an explicitform

Nn(, ) =

2nNm

~N!n!

1/2

Nne||

2

. (159)

We determine then the coherent state as the eigenstate for the op-erators a1 anda2

a1= LH

, (160)

a2=

LH. (161)

In the coordinate representation these equations take form of the differ-ential equations

+

1

4L2H

=

2L2H, (162)

+

1

4L2H=

2L2H. (163)

The normalized solution of Eqs.(162) and (163), as one can check by thedirect substitution, is

37


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= 1

LH

2exp

1

4L2H

| |2 + | |2

14L2H

( 2) ( 2) + 2L2H

.

(164)We see that is the Gauss-like wave function which means that

minimize the uncertainty principle for the coordinate and momentum

xp= ~

2. (165)

and, hence, is the coherent wave function in the sense of the Schrdingerdefinition.

7 Electron in the field of a thin solenoid. The

Aharonov-Bohm effect

Consider an electron moving in the magnetic field of an infinitesimally thinand infinitely long solenoid whose magnetic field is oriented along the z-axis of the Cartesian coordinates. Let the flux through this solenoid be ,so that the magnetic field of the solenoid depends only on the coordinater=xi + yjwithin the plane perpendicular to the magnetic field

B (r) = (r) . (166)

Since the Schrdinger equation depends on the vector potential A (r),we have to findA (r)such that

rotA (r) = (r) . (167)

Equation (167) implies thatI A (r) dl=

I I rotA (r) dS= (168)

for any contour around the singular point r= 0.It is easy to see that

A (r) =

2

k r

r2 =

2

yi + xjx2 + y2

. (169)

satisfies the required conditions. (i,j, k are the unit vectors along the X, YandZ axes correspondingly).

We can introduce an angle (r) between the r and X-axis by

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(r) = arctan

yx

. (170)

In terms of (r) the vector potential (169) reads

A (r) =

2

(r)

r . (171)

The Schrdinger equation for an electron in the field of the solenoid isgiven by

1

2m

~

i

r e

cA (r)

2 (r) = E (r) (172)

with the vector potential determined by Eqs.(169) and (171).In view of the axial symmetry of the problem the wave function should

be periodic under rotations on the angle = 2 around thez-axis.Consider a phase transformation of the wave function

0 (r) = e

i 0

(r) (r) , (173)

where 0 = 2~c

e is the flux quantum. Since |0 (r)|2 = | (r)|2 both 0 (r)

and (r)belong to the same quantum state. On the other hand, the primedwave function satisfies the Schrdinger equation without vector potential

~

2me

2

r20

(r) = E0

(r) (174)

because of the relation

~

i

r e

cA (r)

e

i 0

(r)

0 (r) = ei 0

(r)

~

i

r e

cA0 (r)

0 (r) (175)

in which

A0 (r) = A (r) 2

(r)

r = 0 (176)

in view of Eq. (171).We must supply Eq. (174) with the boundary conditions under rotation

which are rather unusual and depending on the flux . Indeed, since (r)is the2-periodic, we have from Eq. (173)

0 (+ 2) = e

i2 0

0 () . (177)

39


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In the polar coordinatesr and Eq. (174) reads

~2

2m

1

r

r

r

r+

1

r22

2

0 (r, ) = E0 (r, ) (178)

and one can separate variables in (r) (r, )

(r, ) = f(r) um() , (179)

where um() is the eigenfunction of the operator Lz = ~

i because of the

axial symmetry of the problem in question

um() = 1

2eim. (180)

Note first that in the polar coordinates x = r cos , y = r sin and theangle in Eq. (170), in fact, does not depend on coordinates.

Thus, the primed wavefunction becomes

0 (r, ) = f(r)

eim

0

2

, (181)

where m is an integer (the Lz quantum number).We see therefore that one can interpret the angular dependence of the

wavefunction (181) in terms of the angular momentum projection on the

solenoid axis, i.e.12

eiLz

~ = 1

2ei

m 0

. (182)

This implies that the angular momentum is now depends on the fluxthrough the solenoid:

Lz = ~

m

0

. (183)

Substituting then 0 (r, )(181) into Eq. (178) we arrive at the equationfor the radial wavefunction f(r):

d2f

dr2 +

1

r

df

dr+

m

0

2r2

f 2meE~2

f= 0. (184)

This is the Bessel equation and its regular at r= 0solution is given bythe Bessel function J:

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f(r) = CJ(kr) , (185)

where

k=

r2meE

~2 and =

m 0

. (186)

In case when there is an unpenetrable cylinder at r = R, we havef(R) = 0and correspondingly

J(kR) = 0, (187)

which means that kR is one of the roots of the Bessel function (n, )

kn= (n, )

R , (188)

so that the energy spectrum is a discrete and determined by equations (186)and (188)

E(n, ) = ~

2

2mR22 (n, ) . (189)

We see that angular momentumLz (183) and the energy spectrum (189)depend on the magnetic field /0 though electron moves in the regionr > 0, where magnetic field equals zero. This paradoxical phenomenonwhich has no analog in classical electrodynamics was predicted in 1959 byAhronov and Bohm. In classical mechanics the Lorentz force acts locallyand therefore has no impact on a charged particle in the region where B = 0even though the vector potential is nonzero. After its theoretical predictionin 1959 the Ahronov-Bohm effect have been found then experimentally andhas numerous manifestations in the modern physics.

8 The density matrix of a charged particle in quan-

tizing magnetic field

So far we have considered the Landau problem within the Schrdinger equa-tion approach which implies that the charged particle is isolated from theenvironment and only under this assumption a description in terms of thewave function is relevant. In reality electrons in the solids are involved indifferent interactions and move under the action of the atomic forces fromthe crystal lattice. Another words they correlate somehow with the rest

41


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of the sample. This correlation means that the true quantum mechanical

description should be based not on the wave function but rather must bedone in terms of the density matrix . The density matrix approach is analternative to the Schrdinger equation description in case when a system isin contact with the environment. In this section we will give a description ofthe Landau problem in terms of the density matrix in the most simple casewhich assumes a contact between the charged particle and the thermostat(environment) being at the temperature T .

8.1 The density matrix in the Landau problem

The Landau energy spectrum and the wave functions have been calculated in

detail in section 2.4. According to the results of this section we can write thedensity matrix (r, r0, ) in the Landau basis (taken in the Landau gauge)as follows:

r, r0,

= 1

2~

Z

dpy(q, q0, )ei

py~ (yy0)k(z z0, ). (190)

Here the quantityk(zz0, )stands for the longitudinal density matrixfor a free particle moving parallel to the magnetic field

k

(z

z0, ) = 1

2~Z

dpze

p2z2m

ipz~ (zz0) (191)

and the function (q, q0, ) is the density matrix of the Larmor oscillator

centered at the coordinatex0(py) = cpy/eB in the plane perpendicular tothe applied magnetic field:

(q, q0, ) =

Xn=0

e~(n+1

2)n(q)n(q

0). (192)

The Landau basis n(q) is given by the Eq. (81), and dimensionlesscoordinates q and q0 are connected with the xaxes coordinates by therelations

LHq= x x0(py), LHq0 =x0 x0(py). (193)Completing integration in the Eq. (191) we obtain

k(z z0, ) =r

m

2~2e

m

2~2(zz0)2

. (194)

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This quantity, normalized by the condition k(z

z0, 0) = (z

z0), is

exactly the statistical operator for a free particle in a one dimensional space.To calculate the(q, q

0, )is a more tricky business. With this purposein mind, we first derive the differential equation for this quantity. To dothis, note that from the definition of the operators a and a+ by Eq. (70)and Eqs.(84), (85) it follows that

qn(q) = 1

2

nn1(q) +

n + 1n+1(q)

, (195)

qn(q) =

12

nn1(q)

n + 1n+1(q)

. (196)

Using these equations as well as Eq. (192), we have

q(q, q

0, ) = e~f(q, q0) f(q0, q), (197)

where the following function was introduced

f(q0, q) = 1

2

Xn=0

e~(n+1/2)

n + 1n(q0)n+1(q). (198)

With the help of Eqs.(195) and (192) we find a useful relations betweenthe functionf(q0, q) and the perpendicular component of the statistical op-erator:

q(q, q0, ) =e~f(q, q0) + f(q0, q), (199)

q0(q, q0, ) = e~f(q0, q) + f(q, q0). (200)

Combining Eqs.(197)-(200) we arrive at the differential equation for thedensity matrix,which reads

q(q, q

0, ) = q

tanh ~+

q0

sinh ~(q, q0, ). (201)

The solution of this simple equation is trivial an yields

(q, q0, ) = C(q0, )exp

q2

2tanh ~ qq

0

sinh ~

. (202)

43


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According to the definition (192) the quantity (q, q0, ) is symmetric

with respect to the substitution q q0. This condition tells us that theconstant C(q0, )should be taken in the form

C(q0, ) = C0()exp

q02

2 tanh ~

. (203)

It follows also from the Eq. (65) that the function (q, q0, 0)should be

normalized by the condition

(q, q0, 0) =

q q0 . (204)

Choosing then the constantC0()to satisfy the equation (4.28), we have

(q, q0, ) = (2 sinh ~)1/2 exp

(q

2 + q02)

2tanh ~+

qq0

sinh ~

. (205)

Substituting the Eq. (205) into the Eq. (190), we find

r, r0,

= (x, x0, y , y0, )k(z z0, ), (206)

where the perpendicular component of the density matrix(x, x

0, y , y0, ) = (x, x0, y y0, ) is determined by the integral

(x, x0, y , y0, ) =

1

2~Z

dpy(q, q0)ei

py~ (yy0) (207)

and the dependence on the momentum py enters the function (q, q0)

through the dimensionless coordinates

q(py) = [x x0(py)]/LH, q0(py) = [x0 x0(py)]/LH. (208)

We can single out of the Eq. (207)osc(x, x0, ) the statistical operator

of the quantum oscillator of the frequency so that(x, x0, y , y0, )can be

written as a product

(x, x0, y , y0, ) =osc(x, x

0, )G(x, x0, y , y0, ), (209)

where osc(x, x0, ) is given by the formula

osc(x, x0, ) =

m

2~sinh ~

1/2exp

m

2~

x2 + x02

tanh ~ 2xx

0

sinh ~

.

(210)

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TheGfunction is given by the Gauss integral

G(x, x0, y , y0, ) = 1

2

Z

dkyeAk2y+Bky =

1

2

r

AeB2

4A (211)

with

A=L2Htanh

~

2

, B(x, x0, y , y0, ) = tanh

~

2

(x + x0) + i(y y0).

(212)Combining all these equations , we finally have

r, r0,

= k(z z0, )

4L2Hsinh

~2

eS(x,x0,y,y0,), (213)where

S(x, x0, y , y0, ) = (214)

= 1

4L2H

coth

~

2

(x x0)2 + (y y0)2 + 2i(x + x0)(y y0) .

The partition function of the problem in question is given by

Q () =Z

Lx

0dx

Z Ly

0dy

Z Lz

0dz (r, r, ) . (215)

Taking into account that S(x,x,y,y,) 0 we see that (r, r, ) doesnot depend on the coordinate r

(r, r, ) =k(0, )

4L2Hsinh

~2

. (216)Then, after the trivial integration in the Eq. (215) we obtain an explicit

formula for the partition function Q ():

Q() = 0

12sinh

~2

Lzr m2~2

. (217)

The origin of each factor in the equation (217) is absolutely clear: g =LxLy/2L

2H = /0 is the degeneracy of the Landau level on the Lar-

mor orbit center position,h

2sinh

~2

i1is the partition function of

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the quantum oscillator of the cyclotron frequency , and the last factor

Lz(m/2~2)1/2 is the partition function of a free particle in one dimensionassociated with its motion along the z axis (i.e. along the magnetic field).

The free energyF= (1/) ln Q() is given by

F= ~

2 + Tln

1 e ~T

Tln

0Lz

rmT

2~2

!. (218)

The sum of the first two terms in the Eq. (218) is exactly the free energyof the oscillator with the frequency , whereas the last term is due to thedegeneracy of the Landau orbits and because of the free motion of a particlealong the magnetic field.

9 The Greens function of a particle in external

magnetic field

The results of the previous section, as we will show, may be used for thecalculations of the Greens function of the Landau problem because of theformal similarity of the equation of motion in both cases. We start from theequation of motion for the Greens function which in a general form readsas follows

i~

t H(r)G r, t, r0, t0=i~(r r0)(t t0), (219)where H(r) is the Hamiltonian of the system.

If the eigenvalue equation is solved

H(r)n(r) = Enn(r) (220)

so that the energy spectrum En and the wave functions n(r) are foundexplicitly, then it is straightforward to check that the Greens function canbe calculated as a sum over the quantum spectrum:

G r, t, r0, t0= (t t0)X

n

ei~En(tt0)n(r)

n(r0), (221)

where

() =

1, i f 00, if


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Puttingt0 = 0in Eq. (221) and compare it with equation, describing the

coordinate representation for the statistical operator (r, r0, ) ,we found asimple relation between the Greens function and the density matrix:

G

r, t, r0, 0

=

r, r0,

|= it~

. (222)

Since we have calculated above the density matrix for the Landau prob-lem, the Greens function for a charged particle in the magnetic field followsimmediately from Eqs.(213) and (194)

G

r, t, r0, 0

=Gk

z z0, tG , t,0, 0 . (223)HereGkis the Greens function of a free particle moving along the z-axis

(i.e. along the magnetic field B) Gk(z z0, t) = k(z z0, ) |= it~

,

Gk

z z0, t= m2i~t

1/2exp

im

2~t(z z0)2

(224)

andG stands for the Greens function of a charged particle moving withinthe plane perpendicular quantizing magnetic field

G,0, t

=

1

4iL2Hsint2

eiS(x,x0,y,y0,t), (225)where

S= 1

4L2H

cot

t

2

hx x02 + y y02i + 2 x + x02 y y02 .

(226)The above equations for the Greens function G (r, t, r0, 0) have been

obtained in the Landau gauge A= (0,By, 0) . A natural question arises inthis connection how the gauge transformations may influence the shape ofthe Greens function determined by Eq. (221). To answer this question letus rewrite the Hamiltonian in the eigenvalue equation (220) in the followingform

H= 12m

(DA)2 , (227)

where

DA= (~

ie

cA). (228)

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Consider now the gauge (r)transformations given by two simultaneous

relations: A0 = A+ f(r) and 0n(r) = n(r)ei(r). If the phase (r)in these transformations satisfies the condition ~ (r) = f(r)e/c it isstraightforward to see that a following equation holds

DA00n(r) = e

i(r)DAn(r). (229)

The latter means that changes in the vector potential due to the gradientterm f(r) may be compensated by the gauge transformation 0n(r) =n(r)ei(r) with (r) =

20

f(r) + Cand the theory became gauge invariantunder these transformations. Since the function 0n(r) is the eigenfunctionof the Schrdinger equation (220) belonging to the same eigenvalue En,

as the wave function

n(r), the Greens function (221) under the gaugetransformation A0 =A + f(r) acquire an additional factor g:

g= exp

2

0i

f(r) f(r0) . (230)In particular case of the symmetric gauge A= 12[Br] the function f(r)

should be tacking in the form f = 12Bxy, so that the gauge factor is givenby

g= exp

B

0i

xy x0y0

. (231)

10 The supersymmetry of the Landau problem

The supersymmetry of a system as the invariance of its Hamiltonian un-der the transformations of bosons into fermions and vice versa has beenconsidered first in the quantum field theory. This notion appeared to be ex-tremely creative both from physical and mathematical points of view. Forthe first time a matter (fermions) and carriers of interactions (bosons) havebeen involved into a theory on the equal footing. It was novel also thatcommuting and anticommuting variables have been incorporated into a newtype of mathematics - the superalgebra. The basic property of the super-

symmetry is that it unities in a nontrivial way the continuous and discretetransformations. Except the quantum field theory the ideas and methodsof the supersymmetry have been spread wide over the different branches ofphysics: the statistical physics, the nuclear physics, the quantum mechanicsand so on.

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In this section we will show that incorporation of the discrete spin vari-

able into the Landau problem makes the latter belonging to the so calledsupersymmetric quantum mechanics. Because the supersymmetric quantummechanics so far is not a common textbook knowledge, we have to considerfirst some fundamentals of the supersymmetry in the nonrelativistic quan-tum mechanics. After that we will go ahead with the consideration of thesupersymmetry in the Landau problem.

The simplest way to introduce operators transforming bosons into fermi-ons and vice versa is as follows

Q+| NB, NFi | NB 1, NF+ 1i, (232)

Q| NB, NFi | NB+ 1, NF 1i, (233)| NB, NFi is the state vector with fixed number of bosons NB and fermionsNF. The integersNB andNFcan take the following values: NB = 0, 1, 2...,butNFtakes only two values NF = 0, 1.

The operator Q+ transforms bosons into fermions (i.e. annihilates theboson and creates the fermion) and Q contrary destroys one fermion andcreates one boson. These operators may be presented in an evident formwith the help of the creation and annihilation operators

Q+= qbf+, Q= qb

+f , (234)

where band fsatisfy standard for bosons and fermions commutation rules:

hb, b+

i= 1,

nf , f+

o ff+ + f+f= 1, f2 = f+2 = 0,

hb, f

i= 0.

(235)The nilpotentcy of the fermion operators (i.e. the property f2 = f+2 =

0) makes the operators Q nilpotent too

Q2+= Q2 = 0. (236)

This property is closely related with the anticommutation. Let us intro-

duce two Hermitian operators Q1and Q2 by the relations

Q1= Q++ Q, Q2 = i

Q+ Q

. (237)

It is easy to check that these operators are anticommuting

49


46/51


47/51

On the other hand, since hH, Q2i= 0, we obtainH2= HQ21= Q2H1= q

22. (246)

Thus, if q 6= 0 then both 1 and 2 belong to the eigenvalue E = q2

which means a double degeneracy, whereas the energy level E = 0 (q =0) is nondegenerated. This properties are the direct consequence of thesupersymmetry of the Hamiltonian (240).

It is instructive for further consideration to express H in terms of theoperators band f. Using Eqs.(240) and (234) we have

H= nQ+,Qo=

HB+ HF. (247)

Therefore, we see that H is a sum of two Hamiltonians quadric in oper-ators, which we will call the bosonic and fermionic oscillators:

HB = q2

b+b +

1

2

, EB =q

2

NB+

1

2

, NB = 0, 1, 2..., (248)

HF =q2

f+f 1

2

, EF=q

2

NF

1

2

, NF = 0, 1. (249)

The frequencies of these two oscillators are the same = q2 which makesthe Hamiltonian of Eq. (240) supersymmetric.

The eigenvalues of the Hamiltonian Hare positive and given by the sum

ENBNF =EB+ EF = (NB+ NF) . (250)

The ground state (the vacuum) of the Hamiltonian Hcorresponds to thequantum numbers NB =NF = 0, so that the positive energy of the bosonoscillator ground state, E0B =/2, is exactly compensated by the negativefermion vacuum energy E0F =/2. This is the simplest manifestationof the famous cancellation of the vacuum zero oscillations energy in thesupersymmetric theories.

In the quantum field theory, owing to the infinite degrees of freedom, theenergies of fermionic and bosonic vacua are infinite and have opposite signs.

Thus, the problems with the infinite vacuum energies in nonsupersymmetrictheories are no more than an artifact arising because of the inappropriatedivision of the zero vacuum energy of the unified theory including bosonsand fermions into two (infinite) parts: positive bosonic and negative fermi-onic. We see that in the supersymmetric theory the fermionic and bosonicvacuum energies simply cancel each other.

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The Qoperators may be generalized in a way preserving the supersym-

metric form of the Hamiltonian given by the Eq. (240). For example, if wetake them in the form

Q+= B

b, b+

f+, (251)

Q= B+

b, b+

f , (252)

then it is straightforward to check that the key relationh

H, Qi

= 0holds

for an arbitrary function B

b, b

of the bosonic operatorsbandb+, because

of the nilpotentcy of the operators (251) and (252). On the other hand, thesupersymmetric Hamiltonian (247) after substitution of the operators (251)and (252) describes a system of bosons interacting with themselves andfermions, in contrast to the noninteracting fermionic and bosonic oscillatorsin case when operators Q are determined by the Eq. (234).

In as much as the fermion filling number NFmay take only two valuesNF = 0, 1it is convenient to use a two-component wave vector in the form

=

1

0

(253)

with 1 corresponding toNF = 1, and 0 toNF = 0. The Fermi operatorsf and f+ in this representation are given by the 2 2matrix

f+ = + =

0 10 0

, f= =

0 01 0

, (254)

where = 1/2(1 i2) and symbol j(j = 1, 2, 3) stands for the Paulimatrices.

The Hamiltonian H in the matrix representation takes the form

H=n

Q+, Qo

=1

2

nB, B+

o+

1

2

hB, B+

i3. (255)

We see that fermionic degree of the freedom (given by the term contain-

ing 3) vanishes if the commutatorh

B, B+i

= 0.Taking then B in the form

B= 1

2

hiP+ W(x)

i, P =

~

i

d

dx (256)

where W(x) is an arbitrary function of the coordinate, we have

52


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H=12

P2 + W2(x) + 3+ ~dW(x)

dx

. (257)

This differential operator is known as the Hamiltonian of the super-symmetric quantum mechanics of Witten. It takes the form of the PauliHamiltonian

H=P2

2 + U(x) + 30B(x), (258)

if we adopt the potential energy to be U(x) = W2(x)/2 and associate theZeeman splitting 0B(x) with the last term in the Eq. (258) i.e. choosethe magnetic field according to the relation ~dW(x)/dx=

0B(x). Of

course, it is not true magnetic field since there are no real magnetic field inone dimension.

In the three dimensional case the Pauli Hamiltonian reads

H= 1

2m

p e

cA2

+ U 0Bs, s= ~

23. (259)

The eigenfunctions of this Hamiltonian can be written as a product ofthe coordinate and spin functions

n,Pz ,,sz = nPz(r) (sz) . (260)

The index here depends on the momentum Py which determines theLarmore orbit position in the Landau gauge (A = (0,Bx, 0)), or on theinteger min the case of symmetric gauge (A= 1/2(By,Bx, 0)).

The energy spectrum for U 0 is given by the equations

EnPz(sz) =En(sz) +P2z2m

, (261)

En(sz) = ~

n +

1

2

20Bsz. (262)

We see that the transverse energy (262) possesses the properties of the

supersymmetric quantum mechanics. It becomes clear if we rewrite Eq.(262) in the form

En(sz) EN= ~N, (263)where N=n + sz+ 1/2 is a sum of the two quantum numbers.

53


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The ground state, N = 0, (n = 0, sz =

1/2) has zero energy E0 = 0

and this energy level is not degenerated. Contrary, the levels withN 6= 0are degenerated twice since two states with different quantum numbersn =N, sz = 1/2andn= N1, sz = 1/2belong to the same energy level. Thisis exactly what we should have in the supersymmetric quantum mechanics.

(We do not consider here the degeneracy on the orbit centre position). Inessence, the supersymmetry of the energy spectrum En(sz) stems from thefact that the Bohr magneton equals to 0 = e~/2mcso that the energy 0Bis exactly one half of the cyclotron energy~. Because of that, the transversepart of the Hamiltonian (259) can be written in the supersymmetric form

H= 12m

p ec A2 0Hs= ~b+b +12 + ~f+f 12 ,

(264)where the Fermi operators f+ and fare taken in the form (254) while theBose operators b+ and bare given by

b= (y+ ix) (2m~)1/2 ,

hb, b+

i= 1 (265)

with =

p ec A

.The orbital moment corresponds here the bosonic degree of freedom with

the quantum numbersNB =n(0, 1, 2...)whereas the spin variable plays therole of the fermionic degree of freedom with NF = sz + 1/2. The energyspectrum of the Hamiltonian (264) is equal to

E(NB, NF) = ~(NB+ NF). (266)

The above consideration shows that the supersymmetry of the electron mov-ing in an external magnetic field is not an abstract mathematical construc-tion, since it has a practical realization in the Landau problem.

The authors acknowledge the EuroMagNET of FP6, RII3-CT-2004-506-239.

References[1] L.D. Landau and E.M. Lifshitz, Quantum Mechanics(Pergamon Press,

New York, 1976).

[2] J.M. Ziman,Principles of the Theory of Solids(Cambridge Univ. Press,Cambridge, 1972).

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[3] L.D. Landau, Z. Phys.64, 629 (1930).

[4] S. Flugge, Practical Quantum Mechanics(Springer, 1974).

[5] M.H. Johnson and B.A. Lippmann, Phys. Rev.76, 828 (1949).

[6] A. Feldman and A.H. Kahn, Phys. Rev. B 1, 4584 (1970).

[7] P. Carruthers and M.N. Nieto, Rev. Mod. Phys.40, 411 (1968).

[8] S. Ruschin and J. Zak, Phys. Rev. A 20, 1260 (1979).

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quantum mechanis of strong mag field

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